fundoogyan's Blog

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download free IELTS material and books

November 16, 2009 by fundoogyan

Note: The below mentioned links are external links only and no file is being uploaded on this server. If any of these links violates copyright, please inform us, we will remove that link immediately.

Free TOEFL eBooks/CDs Download links

November 16, 2009 by fundoogyan

Download Free GMAT Material, e-Books, Mock tests, Software

November 16, 2009 by fundoogyan

The below mentioned links are external links only and no file is being uploaded on blogger’s server. If any of these links violates copyright, please inform us, we will remove that link(s) immediately.

Courtesy:- www.esnips.com

Download Top 10 Must Have ebooks for MBA aspirants

November 16, 2009 by fundoogyan

Here is the collection of top 10 e-books for MBA Aspirants or for those who want to be the Manager in future. Just click on the links to watch & download files.

1. The Foundations of Personality

2. Why Should you do MBA

3. The One Minute Manager

4. You Dont Have To Do It Alone How to Involve Others to Get Things Done

5. Things We Wished We Had

6. The Guru Guide to Marketing

7. The God of Small Things

8. MBA In A Day

9. Say It Right the First Time

10. Lasting Leadership : What You Can Learn from the Top 25 Business People of our Times

Prepare IIFT GK

November 11, 2009 by fundoogyan

hi,
here are some documents, which are beneficial for IIFT entrance exam. Please go through it and let me know if u require something more

Attachment: iift gk spl.doc (108.0KB)

Attachment: IIFT-paper1.pdf (41.0KB)

Attachment: IIFT-paper2.pdf (39.0KB)

IRMA GK question Bank

November 3, 2009 by fundoogyan

hi, here are some files attached having questions related to irma entrance.
hope it will help.
tell me if u need more
fundoogyan

Attachment: IRMA-Questions-Bank-2009.pdf (92.0KB)

Attachment: IRMA-Question-Papers.docx (41.0KB)

Attachment: IRMA-Paper2(2).pdf (45.0KB)

IRMA GK material

November 2, 2009 by fundoogyan

hi friends,
here i have attached Gk material useful for IRMA as an attachment. Plus here is a link where u can find out year book 2009, which is very useful for IRMA preparation.
http://www.publicationsdivision.nic.in/others/india_2009.pdf

original snap 2008 paper

October 12, 2009 by fundoogyan

Attachment: SNAP2008.pdf (684.0KB)

know the basic fundas of geometry

July 14, 2009 by fundoogyan

Attachment: Geometry Success in 20 Minutes a Day.pdf (4327.0KB)

writing skill for GRE/GMAT

June 23, 2009 by fundoogyan

Wanna clear the AWA section in GRE/GMAT, just go through the book.

Attachment: Writing_skillsGRE-GMAT.pdf (7124.0KB)

general knowledge questions

June 5, 2009 by fundoogyan

General Knowledge Questions

1. 'Air Asia', is a budget airlines in which of the following countries?

(A) Pakistan (B) Sri Lanka (C) India (D) Malaysia

2. Which of the following organisations uses the ad-line "The world put stock on usî?

(A) NYSE (B) NASDAQ (C) BSE (D) IMF

3. Rasna Private Ltd has recently signed up which of the following Bollywood icon as its

brand ambassador?

(A) Kajol (B) Sushmita Sen

4. Pick the incorrect Corporate CEO pair:

(A) McKinsey and Company - Rajat Gupta

(B) Vodafone - Arun Sarin

(D) Bell Labs - Arun Netravali

5. The controversial book ìFreakonomicsî, by the economist Steven D Levitt, lays the onus

of unexplained drop in crime rate in the US on:

(A) Strong Economy (B) Abortion Laws

6. Business Week's Infotech 100' 2006 published a list of global rankings. How many Indian

telecom and software services firms made it to this list?

(A) Two (B) Ten (C) Eight (D) Six

7. World Tuberculosis Day is observed on which of the following dates?

(A) 22nd May (B) 24th March (C) 23rd June (D) 3rd April

8. Which country tops the list of the world's most corrupt countries, issued by Transparency

International:

(A) Mexico (B) Iraq (C) Afghanistan (C) Haiti

9. Who updates a blog known as the ìGadgets, Google & SEOî?

(A) Eric Schmidt (B) Larry Page

10. What is the main attribute of a Gilt-Edged stock?

(A) Bonds issued by the government whose likelihood of default is zero

(B) A stock backed by gold.

(D) Stocks doing well in the capital market

11. Which parameter defines the poverty line in India?

(A) housing and clothing facilities (B) income of the family

12. Sharad Pawar was recently elected as the BCCI President, he replaced:

(A) Ranbir Singh Mahendra (B) Jagmohan Dalmiya

13. Which of the following countries does not have 'dinar' as its currency unit?

(A) Iraq (B) Bahrain (C) Iran (D) Jordan

14. Which among the following facts is not true about the life expectancy of females in India?

(A) The life expectancy at birth among females has been steadily improving over the years from 23.3 in 1901 to 61.8 in 1997.

(B) In India life expectancy of women is more than men.

(C) According to the 2001 census figure Madhya Pradesh had the lowest life expectancy among female while Delhi had the highest life expectancy.

(D) The urban female life expectancy is higher at 68.

15. The MPA (Members Participations Agreement), is the bone of contention between which

two organizations?

(A) ICC & ICICI (B) ICBC and ICC

16. Bollgard is a/an

(A) New variety of Bt. Cotton.

(B) A new insurance scheme for spice corps.

(D) A new trade initiative to raise the output of cotton in India.

17. Which among the following organisations/firms has recently took up the task of

rediscovering the river Saraswati?

(A) Archeological Survey of India (B) Indian Water Commission

18. Which among the following states is India's largest producer of eggs?

(A) Maharashtra (B) Andhra Pradesh

(C)Karnataka (D) Uttar Pradesh

19. What is India's rank in the global production of fruits?

(A) First (B) Fourth (C) Third (D) Second

20. 2006 Nobel Peace Prize winner Mohammad Yunus shared his prize with the bank that he

created. The bank is named:

(A) Gram Bank (B) Grameen Bank (C) Prabha Bank (D)Bangla Bank

21. While calculating the development indexes a term called GDP at Purchasing Power

Parity is always used. Which among the following facts is not true about Purchasing Power Parity index?

(1) Purchasing Power parity (PPP) is a theory, which says that the long-run equilibrium
exchange rate of two currencies is the rate that equalizes the currencies' purchasing
power.

(2) These special exchange rates are often used to compare the standards of living of two or more countries.

(3) In works on the basis of the law of one price which says: ìIn an efficient market all identical goods must have only one price.î

Which of the above statements is/are true?

(A) Only (1) and (2) (B) Only (2) and (3)

22. Malnutrition may cause a situation called Cachexia. In general terms it refers to:

(A) Loss of weight

(B) Inability to adapt to increased food intake.

(D) Depletion of bones

23. For his ìanalysis of intertemporal tradeoffs in macroeconomic policyî this Columbia

University Professor won the 2006 Nobel Prize for Economics. He is

(A) Edmund S Phleps (B) Robert J Aumann

24. Omimex De Columbia, a Columbian oil firm, has recently been acquired by ONGC

Videsh Limited (OVL) in partnership with

(A) Reliance Petrochemical (B) Sinopec

25. Indira Nooyi sometimes brings her kids to office so that they can do their homework

under her supervision. Which company elevated her to the rank of the CEO?

(A) Coca Cola (B) Proctor and Gamble

26. Google is in the process of $1.65 billion takeover of an online media website called:

(A) YouTube (B) WeTube (C) SeeTube (D) TVTube

27. Which among the following ministries has been identified as the nodal ministry for

National Mission on Bio-diesel?

(A) Ministry for Rural Development

(B) Ministry of Environment and Forest

(D) Ministry of Petroleum and Oil resources

28. Which philanthropist organization has been established by the Chairman of the Microsoft

Corporation? It specifically focuses on lending grant for AIDS research.

(A) Bill Gates Foundation (B) Melinda Gates Foundation

29. Brazil is the world's biggest producer of Coffee, which among the following is the second

largest producer of Coffee?

(A) Vietnam (B) India (C) Indonesia (D) Colombia

30. The recent bid of TATA Steel, to acquire UK based Anglo-Dutch steel venture named

ìCorusî, is worth:

(A) $5.1 bn (B) $6.1 bn (C) $7.1 bn (D) $8.1 bn

31. USA ranks at the top of India's list of export destinations. Which of the following regions

ranks second?

(A) European Union (B) Middle East (C) ASEAN (D) CWIS

32. Consider the following statements:-

(1) Two agreements were recently signed between the Government and the GMR and
GVK groups to set up joint venture companies involving the public sector Airports
Authority of India for handing over the Delhi and Mumbai airports for modernisation.

(2) K Ramalingam is Chairman of AAI.

(3) The headquarter of AAI is situated in Jamshed G. Tata House, Navi Mumbai. Which among the above statements is/are true?

(A) Only (1) is true. (B) Only (2) and (3) are true.

33. According to a recent report in Forbes Magazine, India's youngest Billionaire is Vikrant

Bhargarva, the oldest is

(A) Azim Premji (B) Narayana Murthy

34. Which state recently topped the list of best performing states Panchayati raj by the Union

Panchayati Raj ministry for the year 2004-05?

(A) Madhya Pradesh (B) Kerala (C) Gujarat (D) Sikkim

35. The Governor draws a monthly salary of

(A) Rs. 5,000 (B) Rs. 11,000 (C) Rs. 7,500 (D) Rs. 10,000

36. Which was the first public sector bank to launch the Visa card in India?

(A) SBI (B) Andhra Bank

37. Which among the following is name of only airlines operating form Afghanistan to

International destinations?

(A) Air Afghania (B) Air Afghan

38. Which company confers the Red & White Bravery Award?

(A) ITC (B) Godfrey Philips

39. ìBe the first to knowî is the punchline associated with which of the following news

channel?

(A) CNBC (B) Star News (C) BBC (D) CNN

40. Central-State financial distribution takes place following recommendations made by the

(A) Finance Minister (B) Finance Commission

41. Why the Industrial and Commercial Bank of China (ICBC), has recently hit World's

business headlines?

(A) Making a largest acquisition in China

(B) Undergoing largest public sector merger in China

(D) Setting a record for banking turnover in China

42. Which auto major has launched cars in India with VTEC engine technology?

(A) Hyundai (B) Honda (C) Mercedes (D) Opel

43. Which among the following awards cannot be given to a non-Indian?

(A) Indira Gandhi Peace Award (B) Padam Shree

44. With which company would you associate the slogan, "We make the things that make

India proud"?

(A) ACC (B) L&T

45. Anurag Dixit, an IIT Alumni, runs an online company with major operations based at

Gibraltar. This company suffered considerable losses according to the recent financial

results. Its named:

(A) Partygaming.com (B) Pokergaming.com

46. Which airline in India serves Domino's Pizzas?

(A) Sahara (B) Jet Airways (C) Indian Airlines (D) Air-India

47. The world's first water-powered car has been manufactured by

(A) BMW (B) Mitsubishi

48. The print ad of which brand carried the headline-"So plush, so comfortable, so

depressing for other cars"?

(A) Hyundai (B) Scorpio

49. The India Golf Tour is sponsored by which two-wheeler manufacturer

(A) Hero Honda (B) Bajaj Auto

50. Which of the following banks is planning to become a partner in a joint venture in

Malaysia along with Bank of Baroda and Oriental Bank of Commerce?

(A) Bank of Rajasthan (B) ICICI Bank

Answer Key

1. (D) 2. (A) 3. (D) 4. (C) 5. (B)

6. (D) 7. (B) 8. (C) 9. (B) 10. (A)

11. (D) 12. (A) 13. (D) 14. (C) 15. (D)

16. (A) 17. (C) 18. (B) 19. (D) 20. (B)

21. (C) 22. (A) 23. (A) 24. (B) 25. (C)

26. (A) 27. (A) 28. (D) 29. (A) 30. (D)

31. (B) 32. (D) 33. (C) 34. (B) 35. (B)

36. (B) 37. (D) 38. (B) 39. (D) 40. (C)

41. (B) 42. (B) 43. (C) 44. (B) 45. (A)

46. (B) 47. (A) 48. (B) 49. (A) 50. (C)

LOGICAL REASONING

May 29, 2009 by fundoogyan

Direction for question 1 to 4: Answer the questions based on the following

information.

Four people of different nationalities live on the same side of a street in four houses, each of a different colour. Each person has a different favorite drink. The following additional information is also given.

I. The Englishman lives in red house.

II. The Italian takes tea.

III. The Norwegian lives in the first house on the left.

IV. In the second house from the right, they drink milk.

V. The person living adjacent to blue house drinks cocoa.

VI. The Spaniard drink fruit juice.

VII. Tea is taken in blue house.

VIII. The white house is to the right of the red house.

IX. No other description of cocoa is available.

Example 1: Milk is drunk by

(A) the Norwegian (B) the Englishman

Example 2: The Norwegian drinks

(A) milk (B) cocoa

Example 3: The colour of the Norwegian’s house is

(A) white (B) red

Example 4: Which of the following is not true?

A. Milk is drunk in the red house

B. The Italian lives in blue house

C. The Spaniard lives in a corner house

D. The Italian lives next to the Spaniard

Using the given information, try to draw a diagram. In this case, a table would be more apt and it might look like,

Nationality	Norwegian	Italian	Englishman	Spaniard
Colour of the house	---	Blue	Red	White
Drink	Cocoa	Tea	Milk	Fruit juice

We would see now, how easy it is solve the questions.

Solution1: Ans (B) (Straight from table)

Solution2: Ans (B) (Straight from table)

Solution: Ans (D) (Fourth colour is not mentioned in the data)

Solution: Ans (D) ((Straight from table))

As seen above, logical reasoning problems don’t involve any complex mathematical calculations. If we are able to visualize the problem in any suitable figure, 99% job is done. Rest is to read the question carefully and answer them

Direction for the question 5 to 8: Answer the questions based on the following information.

Four friends – Manas, Kailash, Shashidev and Bhagwan – have different preferences for watches and bike. Each person prefers one exclusive watch and bike. Bikes are Passion, Pulsar, Enticer and Fiero. Watches are radio, Omega, Cartier and Tissot. One who likes Enticer like Cartier, and one who likes Tissot also likes Pulsar. Bhagwan likes Fiero and Rado. Manas does not like either Omega or Enticer.

Example 5: Which of the following are Manas choices?

(A) Tissot and Pulsar (B) Pulsar and Cartier

Example 6: The one who likes Omega likes

(A) Passion (B) either Passion or Fiero

Example 7: What is Shashidev’s choice for the watch?

(A) Omega (B) Cartier

Example 8: What is Kailash’s preference for the bike?

(A) Pulsar (B) Passion

Using the given information, try to draw a diagram. In this case, a table would be more apt and it might look like,

Person	Bike	Watch
Shashidev/ Kailash	Passion	Omega
Shashidev/ Kailash	Enticer	Cartier
Manas	Pulsar	Tissot
Bhagwan	Fiero	Rado

Now, all the questions can be cracked easily.

Solution5: Ans (A) (Straight from table)

Solution6: Ans (D) (As seen from table, it can be Shashidev or Kailash, so can’t be determined)

Solution7: Ans (D) (As seen from table, it can be Omega or Cartier, so can’t be determined)

Solution8: Ans (D) ((As seen from table, it can be passion or enticer, so can’t be determined)

Directions for Examples 9-10: Study the following information: (CAT 2001)

Elle is three times Yogesh

Zahir is half Wahida

Zahir is younger than Yogesh.

Example 9: Which of the following are necessary to find the age of each?

(A) Wahida is same age as that of Yogesh

(B) Age of Zahir is ten

(D) None of these.

Solution9: Given , E=3Y, Z=1/2W, Z<Y

Now, From statement II, Z = 10 years

So, W = 2Z = 20 years

From Statement 1, Y=W=20

Then E = 3Y=3 x 20 = 60 years.

Hence, both the statements are necessary to find the age of each person.

Ans=(C)

Example 10: Which of the following is true?

(A) Elle is the eldest

(B) Wahida can be elder to Elle

(D) None of these.

Solution10: Given that Z < Y

so that 2Y < 2Y or W < 2Y < 3Y = E

Hence, Elle is the eldest.

Ans=(A)

Directions for Examples11-13: Read the following passage and answer the questions which follow.

Two union representative and one management representative are seated together at an octagonal table with only one seat to a side of the table. No pair of either union or management representatives may be seated together. Two additional management representatives are seated. (CAT 2001)

Example 11: Seated between the two union reps are

(A) at most two management people.

(B) only two management people.

(D) three management people.

Solution: At most, two management people can be seated between the two union representatives.

<<Logical Reasoning Solved 1>>

Ans=(A)

Example 12: Seated opposite the first management representative

(A) must be a union person.

(B) may be a union person.

(D) must be a management person.

Solution: Seated opposite the first management representative may be a

management person. (The only other alternative is that the seat

would be empty).

Ans=(C)

Example 13: If two more union reps are seated without causing any changes of

seats, then there is (are)

(A) no empty seat next to a union person.

(B) no empty seat next to a management person.

(D) at most one empty seat between union reps.

Solution: If two more union representatives are seated without causing any

changes of seats, then there is at most one empty seat between union

representatives.

<<Logical Reasoning Solved 2>>

Ans=(D)

Example 14: There are six houses in a row. A has B and C as neighbours. D has E and F neighbours. E’s house is not next to B or C, and F does not be next to C. Who are B’s next door neighbours?

(A) A and F (B) C and F

Solution14: According to given conditions:

A is in between B and C and D is in between E and F. But E is not

neighbour of B and F is not neighbour of C. So the sequence will be

EDFBAC. Hence B neighbours are F and A

Ans=(A)

Directions for example 15 to 19: Refer to the data below and answer the questions that follow.

There are five events A, B, C, D and E that can happen. The occurrence of every event is governed by few rules, which are:

- If A occurs then either of B or C or both must occur.

- If B occurs then D cannot occur.

- If C occurs then E must occur.

- If D occurs then C must occur.

- If E occurs then A must occur and B cannot occur.

- If D has not occurred then A will also not occur.

Example 15: If C has occurred, then which of the events must happen?

(A) A (B) B

Example 16: If E has not occurred, then which of the statements must be true?

I. C has not occurred. II. B has occurred.

III D has not occurred. IV. A has not occurred.

(A) I and II (B) III and IV

Example 17: If B has occurred, then which statement will be definitely false?

(A) D has not occurred. (B) C has not occurred.

Example 18: If A has occurred, then which event(s) will definitely occur?

(A) B (B) C and D

Example 19: If D occurs, then any of the events can occur except:

(A) A (B) B

Solution

Let us first draw the figure of the information given. As this is a logical problem, an arrow diagram would best describe. Let us symbolize the given conditions..

1.) A > B or C or both

When A occurs, then either B., C or both would occur so, If both of B or C not occurred then A will not occur. (Logical inference) But it may happen that B or C has occurred and still A hasn’t occurred.

2)B > - D (- sign means can’t occur)

3) C > E

4) D > C

5) E > A

6) E > - B

7) A > D (Last statement actually means that if A happens then D must happen)

Now read all the questions and check all the options for the information given.

Solution 15:

<<Logical Reasoning Solved 3>>

If C happens, then E must happen. If E happened then A must happen and B would not happen and If A happens then D must happen Also either B or C should happen. As B can’t happen then C would happen and it is happening.Thus E, A and D must occur.

Ans=(D)

Solution 16:

<<Logical Reasoning Solved 4>>

If E doesn’t occurs, then C can’t occur. If C doesn’t occur then D can’t occur. If D doesn’t occur, then A can’t occur. So if E doesn’t occur, then C, D and A can’t occur. We don’t know about B. So

Ans=(C)

Solution 17:

<<Logical Reasoning Solved 5>>

If B occur, then D can’t occur. If D doesn’t occur then A can’t occur.

Ans=(D)

Solution 18:

<<Logical Reasoning Solved 6>>

Ans=(D)

Solution 19:

<<Logical Reasoning Solved 7>>

Ans=(B)

Example 20: There are 3 families…. Bannerjees, Guptas and Sharmas. Each family has a feast every Sunday at different timings of 12:00, 1:00 and 2:00. Each family eats different dishes and uses different coloured dinner sets. (CAT 2001)

The Bannerjees eat sambhar but not in the red dinner set.

The last family does not eat karela or brinjal.

The other dinner sets are yellow and blue in colour.

Which of the following is true?

(A) The Bannerjees eat at 12

(B) The last family eats sambhar in the blue dinner set

(D) None of these

Solution20: Given information may be put in the tabular form as

Time

Family

Dish

Colour of Dinner Set

12.00

1.00

2.00

Gupta or sharma)

Sharma or Gupta

Bannerjee

Karela (or Brinjal)

Brinjal (or Karela)

Sambhar

Any

Not Red

So, option (A) of definitely false. Both options B and C can be true but we are not sure.

Ans=(D)

Direction for examples 21-24 :Answer the questions on the basis on the following information.

