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NUMBER SYSTEM

Categories > Fundoo Unsolved Examples

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fundoogyan

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Q1. Which of following numbers are divisible by 12?

(a) 66312 (b) 17525

Q2. What maximum power of 2 would perfectly divide 40!?

(a) 48 (b) 28

Q3. How many zeros will be there in the value of 50!

(a) 12 (b) 18

Q4. What is the unit’s digit in 1896?

(a) 2 (b) 8

Q5. Which of the following number is prime?

(a) 593 (b) 667

Q6. Which of the following is the HCF of 440 , 1800 and 2800.

(a) 400 (b) 40

Q7. What least number must be added to 11661 so that it is exactly divisible by 22?

(a) 1 (b) 41

Q8. Which of the following is the LCM of 299, 221 and 759?

(a) 167739 (b) 166640

Q9. Which is the largest three digit number which when divided by 6 leaves the remainder 5 and when divided by 5 leaves remainder 3?

(a) 990 (b) 983

Q10. What is the remainder when 1559 × 1163 is divided by 20?

(a) 17 (b) 13

Q11. What is the total number of different divisors including 1 and the number that can divide the number 12800?

(a) 20 (b) 27

Q12. The largest number amongst the following that will perfectly divide 101100 – 1 is

(a) 100 (b) 10000

Q13. Which is greater of the two 2300 or 3200

(a) 2300 (b) 3200

(d) cannot be determined

Q14. log6216 v6 is: (CAT 1994)

(a) 3 (b)3/2

Q15. Three bells chime at intervals of 18 min, 24 min and 32 min respectively. At a certain time, they begin together. What length of time will elapse before they chime together again? (CAT 1995)

(a) 2 hr and 24 min

(b) 4 hr and 48 min

(d) 5 hr

Q16. Two positive integers differ by 4 and sum of their reciprocals is 10/21. then one of the numbers is: (CAT 1995)

(a) 3 (b) 1

Q17. The remainder obtained when a prime number greater than 6 is divided by 6 is :

(a) 1 or 3 (b) 1 or 5

Q18. For the product n(n + 1)(2n +1 ) , n e N, which one of the following is not necessarily true? (CAT 1995)

(a) It is even

(b) Divisible by 3

(d) Never divisible by 237

Q19. The value of is :

(CAT 1995)

(a) 100 (b) 105

Q20. 72 hens cost Rs …96.7 … then what does each hen cost, where two digits in place of ‘…’ are not visible written in illegible hand writing? (CAT 1995)

(a) Rs. 3.23 (b) Rs. 5.11

Q21. If n is any odd number greater than 1 , then n(n2 – 1) is : (CAT 1996)

(a) Divisible by 96 always

(b) Divisible by 48 always

(d) None of these

Q22. If a number 774958A96B is divisible by 8 and 9, the respective values of A and B will be: (CAT 1996)

(a) 7 and 8 (b) 8 and 0

Q23. ABC is a three digit number in which A > 0. The value of ABC is equal to the sum of the factorials of its three digits. What is the value of B? (CAT 1997)

(a) 9 (b) 7

Q24. P,Q and R are three consecutive odd numbers in ascending order. If the value of three times P is 3 less than two times R, find the value of R. (CAT 1997)

(a) 5 (b) 7

Q25. Which of the following is true? (CAT 1997)

(a) 7^3^2 = ( 73)2 (b) 7^3^2 > ( 73)2

Q26. If m and n are integers divisible by 5, which of the following is not necessarily true?

(a) (m – n) is divisible by 5

(b) (m2 – n2) is divisible by 25

(d) none of these (CAT 1997)

Q27. log2 [ log7 (x2 – x + 37 )] = 1, then what could be the value of x ? (CAT 1997)

(a) 3 (b) 5

Q28. P and Q are two positive integers such that PQ = 64. Which of the following cannot be the value of P + Q ? (CAT 1997)

(a) 20 (c)16

(b) 65 (d) 35

Q29 A student instead of finding the value of 7/8 of the number, found the value of 7/18 of the number. If his answer differed from the actual one by 770, find the number.

(a) 584 (b) 2520

Q30. If n is an integer, how many values of n will give an integral value of (16n2 + 7n + 6)/n ?

