Question 1

The smallest number which when diminished by 7, is divisible by 12, 16, 18, 21 and 28 is

Accepted Answer

Required Number = (L.C.M of 12, 16, 18,21,28)+7 = 1008 + 7 = 1015

Question 2

L.C.M of two prime numbers x and y (x>y) is 161. The value of 3y-x is :

Accepted Answer

H. C. F of two prime numbers is 1. Product of numbers = 1 x 161 = 161. Let the numbers be a and b . Then , ab= 161. Now, co-primes with product 161 are (1, 161) and (7, 23). Since x and y are prime numbers and x >y , we have x=23 and y=7. Therefore, 3y-x = (3 x 7)-23 = -2

Question 3

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

Accepted Answer

L.C.M. of 6, 9, 15 and 18 is 90. Let required number be 90k + 4, which is multiple of 7. Least value of k for which (90k + 4) is divisible by 7 is k = 4. =>Required number = (90 x 4) + 4 = 364.

Question 4

If the sum of two numbers is 55 and the H.C.F. and L.C.M. of these numbers are 5 and 120 respectively, then the sum of the reciprocals of the numbers is equal to:

Accepted Answer

Let the numbers be a and b. Then, a + b = 55 and ab = 5 x 120 = 600. The required sum = 1 a + 1 b = a + b a b = 55 600 = 11 120

Question 5

The ratio of two numbers is 3 : 4 and their H.C.F is 4. Their L.C.M is

Accepted Answer

Let the numbers be 3x and 4x . Then their H.C.F = x. So, x=4 Therefore, The numbers are 12 and 16 L.C.M of 12 and 16 = 48

Question 6

The H.C.F and L.C.M of two numbers are 11 and 385 respectively. If one number lies between 75 and 125 , then that number is

Accepted Answer

Product of numbers = 11 x 385 = 4235 Let the numbers be 11a and 11b . Then , 11a x 11b = 4235 => ab = 35 Now, co-primes with product 35 are (1,35) and (5,7) So, the numbers are ( 11 x 1, 11 x 35) and (11 x 5, 11 x 7) Since one number lies 75 and 125, the suitable pair is (55,77) Hence , required number = 77

Question 7

Three numbers are in the ratio 1 : 2 : 3 and their H.C.F is 12. The numbers are

Accepted Answer

Let the required numbers be x, 2x, 3x. Then, their H.C.F =x. so, x= 12 Therefore, The numbers are 12, 24, 36

Question 8

H.C.F of 4 x 27 x 3125, 8 x 9 x 25 x 7 and 16 x 81 x 5 x 11 x 49 is

Accepted Answer

180

Question 9

The product of two numbers is 4107. If the H.C.F of these numbers is 37, then the greater number is

Accepted Answer

Let the numbers be 37a and 37b. Then , 37a x 37b =4107 => ab = 3. Now co-primes with product 3 are (1,3) So, the required numbers are (37 x 1, 37 x 3) i.e, (37,111). Therefore, Greater number = 111.

Question 10

The H.C.F of two numbers is 11 and their L.C.M is 7700. If one of the numbers is 275 , then the other is

Accepted Answer

Other number = 11 × 7700 275 = 308

Question 11

If HCF of two numbers is 8,which of the following can never be their LCM?

Accepted Answer

60 is not a multiple of 8

Question 12

The least number which when divided by 5, 6, 7 and 8 leaves a remainder 3, but when divided by 9 leaves no remainder, is

Accepted Answer

L.C.M of 5, 6, 7, 8 = 840 Therefore, Required Number is of the form 840k+3. Least value of k for which (840k+3) is divisible by 9 is k = 2 Therefore, Required Number = (840 x 2+3)=1683

Question 13

A heap of coconuts is divided into groups of 2, 3 and 5 and each time one coconut is left over. The least number of Coconuts in the heap is ?

Accepted Answer

To get the least number of coconuts : LCM = 30 => 30 + 1 = 31

Question 14

LCM of 80, 85, 90

Accepted Answer

LCM of (80, 85, 90) can be found by prime factorizing them. 80 → 2 × 2 × 2 × 2 × 5 85 → 17 × 5 90 → 2 × 3 × 3 × 5 L.C.M of (80,85,90) = 2 × 2 x 2 × 2 × 3 × 3 × 5 × 17 = 16 x 9 x 85 = 144 x 85 = 12240 L.C.M of (80,85,90) = 12240.

Question 15

A drink vendor has 368 liters of Maaza, 80 liters of Pepsi and 144 liters of Sprite. He wants to pack them in cans, so that each can contains the same number of liters of a drink, and doesn't want to mix any two drinks in a can. What is the least nu…

Accepted Answer

The number of liters in each can = HCF of 80, 144 and 368 = 16 liters. Number of cans of Maaza = 368/16 = 23 Number of cans of Pepsi = 80/16 = 5 Number of cans of Sprite = 144/16 = 9 The total number of cans required = 23 + 5 + 9 = 37 cans.

Question 16

The maximum number of students among them 1001 pens and 910 pencils can be distributed in such a way that each student gets the same number of pens and same number of pencils is:

Accepted Answer

Required number of students = H.C.F of 1001 and 910 = 91

Question 17

Find the Greatest Number that will devide 43, 91 and 183 so as to leave the same remainder in each case

Accepted Answer

Required Number = H.C.F of (91- 43), (183- 91) and (183-43) = H.C.F of 48, 92, and 140 = 4

Question 18

LCD of 12 and 18

Accepted Answer

LCD is nothing but Lowest or Least Common Denominator Here LCD of 12 and 18 means LCD of two fractions with denominators 12 and 18 respectively. Therefore, LCM of 12 & 18 = 6 x 3 x 2 = 36 How to calculate LCD :: The lowest common denominator or least common denominator (LCD) is the least common multiple (LCM) of the denominators of a set of fractions.

Question 19

Which of the following has the most number of divisors?

Accepted Answer

99 = 1 x 3 x 3 x 11 101 = 1 x 101 176 = 1 x 2 x 2 x 2 x 2 x 11 182 = 1 x 2 x 7 x 13 So, divisors of 99 are 1, 3, 9, 11, 33, .99 Divisors of 101 are 1 and 101 Divisors of 176 are 1, 2, 4, 8, 11, 16, 22, 44, 88 and 176 Divisors of 182 are 1, 2, 7, 13, 14, 26, 91 and 182. Hence, 176 has the most number of divisors.

Question 20

54, 60 LCM is

Accepted Answer

Here the LCM of 54, 60 is => 2 x 3 x 3 x 3 x 10 = 6 x 9 x 10 = 54 x 10 = 540.

HCF and LCM Questions