Four families decided to attend the marriage ceremony of one of their colleagues. One family has no kids, while the others have at least one kid each. Each family with kids, has at least one kid attending the marriage. Given below is some information about the families, and who reached when to attend the marriage. (CAT 2003)

The family with 2 kids came just before the family with no kids.

Shanthi who does not have any kids reached just before Sridevi’s family.

Sunil and his wife reached last with their only kid.

Anil is not the husband of Joya.

Anil and Raj are fathers.

Sridevi’s and Anita’s daughters go to the same school.

Joya came before Shanthi and met Anita when she reached the venue.

Raman stays the farthest from the venue.

Raj said his son could not come because of his exams.

Example 21:Which woman arrived third?

A. Shanthi

B. Sridevi

C. Anita

D. Joya.

Example 22:Name the correct pair of husband and wife.

A. Raj and Shanthi

B. Sunil and Sridevi

C. Anil and Sridevi

D. Raj and Anita.

Example 23:Of the following pairs, whose daughters go to the same school?

A. Anil and Raman

B. Sunil and Raman

C. Sunil and Anil

D. Raj and Anil.

Example 24:Whose family is known to have more than one kid for certain?

A. Raman’s

B. Raj’s

C. Anil’s

D. Sunil’s

Solution for Examples 21-24

The key to cracking this question is to follow the simple fundamentals in logical

reasoning. Read all the data and try to draw a picture. In this question, a table would best represent the data.

Let us interpret all the data one by one:

Sentence 1- Family with 2 kids came just before no kids. ( they should be together)

Sentence 2 – Shanthi with no kids came just before Sridevi

Sentence 3 - Sunil and wife came last with only kid

Sentence 4 – Anil and Joya not husband and wife.

Sentence 5 – Anil and Raj are fathers – hence cannot be the family with no kids.

Sentence 6 – Sridevi and Anita cannot be the persons with no kid

Sentence 7 – Joya came before Shanthi and Anita was already present.

Sentence 8-- Raman stays the farthest from the venue. (Useless information)

Sentence 9- Raj said his son could not come because of his exams.

Using the above into – Anil and Raj cannot be married to Shanthi as Shanthi has no kids whereas Anil and Raj are fathers. Also, Sunil and wife came last but Shanthi can’t come last as she came before Sridevi so she can’t be wife of Sunil.

So, Shanthi is married to Raman. Also from Sentence 7 and Sentence 2, Sridevi has to come last and is wife of Sunil.

As Anil and Joya are not spouses, so Anil is married to Anita and Raj is married to Joya.

Now, combing all the information, the data can be summarized as

Arrival	Husband	Wife	Kids
1	Anil	Anita	1
2	Raj	Joya	2
3	Raman	Shanthi	0
4	Sunil	Sridevi	1

Solution21 : Straight from table

Ans=(A)

Solution 22 : Straight from table

Ans=(B)

Solution 23: Straight from table

Ans=()

Solution 24: As Raj said that his son can’t come and he came with atleast one kid,

he surely have two or more kids.

Ans=(B)

Directions. for Examples 25-28: Answer the questions on the basis of the following information.

The plan above shows an office block for six officers, A, B, C, D, E, and F. Both B and C occupy offices to the right of the corridor (as one enters the office block) and A occupies an office to the left of the corridor. E and F occupy offices on opposite sides of the corridor but their offices do not face each other. The offices of C and D face each other. E does not have a corner office. F’s office is further down the corridor than A’s, but on the same side. (CAT 2003)

<<Logical Reasoning Solved 8>>

Example 25: If E sits in his office and faces the corridor, whose office is to his

left?

A. A

B. B

C. C

D. D.

Example 26: Whose office faces A’s office?

A. B

B. C

C. D

D. E.

Example 27: Who is/are F’s neighbour(s)?

A. A only

B. A and D

C. C only

D. B and C.

Example 28: D was heard telling someone to go further down the corridor to the

last office on the right. To whose room was he trying to direct that

person?

A. A

B. B

C. C

D. F.

Solutions to Examples 25-28

Like last example, let us comprehend the data and try to draw a suitable diagram. As the examiner himself has given the diagram let us try to fill it by placing different people in appropriate positions.

<<Logical Reasoning Solved 9>>

This is the only combination possible given all the constrains. Why, let us explain

F is further down the corridor from A, so F can take 2 or 3 position on left. As F is on left, so E would be on right. As E can’t be take corner office, he would take middle one on right. So E is fixed. Now F can’t face E, F would take 3 on Left as 2 on left would face E. So Now E and F are fixed. C and D have to face each other that are possible now only at poison 1. Hence, C takes 1 position on right and D takes 1 position on left. Now, A has to take 2 position on left and B would take 3 position on right.

Understood, As said many times before, logical reasoning don’t involve any mathematical calculations or formulas to remember. Simple logic application.

Solution 25 : From the figure, If E faces the corridor, person to his left is C.

Ans=(C)

Solution 26 : According to figure, E face A’s office.

Ans=(D)

Solution 27 :According to figure, F’s neighbour is A.

Ans=(A)

Solution 28 : According to figure, B’s room is last on the right.

Ans=(B)

This Question appeared in CAT’05. (In Quantitative section)

But we believe it should be in Data Analysis Section.

Question: A telecom service provider engages male and female operators for

answering 1000 calls per day. A male operator can handle 40 calls per day whereas a female operator can handle 50 calls per day. The male and the female operators get a fixed wage of Rs. 250 and Rs. 300 per day respectively. In addition, a male operators get a male operator gets Rs.15 per call he answers and a female operator gets Rs. 10 per call she answers. To minimize the total cost, how many male operators should the service provider employ assuming he has to employ more than 7 of the 12 female operators available for the job?

(1) 15 (2) 14

(3) 12 (4) 10

Solution: By the condition given in question.

Each male operator get Rs. 250/40 = Rs. 6.25 for one call as fixed wage

Similarly, Each female operator fixed cost is Rs. 300/50 = Rs. 6 for one call.

Also the variable cost for male is Rs 15 per call amd for female is Rs 10 per call. So total cost per call

For Male=6.25+15=Rs 21.25

For female=6+10=Rs 16

So, female operator is cheaper than man to minimize one should use the maximum possible number of female operators. The telecom service

provider engages maximum 12 female operator, which will answer 12 x 50 = 600 calls. The remaining 400 calls will be answered by 10 operators.

Ans (4)

The Answer was rather simple. Only common sense and logic was required to answer the above question.

This Question appeared in CAT’05 (In Quantitative section)

But we believe it should be in Data Analysis Section.

Question: Three Englishmen and three Frenchmen work for the same company. Each of them knows a secret not known to others. They need to exchange these secrets over person-to-person phone calls so that eventually each person knows all six secrets. None of the Frenchmen knows English, and only one Englishman knows French. What is the minimum number of phone calls needed for the above purpose?

(1) 5 (2) 10

(3) 9 (4) 15

Solution: The key to answer lies in drawing the below figure.

<<Logical Reasoning Solved 10>>

For min number of phone calls let

E2 & E3 converse to E1 --2 calls

F2 & F3 converse to F1 --2 calls

E1 & F1 interchange their code --1 call

Now F1 calls F2 & F3 --2 calls

& E1 calls to E2 & E3 -- 2 calls

Total calls = 2 + 2 + 1 + 2 + 2 = 9

Ans (3)

Once again, No permutation-combination was required. Just draw the diagram and the logic would flow.

LOGICAL REASONING

May 29, 2009 by fundoogyan

Directions for problems 1-4: In a group of 7 people, there are two ladies Fiza and Kavita and five men Ram, Rahim, Peter, David and Shyam. (CAT 2001)

A group of 3 or 4 has to be selected.

If Fiza goes, she wants David.

David wants Kavita.

Shyam and Rahim insist on going together.

David and Peter do not go together.

Ram and Shyam do not go together.

Problem 1: Can both the females be there is a group of four?

(A) Yes

(B) No

(D) Cannot be determined.

Problem 2: Which is a feasible group of three

(A) Ram, David, Fiza

(B) Shyam, David, Peter

(D) None of these.

Problem 3: Which is a feasible group of four?

(A) Ram, David, Fiza, Shyam

(B) Shyam, David, Kavita, Rahim

(D) None of these.

Problem 4: Which of the following will be true ?

(A) A group of four can be formed with only the men

(B) Both women can be included in a group of four

(D) None of these.

Directions for problem 5: Amti wants to see some plays. There are six plays going on. Amit wants to see all to them, as well as take a lunch break for one hour from 12.30 p.m. to 1.30 p.m. The names of the plays, their durations and timings are all mentioned in the following table. (CAT 2001)

No.	Play	Duration	Timings
1.	Sati Savitri	1 hour	9:00 a.m. 2:00 p.m.
2.	Tipu Sultan	1 hour	10:00 a.m. 11:00 a.m.
3.	Sunder Kand	30 min.	10:30 a.m. 12:00 p.m.
4.	Hayavardhana	1 hour	10:00 a.m. 11:00 a.m.
5.	Nagamandala	1 hour	11:00 a.m. 2:00 p.m.
6.	Jhansi ki Rani	30 min.	10:30 a.m. 1:30 p.m.

Problem 5 : Which is the best possible plan for Amit.

(A) Sundar Kand first, Jhansi ki Rani third, Tipu Sultan fifth.

(B) Sati Savitri first, Nagamandala third, Sundar Kand fifth.

(D) None of these.

Direction for Problem 6 to 9: Answer the questions based on the following information. A parking space has seven parallel rows that is occupied by cars of different models. These rows are numbered consecutively from 1 to 7. In each row, a different model of car – Palio, Siena, Baleno, Escort, Accent, Astra and Indica – is to be parked according to the following conditions.

I. If Astra is next to only one other row of cars, that row must be Siena.

II. Neither Palio nor Siena can be in a row next to the row Indica is in.

III. Baleno must be parked in a row next to a row containing either Accent or Indica or in a row that is the only row between the rows containing Accent and Indica.

IV. Escort must be parked in either row 1 or row 7.

Problem 6: If Astra is in row 1 and Palio is in 6 row, which car must be parked in row 4?

(A) Indica (B) Siena

Problem 7: If Indica is next to Escort and Astra is in row 7, then in which row must Indica be placed?

(A) 6 (B) 4

Problem 8: If Palio, Baleno, Siena, Accent and Indica are parked in the inner five rows, then which of the following is the correct and possible order of arrangment of these cars in the row?

A. Palio, Siena, Baleno, Indica, Accent

B. Palio, Accent, Baleno, Indica, Siena

C. Siena, Accent, Indica, Baleno, Palio

D. Baleno, Palio, Sina, Accent, Indica

Problem 9: Which of the following is possible order for the parking of these cars in the parking space, beginning with row1?

A. Accent, Indica, Baleno, Palio, Escort, Astra, Siena

B. Escort, Indica, Baleno, Accent, Palio, Siena, Astra

C. Escort, Accent, Baleno, Indica, Siena, Palio, Astra

D. Baleno, Palio, Siena, Accent, Indica, Astra, escort

Problem 10: Consider the facts given below.

I. Sushma finished her graduation five years ago form today.

II. Geeta did her graduation three years before Sushma and joined the teaching profession the same year that Sushma entered college.

III. Aarti did her graduation three years after Sushma.

IV. Geeta taught both Sushma and Aarti during their graduation. If Geeta is still teaching, for how many years has Geeta been a teacher?

(A) 3 years (B) 5 years

Direction for the Problem 11 to 14: Answer the question based on the following on the following information.

A group of three persons must be selected from six individuals – Keshto, Omprakash, Sanjiv Kumar, Tuntun, Vijayendra and Wahid, according to the following conditions.

I. Either Omprakash or Vijayendra must be selected but neither Vijayendra nor Sanjiv Kumar can be selected with Omprakash.

II. Either Sanjiv Kumar or Keshto or both must be selected.

Problem 11: If Vijayendra is not selected, which pair of individuals must be among those selected?

(A) Omprakash and Wahida (B) Tuntun and Keshto

Problem 12: Which of the following is an acceptable selection of persons?

A. Omprakash, Vijayendra and Tuntun

B. Keshto, Tuntun and Sanjiv Kumar

C. Keshto, Omprakash and Snajiv Kumar

D. Keshto, Sanjiv Kumar and Vijayendra

Problem 13: If Sanjiv Kumar is selected, which of the following individuals must also be among the people selected?

(A) Wahida (B) Vijayendra

Problem 14: Which of the following pairs of persons cannot both be among the persons selected?

(A) Vijayendra and Wahida (B) Omprakash and Wahida

Direction for the Problem 15 to 19: Answer the question based on the following on the following information.

In a typical college day at IIM, Bangalore, exactly seven lecturers A, B, C, D, E, F and G were to give their lecturers in a first year class. In the schedule for the day, seven-time slots are available for the speakers and they are numbered from 1 to 7. Only one speaker is assigned one time slot, according to the following conditions.

I. C must speak in the either time slot 1 or time slot 7

II. A must speak immediately before or immediately after D speaks.

III. F must speak in the fourth time slot.

IV. D must speak sometime before B speaks.

Problem 15: If G speaks at position 7, any of the following pairs of speakers could speak in time slots immediately adjacent to each other except

(A) F and E (B) C and A

Problem 16: Which of the following must be true?

A. G speaks sometime before F speaks

B. C speaks sometime before D speaks

C. A speaks sometime before F speaks

D. A speaks sometime before B speaks

Problem 17: If E wants to speak in the second time slot, then there will be a total of how many scheduling possibilities from which to select the schedule of speakers?

(A) 6 (B) 1

Problem 18: If B speaks immediately before F speaks, which of the following could be true?

A. D speaks in the third time slot

B. G speaks in the sixth time slot

C. C speaks in the first time slot

D. A speaks in the fifth time slot

Problem 19: If E speaks sometime before A speaks, which of the following must be true?

A. G speaks sometime before C speaks

B. E speaks sometime before G speaks

C. D speaks sometime before F speaks

D. F speaks sometime before B speaks

Problem 20: Peter owned a butcher’s shop. In Peter’s absence, a dog ran away with a piece of meat. When Peter returned, the other shopkeepers, who were jealous of him, gave two statements each, one of which was a lie. (CAT 2001)

1^st shopkeeper: The dog was black. It had no collar.

2^nd shopkeeper: The dog was black. It had a short tail.

3^rd Shopkeeper: The dog was white. It had a collar.

Therefore the dog was

(A) white with a short tail and no collar.

(B) black with a long tail and a collar.

(D) white with a long tail and no collar.
Solutions

Solutions 1-4

Let us put together all the constrains in a diagram

Fi>Da (if Fiza goes, she wants David)

Da>Kv (if David goes, he wants Kavita)

Sh<>Rh (Shyam and Rahim want to go together)

Ra>Pe (if Ram goes, he wants Peter)

Da><Pe (David and Peter would not go together)

Ra><Sh (Ram and Shyam don’t go together)

Solution 1: Both the ladies – Fiza and Kavita can go only with David. But when

David goes then none of Ram, Shyam, Rahim or Peter can go. Hence, such a combination is not possible.

Ans=(B)

Solution2: Checking the options, all the groups fail one or more constraints.

Ans=(D)

Solution3: Checking the options ,Only feasible group of four Shyam, David, Kavita and Rahim.

Ans=(B)

Solution4: Checking the options , option (C) is true.

Ans=(C)

Solution 5: The best possible plan for Amit is

III

Sati

Savitri

Tipu

Sultan

Hayavar-

dhana

Sundar

Kand

Lunch

Jhansi

Ki Rani

Nagaman-

dala

(9-10)

(10-11)

(11-12)

(12-12.30)

(12.30-1.30)

(1.30-2.00)

(2-3)

Ans=(D)

Solutions 6 to 9

We can put a linear arrangement like below based on data given.(From I and II)

1	2	3	4	5	6	7
As/Es	Si				Si	As/Es

Now see the information in each question

Solution6: Under the given conditions, the arrangement would be

1	2	3	4	5	6	7
As	Si	Ba/Acc	Indica	Ba/Acc	pa	Es

So Indica would be in Row 4.

Ans=(A)

Solution7: Using the given conditions, see main table. If Astra is in Row 7 then, Escort would be in Row 1 and Indica in Row 2.

Ans=(C)

Solution 8: In this question, remember first not to use any information from Problem 6 or 7. Use only the basic common data in problem, thus the main table above. As no other info is given, the best way is to check the various options and see which one satisfy the constrains of the problem

So, According to the given conditions, the only valid possible option would be Siena, Accent, Indica, Baleno and Palio

Ans=(C)

Solution9: Following same approach as in Problem 8, Undre the given conditions, the only valid option out of the given ones is: Escort, Indica, Baleno, Accent, Palio, Siena, Astra.

Ans=(B).

Solution10: Sushma finished her graduation 5 years back from today. Geeta started teaching from the year Sushma entered college, i.e. 3 years before Sushma finished her graduation. So Geeta started teaching 8 years back from today.

Ans=(C)

Did you notice one thing? Data in points 3 and 4 was totally useless and is called Superfluous data. The idea is to fool student and make him think where to use it.

Solutions 11 to 14

Using abbreviations for various names and classifying the given conditions, we get the following.

I. Either O or V must be selected but neither V or S with O, i.e. VO and SO is possible.

II. Either S or K or both SK must be selected.

Solution 11: Using the given conditions, we get:

If V is not selected, then O must be selected (according to condition I).

Now using condition II V, S or K or SK must be selected along with O, but as given S cannot be with O. Therefore, K has to be there with O. Hence, Om Prakash and Keshto must be selected.

Ans=(D)

Solution12: In type of questions, we have to explore all the choices with the conditions given. So, Using the given conditions, the combination of K, V and S is the only acceptable combination form the given choices.

Ans=(D)

Solution13: According to the given conditions, if S is selected, then V must be selected and not Q. Therefore, V must be selected if S is selected

Ans=(B).

Solution 14: As, one from O and P have to be selected and one from S and K have to be selected, we can never select T and W together.

Ans=(D)

Solution15: As G speaks in time slot 7, and then C will be first slot (according to condition I). and F always speak in slot 4. So 1-4-7 is taken.

Now, as A has to speak adjacent to D, and D has to speak before B, AD would speak in 2-3slots. Now we have to check the options.

In any case, AB can never speak together.

Ans=(C)

Solution16: Remember not to take any info from problem 15. We know only what is given in common directions.

We have to check the options first.

In every combination, A speaks sometimes before B speaks, will always be true. (As A and D speak adjacent and D speaks before B)

Ans=(D)

Solution17: If E wants to speak in the second time slot, then according to the given conditions, we have these restrictions.

As A and D have to speak together, they can speak in 5-6 or 6-7 positions. But As D have to speak before B, A and D Have share 5-6. So B would speak at 7 and thus C would speak at 1.

So we have C at 1, E at 2, G at 3, F at 4, A and D at 5 or 6 and B at Seven So, only two combinations are possible.

(i) C/1, E/2, G/3, F/4, D/5, A/6, B/7.

(ii) C/1, E/2, G/3, F/4, A/5, D/6, B/7.

Ans=(C)

Solution18: If B speaks before F (slot 4), we have these restrictions.

As D has to speak before B and A and D are adjacent, B would speak at a slot 3. A and D at 1 and 2 slots and so C at 7.

Now, check for various options,

then only statement which would be true out of the given options is G speaks in the sixth time slot and one of that arrangement can be

A/1, D/2, B/3, F/4, E/5, G/6, C/7.

Ans=(B)

Solution19: Using the given conditions that E speaks sometime before A speaks, we get the condition, F speaks sometime before B speaks, must be true.

Ans=(D)

Solution20: Based on the statements, only two possibilities are there.

If 1^st shopkeeper is correct about color, then dog would be Black with collar.

if Ist shopkeeper lies about color, then dog would be white with no color and short tail

Checking the options now, Option A is correct

Ans=(A)

LOGICAL REASONING

May 29, 2009 by fundoogyan

Directions for problems 1-4: In a group of 7 people, there are two ladies Fiza and Kavita and five men Ram, Rahim, Peter, David and Shyam. (CAT 2001)

A group of 3 or 4 has to be selected.

If Fiza goes, she wants David.

David wants Kavita.

Shyam and Rahim insist on going together.

David and Peter do not go together.

Ram and Shyam do not go together.

Problem 1: Can both the females be there is a group of four?

(A) Yes

(B) No

(D) Cannot be determined.

Problem 2: Which is a feasible group of three

(A) Ram, David, Fiza

(B) Shyam, David, Peter

(D) None of these.

Problem 3: Which is a feasible group of four?

(A) Ram, David, Fiza, Shyam

(B) Shyam, David, Kavita, Rahim

(D) None of these.

Problem 4: Which of the following will be true ?

(A) A group of four can be formed with only the men

(B) Both women can be included in a group of four

(D) None of these.

No.	Play	Duration	Timings
1.	Sati Savitri	1 hour	9:00 a.m. 2:00 p.m.
2.	Tipu Sultan	1 hour	10:00 a.m. 11:00 a.m.
3.	Sunder Kand	30 min.	10:30 a.m. 12:00 p.m.
4.	Hayavardhana	1 hour	10:00 a.m. 11:00 a.m.
5.	Nagamandala	1 hour	11:00 a.m. 2:00 p.m.
6.	Jhansi ki Rani	30 min.	10:30 a.m. 1:30 p.m.