(CAT 1997)

(a) 2 (b) 3

Q31. A , B , C…..Y, Z are the players who participated in a tournament. Everyone played with every other player exactly once. A win scores 2 points , a draw scores 1 point and a lose scores none. None of the matches ended in a draw. No two players scored the same score. At the end of the tournament, a ranking list is published which is in accordance with the alphabetical order. Then : (CAT 1998)

(a) M wins over N

(b) N wins over M

(d) none of these

Q32. A hundred digit number is formed by writing first 54 natural numbers in front of each other as 12345678910111213…… find the remainder when this number is divided by 8. (CAT 1998)

(a) 1 (b) 7

Q33. A is a set of positive integers such that when divided by 2,3,4,5,6 leaves the remainders 1,2,3,4,5 respectively. How many integers between 1 and 100 belong to set A. (CAT 1998)

(a) 0 (b) 1

Q34. A certain number when divided by 899 leaves the remainder 63. find the remainder when the same number is divided by 29: (CAT 1998)

(a) 5 (b) 4

Q35. Three wheels can complete 60,36,24 revolutions per minute respectively. There is a red spot on each wheel that touches the ground at time zero. After how much time, all these spots will simultaneously touch the ground again? (CAT 1998)

(a) 5/2s (b) 5/3s

Q36. Number of students who have opted for the subjects A,B and C are 60 , 84 and 108 respectively. The exam to be conducted for these students such that only the students of the same subject are allowed in one room. Also the number of students in each room must be same. What is the minimum number of rooms that should be arranged to meet all these conditions? (CAT 1998)

(a) 28 (b) 60

Q37. (BE)2 = MPB, where B,E,M and P are distinct integers, then M = ? (CAT 1998)

(a) 2 (b) 3

Q38. n3 is odd. Which of the following statements is (are) true? (CAT 1998)

A. n is odd B. n2 is odd C. n2 is even

(a) A only (b) B only

Q39. For two positive integers a and b, define the function h(a,b) as the greatest common factor (GCF) of a,b. let A be a set of n positive integers G (A) the GCF of the elements of set A is computed by repeatedly using the function h. the minimum number of times h is required to be used to compute G is: (CAT 1999)

(a) 1/2n (b) (n – 1 )

Q40. The remainder when 784 is divided by 342 is: (CAT 1999)

(a) 0 (b) 1

Q41. Let a, b, c be distinct digits. Consider a two digit number ‘ab’ and a three digit number ‘ccb’, both defined under the usual decimal number system, if (ab)2 = ccb > 300 , then the value of b is :

(CAT 1999)

(a) 1 (b) 0

Q42. If n2 = 12345678987654321, what is n? (CAT 1999)

(a) 12344321 (b) 1235789

Q43. Convert the number 1982 from base 10 to base 12. the result is: (CAT 2000)

(a) 1182 (b) 1912

Q44. ABCDEFGH is a regular octagon. A and E are opposite vertices of the octagon. A frog starts jumping from vertex to vertex, beginning from A. from any vertex of the octagon except E, it may jump to either of the two adjacent vertices. When it reaches E, the frog stops and stays there. Let an be the number of distinct paths of exactly n jumps ending in E. then, what is the value of a2n-1? (CAT 2000)

(a) 0 (b) 4

Q45. Let N = 553 + 173 – 723. N is divisible by : (CAT 2000)

(a) both 7 and 13 (b) both 3 and 13

Q46. Let S be the set of integers x such that (CAT 2000)

I 100< x < 200 II x is odd

III x is divisible by 3 but not by 7

How many elements does S contain? (CAT 2000)

(a) 16 (b) 12

Q47. The integers 34041 and 32506, when divided by a three digit integer n, leave the same remainder. What is the value of n? (CAT 2000)

(a) 289 (b) 367

Q48. Let N = 1421 × 1423 × 1425. What is the remainder when N is divided by 12? (CAT 2000)

(a) 0 (b) 9

Q49. Consider a sequence of seven consecutive integers. the average of the first five integers is n. the average of all the seven integers is: (CAT 2000)

(a) n

(b) n + 1

(d) n + ( 2/7 )

Q50. What is the value of the following expression? (CAT 2000)

[1/(22–1)] + [1/(42–1)] + [1/(62–1) ] + ……[1/(202–1)]

(a) 9/19 (b) 10/19

Q51. Let D be a recurring decimal of the form D = 0.a1a2a1a2a1a2….., where digits a1 and a2 lie between 0 and 9. Further at most one of them is zero. Which of the following numbers necessarily produces an integer, when multiplied by D? (CAT 2000)

(a) 18 (b) 108

Q52. In a number system, the product of 44 and 11 is 2124. the number 3111 of this system, when converted to the decimal number system becomes: (CAT 2001)

(a) 406 (b) 1086

Q53. Three friends, returning from a movie, stopped to eat in a restaurant. After dinner, they paid their bill and noticed a bowl of mints at the front counter. Sita took 1/3 of the mints, but returned 4 because she had a monetary pang of guilt. Fatima then took 1/4 of what was left but returned 3 for the similar reasons. Eswari then took half of the remainder but threw two back into the bowl. The bowl had only 17 mints left when the raid was over. How many mints were originally in the bowl?

