fundoogyan's Blog

iift 2008 paper with answer key

Attachment: iift2008 paper wih key.pdf (255.0KB)

IRMA GK question Bank

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

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

Attachment: SNAP2008.pdf (684.0KB)

know the basic fundas of geometry

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

writing skill for GRE/GMAT

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

Attachment: Writing_skillsGRE-GMAT.pdf (7124.0KB)

general knowledge questions

(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

(C) Maruti-Suzuki                                               (D) General Motors

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

depressing for other cars"?

(A) Hyundai                                                       (B) Scorpio

(C) Maruti 800                                                    (D) Ford IKON NXT

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

(A) Hero Honda                                                  (B) Bajaj Auto

(C) Kinetic Honda                                               (D) LML Industries

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

(C) Bank of Maharashtra                                     (D) HDFC

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

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

(C) the Italian (D) None of these

Example 2: The Norwegian drinks

(A) milk (B) cocoa

(C) tea (D) fruit juice

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

(A) white (B) red

(C) blue (D) Cannot be determined

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,

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

(C) Passion and Cartier (D) None of these

Example 6: The one who likes Omega likes

(A) Passion (B) either Passion or Fiero

(C) Fiero (D) Cannot be determined

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

(A) Omega (B) Cartier

(C) Tissot (D) Cannot be determined

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

(A) Pulsar (B) Passion

(C) Enticer (D) Either (b) or (c)

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

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

(C) Both are required

(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

(C) Wahida is elder to Yogesh

(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.

(C) no more than one management person.

(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.

(C) may be a management 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.

(C) two empty seats between union people.

(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

(C) D and E (D) F and D

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

(C) D (D) A and D

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

(C) I, III and IV (D) I 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.

(C) E has not occurred. (D) A must occur.

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

(A) B (B) C and D

(C) B and E (D) C, D and E

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

(A) A (B) B

(C) A and E (D) Cannot say

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

(C) The Guptas eat karela in the yellow dinner set

(D) None of these

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

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

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

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

(C)   Uncertain

(D)   Cannot be determined.

Problem 2:        Which is a feasible group of three

(A)   Ram, David, Fiza

(B)   Shyam, David, Peter

(C)   Kavita, David Shyam

(D)   None of these.

Problem 3:        Which is a feasible group of four?

(A)   Ram, David, Fiza, Shyam

(B)   Shyam, David, Kavita, Rahim

(C)   Fiza, David, Ram, Peter

(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

(C)   Feasible groups of three and four cannot be made simultaneously.

(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)

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.

(C)   Jhansi ki Rani, 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

(C) Baleno                                (D) Escort

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

(C) 2                                        (D) 5

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

(C) 8 years                               (D) 10 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

(C) Tuntun and Omprakash                                (D) Omprakash 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

(C) Omprakash                                     (D) Tuntun

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

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

(C) Keshto and Omprakash                   (D) Tuntun 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

(C) A and B                              (D) C and D

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

                        (C) 2                                        (D) 4

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.

(C)   black with a short 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

Ans=(D)

Solutions 6 to 9

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

Now see the information in each question

Solution6:          Under the given conditions, the arrangement would be

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

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

(C)   Uncertain

(D)   Cannot be determined.

Problem 2:        Which is a feasible group of three

(A)   Ram, David, Fiza

(B)   Shyam, David, Peter

(C)   Kavita, David Shyam

(D)   None of these.

Problem 3:        Which is a feasible group of four?

(A)   Ram, David, Fiza, Shyam

(B)   Shyam, David, Kavita, Rahim

(C)   Fiza, David, Ram, Peter

(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

(C)   Feasible groups of three and four cannot be made simultaneously.

(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)

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.

(C)   Jhansi ki Rani, 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

(C) Baleno                                (D) Escort

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

(C) 2                                        (D) 5

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

(C) 8 years                               (D) 10 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

(C) Tuntun and Omprakash                                (D) Omprakash 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

(C) Omprakash                                     (D) Tuntun

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

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

(C) Keshto and Omprakash                   (D) Tuntun 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

(C) A and B                              (D) C and D

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

                        (C) 2                                        (D) 4

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.

(C)   black with a short 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

Ans=(D)

Solutions 6 to 9

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

Now see the information in each question

Solution6:          Under the given conditions, the arrangement would be

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)

PROBABILITY

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

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 = 8C₃ x 5³ x 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 = 6C₃ x 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 - 6C₅ is 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 = 11C₆ x 6C₄ x 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

(c) Three circles and three 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