Problem 5 : Which is the best possible plan for Amit.

(A) Sundar Kand first, Jhansi ki Rani third, Tipu Sultan fifth.

(B) Sati Savitri first, Nagamandala third, Sundar Kand fifth.

(D) None of these.

I. If Astra is next to only one other row of cars, that row must be Siena.

II. Neither Palio nor Siena can be in a row next to the row Indica is in.

III. Baleno must be parked in a row next to a row containing either Accent or Indica or in a row that is the only row between the rows containing Accent and Indica.

IV. Escort must be parked in either row 1 or row 7.

Problem 6: If Astra is in row 1 and Palio is in 6 row, which car must be parked in row 4?

(A) Indica (B) Siena

Problem 7: If Indica is next to Escort and Astra is in row 7, then in which row must Indica be placed?

(A) 6 (B) 4

Problem 8: If Palio, Baleno, Siena, Accent and Indica are parked in the inner five rows, then which of the following is the correct and possible order of arrangment of these cars in the row?

A. Palio, Siena, Baleno, Indica, Accent

B. Palio, Accent, Baleno, Indica, Siena

C. Siena, Accent, Indica, Baleno, Palio

D. Baleno, Palio, Sina, Accent, Indica

Problem 9: Which of the following is possible order for the parking of these cars in the parking space, beginning with row1?

A. Accent, Indica, Baleno, Palio, Escort, Astra, Siena

B. Escort, Indica, Baleno, Accent, Palio, Siena, Astra

C. Escort, Accent, Baleno, Indica, Siena, Palio, Astra

D. Baleno, Palio, Siena, Accent, Indica, Astra, escort

Problem 10: Consider the facts given below.

I. Sushma finished her graduation five years ago form today.

II. Geeta did her graduation three years before Sushma and joined the teaching profession the same year that Sushma entered college.

III. Aarti did her graduation three years after Sushma.

IV. Geeta taught both Sushma and Aarti during their graduation. If Geeta is still teaching, for how many years has Geeta been a teacher?

(A) 3 years (B) 5 years

Direction for the Problem 11 to 14: Answer the question based on the following on the following information.

A group of three persons must be selected from six individuals – Keshto, Omprakash, Sanjiv Kumar, Tuntun, Vijayendra and Wahid, according to the following conditions.

I. Either Omprakash or Vijayendra must be selected but neither Vijayendra nor Sanjiv Kumar can be selected with Omprakash.

II. Either Sanjiv Kumar or Keshto or both must be selected.

Problem 11: If Vijayendra is not selected, which pair of individuals must be among those selected?

(A) Omprakash and Wahida (B) Tuntun and Keshto

Problem 12: Which of the following is an acceptable selection of persons?

A. Omprakash, Vijayendra and Tuntun

B. Keshto, Tuntun and Sanjiv Kumar

C. Keshto, Omprakash and Snajiv Kumar

D. Keshto, Sanjiv Kumar and Vijayendra

Problem 13: If Sanjiv Kumar is selected, which of the following individuals must also be among the people selected?

(A) Wahida (B) Vijayendra

Problem 14: Which of the following pairs of persons cannot both be among the persons selected?

(A) Vijayendra and Wahida (B) Omprakash and Wahida

Direction for the Problem 15 to 19: Answer the question based on the following on the following information.

I. C must speak in the either time slot 1 or time slot 7

II. A must speak immediately before or immediately after D speaks.

III. F must speak in the fourth time slot.

IV. D must speak sometime before B speaks.

Problem 15: If G speaks at position 7, any of the following pairs of speakers could speak in time slots immediately adjacent to each other except

(A) F and E (B) C and A

Problem 16: Which of the following must be true?

A. G speaks sometime before F speaks

B. C speaks sometime before D speaks

C. A speaks sometime before F speaks

D. A speaks sometime before B speaks

Problem 17: If E wants to speak in the second time slot, then there will be a total of how many scheduling possibilities from which to select the schedule of speakers?

(A) 6 (B) 1

Problem 18: If B speaks immediately before F speaks, which of the following could be true?

A. D speaks in the third time slot

B. G speaks in the sixth time slot

C. C speaks in the first time slot

D. A speaks in the fifth time slot

Problem 19: If E speaks sometime before A speaks, which of the following must be true?

A. G speaks sometime before C speaks

B. E speaks sometime before G speaks

C. D speaks sometime before F speaks

D. F speaks sometime before B speaks

1^st shopkeeper: The dog was black. It had no collar.

2^nd shopkeeper: The dog was black. It had a short tail.

3^rd Shopkeeper: The dog was white. It had a collar.

Therefore the dog was

(A) white with a short tail and no collar.

(B) black with a long tail and a collar.

(D) white with a long tail and no collar.
Solutions

Solutions 1-4

Let us put together all the constrains in a diagram

Fi>Da (if Fiza goes, she wants David)

Da>Kv (if David goes, he wants Kavita)

Sh<>Rh (Shyam and Rahim want to go together)

Ra>Pe (if Ram goes, he wants Peter)

Da><Pe (David and Peter would not go together)

Ra><Sh (Ram and Shyam don’t go together)

Solution 1: Both the ladies – Fiza and Kavita can go only with David. But when

David goes then none of Ram, Shyam, Rahim or Peter can go. Hence, such a combination is not possible.

Ans=(B)

Solution2: Checking the options, all the groups fail one or more constraints.

Ans=(D)

Solution3: Checking the options ,Only feasible group of four Shyam, David, Kavita and Rahim.

Ans=(B)

Solution4: Checking the options , option (C) is true.

Ans=(C)

Solution 5: The best possible plan for Amit is

III

Sati

Savitri

Tipu

Sultan

Hayavar-

dhana

Sundar

Kand

Lunch

Jhansi

Ki Rani

Nagaman-

dala

(9-10)

(10-11)

(11-12)

(12-12.30)

(12.30-1.30)

(1.30-2.00)

(2-3)

Ans=(D)

Solutions 6 to 9

We can put a linear arrangement like below based on data given.(From I and II)

1	2	3	4	5	6	7
As/Es	Si				Si	As/Es

Now see the information in each question

Solution6: Under the given conditions, the arrangement would be

1	2	3	4	5	6	7
As	Si	Ba/Acc	Indica	Ba/Acc	pa	Es

So Indica would be in Row 4.

Ans=(A)

Solution7: Using the given conditions, see main table. If Astra is in Row 7 then, Escort would be in Row 1 and Indica in Row 2.

Ans=(C)

So, According to the given conditions, the only valid possible option would be Siena, Accent, Indica, Baleno and Palio

Ans=(C)

Solution9: Following same approach as in Problem 8, Undre the given conditions, the only valid option out of the given ones is: Escort, Indica, Baleno, Accent, Palio, Siena, Astra.

Ans=(B).

Ans=(C)

Did you notice one thing? Data in points 3 and 4 was totally useless and is called Superfluous data. The idea is to fool student and make him think where to use it.

Solutions 11 to 14

Using abbreviations for various names and classifying the given conditions, we get the following.

I. Either O or V must be selected but neither V or S with O, i.e. VO and SO is possible.

II. Either S or K or both SK must be selected.

Solution 11: Using the given conditions, we get:

If V is not selected, then O must be selected (according to condition I).

Now using condition II V, S or K or SK must be selected along with O, but as given S cannot be with O. Therefore, K has to be there with O. Hence, Om Prakash and Keshto must be selected.

Ans=(D)

Solution13: According to the given conditions, if S is selected, then V must be selected and not Q. Therefore, V must be selected if S is selected

Ans=(B).

Solution 14: As, one from O and P have to be selected and one from S and K have to be selected, we can never select T and W together.

Ans=(D)

Solution15: As G speaks in time slot 7, and then C will be first slot (according to condition I). and F always speak in slot 4. So 1-4-7 is taken.

Now, as A has to speak adjacent to D, and D has to speak before B, AD would speak in 2-3slots. Now we have to check the options.

In any case, AB can never speak together.

Ans=(C)

Solution16: Remember not to take any info from problem 15. We know only what is given in common directions.

We have to check the options first.

In every combination, A speaks sometimes before B speaks, will always be true. (As A and D speak adjacent and D speaks before B)

Ans=(D)

Solution17: If E wants to speak in the second time slot, then according to the given conditions, we have these restrictions.

As A and D have to speak together, they can speak in 5-6 or 6-7 positions. But As D have to speak before B, A and D Have share 5-6. So B would speak at 7 and thus C would speak at 1.

So we have C at 1, E at 2, G at 3, F at 4, A and D at 5 or 6 and B at Seven So, only two combinations are possible.

(i) C/1, E/2, G/3, F/4, D/5, A/6, B/7.

(ii) C/1, E/2, G/3, F/4, A/5, D/6, B/7.

Ans=(C)

Solution18: If B speaks before F (slot 4), we have these restrictions.

As D has to speak before B and A and D are adjacent, B would speak at a slot 3. A and D at 1 and 2 slots and so C at 7.

Now, check for various options,

then only statement which would be true out of the given options is G speaks in the sixth time slot and one of that arrangement can be

A/1, D/2, B/3, F/4, E/5, G/6, C/7.

Ans=(B)

Solution19: Using the given conditions that E speaks sometime before A speaks, we get the condition, F speaks sometime before B speaks, must be true.

Ans=(D)

Solution20: Based on the statements, only two possibilities are there.

If 1^st shopkeeper is correct about color, then dog would be Black with collar.

if Ist shopkeeper lies about color, then dog would be white with no color and short tail

Checking the options now, Option A is correct

Ans=(A)

PROBABILITY

May 29, 2009 by fundoogyan

Q1. A dice is thrown once, what is the probability of 3 showing up?

Ans1. Total outcomes = 6, favorable outcomes = 1

Probability = 1/6

Q2. Two dice are thrown, which event is more probable a score of 11 or 4?

Ans2. When two dice are thrown total outcomes are 36

For 11, favorable outcomes = 2 (6, 5) (5, 6),

Therefore probability = 2/36 = 1/18

For 11, favorable outcomes = 2 (6, 5) (5, 6),

Therefore probability = 2/36 = 1/18

Q3. Two dice are thrown, what is the probability of getting two 6’s?

Ans3. Here favorable outcome is one (6,6)

Total outcomes = 36

Probability = 1/36

Q4. Two coins are tossed. What is the probability of having 2 heads?

Ans4. The Total outcomes are 4 (HH , HT , TH , TT)

The favorable outcomes is 1 (HH)

Probability = ¼

Q5. A card is selected from a pack of 52 cards. Find the probability that it is an ace or a spade?

Ans5. There are 13 spades which includes one ace, and another 3 aces in the deck

Therefore, favorable outcomes = 16

Total Outcomes = 52

Probability = 16/52 = 4/13

Q6. A card is selected from a pack of 52 cards .find the probability that it is a spade or a ace or a king.

Ans6. Number of spades = 13, number of aces = 4, number of kings = 4

Now spades have one king and one ace, so

Number of aces left = 3, Number of kings left = 3

Probability = 13/52 + 3/52 + 3/52 = 19/52

Q7. A bag contains 2 violet, 3 black and 4 green balls. Find the probability that a ball drawn at random will be violet or green.

Ans 7. Total balls are 9, with violet = 2, Black = 3 and Green = 4

Probability of violet ball = 2/9

Probability of green ball = 4/9

P (violet or green) = 6/9

Q8. A bag contains 3 violet, 3 black and 3 green balls. If three balls are taken out with replacement. Find the probability of:

1. All are green

2. None is green

3. All are of same colour

Ans8. Total balls are 10, with violet = 3, Black = 3 and Green = 3

The experiment is happening with replacement, which means each time a ball is taken out it is being replaced, so total balls and respective number of balls stay the same

Probability of violet ball = 3/9 = 1/3

Probability of black ball = 3/9 = 1/3

Probability of green ball = 3/9 = 1/3

1. All are green

P(all green) = 1/3 x 1/3 x 1/3 = 1/27

2. None is green

Now P(not green) = 6/9 = 2/3

And P(none are green) = 2/3 x 2/3 x 2/3 = 8/27

3. All are same colour

All are same colour means = 3 violets or 3 blacks or 3 greens

= (1/3 x 1/3 x 1/3) + (1/3 x 1/3 x 1/3) + (1/3 x 1/3 x 1/3)

= 3/27 = 1/9

Q9. A bag contains 3 violet, 3 black and 3 green balls. If three balls are taken out without replacement. Find the probability of:

1. All are green

2. None is green

3. All are of same colour

Ans9. Total balls are 10, with violet = 3, Black = 3 and Green = 3

The experiment is happening without replacement, which means each time a ball is taken out it is not being replaced, so total balls and respective number of balls are reducing as balls are being taken out

Probability of violet ball = 3/9

Probability of black ball = 3/9

Probability of green ball = 3/9

1. All are green

P(all green) = 3/9 x 2/8 x 1/7 = 6/504

2. None is green

Now P(not green) = 6/9

And P(none are green) = 6/9 x 5/8 x 4/7 = 120/504

3. All are same colour

All are same colour means = 3 violets or 3 blacks or 3 greens

= (3/9 x 2/8 x 1/7) + (3/9 x 2/8 x 1/7) + (3/9 x 2/8 x 1/7)

= 18/504

Q10. Ravi is going for a blood test; his chance of being positive for a disease is 0.1, to confirm again and again, he takes three tests. What is the probability that

1. He will be positive in all the tests

2. He is be positive at least once

Ans10. Chance of being positive = 0.1

Chance of being positive in three tests = 0.1 x 0.1 x 0.1 = 0.001

Chance of being negative = 1 – 0.1 = 0.9

Chance of being negative in three tests = 0.9 x 0.9 x 0.9 = 0.729

Chance of being positive at least once = 1 – Chance being negative in all tests

= 1 – 0.729 = 0.271

Q11. There are 6 blue marbles and 4 red marbles. What is the probability of your drawing a blue marble and then my drawing a red one?

Ans11. The language is everything in a question; student should be able to understand that this is a question of conditional probability. Here, what is chance of drawing a red ball, when a blue ball has already been drawn?

Which is P(R/B), which is P(R and B)/P(R)

Now P(R and B) = 6/10 x 4/9 = 24/90 = 4/15

And P(R) = 6/10, therefore P(R/B) = (4/15)/(6/10) = 4/9

Q12. Assume that a test to detect a disease whose prevalence is (1/1000) has a false positive rate of 5% and a true positive rate of 100%. What is the probability that a person found to have a positive result actually has the disease assuming that you know nothing about the person’s symptoms?

Ans12. H = has the disease, P = Test result is positive

As prevalence is 1/1000, therefore P(H) = .001, therefore P(Not H) = .999

As false rate is 5%, P(P/not H) = 0.05

And true positive is 100%, therefore P(T/H) = 1.00

Now we have to find if person has the disease in case he has tested positive already, which is P(H/P)

P(H/P) = P(H and P)/P(P)

Now P(H and P) = 0.001 x 1 = 0.001

P(P) = P(P and H) + P(P and not H)

= 0.001 + (.999)x (.05)

= .05095

P (H/P) = .001/.05095 = .019627

Q13. A box contains 6 white balls and 4 black balls and another box contains 4 white balls and 6 black balls. Find the probability that a ball selected from one of the box is a white ball.

Ans 13. Now, In Box I, 6 white and 4 black balls are there

Probability of white ball from box I = 6/10

Probability of selecting box I = ½

P(white ball from box I) = ½ x 6/10 = 6/20

Now, In Box II, 4 white and 6 black balls are there

Probability of white ball from box I = 4/10

Probability of selecting box I = ½

P(white ball from box II) = ½ x 4/10 = 4/20

Therefore, P(white ball from box I or box II) = 4/20 + 6/20 =1/2

Q14. A box contains 6 white balls and 4 black balls and another box contains 4 white balls and 6 black balls. Find the probability of selecting a white ball from box I.

Ans14. Students should be able to understand the difference in this question and last question, in this question the chance of drawing a white ball from Box I has been asked, which is clear case of conditional probability.

Now, In Box I, 6 white and 4 black balls are there

Probability of white ball from box I = 6/10

Probability of selecting box I = ½

P(white ball from box I) = ½ x 6/10 = 6/20

Now, In Box II, 4 white and 6 black balls are there

Probability of white ball from box I = 4/10

Probability of selecting box I = ½

P(white ball from box II) = ½ x 4/10 = 4/20

Therefore, P(white ball from box I or box II) = 4/20 + 6/20 =1/2

Here P(Box I/white ball) = P(Box I and white ball) / P(white ball)

= (6/20) / (1/2) = 6/40 = 3/20

Q15. From a deck of 52 cards, one card is lost; the next two cards drawn are spades, what is the probability that the lost card was a spade?

Ans15. There are 52 cards, In case the card lost was a spade, then its probability at that time would be 13/52, the other cards to be spades will be 12/51 and 11/50. In case that card is not spade, the probability of other cards being spade is 13/52 and 12/51 (students may think how this is possible). From here the probability of third card to be spade is 11/50

Alternatively, assume you got the third card lying some where around, you added the card to the deck without seeing which card it is, now what is the probability of card being spade, again 11/50.

Q16. In GMAT verbal section there are 41 questions. Each question has 5 options out of which only one is correct. If someone clicks answers at random, what is the probability that he will get 20 out of the 41 questions correct ?

Ans16. P(x) = nCx × px × q(n-x)

41C20 × (1/5)20 × (4/5)21

Q17.The probability that a graduate student being male is 0.25 and that being female is 0.75. The probability that a male student passes the course is 0.7 and that a female student does it is 0.80. A student selected at random is found to have completed the course. What is the probability that the student is (i) male and (ii) female?

Ans17. Probability of being a male = 0.25

Probability of being a Female = 0.75

Probability for female to complete the course = 0.80

Probability for male to complete the course = 0.70

Probability for a male who has done the course = 0.25 × 0.7 = 0.175

Probability for a female who has done the course = 0.75 × 0.8 = 0.6

Probability for a person who has done the course is either a male finishing the course or a female finishing the course = 0.25 × 0.7 + 0.75 × 0.8 = 0.775

Probability a student being male who has finished the course = 0.175/0.775 = 0.225

Probability a student being Female who has finished the course = 0.6/0.775 = 0.774

Q18. A, B and C in order cut a pack of cards replacing them after each cut, on condition that first who cuts a spade shall win a prize. find their respective chances?

Ans18. Probability of A winning = 1/4
Probability of B winning = Probability of A losing * Probability of B winning = ¾ X 4/4 = 3/16
Probability of C winning = Probability of B losing * Probability of C winning = 13/16 X 1/4 = 13/64

Q19. A standard deck of 52 cards is shuffled and the cards are dealt face up one at a time until an ace appears. Show that the probability of getting the first ace on or before the ninth card is greater than 50%.

Ans19. The probability of getting no aces in the first nine cards is 3/8…… = 40/44 = 43/92

Now 43/92 is less than 1/2, therefore getting first ace will be 1 – no ace, so greater than 50%

Q20. To the nearest percent, the probability that any one person selected at random was born on a Monday is 14 percent. What is the probability, to the nearest percent, that of any seven persons chosen at random, exactly one was born on a Monday?

Ans20. The probability of getting a Monday is 1/7 and the probability of getting 6 “non-Mondays” is (6/7)⁶. The number of ways this can happen is a combination of “7 picking 1,” i.e. 7C1 = 7.

Therefore,

P(exactly 1 Monday out of 7 people) = 7C1 * (1/7)1 *(6/7)6 = .3966, so 40 %

Q21. Mr. Ryan kept two matchboxes, one in each pocket. Each box contained exactly n matches. Whenever he wanted a match he reached at random into one of his pockets. When he found that the box he picked was empty, what is the probability that the other one has exactly k matches (k = < n)?

Ans21. If k matches remain in the other box, then n–k matches have been selected from that box. Suppose Mr. Ryan attempts to select the (n + 1)st match from the box in his left pocket. Then a total of 2n – k + 1 selections have been made; thus we have 22n – k+1 ways in which the matches can be selected. Of these, 2n – kCn (where mCr = m!/[r! (m – r)!] is the number of ways of choosing r outcomes out of m possibilities, ignoring order) combinations are such that the (n + 1)st selection is from his left pocket.

Therefore the probability the professor will open an empty box from his left pocket is 2n – kCn/22n–k+1. Of course, there is an equal probability that he will open an empty box from his right pocket.

Therefore the probability that the other box currently contains k matches is 2n – kCn/22n-k.

Q22. Which is more likely, to get at least one double six in 24 throws of a pair of dice or to get at least one six in 4 throws of a die?

Ans22. P (no double sixes in 24 throws) = (35/36)24 = 0.509

P (at least one double six) = 1 – 0.509 = 0.491

P (No sixes in four throws) = (5/6)4 = 0.482

P (at least one six in four throws) = 1 – 0.482 = 0.518, this is more likely.

PERMUTATION AND COMBINATION

May 29, 2009 by fundoogyan

Q1. Find the number of even natural numbers, which have three digits?