(CAT 2001)

(a) 38 (b) 31

Q54. M is the smallest positive integer such that for any integer n > m, the quantity n3 – 7n2 + 11n – 5 is positive. What is the value of m? (CAT 2001)

(a) 4 (b) 5

Q55. Anita had to do a multiplication. Instead of taking 35 as one of the multipliers, she took 53. As a result, the product went up by 540. What is the new product? (CAT 2001)

(a) 1050 (c) 1040

(b) 540 (d) 1590

Q56. In a 4 digit number, the sum of the first two digits is equal to that of the last two digits. The sum of the first and last digits is equal to the third digit. Finally, the sum of the second and fourth digits is twice the sum of the other two digits. What is the third digit of the number? (CAT 2001)

(a) 5 (b) 8

Q57. A red light flashes 3 times per minute and a green light flashes 5 times in two minutes at regular intervals. If both lights start flashing at the same time, how many times do they flash together in each hour? (CAT 2001)

(a) 30 (b) 24

Q58. Three pieces of cakes of weight 4 1/2 lbs , 6 3/4 lbs and 7 1/5 lbs respectively are to be divided into parts of equal weights. Further, each part must be as heavy as possible. If one such part is served to each guest, then what is the maximum number of guests that could be entertained? (CAT 2001)

(a) 54 (b) 72

Q59. 76n – 66n, where n is an integer > 0 , is divisible by:

(CAT 2002)

(a) 13 (b) 127

Q60. After the division of a number successively by 3, 4 and 7, the remainders obtained are 2, 1 and 4 respectively. What will be the remainder if 84 divides the same number? (CAT 2002)

(a) 80 (b) 75

Q61. When 2256 is divided by 17, the remainder would be : (CAT 2002)

(a) 1 (b) 16

Q62. A child was asked to add first few natural numbers ( that is 1 + 2 + 3….) so long as his patience permitted. As he stopped, he gave the sum as 575. When teacher declared the result wrong, the child discovered he had missed one number in the sequence during addition. The number he missed was:

(CAT 2002)

(a) less than 10 (b) 10

Q63. A rich merchant had collected many gold coins. He did not want anybody to know about him. One day is wife asked, “How many gold coins do we have”? After pausing a moment, he replied, “well! If I divide the coins into two unequal numbers, then 48 times the difference between the two numbers equals the difference between the squares of the two numbers”. The wife looked puzzled. Can you help the merchant’s wife by finding out how many gold coins the merchant has? (CAT 2002)

(a) 96 (b) 53

Q64. The owner of a local jewellery store hired 3 watchmen to guard his diamonds, but a thief still got in and stole some diamonds. On the way out, the thief met each watchman, one at a time. To each he gave ½ of the diamonds he had then and 2 more besides. He escaped with one diamond. How many did he steal originally? (CAT 2002)

(a) 40 (b) 36

Q65. Number S is obtained by squaring the sum of digits of a two digit number D. if difference between S and D is 27, then the two digit number D is:

(CAT 2002)

(a) 24 (b) 54

Q66. Let x and y be positive integers such that x is prime and y is composite. Then, (CAT 2003)

(a) y – x cannot be an even integer

(b) xy cannot be an even integer

(d) none of these

Q67. Let n(>1) be a composite integer such that vn is not an integer. Consider then following statements

A : n has a perfect integer- valued divisor which is greater than 1 and less than vn

B : n has a perfect integer- valued divisor which is greater than vn but less than n (CAT 2003)

(a) both A and B are false

(b) A is true but B is false

(d) both A and B are true

Q68. What is the remainder when 496 is divided by 6?

CAT 2003)

(a) 0 (b) 2

Q69. What is the sum of all two digit numbers that give a remainder of 3 when they are divided by 7 ?

