Question 1

A rectangular courtyard 3.78 meters long 5.25 meters wide is to be paved exactly with square tiles, all of the same size. what is the largest size of the tile which could be used for the purpose?

Accepted Answer

3.78 meters =378 cm = 2 × 3 × 3 × 3 × 7 5.25 meters=525 cm = 5 × 5 × 3 × 7 Hence common factors are 3 and 7 Hence LCM = 3 × 7 = 21 Hence largest size of square tiles that can be paved exactly with square tiles is 21 cm.

Question 2

The L.C.M of 22, 54, 108, 135 and 198 is

Accepted Answer

22 = 2 x 11 54 = 2 × 3 3 108 = 2 2 × 3 3 135 = 3 3 × 5 198 = 2 × 3 2 × 11 ∴ L . C . M = 2 2 × 3 3 × 5 × 11 = 5940

Question 3

Product of two co-prime numbers is 117. Their L.C.M should be

Accepted Answer

H.C.F of co-prime numbers is 1. So, L.C.M = 117/1 =117

Question 4

The L.C.M of two numbers is 495 and their H.C.F is 5. If the sum of the numbers is 100, then their difference is

Accepted Answer

Let the numbers be x and (100-x). Then, x 100 - x = 5 * 495 => x 2 - 100 x + 2475 = 0 => (x-55) (x-45) = 0 => x = 55 or x = 45 The numbers are 45 and 55 Required difference = (55-45) = 10

Question 5

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 6

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 7

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 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

To find the HCF (Highest Common Factor) of the given numbers, we begin by breaking down each number into its prime factorization. Here are the steps: Prime factorization of 4 × 27 × 3125: 4 = 2² 27 = 3³ 3125 = 5⁵ So, the prime factorization of 4 × 27 × 3125 is 2² × 3³ × 5⁵. Prime factorization of 8 × 9 × 25 × 7: 8 = 2³ 9 = 3² 25 = 5² 7 = 7¹ So, the prime factorization of 8 × 9 × 25 × 7 is 2³ × 3² × 5² × 7¹. Prime factorization of 16 × 81 × 5 × 11 × 49: 16 = 2⁴ 81 = 3⁴ 5 = 5¹ 11 = 11¹ 49 = 7² So, the prime factorization of 16 × 81 × 5 × 11 × 49 is 2⁴ × 3⁴ × 5¹ × 7² × 11¹. Find the common prime factors and their lowest powers: For 2, the lowest power is 2². For 3, the lowest power is 3². For 5, the lowest power is 5¹. For 7, the lowest power is 7⁰ (since 7 is absent in the first number). For 11, the lowest power is 11⁰ (since 11 is absent in the first two numbers). Calculate the HCF: HCF = 2² × 3² × 5¹ = 4 × 9 × 5 = 180. Final answer: The HCF of the given numbers is 180.

Question 9

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 10

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 11

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 12

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 13

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 14

H.C.F of 3240, 3600 and a third number is 36 and their L.C.M is 2 4 x 3 5 x 5 2 x 7 2 . What is the third number?

Accepted Answer

47628

Question 15

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.

Question 16

If the HCF of 210 and 55 is expressible in the form of 210 x 5 + 55P, then value of P = ?

Accepted Answer

HCF of 210 and 55 is 5 Now, 210x5 + 55P = 5 => 1050 + 55P = 5 => 55P = -1045 => P = -1045/55 => P = -19.

Question 17

The difference of two numbers is 14. Their LCM and HCF are 441 and 7. Find the two numbers ?

Accepted Answer

Since their HCFs are 7, numbers are divisible by 7 and are of the form 7x and 7y Difference = 14 => 7x - 7y = 14 => x - y = 2 product of numbers = product of their hcf and lcm => 7x * 7y = 441 * 7 => x * y = 63 Now, we have x * y = 63 , x - y = 2 => x = 9 , y = 7 The numbers are 7x and 7y => 63 and 49

Question 18

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. Therefore, Required sum = 1 a + 1 b = a + b a b = 55 600 = 11 120

Question 19

A drink vendor has 80 liters of Mazza, 144 liters of Pepsi, and 368 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 n…

Accepted Answer

If we want to pack the drinks in the least number of cans possible, then each can should contain the maximum numbers of liters possible.As each can contains the same number liters of a drink, the number of liters in each can is a comman factor for 80,144 and 368; and it is also the highest such factor, as we need to store the maximum number of liters in each can. So, the number of liters in each can = HCF of 80,144 and 368 = 16 liters. Now, number of cans of Maaza = 80/16 = 5 Number of cans of Pepsi = 144/16 = 9 Number of cans of Sprite = 368/16 = 23 Thus, the total number of cans required = 5 + 9 + 23 = 37

Question 20

The H.C.F and L.C.M of two numbers are 84 and 21 respectively. If the ratio of the two numbers is 1 : 4 , then the larger of the two numbers is

Accepted Answer

Let the numbers be x and 4x. Then, x × 4 x = 84 × 21 ⇔ x 2 = 84 × 21 4 ⇔ x = 21 Hence Larger Number = 4x = 84

HCF and LCM Questions