Ans1. The total number of digits are 10 (0 to 9), The three digit number has three places to fill. The first place can be filled by 9 (excluding zero), the second by all 10 and to make it even the third has can only be filled by 5 (0, 2, 4, 6, 8).

9 10 5

Total three digit even numbers = 9x10x5 = 450

Students should not get confused here with the concept of repeat and non-repeat usage of digits, since the question asks for the three digits numbers present in the natural numbers. Now in natural numbers, numbers are formed from all digits with repetitions, so there should be no confusion.

Q2. In how many different ways six questions of true false type can be answered?

Ans2. Each question can be answered in two ways – true or false, which is 2 ways, as total number of questions is six, so all six can be answered in 2x2x2x2x2x2 = 64 ways

Q3. In how many different ways six questions of true false type can be answered incorrectly?

Ans3. Here incorrectly means, the options which are other than where all the answers are correct. The option where all the answers are correct is only 1. Since total number of ways of answering is 64(from last example). So total no. of ways are 64 - 1 = 63 ways

Q4. Find the number of permutations of the letters of the word custom such that no repetitions are there. How many words beginning with M? How many digits begin with M and end with S?

Ans4. The word “custom” has all distinct alphabets, no repeats. The permutation of 6 objects to be placed in six places is 6! = 720 ways

If the word has to begin with M, then first place is locked with M, so there are five places left to be filled with five alphabets = 5! = 120 ways

If the word has to begin with M, and end with S, then these two positions are locked, so there are four places to be filled with four alphabets = 4! = 24 ways

Q5. Find the number of ways in which the letters of the word “epidemic” can be arranged?

Ans5. The word “epidemic” has total 8 alphabets with “i” repeating two times and “e” repeating two times, so by the formula n! / (p! q! r!..), since there are two repeats, the number of ways is

8!/(2!2!) = 10080

Q6. Find the number of ways in which the letters of the word India can be arranged?

Ans6. The word “India” has total 5 alphabets with “i” repeating two times, so by the formula n! / (p! q! r!..), since there is one repeat, the number of ways is

5!/2! = 60

Q7. In how many ways can the six letters A, B, C, D, E, F can be arranged such that B and C always come together?

Ans7. Since B and C have to together, therefore it becomes set of 5 things

5 places 5 things is given by 5P₅ ways = 120 ways

Now B and C can arrange themselves in two ways BC and CB

Therefore total ways = 120 x 2 = 240 ways

Q8. If Nirula’s offers 31 flavours of ice cream cones in sizes small, medium and large, how many different selections of cones are possible? if 4 toppings are available to put on a cone, and a cone can be bought without a topping, how many different selections of ice cream cones are available?

Ans 8 Since there are 31 flavours and three sizes, the total number of selections will be = 31 x 3 = 93.

If there are four toppings, they can be taken in 2⁴, since there are 93 ways of taking the ice-cream itself total number of ways = 93 x 2⁴= 1488

Q9. How many 8 letter words can be constructed using 26 letters of the alphabet if each word contains 3 vowels?

Ans9. Out of eight spaces 3 are to be reserved for vowels, they can be selected in 8C₃ ways. Now there are 5 vowels and each can take any of the three positions = 5³ and there are 21 consonants which can take rest of the 5 positions = 25⁵

Total number of ways = 8C₃x 5³x 25⁵= 28588707000 ways

Q10. In how many ways can 5 boys and 3 girls be made to stand in a row such that no two girls are together?

Ans10. Since no two girls can stand together they take places between the boys, before after and between, there are six positions around 5 boys, where three girls have to fit, therefore it is 6P₃. Now boys are left to take 5 positions (for 5 places) which is 5!

Total ways = 6P₃+ 5! = 14400 (why add, as both the things are independent)

(permutationcombination theory Fig 3)

Q11. In how many ways can a committee of 3 men and 3 ladies be appointed from 6 men and 4 ladies?

Ans11. 6 Men, three to be selected, which is 6C₃

4 Ladies, two to be selected, which is 4C₂

Total ways of selection = 6C₃x 4C₂ = 120

Q12. From 6 men and 4 ladies five people have to be chosen with at least one lady?

Ans12. 6 Men + 4 Ladies, 10 people, five to be selected, which is 10C₅

The scenario in which no women is selected, 6 men, 5 to be selected 6C₅

Here 10C₅- 6C₅is where at least one lady is selected = 246 ways

Q13. In a conference of 9 schools, how many intra conference football games are played during the season if the teams all play each other exactly once?

Ans13. There are total of nine teams and two teams required to play a match, so the total number of ways is 9C₂= 36 games

Q14. How many different signals can be made by using at least three distinct flags if there are five different flags from which to select?

Ans14. At least three distinct flags are to be chosen from give 5 flags, which is:

5P₃ + 5P₄ + 5P₅= 5!/2! + 5!/1! + 5!/0! = 300 signals

Q15. In how many ways 11 cricketers can be chosen from 6 bowlers , 4 wicket keepers and 11 batsmen to give a majority of batsmen if at least 4 bowlers are to be included and there is one wicket keeper?

Ans 15. One wicketkeeper has to be choose out of 4 which is 4C₁ = 4

Now since 11 players have to be selected, post wicketkeeper selection only 10 are left, out of which at least 4 bowlers have to be selected and batsman should have majority, so they will be 6, and that’s the only way.

Ways of selecting four bowlers out of 6 = 6C₄

Ways of selecting Six batsmen out of 11 = 11C₆

Total ways = 11C₆x 6C₄x 4 = 27720

Q16. If 6 balls of different colors - black, white, yellow, green, blue, violet are to be arranged in a row that the black and white balls may never come together?

Ans16. There are 6 balls, in case black and while ball come together, then the scenario will be of 5 objects and 5 places = 5P₅ x 2 (since black and white can change positions) = 240

Total arrangements = 6P₆ = 6! = 720

Arrangements where black and white ball are not together = total arrangements – the case black and while ball come together = 720 – 240 = 480

Q17. A Teacher with 8 students takes three at a time to the computer room, as often as he can without taking the same three students together more than once. How often will he go, and how often will each student go?

Ans17. The visits of the teacher = number of groups, as he visits with each group

Which is three students out of eight = 8C₃ = 56

The student cannot be paired with the other two of his group, therefore he will go as many times as the other two in his group can be taken out of the remaining 7 students other than him = 7C2 = 21

Think about the second point, you will get it.

Q18. A team consists of 8 men, 3 of whom can only work in one city and 2 only in the other. Find the number of ways in which the team can be formed. There have to be equal number of men in both cities. Also find the number of ways in which the team can be formed and internally rearranged.

Ans18. If 3 are fixed in one city and 2 are fixed in other, out of eight five positions are closed. Since teams have to be of 4 people each, one position is open in one city and two positions are open in two cities. So there are three positions and three people, since there is only one position in one city, only one can fitted there, and all three people can be fitted there, those are the only three ways. Therefore there are 3 ways of making the teams.

The two teams have four members each and both can be arranged in 4! Ways = 24

Total ways = 24 x 24 x 3 = 1728.

Q19. A Bus goes from Delhi to Udaipur(last stop) stops at 8 intermediate terminals. 5 persons board the bus during the journey with 5 different tickets. How many different sets of tickets may they have had?

Ans19. There are 8 intermediate terminals, if a person boards from first terminal after Delhi, he can buy 8 type of tickets up to Udaipur. Similarly the next person boarding from next terminal can buy 7 types of tickets and so on.

So type of tickets = 8+7+6+5+4+3+2+1 = 36

Now the five people who boarder the bus may have any of the 36 type of tickets = 36C₅ = 376992 sets

Q20. Find the number of diagonals in a decagon.

Ans20. The number of sides in a decagon is 10; the number of ways two sides can meet are 10C₂ = 45, there are 10 outer sides, therefore 45 -10 = 35

Q21. Find the number of triangles in an octagon.

Ans21. The number of sides in an octagon is 8; the number of ways three sides can meet to form a triangle are 8C₃ = 56, there are 56 triangles

Q22. There are 20 points in which 6 points which are collinear. How many straight lines can be formed by joining them?

Ans22. Total lines (meeting of two points) for 20 points = 20C₂

Collinear points = 6, therefore only one line for 6C₂

Therefore total lines = 20C₂- 6C₂ + 1 = 176 lines

Q23. If you have 100 people at a party, and everyone shakes hands with everyone else, how many handshakes take place?

Ans23. There are two people required for a handshake therefore 100C₂= 4950

Q24. Four figures are to be inserted into a six-page essay, in a given order. One page may contain at most two figures. How many different ways are there to assign page numbers to the figures under these restrictions?

Ans24. In the case that one page may contain 1 figure, the number of ways are 6C4. In the case where 1 page only can contain 2 figures, the ways are 6C3 Consider 2 pages can contain 2 figures then 6C2. The total number of ways are 6C4 + 6C3 + 6C2 = 50 ways.

Q25. There are n Railway stations. Ticket facility is 10 available between every two stations. Recently, m new stations are built so that 42 new tickets are to be printed. How many stations were there? How many are newly Constructed.

Ans25. For the m new stations you will need to print 42 new tickets = 21 new tickets one way. Now, 21 = trains between 1 each of n stations and 1 of m stations = m × n + trains between each of the m new stations = mC2

21 = m*n + mC2 = m(2n + m – 1)/2

3*7 = m(2n + m–1)/2

Taking m = 3

7 = (2n + 3–1)/2

14 = 2n + 2

n = 6

If we take m = 7, (2n + 6) = 6  n = 0

Therefore, there were no stations initially and 7 new stations were added or 6 originally, 3 new were added.

Q26. How many ways can you buy a dozen donuts from an unlimited supply of 5 types of donuts?

Ans26. The way here is to think of how many ways can you line up 12 x’s and 4/’s. Why? There is a one to one correspondence between such lineups and possible purchases — xx/xxx//xxxxxx/x corresponds to 2 of type 1, 3 of type 2, 0 of type 3, 6 of type 4 and 1 of type 5 etc. Thus we need to count the number of such lineups. Equivalently, how many ‘words’ can we make from 4/’s and 12 x’s? This is fairly easy as we have 16 spots to fill and 4 of them have to be chosen to be occupied by a/. Thus there are 16C4*12C12 = [16]C[4] = 1820 ways to buy the donuts.

What if the question says that you must purchase at least one of each type? Then the answer is only 11C4*7C7 = 11C4 ways.

Q27. A company president is deciding how to fill three vice-presidencies in the company: VP-Marketing, VP-Finance, and VP-Production. Twelve executives are eligible and qualified for promotion, and each could fill any of the three positions. In how many ways can the positions be filled?

Ans27. The decision-maker first selects three people from among the twelve, not yet thinking about their job assignments (order not important). This can be done C(12,3) = 220 ways. Then the decision-maker assigns the three chosen people to the three jobs (order important). This can be done P(3,3) = 6 ways. So the total number of ways is 220 x 6 = 1,320.

Q28. The Lottery Commission is considering a new game in which five balls would be withdrawn from a box containing 10 balls, numbered 0 to 9. The five balls would come out of the box at nearly the same time, as they do in the current Lotto game, in which six balls come out of a box into a tube at nearly the same time. In this new game, however, the winning ticket must have the five lucky numbers in the same order as they came out of the box. What is the chance of winning with a single five-number ticket?

Ans28. The order is important, duplicates are not possible. Therefore 10P5 = 30,240.

Formula-free sequential method: the first number has 10 possibilities, the second number has 9, the third number has 8, the fourth number has 7, and the fifth number has 6. 10 × 9 × 8 × 7 × 6 = 30, 240. So the probability of winning is 1/30,240 = 0.000033069.

Q29. A new flag is to be designed with six vertical stripes using some or all of the colors yellow, green, blue and red. Then, find the number of ways this can be done such that no two adjacent stripes have the same color. (CAT 2003).

Ans29. The first stripe can be chosen as 4C1 = 4 ways, as one out of four colors is to be taken, rest all are 3C1 = 3 ways as the adjacent color cannot be repeated, which leaves one color to be chosen from three colors. Therefore total ways = 4 × 3 × 3 × 3 × 3 × 3 = 972 ways

Q30. What is the maximum number of points of intersections of:

(a) Five circles

(b) Five straight lines

Ans30. (a) For circles permutation will be used, as arrangement is important 5P2 = 20

(b) For lines arrangement is irrelevant, therefore use of combination 5C2 = 10

(d) Three circles = 3P2 = 6, three line = 3C2 = 3, Now three lines can maximum cut three circle in 18 times(as one line can cut three circles 6 times), so total intersections = 18 + 6 + 3 = 27

Q31. In a Ranji Cup final, team A and team B play until one team wins 4 games. The sequence of game winners is designated by letter; for example, ABBBB means team A won the first game and team B won the next four games. How many different Ranji Cup finals are possible?

Ans31. The number of ways with B winning is shown in the table.

Combinations Number of Sequences BBB|B 1

ABBB|B 4!/[1!3!] = 4

AABBB|B 5!/[2!3!] = 10

AAABBB|B 6!/[3!3!] = 20

Total = 35

we get a further 35 possible sequences with A winning, so the total number of sequences is 35 + 35 = 70 There are 70 possible sequences for the Ranji Cup finals.

Q32. Ravi’s family consists of a grandfather, ‘X’ sons and daughters and ‘Y’ grand children. They are to be seated in a row for dinner. The grandchildren wish to occupy the Y seats at each end and the grandfather refuses to have a grandchild on either side of him., In how many way’s can the Ravi’s family be made to sit?

Ans32. Total Seats = X + Y + 1

Grandchildren (Y) sit together on Y seats in Y! ways

Since grandfather does not want to sit with any grandchildren, he cannot take any of the Y seats and 2 seats on either side of Y seats. Which is X + 1 – 2 = X – 1 seats left for him to sit, so one person to be accommodated in X – 1 seats is done in X – 1 ways. The rest X people can sit on X seats in X!

So total ways = X! Y! (X-1)

Q33. There are 720 (or 6!) permutations of the digits 1, 2, 3, 4, 5, 6. Suppose these permutations are arranged from smallest to largest numerical value, beginning with 123456 and ending with 654321, then what number falls on the 124th position?

Ans33. Among the 6! Permutations, there are 5! = 120 beginning with 1. They would be numbered from 1 to 120. The next 120, including the 124th, would begin with 2. Among the 5! Permutations beginning with 2, there are 4! = 24, including the 124th, which have second digit 1. Among the 4! Permutations beginning with 21, there are 3! = 6, including the 124th, which have third digit 3. Among the 3! Permutations beginning with 213, there are 2! = 2 with fourth digit 4 (numbers 121 and 122), 2 with fourth digit 5 (numbers 123 and 124), and 2 with fourth digit 6 (numbers 125 and 126). The 124th has fourth digit 5. Finally, among the 2! Permutations beginning with 2135, there is one with 5th digit 4 (number 123) and one with 5th digit 6 (number 124). The 124th is the latter one. Thus the 124th number is 213564.

Q34. Find the number of ways’ in which 9 identical balls can be placed in three identical drums.(assuming each drum can take all nine balls)

Ans34. The balls can go in any drum, and their being in one drum or another does not make a difference since both balls and drums are identical. This is the important point to remember. Now, one situation is that all nine balls in one drum, again remember it does not matter which drum they are in, that’s one way, in this way the other two drums are empty. Now, if one drum has no ball, then other drums could have four combinations to make nine (as total is 9) out of two digits (8, 1), (7, 2), (6, 3), (5, 4), now these are four ways. Now, if one drum has one ball, then other drums could have four combinations to make eight (as total is 9) out of two digits(7, 1), (6, 2), (5, 3), (4, 4), now these are four ways. Now, if one drum has two balls, then other drums could again have three combinations to make seven (as total is 9) out of two digits (6, 1), (5, 2), (4, 3), out of which (6, 1) is a repetition of one of the previous cases, so in all two ways. Finally the only way in which all drums have three balls. So total cases = 1 + 4 + 4 + 2 + 1 = 12 ways

Q35. In how many ways a cricketer can make a century with fours and sixes only?

Ans35. Taking the extreme case, with all fours(25) he can make a century, with only sixes he cannot do, as 100 is not a multiple of 6, so with 16 sixes and one four he can do it. Now from here keep reducing six and see if the number left is a multiple of four that’s a case. The next case is 14 sixes and 4 fours, the next is 12 sixes and 7 fours, and so on, you will find that there are 9 ways in all.

Q36. At a party, the superintendent of an apartment building tells you that his building has seven elevators. Each elevator stops on at most 3 floors. He also tells you that if you take the right elevator, you can get to any one floor from any other floor without changing elevators. What is the greatest number of floors that the building could have?

Ans36. Since the elevator stops at three floors, we can make a combination of the same:

1 2 3

1 4 5

1 6 7

2 4 6

2 5 7

3 4 7

3 5 6

Therefore for one elevator there could be 7 floors, so for seven elevators there could be maximum of 21 floors (7 × 3)

TIME SPEED AND DISTANCE

May 29, 2009 by fundoogyan

Q1. Ram and Ravi are 100 km apart and started to walk towards each other at 10 am. Ram walked at the rate of 5 km/hr and Ravi at 10 km/hr. at what time will they meet?

Ans. Relative speed = 5 + 10 = 15 km/hr

Time to walk 100 km = 100/15 = 6.66 hours, which is 6 hours and 40 min(approx)

As they started at 10 am, therefore they meet after
10 + 6 hours and 40 min = 4:40 pm

Q2. Hari runs after Sam who is 200 m ahead of him. If speed of Hari is 15 km/hr and that of Sam is 10 km/hr, what will be the distance covered by Hari when he catches Sam?

Ans. Relative speed = 15 × 10 = 5 km/hr

Distance to be covered = 200 m = .2 km

Time taken = .2/5= 0.04 hrs

Distance covered = 15 × 0.04 = 0.6 km = 600 meters

Q3. Two trains, 200 and 160 meters long take a minute to cross each other while traveling in the same direction and take only 10 seconds when they cross in opposite directions. What are the speeds at which the trains are traveling?

Ans. Distance covered (sum of lengths of the train)

= 200 + 160 = 360 meters.

Let Train 1 be traveling at X m/sec and Train 2 be traveling at Y m/sec. (considering X > Y)

Now while traveling in same direction Time = (L₁ + L₂)/(X – Y)

Therefore 60 = 360/(X – Y), 60X – 60Y = 360

Now while traveling in opposite direction Time

= (L₁ + L₂)/(X + Y)

Therefore 10 = = 360/(X + Y), 10X +10Y = 360

Solving the two equations X = 21 m/sec

Y = 15 m/sec

Q4. A train traveling at 72 km/hr crosses a platform in 30 seconds and a man standing on the platform in 18 seconds. What is the length of the platform in meters?

Ans. Students may recall from the theory of the chapter, When the train crosses a man standing on a platform, the distance covered by the train is equal to the length of the train and when the same train crosses a platform, the distance covered by the train is equal to the length of the train plus the length of the platform.

Train takes 18 seconds to cross a man, which is basically to cover length of the train

Train takes 30 seconds to cross the platform, which is to cover length of the train and length of the platform. Therefore the extra 12 seconds are to cover the length of the platform. Therefore Length of platform

= (72 x 12)/3600 = 0.24 km = 240 meter.

Q5. Two trains starting from the same station and traveling in opposite directions are 228 km apart in 3 hours. Had they been traveling in same direction they would have been 33 km apart in the same time. What are their speeds?

Ans. They are 228 km apart in 3 hrs traveling in opposite direction

In one hour they will be 228/3 = 76 km = Sum of speeds of two trains

Assume speeds to be X km/hr and Y km/hr

Therefore X + Y = 76 km/hr

In the other case X – Y = 33/3 = 11 km/hr

Solving X = 43.5 km/hr, Y = 32.5 Km/hr

Q6. Two trains traveling in the opposite directions pass each other in 8 seconds. But when they travel in same direction at the same rates, the man in the faster train passes the other in 30 seconds. Find the lengths of the trains when their speeds are 45 km/hr and 35 km/hr.

Ans. When trains travel in opposite direction,

Time = (L₁ + L₂)/(X + Y)

Therefore, 8/3600 = (L₁ + L₂)/(45 + 35)

Lengths of train, L1 + L2 = 0.17778 km

= 177.78 meter

When trains travel in same direction, the man in the faster train passes the other in 30 seconds, the man only passes the length of the smaller train.

Therefore distance traveled by trains in 30 seconds = length of the smaller train

Relative speed = 45 – 35 = 10 km/hr

Distance traveled = 30/3600× 10

= 0.08334 km = 83.34 meter

Therefore length of longer train

= 177.78 – 83.34

= 94.44 meters

Q7. Two trains 150 miles distant travel towards each other along the same track, the first train at 60 km/hr, the second at 90 km/hr. A fly buzzes back and forth between the two trains until they collide. If the fly’s speed is 110km/hr, how far will it travel and how many rounds will it take?

Ans. Since the trains start 150 km apart and have a relative speed of 90 + 60 = 150 km/hr, they will meet in exactly 1 hour. The bird is flying at a speed of 110 km/hr, so in 1 hour it flies 110 km. The interesting part remains the number of rounds, which actually is an interesting physics problem with a infinite series being formed and the answer is infinity.