(CAT 2003)

(a) 666 (b) 676

Q70. Each family in a locality has at most two adults and no family has fewer than 3 children. Considering all the families together, there are more adults than boys, more boys than girls and more girls than families. Then the minimum possible number of families in the locality is : (CAT 2004)

(a) 4 (b) 5

Q71. On January 1, 2004 two new societies S1 and S2 are formed, each with n members. On the first day of each subsequent month S1 adds b members while S2 multiples its current number of members by a constant factor r. both the societies have the same number of members on July 2, 2004. If b = 10.5n, what is the value of r? (CAT 2004)

(a) 2.0 (b) 1.9

Q72. N persons stand on the circumference of a circle at distinct points. Each possible pair of persons, not standing next to each other, sings a two minute song one pair after the other. If the total time taken for singing is 28 minutes, what is N?

(CAT 2004)

(a) 5 (b) 7

Q73. Let S be the set of prime numbers greater than or equal to two and less than 100.

All such existing terms are then multiplied, then how many consecutive zeros would the final digit end with.

(1) 1 (2) 4

(3) 5 (4) 10

Solutions

Ans1. (a) Divisibility rule of 12, number has got to be divisible by 3 and 4.

66312 = 18, divisible by 3 and last two digits divisible 4.

Ans2. (d) It will be the number of 2’s in 40!

No. of 2’s(leave decimals)

= 40/2 + 40/22 + 40/23 + 40/24 + 40/25

onwards powers of 2 will exceed 40,

= 20 + 10 + 5 + 2 + 1 = 38

The maximum power will be 38.

Ans3. (a) The number of zeros are number’s of (5 × 2)’s

No. of 5’s(leave decimals)

= 50/5 + /52 onwards powers of 5 will exceed 25,

= 10 + 2 = 12, as twos are plenty, but 5’s are 12 therefore number of zeros is 12

Ans4. (c) 1896 = (1824) (1824) (1824) (1824)

Since we know multiple of 4 of power of 8, last digit is 6

Last digit = 6 × 6 × 6 × 6 = 6

Ans5. (a) try dividing by all prime numbers less than the rough square root of given number

Ans6. (b) 10 × 4 = 40

Ans7. (a) Least number to be added is the remainder when divided by the given number

Here when we divide 11661 by 22, quotient is 530 and remainder is 1, so the least number is 22 – 1 = 21

Ans8. (d) 299 = 13 × 23, 221 = 13 × 17,

759 = 23 × 3 × 11

LCM = 13 × 23 × 17 × 3 × 11 = 167739.

Ans9. (a) You can directly check the answer by using the given operator

Alternatively, Let A be the required number

Then A = 6x + 5 and A = 5y + 3 (with x and y as respective multiples)

The first such number = 23.

The largest number less than 1000 which is a multiple of 30 (LCM of 6,5) is 990 and the penultimate number is 960.

Hence N = 960 + 23 = 983.

Ans10. (a) Here when 20 divides 1559, remainder is 19.

And when 20 divide 1163 it is 3

Now the rule is the remainders multiplication has to be divided by divisor to get the remainder

Therefore 19 × 3 = 57. Dividing by 20 remainder is 17, which is the answer.

Ans11. (d) The number of divisors (including 1 and itself) of a given number N where

N = Am * Bn * Co … where A,B,C are prime numbers are (1 + m)(1 + n)(1 + o)…

Now, 6400 = 29 * 52. Here m = 8 and n = 2
Therefore number of divisors are (9 + 1)(2 + 1) = 10*3 = 30.

Ans12. (b) Since 1012 = 10201 and 1012 – 1 = 10200. This is divisible by 100.

And 1013 – 1 = 1030301 – 1 = 1030300. This is also divisible by 100.

Now 10110 will be divisible by 1000, since a zero will increase.

Therefore 101100 will be divided by 10000

Ans13. (b) As both can be expressed in powers of 100 with base 8 and 9, 2300 = 8100 and 3200 = 9100, therefore it is higher.

Ans14. (c) Let log6 216Ö6 = x, as logax = y Þ x = (a)y

Then 216Ö6 = (6)x

(6)3 × (6)1/2 = (6)x thus (6)7/2 = (6)x so x = 7/2

Ans15. (b) It is like the question of three people running around a circular track and when will they be together, which will be their LCM. Therefore the bells will chime together again after a time that is equal to the LCM of 18,24 and 32 = 288 min = 4 hr and 48 min.

Ans16. (a) Let one number be x, then second number will be (x + 4)

1/x + 1/(x+4) = 10/21 or x + x + 4/x(x + 4) = 10/21

2x + 4/x(x + 4) = 10/21,

10x2 – 2x – 84 = 0

Solving x = 3.

Ans17. Prime numbers greater than 6 are 7, 11, 13, 17, 23, 29 etc. If these numbers are divided by 6, the remainder is always either 1 or 5.

