Question 1

The least multiple of 7, which leaves a remainder of 4, when divided by 6, 9, 15 and 18 is?

Accepted Answer

Let the least multiple of 7 be x, which when divided by 90 leaves the remainder 4. Then, x is of the form 90k + 4 Now, the minimum value of k for which 90k + 4 divisible by 7 is 4 ∴ x = 90 x 4 + 4 = 364

Question 2

The largest natural number, which exactly divides the products of any four consecutive natural numbers, is?

Accepted Answer

In any 4 consecutive natural numbers: Two numbers will be divisible by 2, one of which will be divisible by 4, at least one number will be divisible by 3. These facts give us a lower bound of 2 x 4 x 3 = 24 for the solution. But 24 is also the product of the first 4 consecutive natural numbers, so it is also an upper bound.

Question 3

The total number of prime factors of the products (8) 20, (15) 24, (7) 15 is?

Accepted Answer

Since 2, 3, 5, 17 are prime numbers and The given expression = (23)20 x (3 x 5)24 x (17)15 = 260 x 324 x 524 x 1715 So the total number of prime factors in the given expression is (60 + 24 + 24 + 15 ) = 123.

Question 4

The traffic lights at three different road crossings changes after every 48 sec., 72 sec. and 108 sec. respectively. If they all change simultaneously at 8 : 20 : 00 hrs, then they will again change simultaneously at?

Accepted Answer

Interval of changes = (L . C . M of 48, 72, 108) sec. = 432 sec. So, the lights will simultaneously changes after every 432 seconds i,e 7 min. 12 sec. So, the next simultaneously changes will take place at 8 : 27 : 12 hrs.

Question 5

Four metal rods of lengths 78 cm, 104 cm, 117 cm and 169 cm are to be cut into parts of equal length. Each part must be as long as possible. What is the maximum number of pieces that can be cut?

Accepted Answer

Given, length of four metal roads are 78, 104, 117 and 169 cm. Now, 78 = 13 x 2 x 3, 104 = 13 x 2 x 2 x 2, 117 = 13 x 3 x 3, 169 = 13 x 13 Length of each piece of rod as long as possible, HCF = 13 CM ∴ Number of piece = 6 + 8 + 9 + 13 = 36

Question 6

There are five hobby clubs in a college viz. photography, yachting, chess, electronics and gardening. The gardening group meets every second day, the electronics group meets every third day, the chess group meets every fourth day, the yachting group…

Accepted Answer

Gardening group meets once in 2 days, electronics group meets once in 3 days, chess group meets once in 4 days, yachting group meets once in 5 days and the photography group meets once in 6 days. If they meet on the same day at one time, then the next time they will meet on same day again will be the LCM of 2, 3, 4, 5 and 6 which is equal to 60. Hence, within 180 days all the five groups will meet on the same day = 180/60 = 3 times.

Question 7

A, B and C start at the same time in the same direction to run around in 252s, B in 308s and C in 198s, all starting at the same point. After what time will they meet again at the starting point?

Accepted Answer

The time at which all the three persons meet will be the LCM of the time taken by each person individually to complete one round. 252 = 22 x 71 x 91 308 = 22 x 71 x 111 198 = 21 x 91 x 111 ∴ LCM of 252, 308 and 198 = 22 x 71 x 91 x 111 = 2772 So, A, B and C will again meet at the starting point in 2772s i.e., 46 min and 12s

Question 8

What is the LCM of (x 2 - y 2 - z 2 - 2yz), (x 2 - y 2 + z 2 + 2xz) and (x 2 + y 2 - z 2 - 2xy)?

Accepted Answer

x2 - y2 - z2 - 2yz = x2 - (y + z)2 = (x+ y + z)(x - y - z) x2 - y2 + z2 + 2yz = (x + z)2 - y2 = (x + y + z) (x + y - z) x2 + y2 - z2 - 2xy = (x - y)2 - z2 = (x - y + z) (x - y - z) ∴ LCM = (x + y + z) (x - y - z) (x - y + z)

Question 9

A man has four copper rods whose lengths are 52 m, 65 m, 78 m, 91 m, respectively. This man wants to cut pieces of same length from each of four rods. what is the least number of total pieces if he is to cut without any wastage?

Accepted Answer

The pieces cut from four rods is least, so length of the pieces is the HCF of 52, 65, 78 and 91 which is 13. ∴ Number of pieces cut =(52 + 65 + 78 + 91)/13 = 22

Question 10

A, B, and C start at the same time in the same direction to run around a circular stadium. A completes a round in 252s, B in 308s and C in 198s, all starting of the same point. After what time will they meet again at the string point?