Q8. Two trains A and B each of length 100m travel in opposite directions in parallel tracks. The speeds are 20m/s and 30m/s respectively. A boy sitting in the front end of train A throws a ball to a boy sitting in the front end of train B when they are at the closest distance. The speed of the ball is 2m/s. The ball, instead of reaching the boy, hits the rear end of the train. Find the distance between the parallel tracks.

Ans. The boy throws the ball when trains are at closest distance, which is when front ends of trains meet; therefore boy throws perpendicular to the train, in line with the distance between the tracks.

Suppose from the time ball leaves boy’s hands to hitting the train’s rear is T

In this time T, Train A has traveled 20T

Train B has traveled 30T

Since ball hits the rear of train B, with length of trains being 30T + 20 T = 100, therefore T = 2 seconds

Since the ball travels at 2 m/s, and ball hitting the other train means that ball has traveled the distance between two tracks, assuming distance between tracks is D

Therefore 2T = D, D = 4 meters

Q9. Ram and Mohan start at the same time from A to B to go to B and A, a distance of 42 km at the rates of 4 km/hr and 3 km/hr. They meet at X, then go to B and A and return immediately and meet again at Y. find the distance XY

Ans. First Ram and Mohan meet at X, Distance = 42 km

First Ram and Mohan meet at Y,

Distance = 42 × 3 = 126 km

Speed of Ram = 4 km/hr,

Speed of Mohan = 3 km/hr

Ram travels AX distance and Mohan BX

Relative speed = 4 + 3 = 7 km/hr

Time when they meet first = 42/7 = 6 hrs

In 6 hrs Ram travels 6 × 4 = 24 km = AX

Therefore Mohan travels = 42 – 24 = 18 = BX

Time when they meet second = 126/7 = 18 hrs

In 18 hrs Ram travels 18 × 4 = 72 km

= AX + XB + BY

Since AX + XB = 42, BY = 30 and AY = 12

Since AX = 24, and AY = 12. XY = 12 km

Q10. There are two boats that start out on opposite sides of a river at the same time. Each one is heading across the river to the other side. They each go a constant speed throughout the entire problem (so ignore having to slow down to turn around, and ignore current, etc.), but they are not necessarily the same speed as each other. When each boat reaches its opposite bank, it immediately turns around and heads back to where it started. The boats thus pass each other twice. The first time they pass, they are 700 yards from one of the banks of the river. The second time they pass, they have each turned around after reaching their respective opposite shores and have started back toward where they each began. When they pass the second time, they are 300 yards from the other bank of the river. How wide is the river?

Ans. Assume total distance = X, and speeds of boats being S1 and S2 when the two boats meet for the first time,

Distance travel by boat I, D1 = 700 yards

Distance travel by boat II, D2 = X – 700 yards

Here Time is the same, therefore

T = D₁/S₁ and T = D₂/S₂

Therefore D₁/S₁ = D₂/S₂

Also D₁/D₂ = S₁/S₂ = 700/(700 – x)

Now, Boat I then continues on to the bank, which is (X – 700) yards away, and then it turns around and goes back 300 yards. After the first meeting, boat I travels (X – 700) + 300, which is (X – 400) yards. After the first meeting Boat II travels 700 yards to the bank and then turns around and travels back (X – 300) yards, which is (X + 400) yards, where it then meets Boat I again.

When the two boats meet for the second time after first meeting,

Distance travel by boat I, D1 = X – 400 yards

Distance travel by boat II, D2 = X + 400 yards

Here Time is the same, therefore

T = D₁/S₁ and T = D₂/S₂

Therefore D₁/S₁ = D₂/S₂

Also D₁/D₂ = S₁/S₂ = (x – 400)/(x + 400)

The first and second time ratios are the same, therefore

700/(x – 700) = (x – 400)/(x + 400)

Solving X = 1800 meters

Q11. Ravi can swim with the stream at the rate of 10 km/hr and 5 km/hr against the stream; find his speed in still water.

Ans. Upstream speed = 10 km/hr, Downstream speed is 5 km/hr

Using direct formula 1/2 (10 + 5) = 7.5 km/hr

Q12. Hari swims 20 km downstream a river in 5 hours and returns in 10 hours. What is his speed and speed of the stream?

Ans. Speed downstream

= 20/5 = 4 km/hr

= Speed of Hari + Speed of stream

Speed downstream

= 20/10 = 2 km/hr

= Speed of Hari – Speed of stream

From the two equations speed of Hari = 3 km/hr

Therefore speed of stream = 4 – 3 = 1 km/hr

Q13. A boat travels from point A to point B upstream and returns from point B to point A downstream. If the round trip takes the boat 5 hours and the distance between point A and point B is 120 km and the speed of the stream is 10 km/hr, what is the speed of the boat?

Ans. Total distance = 120 × 2 = 240, Total time = 5hrs

Average speed = 240/5 = 48 km/hr

Assuming speed of boat = X

Speed downstream

= Speed of Boat + Speed of stream

= X + 10

Speed downstream

= Speed of Boat – Speed of stream

= X – 10

Since average speed = 48, and using formula of average speed

48 = 2(X + 10 )(X – 10)/(X + 10) + (X – 10)

Solving for X, X = 50 km/hr

Q14. Mohan can beat Ravi by 50 m in a 1700 m race; Ravi can beat Shyam by 20 m in a 1700 m race. If Mohan and Shyam run 1700 m, by how much will Mohan win?

Ans. The students should be able to do this faster using simple logic:

Mohan can beat Ravi by 50 m, Mohan travels 1700, when Ravi travels 1650

Ravi can beat Shyam by 20m, Ravi travels 1700, when Shyam travels 1680

Therefore when Ravi travels 1 Shyam travels

Therefore when Ravi travels 1650 Shyam travels 1680/1700

1680/1700 × 1650 = 1630.5

Mohan beats Shyam by 1700 – 1630.5 = 69.5 meters

Q15. In a kilometer race, A can give B a start of 100 m or 15 seconds. How long does A take to complete the race?

Ans. In a 1000 meter race A gives B a start of 100 m or 15 seconds. This means that B takes 15 seconds to run 100 m. Therefore, B will take 150 (1000/100 x 15) seconds to run the stretch of 1000 meters. As A takes 15 seconds less than B, he will take 135 seconds to run the 1000 m.

Q16. Ravi can give anand a start of 20 seconds in a kilometer race. Ravi can give ranjan a start of 200 meters in the same kilometer race. And anand can give ranjan a start of 20 seconds in the same kilometer race. How long does Ravi take to run the kilometer?

Ans. Ravi can give anand a start of 20 seconds in a km race.

Let anand takes X seconds to run a km, then Ravi will take X – 20 seconds

Now, Anand can give ranjan a start of 20 seconds in a km race.

Let ranjan takes Y seconds to run a kilometer, then anand will takeY – 20 seconds

From the two X = Y – 20

Therefore Ravi can give ranjan a start of 40 seconds

Now, Ravi can give ranjan a start 200 meters or 40 seconds in a km race.

This essentially means that ranjan runs 200 meters in 40 seconds.

Therefore, ranjan will take 200 seconds to run a km.

Then ravi will take 200 – 40 = 160 seconds

Q17. Rohit and Ravi walk around a circular path of circumference 1000 meters. Rohit walks at speed of 100 m/min and Ravi at a speed of 50 m/min. if they start at the same point and walk in the same direction, when will they be together again? When will they meet at the starting point?

Ans. Rohit gains 50 meters in a minute over Ravi, so it will gain 1000 meters in 1000/50 = 20 minutes, so they will meet after 20 minutes

Time taken by Rohit to complete one round

= 1000/100 = 10 min

Time taken by Ravi to complete one round

= 1000/50 = 20 min

They meet again at starting point, LCM of 10 and 20 = 20 minutes

Q18. Three cyclists Raman, Mohan and Nitin ride around a circular course 85 km around at the rate of 8, 12 and 20 km an hour. Raman and Mohan ride in the same direction and Nitin in the opposite direction. In how many hours will they meet again?

Ans. Mohan will meet Raman in every

85/(12 – 8) = 85/ 4 hrs

Mohan and Nitin will meet in every

85/(12 + 20 ) = 85/32 hrs

Raman and Nitin will meet in every 85/(8 + 20)

= 85/28 hrs

They all will meet at the LCM of the three 85/4, 85/32 and 85/28

LCM is 85/4 hrs

Q19. At what time between 3 and 4 will the hands of a watch coincide?

Ans. At 3, the hour hand and the minute hand are at right angle and are 15 minute away, the minute hand gains 1 in 12/11 minute, it will gain 15 in 12/11 × 15 = 16.3 minutes.

Therefore hands will coincide 16.3 min past 3.

Q20. A clock loses 1% time during the first week and then gains 2% time during the next one week. If the clock was set right at 12 noon on a Sunday, what will be the time that the clock will show exactly 14 days from the time it was set right?

Ans. The clock loses 1% time during the first week. In a day there are 24 hours and in a week there are 7 days. Therefore, there are 7 × 24 = 168 hours in a week. Therefore time lost from noon to noon = 1/100 × 168 = 1.68 hrs (behind)

The clock gains 2% time during the first week. Therefore time gained from noon-first week to noon-second week = 2/100 × 168 = 3.36 hrs (Ahead)

Net change = 3.36 – 1.68

= 1.68 (ahead of 12 noon)

1.68 hours = 1 hour + 40 minutes + 48 seconds. Therefore time is 1:40:48 P.M.

Q21. A truck travels 15 mph for the first half of the distance of a trip. The driver wants to average 30 mph for the whole trip. How fast must he travel for the second half of the trip?

Ans21. This is more of a puzzle, assume any distance for the trip. If you said 30 miles, then the first half took 1 hour. But to average 30 mph for the total trip would require an hour. Therefore it has no solution and it is impossible!

Q22. A ship went on a voyage. After it had traveled 180 miles a plane started with 10 times the speed of the ship. Find the distance when they meet from starting point.

Ans22. Let’s say the ship is travelling x miles/hr and plane took t hours to meet the ship

Then, xt is the distance ship traveled after plane started, Let 10xt is the distance plane traveled. So we have, xt + 180 = 10xt

Solving xt = 20, therefore distance traveled by the ship = 180 + 20 = 200 miles

Q23. A can complete a piece of work in 4 days, B takes double the time taken by A, C takes double that of B, and D takes double that of C, to complete the same task. They are paired in-groups of two each, one pair takes 2/3rds the time needed by the second pair to complete the work. Which is the first pair? (CAT 2001)

Ans23. A takes 4 days, B takes 8, C takes 16, D takes 32, In one day A does 1/4th, B does 1/8th, C does 1/16th, D does 1/32th of work.

Now A and D work in a day = ¼ + 1/32 = 9/32 , so finish work in 32/9 days

A and C work in a day ¼ + 1/16 = 5/16 , so finish work in 16/5 days

A and B work in a day ¼ + 1/8 = 3/8, so finish work in 8/3 days

B and C work in a day 1/8 + 1/16 = 3/16, so finish work in 16/3 days

We can already see a relationship between pair of A and D, and B and C,

16/3 X 2/3 = 32/9, so these are the pairs and B and C finish work faster.

Q24. Three small pumps and a large pump are filling a tank. Each small pump works at 2/3rd the rate of the large pump. If all four work at the same time, they should fill the tank in what fraction of time, it would have taken the large pump alone.(CAT 2001)

Ans24. Let the rate of large pump is 1 liter/hr, therefore the rate of small pump is 2/3 liter/hr. Assume there is a 3 liter tank, all four will fill it in 1 hour. Alone large pump will take 3 hours, ratio = 1/3

Q25. Six technicians working at the same rate complete work of one server in 10 hours. If they start at 11 am and one additional technician per hour been added, beginning at 5 PM, at what time the server will be complete? (CAT 2002)

Ans25. The work requirement is 10 × 6 = 60 Man hours

Man hours till 5 pm = 6 × 6 = 36, Man hours required more = 24. Next hour will contribute 7, next will 8, and next 9, making for the entire 24(9 + 8 + 7), therefore work will finish by 8:00 PM.

Q26. Two boats travelling at 5 and 10 km/hr respectively, head directly towards each other. They begin at a distance of 20 km from each other. How far apart are they (in km) 1 minute before they collide? (CAT 2004)

Ans26. Relative speed = 15(10 + 5) km/hr, time to travel 20 km = 20/15 = 4/3 hours = 80 minutes

So in 80 minutes they will travel 20 km

In 1 minute they will travel 20/80 = 4 km

In 79 minutes they will travel 79/4 = 19.75 km

Since total is 20, they were apart by 20 – 19.75 = 0.25 km

Q27. A sprinter starts running on a circular path of radius r meters. Her average speed (in meter/min) is Pr, during the first 30 seconds, πr/2, during next one minute, πr/4, during next two minutes, πr/8 , during next four minutes and so on. What is the ratio of the time taken for the nth round trip that for the previous round? (CAT 2004)

Ans27. The circumference of circle is 2Πr, so distance traveled in first 30 seconds in , in next one minute is again Πr/2(Πr/2 for one minute), in the next two minutes is also Πr/2(Πr/4 for two minutes), and so on as per question. Now for the 1st round the time is 71/2 minutes, and basically four intervals, for 2nd round the intervals will be 16 times the intervals as it will be Πr/2(Πr/16, in next 8 minutes), so the ratio will be 16.

Q28. A company has a job to prepare certain number of cans and there are three machines A, B and C for this job. A can complete the job in three days. B can complete the job in four day and C can complete the job in six days. How many days the company will take to complete the job if all the machines are used simultaneously? (CAT 1998)

Ans28. A can do the job in 3 days, which means 1/3rd job in a day

B can do the job in 4 days, which means 1/4^th job in a day.

C can do the job in 6 days, which means 1/6th job in a day.

In one day they all can do 1/3 + ¼ + 1/6 = 3/4th of job, so they will do the job in 4/3 days.

Q29. Two full tanks, one shaped like a cylinder and other like a cone contains jet fuel. The cylindrical tank holds 500 liters more than the conical tank. After 200 liters of fuel has been pumped out from each tank, the cylindrical tank contains twice the amount of fuel in the conical tank. How many liters of fuel did the cylindrical tank had when it was full? (CAT 2000)

Ans29. Let conical tank hold X liter of fuel, then cylindrical holds X + 500, also after 200 liters from each has been used (X + 300) = 2(X – 200)

X = 700 L, and cylindrical tank has 1200 liters

Q30. A and B walk up an escalator. The escalator moves at a constant speed. A takes 3 steps for every 2 of B’s steps. A gets to the top of escalator in 25 steps, while B takes 20 steps to reach the top. If escalator was turned off, how many steps would they have to take to walk up? (CAT 2001)

Ans30. A is obviously faster than B, let A reaches the top in 1 hour, in 25 steps as given, and let escalator contributed X steps (in one hour), so total steps, T = X + 25. If A takes 25 steps in an hour, B takes 20 steps in 6/5 hrs(A takes 3 steps B takes 2), and escalator contributes 5/6X steps, so total steps, T = 6/5X + 20,

Equating X + 25 = 6/5X + 20

Solving X = 25, therefore, total steps = 50

Q31. At his usual rowing rate Kapil can travel 12 miles downstream in a certain river in 6 hrs less than it takes him to travel the same distance upstream. But if he could double his usual rowing rate for this 24 mile round trip, the downstream 12 miles would then take only 1 hr less than the upstream 12 miles. What is the speed of the current in miles per hour. (CAT 2001)

Ans31. Let Kapils speed in still water = X miles/hr

And the speed of the stream = Y miles/hr

Time difference = 6 = 12/(X – Y) – 12(X+Y)

Time difference for second case = 1 = 12/(2X – Y) – 12(2X + Y).

Solving Y = 8/3 miles per hour

Q32. A car is moving at a rate of 50 miles per hour and the radius of its wheels is 2.5 feet. Find the number of revolutions per minute the wheels are rotating. (1 mile = 5280 feet)

Ans32. We have a car traveling 50 miles an hour with a tire radius of 2.5 ft. We wish to find the rotational velocity of the tires in revolutions per minute. First let’s convert the car’s speed to ft/min, 50 miles/hour * (5280 ft/mile) * (1 hr/60 min) = 4400 ft/min

We know that this is some multiple of the number of circumferences of the tire (since the tire rolls smoothly along). Let’s find out how many circumferences (or revolutions) this is:

C = 2 × pi × r = 2 × pi × 2.5 ft = 5pi ft

Let’s convert this speed to revolutions per minute = 4400/5 X 3.14 = 280.2 revolutions per min.

Q33. A man is walking across a railroad bridge that goes from point A to point B. He starts at point A, and when he is of the way across the bridge, he hears a train approaching. The train’s speed is 60 mph (miles per hour). The man can run fast enough so that if he turns and runs back toward point A, he will meet the train at A, and if he runs forward toward point B, the train will overtake him at B. How fast can the man run?

Ans33. Let M stand for the man’s speed in mph. When the man runs toward point A, the relative speed of the train with respect to the man is the train’s speed plus the man’s speed (60 + M). When he runs toward point B, the relative speed of the train is the train’s speed minus the man’s speed (60 - M). When he runs toward the train the distance he covers is 3 units. When he runs in the direction of the train the distance he covers is 5 units. Therefore (60 + M)/(60 – M) = 5/3, solving M = 15 mph

Special Examples

Q1: A truck travels 15 mph for the first half of the distance of a trip. The driver wants to average 30 mph for the whole trip. How fast must he travel for the second half of the trip?

Ans1: This is more of a puzzle, assume any distance for the trip. If you said 30 miles, then the first half took 1 hour. But to average 30 mph for the total trip would require an hour. Therefore it has no solution and it is impossible!

Q2: A ship went on a voyage. After it had traveled 180 miles a plane started with 10 times the speed of the ship. Find the distance when they meet from starting point.

Ans2: Let's say the ship is travelling x miles/hr and plane took t hours to meet the ship

Then, xt is the distance ship traveled after plane started, Let 10xt is the distance plane traveled. So we have, xt + 180 = 10xt

Solving xt = 20, therefore distance traveled by the ship = 180 + 20 = 200 miles

Q3: A can complete a piece of work in 4 days, B takes double the time taken by A, C takes double that of B, and D takes double that of C, to complete the same task. They are paired in-groups of two each, one pair takes 2/3rds the time needed by the second pair to complete the work. Which is the first pair? (CAT 2001)

Ans3: A takes 4 days, B takes 8, C takes 16, D takes 32, In one day A does 1/4^th, B does 1/8^th, C does 1/16^th, D does 1/32th of work.

Now A and D work in a day = ¼+1/32 = 9/32, so finish work in 32/9 days

A and C work in a day = ¼+1/16 = 5/16, so finish work in 16/5 days

A and B work in a day = ¼+1/8 = 3/8, so finish work in 8/3 days

B and C work in a day = 1/8+1/16 = 3/16, so finish work in 16/3 days

We can already see a relationship between pair of A and D, and B and C,

16/3 x 2/3 = 32/9, so these are the pairs and B and C finish work faster.

Q4: Three small pumps and a large pump are filling a tank. Each small pump works at 2/3^rd the rate of the large pump. If all four work at the same time, they should fill the tank in what fraction of time, it would have taken the large pump alone.(CAT 2001)

Ans4: Let the rate of large pump is 1 liter/hr, therefore the rate of small pump is 2/3 liter/hr. Assume there is a 3 liter tank, all four will fill it in 1 hour. Alone large pump will take 3 hours, ratio = 1/3

Q5: Six technicians working at the same rate complete work of one server in 10 hours. If they start at 11 am and one additional technician per hour been added, beginning at 5 PM, at what time the server will be complete? (CAT 2002)

Ans5: The work requirement is 10 x 6 = 60 Manhours

Manhours till 5 pm = 6 x 6 = 36, Manhours required more = 24

Next hour will contribute 7, next will 8, and next 9, making for he entire 24(9+8+7), therefore work will finish by 8:00 PM

Q6: Two boats travelling at 5 and 10 km/hr respectively, head directly towards each other. They begin at a distance of 20 km from each other. How far apart are they (in km) 1 minute before they collide? (CAT 2004)

Ans6: Relative speed = 15(10+5) km/hr, time to travel 20 km = 20/15 = 4/3 hours = 80 minutes

So in 80 minutes they will travel 20 km

In 1 minute they will travel 20/80 = 4 km

In 79 minutes they will travel 79/4 = 19.75 km

Since total is 20, they were apart by 20 – 19.75 = 0.25 km

Q7. A sprinter starts running on a circular path of radius r meters. Her average speed (in meter/min) is Pr, during the first 30 seconds, Pr/2, during next one minute, Pr/4, during next two minutes, Pr/8, during next four minutes and so on. What is the ratio of the time taken for the nth round trip that for the previous round? (CAT 2004)

Ans7. The circumference of circle is 2Pr, so distance traveled in first 30 seconds in Pr/2, in next one minute is again Pr/2(Pr/2 for one minute), in the next two minutes is also Pr/2(Pr/4 for two minutes), and so on as per question. Now for the 1^st round the time is 71/2 minutes, and basically four intervals, for 2^nd round the intervals will be 16 times the intervals as it will be Pr/2(Pr/16, in next 8 minutes), so the ratio will be 16.