Ans18. (d) This question has to be solved intuitively rather than with using core mathematics, looking at options, the fourth option say never divisible by 237, now for n = 237, the expression n(n + 1) (2n + 1) is divisible by 237, which is not necessarily true, there fourth option is the answer.

Ans19. (a)

(55³ + 45³)/(55² – 55 x 45 + 45²) = (55³ + 45³)/(55² – 55 x 45 + 45²)

[As a3 + b3 = (a + b)(a2 + b2 – ab)]

= (55 + 45) = 100.

Ans20. (c) Here hens are 72, and price of each hen is given as choices, we have to multiply all four with 72 and check, 5.51 × 72 = 396.72 fits, rest do not so third choice is the answer.

Ans21. (c) For n = 3, n(n2 – 1) = 3 × 8 = 24

For n = 5, n(n2 – 1) = 5 × 24 = 120

For n = 7, n(n2 – 1) = 7 × 48 = 336

All divisible by 24, therefore the third choice is correct

Ans22. (b) Using the divisibility rules, For 8, the last three digits have to divisible by 8, therefore the number 774958A96B is divisible by 8 if 96B is divisible by 8. and 96B is divisible by 8 if it is 960 or 968 thus B is either 0 or 8.

For 9, the sum of the number has to be divisible by 9, therefore (55 + A + B) is divisible by 9 if (A + B) is 8. Now either of A or B could be 8, but the other has to be zero. Therefore (b) is the right answer.

Ans23. (c) The number is given to be = A! + B! + C!, Here the only way is to work through given choices, Value of B will be less than 7 because 7! = 5040 which is a four digit number. Therefore B can either be 4 or 2.

Now the value of A and C has to be 6 or 5 as lower than this the factorial with a combination of 4! Or 2! will not make a three-digit number. Now 6! = 720 and 5! = 144.

Taking B = 4, the number fitting is 145(1! + 4! + 5! = 1 + 24 + 120 = 145), with B = 2, nonumber fits, therefore third is the correct choice.

This clearly is not a question that can be solved in one minute by most students; students may like skip these kinds of questions in the first go at questions and like to get back trying to solve them if they get time later.

Ans24. (c) P,Q and R are three consecutive odd numbers therefore Q = P + 2 and R = P + 4

Given 3P = 2(P + 4) – 3

So, P = 5.

Therefore R = 5 + 4 = 9.

Ans25. (b) 7^3^2 = 79 and (73)2 = 76. Here clearly 79 > 76.

Ans 26. (c) Looking at choices (m–n) is divisible by 5, as both m and n are multiples of 5, and five can be taken common out of m and n. Now second choice can be written as (m + n)(m – n) and as per the logic above, 5 can be taken common out of both (m – n) and (m + n), therefore 5 × 5 will be common, which will make it divisible by 25. The third is (m+n) divisible by 10, which may not be always correct, for example 10+15 = 25, not divisible by 10, therefore this is the answer.

Ans27. (c) log2 [ log7 (x2 – x + 37 ) ] = 1

Log7 (x2 – x + 37 ) = (2)1 ( logax = y thus x = (a)y )

(x2 – x + 37) = (7)2

x2 – x + 37 – 49 = 0

x2 – x – 12 = 0 x = 4.

Ans28. (d) (P,Q) can be (1,64),(2,32),(4,16),(8,8). Hence P + Q cannot be 35.

Ans29. (a) Here let the number be x, he took out 7/18x instead of 7/8x

As per the question, 7/8x – 7/18x = 770

Solving, x = 1584.

Ans30. (c) Here (16n2 + 7n + 6 )/n = 16n + 7 + (6/n)

For the entire expression to become an integer (6/n) should be an integer. The values of n could be 1,2,3,6. Therefore n can have four values.

Ans31. (a) Since the ranking is as per the alphabetical order (A to Z). M scores higher than N to be above N, and therefore must win over N to score more victories(thus score) than N.

Ans32. (a) As per the rule of divisibility for 8, the number formed by last three digits has to be divisible by 8.The same rule will be applicable to find the remainder. The last three digits in the hundred digit number are 545. The remainder when 545 is divided by 8 is 1.

Ans33. (b) First we take the LCM of 2,3,4,5,6, which is 60, the number which all can divide. The next number which all of them can divide is 120, which is above 100, so it is rejected. Now for number to leave remainder 1 for 2, it should be 61 or 59, 61 cannot be as it leaves remainder 1 with 3, where we want 2. Checking for all 59 is the number, and it is the only number. Now students should also see that as soon as the LCM is 60, there could be only one number, which can satisfy the problem.