Accepted Answer

The time which all the three persons meet will be the LCM of the time by each person individually to complete one round. 252 = 22 x 71 x 91 308 = 22 x 71 x 111 198 = 21 x 91 x 111 ∴ LCM of 252, 308 and 198 = 22 x 71 x 91 x 111 = 2772 So,A,B and C will again meet at the starting point in 2772 i.e.,46 min and 12s.

Question 11

A person has completely put each of three liquids 403 L of petrol, 456 L of diesel and 496 L of mobile oil in in bottles of equal size without mixing any of the above three types of liquids such that each bottle is completely filled. What is the lea…

Accepted Answer

HCF of 403,456 and 496 = 31 Now, the number of bottles required to put 403 L of petrol = 403/31 = 13 Similarly,the number of bottles required to put 456 L of diesel = 465/31 = 15 The number of bottles required to put 496 L of mobile oil = 496/31 = 16 ∴ Least number of bottles required having 31 L as capacity = 13+15 + 16 = 44

Question 12

30 pineapple trees, 45 orange tree and 60 mango trees have to be planted in rows such that each row contains the same number of trees of one variety only. What is minimum number number of row in which the trees may be planted?

Accepted Answer

Minimum number of rows = Maximum number of trees in each row = HCF of 30,45 and 60 = 15 ∴ Number of rows = (30 + 45 + 60)/15 = 9

Question 13

If HCF of x 3 - 10m 2 + 31x - 30m and x 2 - 8mx + 15 is a linear polynomial, then what is the value of m?

Accepted Answer

Go through options. Put m = 1 in two given expression,we have x3 - 10x2 + 31x - 30 and x2 - 8x + 15 x2 - 8x + 15 = (x - 5) (x - 3) x3 - 10x2 + 31x - 30 = (x - 2) (x - 5) (x - 3) ∴ (x - 5) (x - 3) = x2 - 8x + 15 is the HCF of both expressions. ∴ m = 1

Question 14

What is the um of the digits of the least number which when divided by 54 leaves 35 as remainder, when divided by 80 leaves 61 as remainder and when divided by 119 leaves 100 leaves 100 as remainder?

Accepted Answer

(54 - 35) = 19, (80 - 61) = 19 and(119 - 100) = 19 LCM of 54, 80 and 119 is 257040. Required least number = 257040 + 19 = 257059 Sum of the digits of 257059 = 2 + 5 + 7 + 0 + 5 + 9 = 28

Question 15

What is the sum of digits of the least number,which when divided by 52 leaves 33 as remainder, when divided by 78 leaves 59 as remainder and when divided by 117 leaves 98 as remainder?

Accepted Answer

(52 - 33) = 19, (78 - 59) = 19 and (117 - 98) = 19 LCM of 52, 78 and 117 is 468. Required least number = 468 + 19 = 487 Sum of digits of 487 = 4 + 8 + 7 = 19

Question 16

The ratio of two numbers is 3 : 4 and their HCF is 4. What will be their LCM?

Accepted Answer

According to the question, 1st number = 3M, 2nd number = 4M where, M = HCF But given, M = 4 We know that, LCM = Product of two numbers / HCF = (3M x 4M) / M LCM = 12M = 12 x 4 = 48

Question 17

The sum of two numbers is 1056 and their HCF is 66, Find the Number of such pairs.

Accepted Answer

Let the numbers be 66a and 66b, where a and b are co-primes. According to the question, 66a + 66b = 1056 ⇒ 66(a + b) = 1056 ⇒ (a + b) = 1056/66 = 16 ∴ Possible values of a and b are (a = 1, b = 15), (a = 3, b = 13). (a = 5, b = 11 ), (a = 7, b = 9) ∴ Numbers are (66 x 1, 66 x 15), (66 x 3, 66 x 13), (66 x 5, 66 x 11), (66 x 7, 66 x 9). ∴ Possible number of pairs = 4

Question 18

The HCF and LCM of two natural numbers are 12 and 72, respectively. What is the difference between the two numbers, if one of the number is 24?

Accepted Answer

Second number = (LCM x HCF) / first number = (72 x 12)/ 24 = 36 Difference between the two numbers = 36 - 24 = 12

Question 19

How many numbers are there between 4000 and 6000 which are exactly divisible by 32, 40, 48 and 60?

Accepted Answer

LCM of 32, 40, 48 and 60 = 480 The number divisible by 480 between 4000 and 6000 are 4320, 4800, 5280 and 5760. Hence, required number of numbers are 4.

Question 20

For any integer n, What is HCF (22n + 7, 33n + 10) equal to?

Accepted Answer

HCF of (22n + 7, 33n + 10) is always 1. lllustraction For n = 1, HCF(29, 43) ⇒ HCF = 1 For n = 2, HCF(51, 76) ⇒ HCF = 1 For n = 3, HCF(73, 109) ⇒ HCF = 1

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