Q8: A company has a job to prepare certain number of cans and there are three machines A, B and C for this job. A can complete the job in three days. B can complete the job in four day and C can complete the job in six days. How many days the company will take to complete the job if all the machines are used simultaneously? (CAT 1998)

Ans8: A can do the job in 3 days, which means 1/3^rd job in a day

B can do the job in 4 days, which means 1/4^th job in a day

C can do the job in 6 days, which means 1/6^th job in a day

In one day they all can do 1/3 + ¼ + 1/6 = 3/4^th of job, so they will do the job in 4/3 days

Q9: Two full tanks, one shaped like a cylinder and other like a cone contains jet fuel. The cylindrical tank holds 500 liters more than the conical tank. After 200 liters of fuel has been pumped out from each tank, the cylindrical tank contains twice the amount of fuel in the conical tank. How many liters of fuel did the cylindrical tank had when it was full? (CAT 2000)

Ans9: Let conical tank hold X liter of fuel, then cylindrical holds X+500, also after 200 liters from each has been used (X+300) = 2(X-200)

X = 700 L, and cylindrical tank has 1200 liters

Q10: A and B walk up an escalator. The escalator moves at a constant speed. A takes 3 steps for every 2 of B’s steps. A gets to the top of escalator in 25 steps, while B takes 20 steps to reach the top. If escalator was turned off, how many steps would they have to take to walk up? (CAT 2001)

Ans10: A is obviously faster than B, let A reaches the top in 1 hour, in 25 steps as given, and let escalator contributed X steps (in one hour), so total steps, T = X + 25. If A takes 25 steps in an hour, B takes 20 steps in 6/5 hrs(A takes 3 steps B takes 2), and escalator contributes 5/6X steps, so total steps, T = 6/5X + 20, Equating

X + 25 = 6/5X + 20

Solving X = 25, therefore, total steps = 50

Q11: At his usual rowing rate Kapil can travel 12 miles downstream in a certain river in 6 hrs less than it takes him to travel the same distance upstream. But if he could double his usual rowing rate for this 24 mile round trip, the downstream 12 miles would then take only 1 hr less than the upstream 12 miles. What is the speed of the current in miles per hour. (CAT 2001)

Ans11: Let Kapils speed in still water = X miles/hr

And the speed of the stream = Y miles/hr

Time difference = 6 = 12/(X-Y) - 12(X+Y)

Time difference for second case = 1 = 12/(2X-Y) - 12(2X+Y)

Solving Y = 8/3 miles per hour

Q12: A car is moving at a rate of 50 miles per hour and the radius of its wheels is 2.5 feet. Find the number of revolutions per minute the wheels are rotating. (1 mile = 5280 feet)

Ans12: We have a car traveling 50 miles an hour with a tire radius of 2.5 ft. We wish to find the rotational velocity of the tires in revolutions per minute. First let's convert the car's speed to ft/min, 50 miles/hour * (5280 ft/mile) * (1 hr/60 min)

= 4400 ft/min

We know that this is some multiple of the number of circumferences of the tire (since the tire rolls smoothly along). Let's find out how many circumferences (or revolutions) this is:

C = 2 x pi x r = 2 x pi x 2.5 ft = 5pi ft

Let's convert this speed to revolutions per minute = 4400 / 5 x 3.14 = 280.2 revolutions per min.

Q13: A man is walking across a railroad bridge that goes from point A to point B. He starts at point A, and when he is 3/8 of the way across the bridge, he hears a train approaching. The train's speed is 60 mph (miles per hour). The man can run fast enough so that if he turns and runs back toward point A, he will meet the train at A, and if he runs

forward toward point B, the train will overtake him at B. How fast can the man run?

Ans13: Let M stand for the man's speed in mph. When the man runs toward point A, the relative speed of the train with respect to the man is the train's speed plus the man's speed (60 + M). When he runs toward point B, the relative speed of the train is the train's speed minus the man's speed (60 - M). When he runs toward the train the distance he covers is 3 units. When he runs in the direction of the train the distance he covers is 5 units. Therefore (60 + M)/(60 - M) = 5/3, solving M = 15 mph

PERMUTATION AND COMBINATION

May 29, 2009 by fundoogyan

Q1. How many three digits numbers can be formed using digits 3,6,9?

(a). 22 (b). 27

(c). 20 (d). 25

Q2. How many odd numbers are there less than 10000 using digits 0,2,3,9

(a). 128 (b). 127

(c). 120 (d). 125

Q3. There are 8 people in the bus,3 get off on stop 1, 2 on stop 2, and rest on 3, how many different ways this can happen?

(a). 550 (b). 500

(c). 560 (d). 565

Q4. In how many ways can 7 Irish and 7 Welshmen sit down at a round table, no 2 Welshmen being in consecutive positions?

(a). 3628890 (b). 3628870

(c). 3628820 (d). 3628800

Q5. From 6 men and 5 ladies, in how many ways five people can be chosen with at least one man?

(a). 454 (b). 512

(c). 376 (d). 462

Q6. In how many ways 11 cricketers can be chosen from 6 bowlers, 4 wicket keepers and 11 batsmen to have 5 batsmen, 5 bowlers and 1 wicket keeper?

(a). 11008 (b). 11088

(c). 46200 (d). 44242

Q7. Find the sum of all the four digit numbers which can be formed with the digits 1,2,3,4.

(a). 74000 (b). 74066

(c). 78992 (d). 74088

Q8. An Army officer has 10 soldiers in his platoon; he takes four at a time to the forest trip, as often as he can without taking the same four soldiers together more than once. How often will he go himself to the forest?

(a). 156 (b). 210

(c). 256 (d). 186

Q9. In how many ways can the letters of the word “sincity” be arranged such that the two i’s always come together?

(a). 720 (b). 120

(c). 660 (d). 360

Q10. How many triangles can be formed by joining 12 points, 5 of which are collinear?

(a). 156 (b). 210

(c). 256 (d). 186

Q11. How many signals can be made by raising 4 flags of different colours one above the other when any number of them may be used?

(a). 64 (b). 72

(c). 92 (d). 84

Q12. Ravi is having a party for 10 people. In how many ways can he invite the guests?

(a). 1023 (b). 1024

(c). 720 (d). 1100

Q13. In a chess competition involving some boys and girls of a school, every student had to play exactly one game with every other student. It was found that in 45 games both the players were girls and in 190 games both were boys. The number of games in which one player was a boy and other was a girl is (CAT)

(a). 200 (b). 216

(c). 135 (d). 256

Q14.1 A string of three English letters is formed as per the following rules. The first letter is any vowel. The second letter is m, n or p. If the second letter is m then the third letter is any vowel which is different from the first letter. If the second letter is n then the third letter is e or u….If the second letter is p then the third letter is same as the first letter. How many strings of letters can possibly be formed using the above rules?(CAT)

(a). 40 (b). 45

(c). 30 (d). 35

Q15. There are 12 towns grouped into four zones with three towns per zone. It is intended to connect the towns with telephone lines such that every two towns are connected with three direct lines if they belong to the same zone and with only one direct line otherwise. How many direct telephone lines are required?(CAT)

(a). 72 (b). 90

(c). 96 (d). 144

Q16. An intelligence agency forms a code of two distinct digits selected from 0, 1, 2, ….9 such that the first digit of the code is nonzero. The code, handwritten on a slip, can however potentially create confusion when read upside down – for example, the code

91 may appear as 16. How many codes are there for which no such confusion can arise?(CAT)

(a). 80 (b). 78

(c). 71 (d). 69

Q17. In how many ways is it possible to choose a white square and a black square on a chess board so that the square must not lie in the same row or column?(CAT)

(a). 56 (b). 896

(c). 60 (d). 768

Ans1. (b) Three digit, three places = 3x3x3 = 27

Ans2. (a) Single Digit = 2(3,9)

Two Digit = 6

Three Digit = 24

Four Digit = 96

Total = 128

Ans3. (c) 8!/3!2!3! = 560

Ans4. (d). There are 14 people to sit; let first Irish be seated, first person can sit anywhere, therefore 6! Ways, and then the Welshmen have 7! Ways to take 7 positions

Total ways = 7! X 6! = 3628800

Putting 1 Englishman in a fixed position, the remaining 6 can be arranged in 6!

Ans5. (d). 6 Men + 5 Ladies, 10 people, five to be selected, which is 11C₅

The scenario in which no man is selected, 5 women, all 5 to be selected 5C₅= 1

Here 11C₅- 5C₅= 462 ways

Ans6. (b) One wicketkeeper has to be chosen out of 4 which is 4C₁ = 4

Five batsman have to be chosen out of 11 which is 11C₅

Five bowlers have to be choose out of 6 which is 6C₅ = 6

Total ways = 4 x 11C₅x 6 = 11088

Ans7. (d) Numbers that can be formed from 1,2,3,4 = 4! = 24

The least number is 1234 and the greatest is 4321

Sum = Average x total = (1234+4321)/2 x 24 = 74088

Ans8.(b) The visits of the officer = number of groups of four, as he visits with each group

Which is four soldiers going out of 10 = 10C₄ = 210 times

Ans9.(a) Since two i’s have to come together, total words are 7, so there will be a set of six now(2 i’s and 5 others), which is 6! Ways, now two i’s can arrange themselves in only one way, therefore ways are 6! = 720

Ans10. (d) 3 points required for a triangle therefore 12C_3,since 5 points are collinear, therefore these points will not form a triangle 5C_3.Number of Triangles = 12C₃- 5C₃= 210

Ans11. (a) Flags can be used in any number, so from 1 to 4

Therefore no. signals = 4P₁ + 4P₂ + 4P₃ + 4P₄ = 64

Ans12 (c) 2¹⁰ – 1 = 1023

Ans13. (a) For a match two people are required in case of boys = ⁿC₂ = 45 = n(n-1) / 1

n(n-1) = 90, n = 10 boys

In case of girls = ⁿC₂ = 190 = n(n-1) / 2 = 190

n (n-1) = 380, n = 20 girls

Matches between girls and boys are ¹⁰C₁x ²⁰C₁ = 10 x 20 = 200.

Ans14.1. (d) First place can be taken by 5 vowels

Case I – second place is 1(m) and third is 4(other vowels) = 5x1x4 = 20

Case II – second place is 1(m) and third is 2(2 vowels) = 5x1x2 = 10

Case III – second place is 1(m) and third is 1(same vowel) = 5x1x1 = 5

Total = 35

Ans14.2. (c) First place can be taken by 5 vowels in only case II, In case I first place will be taken by 4 and in Case III it will be taken by 1.

Case I – second place is 1(m) and third is 1(only e) = 4x1x1 = 4

Case II – second place is 1(m) and third is 1(only e) = 5x1x1 = 5

Case III – second place is 1(m) and third is 1(same vowel) = 1x1x1 = 1

Total = 10 ways

Ans15. (b) Total number of lines required for connections in each zone = 9 x 4 = 36

Each town connecting to town in different zone will require 3 x 3 = 9 lines

Selecting 2 out of 4 towns = 4C₂= 6

Lines required for connecting towns of different zones = 6 x 9 = 54

Total number of lines in all = 54 + 36 = 90.

Ans16. (c) Total codes which can be formed = 9 x 9 = 81 (distinct digits)

The digits which can confuse are 1,6,8,9. From these digits we can form the codes

= 4 x 3 = 12

Out of these 12 codes two numbers 69 and 96 will not create confusion. Therefore

( 12 – 2 ) = 10 codes will create confusion

Thus total codes without confusion = 81 – 10 = 71

Ans17.(a) In a chess board , there are 8 rows and 8 columns. If you choose one row or column you have to choose one less, as a box cannot be taken from same row or column, Required number of ways = ⁸C₁ x ⁷C₁ = 8 x 7 = 56.

GEOMETRY AND MENSURATION

May 29, 2009 by fundoogyan

Example1: The sides of a triangle are 13, 14, 15 cm. The in radius of the triangle is :

(1) 8 cm (2) 4 cm

(3) 3 cm (4) 2.43 cm

Example 2: Euclid had a triangle in his mind. The longest side is 20 and the other side is 10. Area of the triangle is 80. The third side is

(CAT 2001)

(1) Ö260 (2) Ö240

(3) Ö250 (4) Ö210

Example3: The radius of the biggest circle which can be fitted in an equilateral

triangle of side 6 cm is:

(1) 3Ö3 cm (2) Ö3 cm

(3) 3 cm (4) 2Ö3cm

Example4: One side of a right triangle is 3/5 times of the hypotenuse and the sum of that side and hypotenuse is 16 cm. The circumradius of the triangle is:

(1) 3 cm (2) 4 cm

(3) 5 cm (4) None of the above

Example 5: Four horses are tethered at four corners of a square plot of side 14 metres so that the adjacent horses can reach one another. There is a small circular pond of area 20 m² at the center. The area let ungrazed is (CAT 2002)

(1) 22 m² (2) 42 m²

(3) 84 m² (4) 168 m²

Example 6 : A square of side 2 cm. is cut from each corner to form a regular

octagon. What is the side of the octagon? (CAT 2001)

(1) Ö2/(Ö2 + 1) (2) Ö2/(Ö2 – 1)

(3) 2/(Ö2 + 1) (4) 2/(Ö2 – 1)

Example7: A spherical ball of radius 8 cm is cut into 4 equal parts. Total surface area of the parts is:

(1) 384 π cm² (2) 512 π cm²

(3) 256 π cm² (4) 488 π cm²

Example8: A cubic meter of silver weighting 900 grams is rolled into a square bar 16 m long. The weight of 1 cm of this bar in grams is:

(1) 5 grams (2) 5.624 gms.

(3) 0.5625 gms (4) None of the above

Example9: The base of a conical tent is 19.5 m in diameter and the height of the tent is 2.8 m. What area of canvas is required to put up such a tent?

(1) 300m²(2) 400 m²

(3) 310.7 m² (4) none of the above

Example10: The radius of a hemisphere is 7 cm. Find the altitude of a right circular cone of the same base radius and same total surface area.

(1) 10.38 cm (2) 12.12 cm

(3) 15.66 cm (4) None of the above

Example11: In a rectangular plot 160 m x 120 m, a square tank of side 40 m is dug out a depth of 8 m. The earth so dug out is evenly spread to cover the remaining area, by how much has the level of plot has been raised?

(1) 0.82 cm (2) 8.2 cm

(3) 0.82 m (4) None of the above

Example12: A cylindrical cistern whose diameter is 21 cm is party filled with water. If a rectangle block of iron measuring 40 cm wholly immersed in the water, then the rise in the water level (in cm) is:

(1) 13 cm (2) 15.24 cm

(3) 18 cm (4) 19.38 cm

Example13: The corners of an equilateral triangle of side 10cm each are cut to form a regular hexagon. The area of the hexagon is:

(1) 20 cm² (2) 28.8 cm²

(3) 24.39 cm² (4) 25.12 cm²

Example14: The surface area of hemisphere is numerically equal to its volume. Its diameter is:

(1) 4.5 units (2) 7.5 units

(3) 9 units (4) 6 units

Example15: Five cylinders of base radius 10 cm and height 1, 2, 3, 4, 5 cm

respectively are melted to form a hemisphere. The surface area of the hemisphere is:

(1) 1615 cm² (2) 1685 cm²

(3) 1729 cm²(4) 1836 cm²

Example16: A circular disc of diameter 9 cm is cut in the center and a circular hole is generated. If the weight of the disc is reduced by one third, then the diameter of the hole is:

(1) 14.3 cm (2) 13.5 cm

(3) 12.4 cm (4) None of the above

Example17: A single pipe of diameter x has to be replaced by six pipes of

diameters 10 cm each. The pipes are used to covey precious liquid in a laboratory. If the speed of the liquid is the same then the value of x is:

(1) 13cm (2) 18 cm

(3) 24.5 cm (4) 24 cm

Example18: The number of edges of a solid having 20 faces and 32 vertices is:

(1) 58 (2) 56 cm

(3) 50 (4) 59

Example19: The percentage change in the total surface area of a cylinder with volume 1000 c.c. & radius 10 cm when it’s cut into two equal parts along its cross section is:

(1) 80% (2) 10%

(3) 30% (4) 75.84%

Example 20: Instead of walking along two adjacent sides of a rectangular field, a boy took a short cut along the diagonal and saves a distance equal to half the longer side. Then the ratio of the shorter side to the longer side is (CAT 2002)

(1) ½ (2) 2/3

(3) ¼ (4) ¾

Example21: Water flows through a pipe of diameter 1/16 meter, at 6 meters/sec. The time taken to fill bath 2 metres by 3 meters by 4 meters is:

(1) 21.8 min (2) 30.33 min

(3) 15.38 min (4) None of these

Example22: A tank 72 cm long, 60 cm wide, 36 cm high contains water to a depth of 18 cm. A metal bloc 48 cm by 36 cm by 15 cm is put into the tank and totally submerged. The rise of water- level is:

(1) 5 cm (2) 5.27 cm

(3) 6.36 cm (4) 6 cm

Example23: The external diameter of a hollow metal sphere is 14 cm and its thickness is 2 cm. Find the radius of a solid sphere containing the same amount of material as the hollow sphere.

(1) 6 cm (2) 5 cm

(3) 4 cm (4) 4.5 cm

Example24: Three solid metal of radii 3 cm, 4 cm, 5 cm respectively, are melted

together. The metal is recast as a single solid sphere. The percentage reduction in the area of surface resulting from this is:

(1) 40% (2) 16%

(3) 12% (4) 28%

Example25: A bucket in the form of a frustum of a right circular cone is 48 cm deep and the radii of the top and bottom are 20 cm and 16 cm respectively. The volume of water it will hold is:

(1) 49049 c.c. (2) 30000 c.c.

(3) 31259 c.c. (4) 24076 c.c.

Example 26: The diameter of a copper sphere is 6 cm. The sphere is melted and is drawn into a long wire of uniform circular cross section. If the length of the wire is 36 cm, The radius is

(1) 1 (2) 0.5

(3) 2 (4) 0.1

Example 27 : A sphere of diameter 6 cm is dropped in a right circular cylindrical vessel, party filled with water. The diameter of the cylindrical vessel is 12cm. If the sphere is completely-submerged in water, by how much will the level of water rise in the cylindrical vessel?

(1) 2 (2) 0.5

(3) 1 (4) 0.1

Example 28: In the given figure AB and BC are 24 and 32 with included angle of 90^o. The other sides are 25 each and none of the other angle is 90^o. Find the area of the fig. (CAT 2001)

<<Geometry Unsolved Fig 1>>

(1) 684 (2) 786

(3) 880 (4) None of these.

Example 29 : Find the number of coins, 1.5 cm in diameter and 0.2 cm thick, to be melted to form a right circular cylinder of height 10 cm and diameter 4.5 cm.

(1) 425 (2) 450

(3) 550 (4) None of these.

Example 30: A solid iron rectangular block of dimensions 4.4 m, 2.6m and 1 m is

cast into a hollow cylindrical pipe of internal radius 30 cm and thickness 5 cm. Find the length of the pipe.

(1) 118 (2) 116

(3) 110 (4) 112

Example31: How many sides have regular polygon whose exterior angle is 1/11 of its interior angle?

(1) 22 (2) 24

(3) 20 (4) 26

Example32: The angles of a pentagon are x^o, x + 20^o, x + 40^o, x + 60^o and

x + 80^o. The smallest angle of the pentagon is:

(1) 65^o (2) 70^o

(3) 68^o (4) 60^o

Example33: In a triangle ABC, the altitudes BD and CE are equal and ÐA = 36^o. What is the value of the angle B?

(1) 65^o (2) 60^o

(3) 75^o (4) 72^o

Example 34: In the following figure

<<Geometry Unsolved Fig 2>>

AE = EF = FB. The ratio of area of triangle CEF and rectangle ABCD is (CAT 2001)

(1) 1/6 (2) 1/8

(3) 1/9 (4) None of these.

Example35: The interior angle of a regular polygon exceeds the exterior angle by 140^o. The sides of the polygon are:

(1) 18 (2) 22

(3) 25 (4) 15

Example36: Two regular polygons have the number of their sides as 3 : 2 and the interior angles as 10 : 9. Find the number of sides of the polygon.

(1) 10, 8 (2) 14, 8

(3) 12, 8 (4) 18, 8

Example 37: In the given figure BC = AC, angle AFD = 40^o and CE = CD. The value of angle BCE = (CAT 2001)

<<Geometry Unsolved Fig 3>>

(1) 100^o (2) 50^o

(3) 60^o (4) None of these.

Example38: In a triangle PQS, R is any point on PS, such that PR = QR and QS = RS. If angle RSQ = 120^o, what is the measure of angle QPR?

(1) 15^o (2) 25^o

(3) 16.5^o (4) 30^o

Example39: O is the centre of the circumcircle of triangle ABC. If ÐA= a and ÐOCB= b. what is the measure of a + b?