Ans34. (a) 899 is 29 × 31, therefore what is divisible by 899 is divisible by 29 also, now when remainder is 63, which can be written as 29 × 2 + 5, Here the component not divisible by 29 is 5, therefore is the answer.

Ans35. (c) For all the three wheels to touch the ground again simultaneously is equal to the LCM of the times taken by the 3 wheels to complete one revolution. Since the answers are given in seconds, lets convert the given data to seconds

Time taken by first wheel = 60/60 = 1 sec.

Time taken by the second wheel = 36/60 = 3/5 sec.

Time taken by the third wheel = 24/60 = 2/5 sec.

LCM of 1,3/5 ,2/5 = LCM of (1,3,2)/HCF of (1,5,5) = 6 sec.

Ans36. (d) Number of students which should be allowed in each room should be the HCF of 60, 84 and 108 which is 12.

Thus number of rooms required for subject A = 60/12 =5, subject B = 84/12 = 7 and subject C = 108/12 = 9 rooms. Total minimum rooms = (5+ 7+ 9) = 21.

Ans37. (b) As ( BE)2 = MPB. A three digit number, from which we can deduce BE should be less than 32, as (32)2 is a four digit number. Since the last digit of number is also B, B can only be 1, for no other number this can be fulfilled. Now for (BE)2 last digit to be 1, E can be 1 or 9, cannot be 1 as digits are distinct BE will become 11 for 1).Therefore it is 9, and BE is 19, (19)2 = 361, M = 3.

Ans38. (c) If n3 is odd then n and n2 has to be odd. One can take an example If n = 5, n2 = 25 ,n3 = 125.

Ans39. (b) If there are n numbers, for first two numbers the operator wil be use once, and then the result with be operated with the next number to give the HCF of three numbers. Thus for three numbers it is used twice, from here for n positive integers function h ( a, b ) has to be used one time less than the number of integers that is ( n – 1) times.

Ans40. (b) Students should understand by now that the closest number to divisor from power of 7 has to be found to solve the question. Therefore (7)84 / 342 = (73)28/342 = (343 )28/ 342, for 343 remainder = 1, as per remainder theorem it will be true for any power of 343.

Ans41. (a) (ab)2 = ccb > 300, The greatest possible value of ‘ab’ to be 31, as 322 is four digit number. As (31)2 = 961, therefore 300 < ccb < 961, Also ab has to be greater than 172, as it is less than 300, therefore 18< ab < 31, since (ab)2 = ccb, the last digit is same, which be possible for a square with last digit 1, 5, 6, so the possible values could be 21, 25, 26 the value of ab which satisfies (ab)2 = ccb is 21.

So (21)2 = 441 thus a = 2 , b = 1 and c = 4.

Ans42. (c) Just looking at last three digits of multiplication of choice one(last two digits (41) and two(last three digits 521), one knows they are incorrect, Now the choices left are third and fourth. From here students can take a guess it is (c) as it has nine digits and the given number has 17. For 8 digits (choice 3) the number will be lower.

Second solution:

It is a standard result on number system. Check this pattern

112 = 121

1112 = 12321

11112 = 1234321 and so on

when you are doing square of 4 one’s, write 1234 and then 321 so 1234321.

Similarly, 11111112 = 1234567654321

Though we don’t encourage remembering a lot of results/methods (they are close to infinite in Number system), this is a very standard result. And I am sure; you anyways would have grasped it for rest of your life by now.

Ans43. (c)

12 1982__

12 165 – 2

12 13 – 9

1 -1

The required number is 1192.

Ans44. (a) The frog can move either clockwise or anticlockwise in order to reach point E. in any case number of jumps required is 4.

For n = 4, distinct ways are 2, therefore a4 = 2, Now a2n–1 for n = 4 is a7, which means 7 jumps, now with seven jumps the frog cannot reach E, for any combination, so for a7, distinct ways are 0.

Ans45. (d) 55 + 17 + (–72) = 0, we have a + b + c = 0

we know when a + b + c = 0, then a3 + b3 + c3 = 3abc

Hence in the given question, 553 + 173 – 723 = 3 × 55 × 17 × 72

We can observer easily that 3 and 17 would be factors of this product.

Ans46. (d) Numbers between 100 to 200, which are divisible by 3 are 102,106 109, ….198. which is 198 = 102 + ( n-1) x 3, from here n = 33. Out of these 33 numbers 17 are even and 16 are odd. Out of these 16 odd numbers there are 3 odd numbers ( = 105 ,147,189) which are divided by the LCM of (7,3) i.e. 21.