(1) a + b = 65^o (2) a + b = 60^o

(3) b + a = 80^o (4) a + b =90^o

Example40: AB and CD are parallel straight lines of lengths 5 cm and 4 cm

respectively. AD and BC intersect at a point O such that AO = 10 cm. Then OD equals to:

(1) 10 cm (2) 8 cm

(3) 6 cm (4) 4 cm

Example41: PSR is a triangle right angled at S and D is the mid point of SP. If the bisector of ÐPSR and perpendicular bisector of SR meet at O, then triangle OSD is:

(1) scalene (2) equilateral

(3) isosceles right angled (4) acute angled

Example42: PR is the diameter of the circle. Find PS where PQ = 6 cm QR = 7 cm and RS = 2 cm?

(1) 7 cm (2) 5 cm

(3) 11 cm (4) 9 cm

Example43: A circle has two parallel chords of lengths 6 cm and 8 cm. If the chord are 1 cm apart and the chord is on the same side of the centre, then the diameter of the circle is of length:

(1) 5 cm (2) 6 cm

(3) 10 cm (4) 12 cm

Example 44: In the given figure, ACB is a right angled triangle. CD is the altitude. Circles are inscribed within the triangles ACD, BCD. P and Q are the centers of the circles. The distance PQ is (CAT 2002)

<<Geometry Unsolved Fig 4>>

(A) 5 (B) Ö50

Example45: In a right angled triangle ABC; ÐA = 90^o and AM is the median of BC. If AB = 6 cm and AC = 8 cm. The length of AM is:

(1) 5 cm (2) 7 cm

(3) 6.5 cm (4) 10 cm

Example46: ACB is a tangent to a circle at C, CD and CE chords such that ÐACE > ÐACD. If ÐACD = ÐBCE = 50^o, then

(1) CD = CE

(2) ED is not parallel to AB

(3) ED passes through the centre of the circle

(4) ÐCDE is a right angled triangle

Example47: A circle with centre O has a chord AB produced so that it meets the line through O in C, such that OB = BC. If angle BCO is 20^o, what is the measure of angle AOD where D lies on circle when CO is produced?

(1) 65^o (2) 60^o

(3) 75^o (4) 72^o

Example48: ABCD is a square E and F are the mid-point of BC and CD. What is the ratio of the area of DAEF to that of the square ABCD?

(1) 3: 8 (2) 5: 8

(3) 8: 3 (4) None of the above

Example49: AB is a chord of a circle whose centre is O. P is a point on the circle such that OP ^ AB and OP intersect AB at the point M. If AB = 8 cm and MP = 2 cm, then the radius of the circle is:

(1) 10 cm (2) 6 cm

(3) 5 cm (4) 4 cm

Example50: The side AB, BC, CD of a regular polygon are such that the measure of angle BAC is 15^o. The number of sides of the polygon are:

(1) 15 (2) 12

(3) 18 (4) 20

Example51: In DABC, D is the mid-point of BC which is 10 cm, AD = 5 cm and AC = 6 cm. The length of side AB is:

(1) 5 cm (2) 7 cm

(3) 6.5 cm (4) 8 cm

Example 52: In the figure given below, ABCD is a rectangle. The area of the isosceles right triangle ABE = 7 cm²; EC = 3 (BE). The area of ABCD (in cm²) is (CAT 2002)

<<Geometry Unsolved Fig5>>

(1) 21 (2) 28

(3) 42 (4) 56

Example53: The sides of the triangle are 5 cm, 4 cm and 3 cm. Find the greatest side of another similar triangle whose area is 16 times greater that the area of the given triangle.

(1) 25 cm (2) 20 cm

(3) 18 cm (4) 15 cm

Example54: P is a point on the base of the equilateral DABC such that

BP=1/3BC.Then AP²/AB²=

(1) 7/9 (2) 5/9

(3) 8/9 (4) 4/9

Example 55: The length of the common chord of two circles of radii 15 cm and 20 cm, whose centers are 25 cm apart, is (in cm) (CAT 2002)

(1) 24 (2) 25

(3) 15 (4) 20

Example 56: In the given figure, BC is the diameter of the circle with the centre O and PAT is the tangent at A. If ÐABC = 38^o, find ÐBAT.

(1) 52^o. (2) 48^o.

(3) 62^o. (4) None of these

<<Geometry Unsolved Fig 6>>

Example 57: In the given figure, O is the centre of the circle and PAQ is the tangent to the circle at A. If ÐPAB = 58^o, find ÐABQ and ÐAQB.

(1) 32^o, 26^o (2) 48^o, 24^o

(3) 42^o, 18^o (4) None of these

<<Geometry Unsolved Fig 7>>

Example 58: In the given figure, DE || BC. If AD = (4x - 3)cm, AE = (8x - 7)cm and BD = (3x - 1)cm and CE = (5x - 3), find the value of x.

(1) 2 (2) 4.

(3) 1 (4) None of these

<<Geometry Unsolved Fig 8>>

Example 59: The perimeters of two similar triangles are 30 cm and 20 cm

respectively. If one side of the first triangle is 15 cm, then find out the

proportional side of the other triangle.

(1) 20 (2) 25

(3) 15 (4) None of these

Example 60 :A vertical stick 12 cm long casts a shadow 8 cm long on the ground. At the same time tower casts a shadow 40 m long on the ground. Determine the height of the tower.

(1) 50 (2) 60

(3) 40 (4) 75

Q61. Find the area of an equilateral triangle whose vertices lie on a circle with radius 2 cm.

(a) 3π cm2 (b) 3√3 cm2

(e) None of these

Q62. The following statements describe a race between: Pat, Chris, Jo, and Sam.

(i) Pat is 12 seconds behind the next runner.

(ii) The leader is 20 seconds ahead of the last person.

(iii) Chris is 1 second ahead of somebody.

(iv) Sam is 19 seconds ahead of Jo.

Which of the following gives the order of the runners. (Runners listed from first to last.)

(a) Sam, Chris, Pat, and Jo

(b) Chris, Sam, Pat, and Jo

(d) Chris, Sam, Jo, and Pat

(e) Sam, Chris, Jo, and Pat

Q63. If a boat travels North for 5 miles then East for 12, then Southeast for 6, approximately how far is it from its starting point?

(a) 13 miles (b 19.01 miles

(e) None of these

Q64. Given the following two circles, find the algebraic equation of the chord they share in common.

Circle I : (x – 1)2 + (y – 2)2 = 9

Circle II: (x + 3)2 + (y – 1)2 = 16

(a) 4x – 3y = 1 (b) 3x + 4y = 1

(e) None of these

Q65. The area of a 300' by 400' rectangle is doubled by adding a strip of width "w" around the perimeter. Approximately how wide is that strip?

<<Geometry Unsolved Fig 9>>

(a) 87.50’ (b) 83.41’

(e) None of these

Q66. Find the length of the perimeter of a right triangle whose area is 30 cm2 and whose hypotenuse is 13 cm.

(a) 23 + √69 cm (b) 16 + 2√24cm

(e) None of these

Q67. Find the area of a triangle whose vertices are (0,0), (12, 9) and (14, 6).

(a) 27 sq units (b) 54 sq units

(e) None of these

Q68. Given that it is 3 o'clock, exactly how long will it take for the minute hand to catch up with the hour hand?

(a) 15 min (b) 16 min

(e) None of these

Q69. If the size of a rectangle's area is twice as large as the size of its perimeter, and the length of one of its sides is 4.5, what is the length of the other side?

(a) 36 (b) 3.6

(e) None of these

Q70. A given circular cylindrical can is made up of a square piece of metal and two circular disks each with diameter c inches. What is the volume of this cylindrical can?

<<Geometry Unsolved Fig 10>>

(a) 2π – c3 in3 (b) π3c3/4 in3

(e) None of these

Q71. Two chords AB and DC intersect each other so that AO = 1.2, OB = 7.5, and OC = DO. How long is DC?

(a) 6 units (b) √8.7 units

(e) None of these

Q72. Find the area of parallelogram ABCD, given that AB = BE = ED = 1, and ÐABE = 90°.

<<Geometry Unsolved Fig 11>>

(a) √2 (b) 1.707

(e) None of these

Q73. Find the solution set for y given that z y x, are natural numbers and the following are true:

(i) If y > 3 then x < 3

(ii) z + y < 7

(iii) x + z > 10

(a) y ε {1, 2} (b) y ε {1, 2,…,5}

(e) y ε {1,2,…,6}

Q74. A rectangle is inscribed in a triangle such that its upper right vertex bisects the triangle's side. What is the ratio of the area of the shaded region to the area of the unshaded region?

<<Geometry Unsolved Fig 12>>

(a) 1:2 (b) 4:3

(e) None of these

Q75. Two 5-12-13 triangles are combined to form a parallelogram. Which of the following statements must be true?

(a) The parallelogram is a rectangle.

(b) The perimeter of the parallelogram is 34 units long.

(d) One of the sides of the parallelogram is 12 units long.

(e) All of these statements must be true.

Q76. To construct a circle that circumscribes a triangle one finds its center by locating the intersection of which two lines.

(a) The perpendicular bisectors of two sides.

(b) The bisectors of two of the angles.

(d) Two altitudes.

(e) It is not possible to construct a circle that circumscribes a triangle.

Q77. Form a triangle by connecting the centers of three circles. Each of these circles is tangent to the other two and their radii are 1, 2, and 3 units long. How large is area of the triangle?

<<Geometry Unsolved Fig 13>>

(a) 12 (b) √5.π

(e) 6

Q78. Given the graph below, how many different paths are there from “S” to “E” if one never visits the same point twice?

<<Geometry Unsolved Fig 14>>

(a) Less than 11 (b) 11

(e) More than 13

Q79. A pentagon is inscribed in a circle of radius 7. How long is the circular arc that connects two neighboring vertices?

(a) 1.4π (b) 2.1π

(e) None of these

Q80. If a rectangle has a diagonal of length c and a perimeter of length p then the expression for its area is?

(a) pc (b) (p² – 4pc²)/8

(e) None of these

Q81. A rectangular solid with a square base has dimensions 5 × 5 × 8. If its volume is quadrupled by doubling the lengths of sides of the base, by what factor is its surface area increased?

(a) 4 (b) 2

(e) None of these

Q82. How long is the room shown in the figure on the right, given that it has a 7.5 foot ceiling, and an infra red beam that starts at one end of the room is bounced off the ceiling then the floor and finally hits the opposite side of the room 5.4 feet above the floor? The point at which it bounces off the ceiling is two feet from the wall.

<<Geometry Unsolved Fig 17>>

(a) 8.6 ft (b) 10.6 ft

(e) None of these

Q83. Find the height of a square pyramid formed by four equilateral triangles whose sides all have length 2.

(a) 1 (b) √6/2

(e) None of these

Q84. If a rectangle whose length is 9 times its width is modified so that its area is doubled but its perimeter is kept constant, what is the ratio of length to width for the new rectangle?

(a) 5 + √5 : 5 - √7 (b) 2 : 1

(e) None of these

Q85. One of the five statements below is false and the other four are true. Who told the falsehood? Ann said, “If the car was not locked then the wallet was stolen.” Bo said, “If the tickets to the game are lost then Ann's statement is false.”

Chris said, “Ann's statement is true”

Dan said, “The tickets to the game are not lost.”

Ed said. “The wallet was stolen but not the tickets to the game.”

(a) Ann (b) Bo

(e) Ed

Q86. A Quadrilateral ABCD is inscribed in a circle. If the size of the angle at vertex A is 36° then the angle at vertex C is:

(a) 72° (b) 54°

(e) None of these

Q87. The following facts are given: a gallon of paint covers 400 square feet of wall space, the room to be painted has an 8 foot ceiling, its dimensions are 20 by 14 feet, it has two doors (36 by 84 inches) and four large windows (72 by 60 inches), and you need two coats of paint. How much paint do you need to paint the walls?

(a) 1.47 gal. (b) 2.06 gal.

(e) None of these

Q88. For the points (0, 2), (6, 6) and (10, 0) which of the following statements are true?

I. The points form the vertices of a right triangle.

II. The points form the vertices of an isosceles triangle.

III. The largest angle is at the vertex located on (6,6).

(a) only III (b) II and III

(e) None are true

Q89. Given you have 4 sticks, two of length 5 and two of length 8, with which you are to form a quadrilateral. If at least one of the angles is a right angle, how many different no congruent quadrilaterals could you form.

(a) 1 (b) 2

(e) More than 4

Q90. A triangle with sides 6, 8, and 10 has its shortest side doubled in length while the other two sides remain the same. What is the area of the new triangle?

(a) 30 (b) 40

(e) None of these

Q91. What is the size of an angle between two adjacent sides of a regular 12 sided polygon?

(a) 144° (b) 120°

(e) None of these

Q92. As shown in the figure on the right six similar triangles are each sharing one side with the next triangle and all are sharing one vertex. All angles at that vertex measure 60°. If the side of the last (smallest) triangle that is adjoining the first triangles 1/6 as large as the longest side of that first triangle, how many times larger is the area of the largest triangle as compared to the smallest?

<<Geometry Unsolved Fig 18>>

(a) 6

(b) <<Geometry Unsolved Fig 19>>

(d) 36

(e) None of these

Q93. Three views of the same block are shown on the right. What letter is on the side parallel to the side with the letter A?

<<Geometry Unsolved Fig 21>>

(a) E (b) O

(e) G

Q94. An arbelos is the region formed by three mutually tangent circles whose centers are collinear, as noted in the image. If the diameters of the two smaller circles are a and b, what is the area of the arbelos.

<<Geometry Unsolved Fig 22>>

(a) abπ. (b) √abπ/4

(e) None of these

Q95. A bicycle has a 70 cm diameter wheel. If you ride in a 120 km race, approximately how many revolutions does the wheel have to make to complete the race?

(a) 54,600 (b) 171,400

(e) None of these

Q96. If the volume of a tetrahedron is doubled without changing its shape, by what factor is the surface area increased?

(a) <<Geometry Unsolved Fig 23>>

(b) <<Geometry Unsolved Fig 24>>

(d) √8

(e) 4

Q97. A semicircle with diameter 12 inches is used to form a conical cup by bending the semicircle so that its two corners are connected and the circle's center forms the point of the cone. How large is the volume?

(a) 9π√3 (b) 27π√3

(e) None of these

Q98. Chord <<Geometry Unsolved Fig 25>> is parallel to the line tangent to the circle at point C. If the distance between the chord and the tangent line is 18, and the radius of the circle is 13, how long is the chord <<Geometry Unsolved Fig 26>> ?

<<Geometry Unsolved Fig 27>>

(a) 12 (b) 2√105

(e) None of these

Q99. A rectangle with dimensions 11 by 13 had its diagonal increased by 50% without lengthening the shorter side. Approximately how big is the area of the new rectangle?

(a) 214.5 (b) 321.75

(e) 253.597

Q100. Nikki sees that the top of a 15 foot lamp, which is 250 feet away, lines up perfectly with the peak of a distant mountain. Nikki knows that the mountain is 15 miles away so she uses the lamp to determine the mountain's height. If Nikki's eyes are 5 feet above the ground, what is the best estimate of the mountain's height relative to Nikki?

(a) 900 feet (b) 3,178 feet

(e) 6,648 feet

Solution1: Area = √[ s(s-a) (s-b) (s-c)]

Here s = 21 cm. so, area = √(21 x 8 x 7 x 6) = 84 cm²

ð r = A/s = 84/21 = 4cm.

Ans. (2)

Solution2: Suppose the three sides are a, b, c,

i.e., a = 20, b = 10

so that s = (a + b + c)/2 = (20 + 10 + c)/2 = (30 + c)/2

By hypothesis,

80 = √[s(s – a) (s – b) (s – c) ]

= √[(30 + c)/2 ((30 + c)/2 – 20) ((30 + c)/2 – 10) x ((30 + c)/2 – c) ]

or 80 = √[ (225 – c²/4)(c²/4 – 25) ]

Write c²/4 = y, so that 80 = (225 – y)(y – 45) or y = 65

Hence, c = √(4 x 65 )= √ 260

Ans=(1)

Solution3: r = A/s, Here A = Ö3/4 x 6² and s = 9 => r = Ö3 cm.

Ans.(2)

Solution4: Forming the equations we have

a = 3/5 h a + h = 16

so 3/5 h + h = 16 or = 5/8 x 16 = 10

For right triangle the circum radius = Hypotenuse/2 = 5 cm

Ans. (3)

Solution 5: 14mtr

<<Geometry Unsolved Fig 28>>

Area of the square field = 14 x 14 = 196 m²

Now, as the horses can reach each other, the rope length is 7 mtrs

Area one horse can graze=90/360 x p7x7 (remember area of sector of a circle)

Area grazed by 4 horses = 4 x p/4r² = 22/7 x 7 x 7 = 154m².

Area of circular pond in the centre = 20m²(Given)

\Remaining area = 196 – 154 – 20 = 22m².

Ans=(1)

Solution6: Let PA = x and B = y. From rt. Ð ed DHPA,

y² = x² + x² = 2x²

or x = y/Ö2 ….(i) <<Geometry Unsolved Fig 29>>

Also x + y + x = 2

or x = (2 – y)/2

Putting it in (i),

(2 – y)/2 = y/Ö2 or Ö2(2 – y) = y

(2 – y) = Ö2y or 2 = (1 + Ö2)y

\ y = 2/(Ö2 + 1)

Ans=(3)

Solution7: 1/4^th part of sphere contains, lateral surface area + 2 semicircle of radius 8. ¼(4pr²) + 2 (1/2 pr²) = 2pr². So the total surface area = 4 x 2pr² = 512p cm².

Ans.(2)

Solution8: Volume of the cube = 100 x 100 x 100

= 10⁶cm³. Volume of the square bar = X² x 1600cm²

=>1000000 = X² x 1600 => X = 25cm

=> Volume of 1 cm length of bar = 1 x 25 x 25 = 625 cm³

=> Required weight = (900 x 625) / (1000000)

=> 0.5625 grams.

Ans.(3)

Solution9: Curved surface area of the tent = prL

since L² = r² + h² so curved surface

= p x 9.75 xÖ(2.8)² + 9.75)² = 310.7 m².

Ans. (2)

Solution10: Surface Area of the hemisphere = 3pR²= 462 cm²

= Total Surface Area of cone = pR² + pR Ö(H² + R²)

ð Putting the value of R = 7 cm, we get H = 12.12 cm.

Ans.(2)

Solution11: Volume of the earth dug out = 40 x 40 x 8 = 12800m3

The area over which this is spread = (160 x 120) – (40 x 40) = 17600 m²

ð The level increases by h = 12800/17600 = 0.82 m

= 82cm.

Ans.(3)

Solution12: Cross sectional area of the cistern = pR²= 346.5 cm².

Volume of the iron block= 40 x 11 x 12 = 5280cm³

ð Increase in height = 5280/346.5 = 15.24 cm.

Ans. (2)

Solution13: Since a side of the triangle is 10cm, when we cut it the side of the hexagon becomes 10/3 = 3.33 cm

= Area of the hexagon = 6 x area of an equilateral triangle

= 6 x Ö3/4 x 3.33² = 28.8 sqcm.

Ans. (2)

Solution14: We have 3pR² = 4/3 pR³ => R = 9/4 or Diameter = 9/2 = 4.5 units.

Ans. (1)

Solution15: Total volume of 5 cylinders = p x 10² x (1 + 2 + 3 +4 +5) = 4710m³.

Volume of hemisphere = 2/3 pR³ = 4710

\ R = 13.1 cm Surface Area = 3pR²

= 1616.5cm² (approx).

Ans.(1)

Solution16: Weight of the disc cut off = 1/3 of the weight of the original disc. Let D be the density => Volume = Weight/Density

r is the radius of the hole and R is radius if original disc.

ð pr²hD = 1/3 x pR²hD => r = 6.75 cm

diameter of hole= 6.75 cm x 2 = 13.5 cm.

Ans.(2)

Solution17: Volume discharge by 1 pipe = Volume discharged by 6 pipes

=> Volume per sec = Area x Speed

ð Area of bigger pipe = Total area of 6 smaller pipes

ð pR² = 6 x p x 5² = R = 12.25 cm =>D = 24.5 cm

Ans.(3)

Solution18: For this question, we require Euler’s formula:

Number of Faces + Number of Vertices = Number of edges +2

ð Number of edges = 50.

Ans.(3)

Solution19: Volume = pR²H => Volume = 1000, R = 10

=> H = 3.18 cm => Curved surface area = 200sq.cm.

Now Old Total Surface Area = 828 cm;

New Total Surface Area = 4pR² + 2pRH = 1456sq.cm.

ð Required percent charged = (1455 – 828)/828 x 100% = 75.84%.

Ans.(4)

Solution20: Let the length and breadth of the rectangular field be a and b respectively.

According to the question,

Diagonal + a/2 = a + b, As diagonal of a rectangle= √(a²+ b² )

=> √( a²+ b²) + a/2 = a + b => a + b – a/2 = √(a² + b²)

=> a/2 + b = √(a² + b²)

On squaring both sides, we get

a²/4 + b² + 2 x a/2 x b = a² + b²

ð a²/4 + ab = a² + b² – b²

ð => a² – a²/4 = ab

ð => 3a²/4 = ab

ð => 3a/4 = b

ð a/b = 4/3

ð => b/a = 3/4

Ans=(4)

Solution21: We have volume of bath = 2 x 3 x 4 = 24 m³

Rate of flow = Area x speed = p/4 x (1/16)² x 6 m x 60 = 1.104m³/min.