Hence in all (16 – 3) = 13 numbers are contained in S.

Ans47. (d) Let the common remainder be x. then numbers ( 34041 – x) and ( 32506 – x) would be completely divisible by n. The difference of these numbers will also be divisible by or (34041 – x –32506 + x) = 1535 will also be divisible by n.

Now using options we find that 1535 is divisible by 307.

Ans48. (c) As per the remainder theorem the remainders for the three numbers could be multiplied and remainder for the number will be the remainder for the original number

Remainder for 1421, 1423, 1425 when divided by 12 is 5, 7, 9, multiplying these numbers = 5

× 7 × 9 = 315, remainder for 315 /12 = 3, which is the answer.

Ans49. (b) Average of first five integers

= 1 + 2 + 3 + 4 + 5 / 5 = 15/5

= 3 = n

And average of first seven integers

= 1 + 2 + 3 + 4 + 5 + 6 + 7/7

= 28/ 7 = 4 = 3 + 1 = n + 1

Ans50. (c)

1/1.3 + 1/3.5 + 1/5.7 + … + 1/19.21

= ½ ( 1 – 1/3 ) + ½ ( 1/3 – 1/5 ) + ½ (1/5 – 1/7) + …. + ½ ( 1/19 – 1/21)

= ½ (1 – 1/21) = ½ × 20/21 = 20/42 = 10/21

Ans51. (c) D = 0.a1a2a1a2a1a2 …

100D – D = a1a2

99D = a1a2

D = a1a2/99

The number should be the multiple of 99. Hence 198 is the required number.

Ans52. (d) The product of 44 and 11 is 484. 3111 is approximately 1.5 times 2124, therefore the required number should be approximately 1.5 times 484 which from choices is 691.

Ans53. (d) Number of mint before Eeswari took them

= (x – x/2) + 2 = 17

x = 30

Number of mint before Fatima took them

= (x – x/4) + 3 = 30

x = 36

Number of mint before sita took them

= (x – x/3 ) + 4 = 36

x = 48

Hence there were 48 mints originally.

Ans54. (b) Let y = n3 – 7n2 + 11n – 5

At n = 1,

y = 0

thus

n3 – 7n2 + 11n – 5 = (n – 1) (n2 – 6n + 5 )

= (n – 1)2 (n – 5)

Now for every value of n, (n – 1)2 is always + ve.(n – 5) is negative for all values < 5.

Hence for n = 6 , (n – 1)2 (n – 5 ) is positive. Thus, the smallest value of m is 5.

Ans55. (d) 53x – 35x = 540

or 18x = 540

or x = 30

The new product = 53 × 30 = 1590.

Ans56. (a) Let the first, second, third and fourth digits be a, b, c and d respectively.

Then, a + b = c + d ...(i)

a + d = c ...(ii)

b + d = 2(a + c) ...(iii)

from (i) and (ii) we get a + b = a + 2d b = 2d

from (iii) we get 2d + d = 2(a + a + d)

3d = 2 (2a + d) d = 4a

or a = d/4

Now from (ii) a + d = d/4 + d = 5d/4 = c

c = 5/4d.

The value of d can be either 4 or 8 . If d = 4, then c = 5. If d = 8 , then c = 10. But the value of c should be less than 10. Hence value of c should be 5.

Ans57. (a) First light blinks after 20 sec and second light blinks after 24 sec

Now they blink together after LCM of 20 and 24 sec = 120 sec = 2min

Hence, the number of times they blink together in an hour = 30.

Ans58. (d) Total weight = (9/2 + 27/4 + 36/5) = 369/20 = 18.45 lb

The weight of a single piece should be HCF of (9/2, 27/4, 36/5)

= HCF of ( 9, 27, 36 )/ LCM of (2, 4, 5) = 9/20 lb.

Number of guests = 18.45/9/20 = 18.45 × 20 /9 = 41

Ans59. (b) For n = 1, 76 – 66 = (73)2 – (63)2

= (73 – 63) (73 + 63) = (343 – 216)(343 + 216 )

= 127 × 559. Clearly it is divisible by 127.

Ans60. (d) The number = 3 {4 (7x + 4 ) + 1} + 2 = 84x + 53

Hence, if the number is divided by 84, the remainder is 53.

Ans61. (a) 2256 can be written as (24)64 = (17 – 1)64

In the expansion of ( 17 – 1 )64 every term is divisible by 17 except (–1)64. Hence the remainder is 1.

Ans62. (d) Sum of n natural numbers = n/2 [2a + (n – 1)d]

575 = n/2 [2 + (n – 1)]

575 = n(n + 1)/2

for n = 33

n(n + 1 )/2 = 551

for n = 34

n(n + 1 )/2 = 595

The difference should be 595 – 572 = 20, which is the number he missed

Ans63. (d) Let the two unequal numbers be x and y.

Then 48(x – y) = (x2 – y2)

x + y = 48.

Ans64. (b) Finally thief is left with one diamond. The number of diamonds before he gave some diamonds to the third watchman = x – (x/2 +2) = 1

= 1

or x = 6.

He had 6 diamonds before he gave 5 to the third watchman. Similarly, number of diamonds before giving to the second watchman = = 6 or x = 16

The number of diamonds before giving to the first watchman = 16

or x = 36.

Ans65. (b) Let D = ab, therefore the condition is (a + b)2 – ab (not a × b) = 27, Using options we find (5 + 4 )2 – 54 = 27.

Ans66. (d) Lets take x = 7 and y = 15 as examples, 15 – 7 = 8, so first is untrue, for second we can take x = 2, and y = 6, where 6 × 2 = 12, so second is untrue. Again = 4, which even, so third is untrue. Therefore fourth is the correct choice.

Ans67. (d) Consider a number

n = 8

then vn = 2.82

In the expansion of ( 17 – 1 )64 every term is divisible by 17 except (–1)64. Hence the remainder is 1.

Ans62. (d) Sum of n natural numbers = n/2 [2a + (n – 1)d]

575 = n/2 [2 + (n – 1)]

575 = n(n + 1)/2

for n = 33

n(n + 1 )/2 = 551

for n = 34

n(n + 1 )/2 = 595

The difference should be 595 – 572 = 20, which is the number he missed

Ans63. (d) Let the two unequal numbers be x and y.

Then 48(x – y) = (x2 – y2)

x + y = 48.

Ans64. (b) Finally thief is left with one diamond. The number of diamonds before he gave some diamonds to the third watchman = x – (x/2 +2) = 1

= 1

or x = 6.

He had 6 diamonds before he gave 5 to the third watchman. Similarly, number of diamonds before giving to the second watchman = = 6 or x = 16

The number of diamonds before giving to the first watchman = 16

or x = 36.

Ans65. (b) Let D = ab, therefore the condition is (a + b)2 – ab (not a × b) = 27, Using options we find (5 + 4 )2 – 54 = 27.

Ans67. (d) Consider a number

n = 8

then vn = 2.82

A : we have a divisor 2 which is greater than 1 and less than v6

B : we have a divisor 4 which is greater than v8 but less than 8

Thus both statements are true.

Ans68. (d) 496 = (42)48 = 1648, for 16 remainder is 4, it will stay that for any power of 16

Ans69. (b) First of all, we have to identify such 2 digit numbers. Obviously, they are 10, 17, 24,….. The required sum = 10 + 17 + …… 94.

Now this is an AP with a = 10, n = 13 and d = 7

13/2 [2 × 10 + (13 – 1) 7] = 676.

Ans70. (d) The number of adults > number of boys > number of girls > number of families. Going by the given choices for 2 families

5 > 4 > 3 > 2

5 adults in two families is not possible (as per “at most two adults”)

Going by 3 families

6 > 5 > 4 > 3

3 families can have a maximum of 9 children and 6 adults. Hence the minimum number of families is 3

Ans71 . (a) Let the number of members as on January 1, 2004 in S1 and S2 be n each. At the beginning of every month, members of society S1 are in arithmetic progression while those of S2 are in geometric progression. Hence on July 1, 2004 the number of members in S1 and S2 are n + ( 7 – 1 )b and nr7–1 i.e. n + 6b and nr6 respectively.

Hence n + 6b = n + 6(10.5n) = 64n = nr6

r6 = 64

or r = 2.

Hence the value of r = 2.

Ans72. (b) Each of the N persons form a pair with (N – 3) persons ( i.e. , excluding the person himself and the adjacent two). So the total number of pairs that can be formed = n(N – 3)/2

thus the total time they sing = N (N – 3)/2 ×2 = 28(given)

N(N –3) = 28
or N2 – 3N – 28 = 0

N = 7 (as N > 0).

Ans73. (1) The number of zero’s in a product depends on number of 2’s and 5’s in the product. As product of 2 and 5 can make a 0

From 2 to 100 in prime numbers, we have only one 2 and one 5.

So number of zero’s = 1

quote

29th May 2009 09:30 a.m.

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