Time = 24/1.104 = 21.8 minutes.

Ans. (1)

Solution22: The rise can be found as –

Vol. increased/area of tank = (48 x 36 x 15)/(72 x 60) = 6cm Ans.(4)

Solution23: Volume of hollow sphere = 4/3 p(7³ – 5³)

volume of solid sphere= 4/3 p R³ = 4/3 p 218 = 6.018 cm.

Ans.(1)

Solution24: Let R be the radius of the recast sphere. 4/3pR³

= 4/3 p(3³ + 4³ + 5³) =>R³ = 216 => R = 6.

Total surface area of 3 spheres

= 4p(3² + 4² + 5²) = 4p x 50 = 200p cm²

Surface area of the recast sphere = 4p x 6² = 144p cm2

% Reduction in area = (200p - 144p)/200p x 100% = 28%

Ans.(4)

Solution25: Let h be the removed part of the cone. Then 16/20 = h/(h + 48)

=>h = 192 cm.

The required volume = 1/3 x p(20)² (48 + 192) – 1/3 x p(16)²(192)

= 1/3p(400 x 240 – 256 x 192) = 49049 c.c.

Ans.(1)

Solution26: Radius of the sphere = 6/2 cm = 3cm.

Volume of the sphere = [4/3 p x (3)³] cm³ = (36p)cm³.

Let the radius of the circular end of the wire be r cm.

Length of the wire = 36 cm.

Volume of the wire = (p² x 36)cm³.

Now, volume of the wire = volume of the sphere

=> 36pr² = 36p => r² = 1 => r = ±1.

But, radius cannot be negative.

Hence, the radius of the wire is 1 cm.

Ans=(1)

Solution 27: Radius of the sphere = 3cm.

Volume of the sphere [4/3 p x (3)³] cm³ = (36p)cm³.

Let the rise in the water level be h cm.

Increase in the volume when the sphere is submerged

= (p x 6 x 6 x h)cm³= (36ph)cm³.

This volume must be equal to the volume of the sphere

\ 36ph = 36p => h = 1 cm.

Hence, rise in the water level is 1 cm.

Ans=(3)

Solution 28: From DABC,

AC² = 24² + 32² = 1600

\ AC = 40 cm

so that AE = EC = 20cm <<Geometry Unsolved Fig 30>>

\ DE = √(25² - 20²)

= √(625 – 400) = √225 = 15

Hence, required area of figure

= Area of DABC + Area of DDAC

= ½ x 32 x 24 + ½ x 40 x 15 = 384 + 300 = 684 cm².

Ans=(1)

Solution 29: Clearly, each coin is a cylinder in which

r = 0.75cm = 75/100cm = 3/4cm,

and h = 0.2 cm = 2/10cm = 1/5cm.

Volume of each coin = pr²h

= [p x (3/4)² x 1/5]cm³ = (9p/80)cm³.

For the required cylinder, we have

Radius of the base (R) = (4.5/2)cm = 2.25cm,

and height (H) = 10cm.

Volume of the cylinder formed

= [p x (2.25)² x 10]cm³ = (405p/8)cm³.

Required number of coins = volume of the cylinder formed/volume

of 1 coin

= (405p/8 x 80/9p) = 450

Ans=(2)

Solution 30: Volume of iron = (440 x 260 x 100)cm³.

Internal radius of the pipe = 30cm.

External radius of the pipe = (30 + 5)cm = 35cm

Let the length of the pipe be h cm. Then, volume of iron in the pipe

= (external volume) – (internal volume)

= [p(35)²h - p(30)²h] cm³ = ph[(35)² – (30)²]cm³.

= (65 x 5) ph cm³ = (325ph)cm³.

\ 325ph = 440 x 260 x 100

ð length = h cm = (440 x 260 x 100 x 7/325 x 22) cm

ð 11200 cm = 112m.

Hence, the length of the pipe is 112m.

Ans=(4)

Solution31: If x be the exterior angle, then according to the question

x = 1/11 (180^o – x) => 11x =180o – x => 12x = 180^o or x = 15^o

Number of sides = 360^o/15^o = 24 (Remember formulae)

Ans.(2)

Solution32: Since the sum of the interior angles of a Pentagon

= (2 x 5 – 4) x 90^o = 540^o

So, x + x + 20 + x + 40 + x + 60 + x + 80 = 540

ð 5x + 200 = 540 => 5x = 340 or 68^o

Ans. (3)

Solution33: D BEC ~= D BDC (Draw the figure yourself)

\ ÐB = ÐC = (180 – 36)/2 = 72^o each.

Ans. (4)

Solution34: Let AB = x, BC = y

so that, AE = EF = FB = x/3

Area of DCEF = Area of DBCE – Area of DBCF

= ½ x 2x/3 x y –1/2 x x/3 x y = xy/6

Also, Area of rect. ABCD = xy

Hence, reqd. ratio = (xy/6)/xy = 1/6.

Ans=(1)

Solution35: Let x be the exterior angle, so (180^o – x) is the interior angle

^ð(180^o – x) – x = 140^o => 2x = 40^o

\ x =20^o

Sum of the exterior angles of n sides polygon is 360^o

\Number of sides = 360^o/20^o = 18.

Ans.(1)

Solution36: Let the sides be 3n and 2n

Now, (6n – 4)/3n x 2n/(4n – 4) = 10/9

so 18n – 12 = 20n – 20 or 2n = 8 or n = 4

\ sides are 12, 8.

Ans.(3)

Solution37: From DAFD,

x + 40^o + y = 180^o

or x + y = 140^o …(i)

Again, from Ds BAC and ECD,

2x + t = 180^o …(ii) <<Geometry Unsolved Fig 31>>

and 2y + l = 180^o …(iii)

Adding (ii) and (iii),

2(x + y) + t + l = 360^o

or 2 x 140 + t + l = 360^o [Using (i)]

or t + l =80^o (iv)

Finally, t + z + l = 180^o or 80^o + z = 180^o [Using (iv)]

or z = 100^o

Ans=(1)

Solution38: Draw the diagram yourself

As RS = SQ (given)

\ ÐQRS = ½ (180^o – 120^o) = 30^o

\ ÐQRP = 180^o – 30^o = 150^o

Hence ÐQPS = ½*(180^o – 150^o) = 15^o (as PR = RQ).

Ans.(1)

Solution39: Draw the diagram yourself

OB = OC (As O is circumcentre), so ÐOBC = b

Also, ÐBOC = 2a (angle subtended by chord BC at centre is twice of in alternate segment)

In D BOC, we have b + 2a + b = 180^o

\ a + b =90^o.

Ans.(4)

Solution40: Draw the diagram yourself

As DAOB and DCDO are similar

\ AB/CD = AO/DO => 5/4 = 10/DO

\ DO = 8 cm.

Ans.(2)

Solution41: Draw the figure yourself

Clearly ÐOSD = ÐSOD => SD = OD

Further ÐSDO = 90^o

\ DOSD is isosceles right angled.

Ans.(3)

Solution42: Draw the figure yourself

PR2 = PQ² + QR²

PR² = 6² – 7² = 85

Again PR2 = PS² + RS²

\ 85 = PS² + 2²

ð PS² = 81 or PS = 9

Ans.(4)

Solution43: Draw the figure yourself

Let OP = x

In DOPB: 4² + x² = r². In DOQD: 3² +√ (1 + x²) = r²

\ 4² + x² = 3² + (1 + x)². i.e. x = 3. \ 2r = 2 4² – x² = 10

Ans.(3)

Solution44:

AB = √ (15² + 20² )= √(225 + 400) = 25

Area of DABC = ½ x 15 x 20 = 150 sq. units.

\1/2 x 25 x CD = 150 => CD = 12 units.

From DADC, AD = √(15² + 12² )= Ö81 = 9 units

\ BD = 16 units.

From DADC, S = AC + CD + DA/2 = 15 + 12 + 9/2 = 18

\ radius = D/S = (½ x 9 x 12)/18 = 3

From DBCD, S = BD + DC + CB/2 = 16 + 12 + 20/2 = 24

\ radius = D/S = (1/2 x 16 x 12)/24 = 96/24 = 4

\ Required distance PQ = 3 + 4 = 7.

Ans=(3)

Solution45: Draw the figure yourself

In a right triangle mid point on hypotenuse is always equidistance from each vertex i.e. BM = MC = AM

But BC = 6² + 8² = 10

\ AM = 5 cm.

Ans.(1)

Solution46: CD = CE. ((Draw the figure first)

Ans.(1)

Solution47 Draw the figure yourself

ÐBCO = 20^o and OB = BC

So ÐBOC = 20^o

ÐABO = 20^o + 20^o = 40^o

(Exterior angle ABO = Sum of two Interior opposite angles)

\ ÐOAB = 40^o (AO = BO)

Thus ÐAOB = 180^o – (40^o + 40^o) = 100^o

\ ÐAOD = 180^o – (100^o + 20^o) = 60^o

Ans.(2)

Solution48: Draw the figure yourself

If a is the side of the square then area of the square = a²…..(i)

DABE + DECF + DADF

½ x a a/2 + ½ x a/2 x a/2 x ½ x a x a/2 = 5a²/8

\ Area of DAEF = a² – 5a²/8 = 3a²/8…….(ii)

From (i) and (ii), ratio of areas of DAEF : square ABCD

3a²/8: a²= 3:8.

Ans.(1)

Solution49 Draw the figure yourself

: OM = OP – MP = r – 2

\ OA² = OM² + AM²

r² = (r – 2)² + 4² or r = 5 cm.

Ans.(3)

Solution50: Draw the figure yourself

Since ABC is an isosceles triangle, so angle ACB is also 15. Hence

ÐCBA = 150^o

and exterior ÐCBA=30^o

\number of sides = 360^o/30^o = 12.

Ans.(2)

Soluition51 Draw the figure you150^o yorrself

AB² + AC² = 2(BD² + AD²) (Remember this formulae)

ð AB² + 36 = 2(25 + 25)

ð AB² = 100 – 36 = 64

\ AB = 8.

Ans.(4)

Solution52: AB = BE (As ABE is isosceles triangle)

\ ½ x AB x BE = 7

ð 1/2AB² = 7

ð AB = Ö14

EC = 3BE (Given)

\ BC = 4BE = 4AB

Now area of ABCD = AB x BC = AB x 4AB = 4AB² = 4 x 14 = 56cm².

Ans=(4)

Solution53: Since the area is 16 times as the original, so the sides are Ö16 =times i.e. 4 times the sides of the given triangle. (Remember properties of similar triangles)

Hence the required length of the side

= 4 x 5 = 20 cm

Ans.(2)

Solution54: Draw the figure yourself

Draw AD perpendicular on BC.

AB=BC=AC (Given)

And BP=1/3BC (given)=1/3 AB

BD=DC=1/2BC (Remember properties of equilateral triangles)=1/2AB

So PD=BD-BP=1/2BC-1/3BC

=1/6BC=1/6AB

Now in DAPD,

AP²=AD²+PD² (Pythagoras theorem)

Or AP²=(AB²- BD²)+(1/6AB)²(Pythagorus theorem in DABD)

Or AP²=(AB²-1/4AB²+1/36AB²

Or AP²=7/9AB²

So AP²/ AB²=7/9

Ans (1)

Solution55:

<<Geometry Unsolved Fig 32>>

Let O and O’ be the centers of the circles of radii 20 cm and 15 cm,

respectively and let PQ be their common chord. We have,

OP = 20cm, O’P = 15cm, OO’ = 25cm.

Let OL = x, \LO’ = 25 - x

Again let PQ = y \ PL = y/2

In right triangle OLP

OL = √ (OP² – LP² )= 20² – (y/2)² => x² = 400 – y²/4 …(i)

In right triangle O’LP, O’L = √(15² – y²/4)

(25 – x)² = 225 – y²/4 …(ii)

By (i) And (ii) we get

x² – 625 + 50x – x² = 400 – y²/4 – 225 + y²/4

=> 50x =625 + 175 => x = 800/50 = 16

From equation (i)

256 = 400 – y²/4 => y²/4 = 400 – 256 = 144 => y² = 144 x 4

ð y = 12 x 2 = 24cm.

Ans=(1)

Solution56: ÐBAC = 90^o [angle in a semicircle].

In DABC, we have

ÐABC + ÐACB + ÐBAC = 180^o

ð 38^o + ÐACB + 90^o = 180^o

ð ÐACB = (180^o – 128^o) = 52^o.

\ ÐBAT = ÐACB = 52^o [angles in alternate segments].

Ans=(1)

Solution57: Let BOQ intersect the circle at C.

Join CA.

ÐACB = ÐPAB = 58^o [Ðs in alternate segments.]

Now, in DACB, we have

ÐCAB + ÐABC + ÐACB = 180^o

ð 90^o + ÐABC + 58^o =180^o

ð ÐABC = (180^o – 148^o) = 32^o.

\ ÐABQ = 32^o [because ÐABQ = ÐABC].

Also, ÐPAB = ÐABQ + ÐAQB [Exterior-Angle Theorem]

ð 58^o = 32^o + ÐAQB

ð ÐAQB = (58^o - 32^o) = 26^o.

Hence, ÐABQ = 32^o and ÐAQB = 26^o.

Ans=(1)

Solution 58: Using BPT theorem, we have

AD/BD = AE/CE => (4x – 3)/(3x – 1) = (8x – 7)/(5x – 3)

ð (4x – 3)/(5x – 3) = (8x – 7)/(3x – 1)

ð 20x² – 27x + 9 = 24x² – 29x + 7

ð 4x² – 2x – 2 = 0 => 2x² – x – 1 = 0

ð (2x + 1)(x – 1) = 0

ð x = - ½ or x = 1.

But, x = -1/2 makes AD = [4 x (-1/2) - 3)] cm = -5 cm.

Since distance can never be negative, therefore x ¹ –1/2.

Hence, x = 1

Ans=(3)

Solution 59: For similar triangles,

Ratio of perimeters = Ratio of Proportional sides

ð 30/20 = 15/X => X =10cm

Ans=(4)

Solution 60 : Use similar triangle property: (Draw the figure yourself)

Height of tower : Height of stick = Shadow of tower : Shadow of

stick

X : 12 cm = 40 m : 8 cm => X = 60 m.

Ans-(2)

61(b). In an equilateral triangle, the centroid, as well as the in center and circumcenter, is located two -thirds of the distance from the vertex to the opposite side.

Then

2(x)2 = x2 + 32 => 3x2 = 9 => x = . This makes the area ½(2√3)³ = 3√3

<<Geometry Unsolved Fig 33>>

62(b). Placing the runners on a number line, we see that there is a spread of 20 seconds from first to last, that Jo was at the end, Sam in second 19 second ahead of Jo and one second behind Chris, the winner. Pat then was in third 12 seconds behind Sam.

63(b). In the figure, we see that

CE = DE = 3√2.

Also FD = BC + CE = 12 + 3√2 and AF = 5 – 3√2.

Thus the distance we want,

AD = <<Geometry Unsolved Fig 34>> = <<Geometry Unsolved Fig 35>> = 16.26

<<Geometry Unsolved Fig 36>>

64(a). Points on both circles must satisfy both equations, so both (1) (x – 1)2 + (y + 2)2 = 9 and (2) (x + 3)2 + (y – 1)2 = 16. Squaring both, and subtracting gives, (x2 + 6x + 9 + y2 – 2y + 1) – (x2 – 2x + 1 + y2 + 4y + 4) = 16 – 9, or simply 8x – 6y = 2 => 4x – 3y = 1.

65(d). The new length and width are 400 + 2w and 300 + 2w, so new area is

(400 + 2w)(300 + 2w) = 2(400)(300) =>

120000 + 1400w + 4w2 = 240000 =>

w2 + 350w – 3000 = 0 =>

w = <<Geometry Unsolved Fig 37>>

66(c). A good guess, since the hypotenuse is 13, would be the 5 – 12 – 13 right triangle. The area is indeed 30 and the perimeter is 30. If you did not guess this, you could use the equations a2 + b2 = 132 = 169 and 1/2 ab = 30 => 2ab = 120. Adding and subtracting the second equation from the first gives a2 + 2ab + b2 = 169 + 120 = 289 and a2 – 2ab + b2 = 169 – 120 = 49. Thus (a + b)2 = 289 => a + b = 17 and (a – b)2 = 49 => a – b = 7. These last two imply that a = 12 and b = 5.

67(a). The simplest way to find this is to put a rectangle around the given triangle and then subtract off the areas of the right triangles that surround it. (See Figure). The area of the entire rectangle is 126, but after subtracting the areas of the 3 right triangles, whose areas are 54, 3, and 42, we are left with an area of 27.

<<Geometry Unsolved Fig 38>>

68(e). We want the two angles, from vertical or the 12 on the clock, going clockwise, to be equal for both hands. The minute hands starts at an angle of zero and rotates at 1/60 revolutions per minute. The hour hand, which begins at 1/4th of a revolution, moves at 1/720 revolutions per minute. (It takes 12 hours times 60 minutes to make one complete revolution.) Setting these equal we have

(1/ω) t= ¼ + 1/720 t => 12t = 180 + t => 1 lt = 180 => t = <<Geometry Unsolved Fig 39>>

69(a). Let the sides have length x and 4.5.

Then we have 4.5x = 2(2x + 9), so 0.5x = 18 => x = 36.

70(b). The circumference of the can us cπ = 2πr, so the radius is r = c/2. This side of the square piece is the same as the circumference. The volume is

V = πr2h = . <<Geometry Unsolved Fig 40>>

71(a). By the power of a point formula, we know that x2 = 1.2 × 7.5 = 9 => x = 3

<<Geometry Unsolved Fig 41>>

72(d). The area of a rectangle is base times height. In the figure we see that the base is √2 + 1 and the height is √2/2 , so the area is A = Bh

= (√2 + 1)√2/2 = (2 + √2)/2

<<Geometry Unsolved Fig 42>>

73(c). z + y < 7 => 0 < z < 7 and 0 < y < 7. x + z > 10 and 0 < z < 7 => x ³ 3. x ³ 3 => y ³ 3. y ³ 3 and 0 < y < 7 => y = 3, 4, 5, 6.

74(e). In the figure we see that the area of the shaded region is x(a + b). The area of the entire triangle is
1/2(2x)(2a +2b) = 2x(a + b).

Since the area of the shaded region is one -half the area of the entire triangle, the ratio of the shaded region to the unshaded region is 1:1.

<<Geometry Unsolved Fig 43>>

75(c). The three configurations are shown here. In each case the area must be the sum of the two areas, since there is no overlap. The other statements are false, depending on the arrangement.

<<Geometry Unsolved Fig 44>>

76(a). The center of the circumscribed circle must be the same distance from each vertex. The perpendicular- bisector of one side is equidistant from two vertices, so the point of intersection of two perpendicular-bisectors is equidistant from all 3 vertices.

77(a). The segments joining the centers will be formed by the radii, so the sides of the triangle are 3, 4, and 5 units long, making it a 3-4-5 right triangle with are 6.

78(a). There are 2 ways to get to point B, so we can start at B and then double our answer. A tree diagram shows there are 8 ways to get from B to S without visiting any point twice.

<<Geometry Unsolved Fig 45>>

79(c). The radius is 7, so the circumference is 14π. Divide this by 5 to get 2.8π for the arc length.

80(e). The perimeter is 2a + 2b = p => a + b = p/2.

Squaring we have (a + b)2 = (p/2)² => a2 + 2ab + b2 = (p/4)².

But the diagonal c2 = a2 + b2, so subtracting we get 2ab = (p/4)² – c² => ab = (p² – 4c²)/8 .

81(d). The volume of the original solid is 200 cubic units. When two sides are doubled, the volume goes to 800. The original surface areas was 2.52 + 4.(5.8) = 210. The new surface area will be 2.102 + 4(10.8) = 520. The ratio of the volumes is 520/210 = 52/21 =2(10/21).

82(b). In the figure was see that triangle MBN is similar to triangle ORN, so BM/BN = OR/RN => 3/2 = 7.5/RN => RN = 5. Similarly, triangle NQO is similar to triangle PDO, so NQ/QO = PD/DO => 7.5/5 = 5.4/Do => DO = 3.6. Thus the length of the rectangle is 10.6.

<<Geometry Unsolved Fig 46>>

83(c). Draw in diagonal AC and look at triangle APC. The length of AC is 2√2. so that makes triangle APC an isosceles right triangle sna the altitude is √2.

<<Geometry Unsolved Fig 47>>

84(a). The old area was 9w2 and the perimeter was 20w. The new area will be 18w2. The perimeter stays the same so the new length and width, L and W are such that L = 10w – W, and (10w – W)W = 18w2. This is quadratic in W, so w2 – 10wW – 18w2 = 0

=> W = 10w ± √(100w² – 72w²)/2. This simplifies to W = (5 ± √7) => L = <<Geometry Unsolved Fig 48>>

. Taking the length

to be the longer side, the ratio is (5 + √7)/(5 – √7).

85(e).