HCF Questions

HCF questions are given here, along with solutions using different methods of finding the HCF of given numbers. All these questions cover the important applications of HCF in an easily understandable way. Practising all these questions will help the students to improve their problem-solving skills, to a great extent. In this article, you will get questions on finding the HCF of the given set of integers, decimal numbers, and fractions, along with suitable formulas to reach the solutions.

What is HCF?

HCF (Highest Common Factor) is the greatest number that divides the given set of two or more numbers. In other words, HCF is the product of the smallest power of each common prime factor in the numbers.

Also, check: HCF

HCF of any given numbers can be calculated using different methods, such as prime factorization, long division, and the short cut method.

Try out: HCF Calculator

HCF Questions and Answers

1. Find the HCF of 12 and 15.

Solution:

Given numbers: 12 and 15

Let us divide 15 by 12.

Therefore, the HCF of 12 and 15 is 3.

2. What is the highest common factor of 96 and 404?

Solution:

Prime factorization of 96 = 2 × 2 × 2 × 2 × 2 × 3 = 2⁵ × 3

Prime factorization of 404 = 2 × 2 × 101 = 2² × 101

HCF(96, 404) = 22 = 4

Therefore, the highest common factor of 96 and 404 is 4.

3. Find the HCF of 14, 36 and 42.

Solution:

Given numbers: 14, 36, 42

First, consider the least number as the divisor and the greatest number as the dividend.

Thus, 14 is the common factor of 14 and 42, so consider 14 as the divisor for 36.

Therefore, 2 is the HCF of 14, 36 and 42.

4. What will be the greatest possible length that can be used to measure exactly the lengths 7 m, 3 m 85 cm, 12 m 95 cm?

Solution:

7m = 7 × 100 cm = 700 cm

3m 85 cm = (3 × 100) cm + 85 cm = (300 + 85) cm = 385 cm

12 m 95 cm = (12 × 100) cm + 95 cm = (1200 + 95) cm = 1295 cm

Prime factorization of 700 = 2 x 2 x 5 x 5 x 7

Prime factorization of 385 = 5 x 7 x 11

Prime factorization of 1295 = 5 x 7 x 37

HCF(700, 385, 1295) = 5 x 7 = 35

Therefore, the greatest possible length that can be used to measure exactly the lengths 7 m, 3 m 85 cm, 12 m 95 cm is 35 cm.

5. Find the numbers if the HCF of two numbers is 29 and their sum is 174.

Solution:

Given that the HCF of two numbers is 29.

Let 29a and 29b be the two required numbers.

According to the given,

29a + 29b = 174

29(a + b) = 174

a + b = 174/29 = 6

The pair of values of co-primes with sum 6 is (1, 5).

So, the possible numbers are:

29 x 1 = 29

29 x 5 = 145

Verification:

Sum of numbers = 29 + 145 = 174

Hence, the required numbers are 29 and 145.

6. Find the HCF of fractions 2/3, 16/81, and 8/9.

Solution:

Given fractions are:

2/3, 16/81, 8/9

We know that,

HCF of fractions = HCF of numerators/LCM of denominators

Here,

Prime factorization of 2 = 2

Prime factorization of 16 = 2⁴

Prime factorization of 8 = 2³

HCF(2, 16, 8) = 2

Prime factorization of 3 = 3

Prime factorization of 81 = 3⁴

Prime factorization of 9 = 3²

LCM(3, 81, 9) = 3⁴ = 81

Therefore, the required HCF = 2/81

7. Simplify the fraction 36/60 using HCF.

Solution:

Given fraction: 36/60

Prime factorization of 36 = 2 × 2 × 3 × 3 = 2² × 3²

Prime factorization of 60 = 2 × 2 × 3 × 5 = 2² × 3 × 5

HCF of 36 and 60 = 2² × 3 = 12

Now, divide the numerator and denominator of the given fraction by the HCF of 36 and 60, i.e. 12.

(36/12)/ (60/12) = ⅗

Hence, the simplified fraction is ⅗.

8. Find the greatest number, which on dividing 1657 and 2037 leaves remainders 6 and 5, respectively.

Solution:

Given,

The number on dividing 1657 and 2037 leaves remainders 6 and 5, respectively.

Let us make the dividend completely divisible by the divisor.

This is possible when we subtract the remainder from the dividend.

So, 1657 – 6 = 1651

2037 – 5 = 2032

Prime factorization of 1651 = 13 x 127

Prime factorization of 2032 = 2 x 2 x 2 x 2 x 127 = 24 x 127

HCF of 1651 and 2032 is 127.

Hence, the required number is 127.

9. What is the largest number, which divides 64, 136 and 238 to leave the same remainder in each case?

Solution:

To find the required number, we need to calculate the HCF of (136 – 64), (238 – 136) and (238 – 64), i.e., HCF (72, 102, 174).

since

136 – 64 = 72

238 – 126 = 102

238 – 64 = 174

Let us write the prime factorization for each of these numbers.

72 = 23 x 32

102 = 2 x 3 x 17

174 = 2 x 3 x 29

Therefore, HCF of 72, 102, and 174 = 2 x 3 = 6

Hence, 6 is the required number.

10. What will be the HCF of 0.63 and 1.05?

Solution:

Given numbers: 0.63 and 1.05

To find the HCF of decimal numbers, first, we need to calculate the HCF of those numbers by removing the decimal point.

So,

Prime factorization of 63 = 3 x 3 x 7 = 32 x 7

Prime factorization of 105 = 3 x 5 x 7

HCF of 63 and 105 = 3 x 7 = 21

Given two numbers have decimal points after two places.

Therefore, the HCF of 0.63 and 1.05 is 0.21.

Practice Questions on HCF

1. Find the side of the largest square slab, which can be paved on the floor of a room 5 meters 44 cm long and 3 meters 74 cm broad.
2. Find the HCF of 68, 76 and 94.
3. The product of two numbers is 6760, and their HCF is 13. How many such pairs can be formed?
4. Three rectangular fields having areas 60 m², 84 m² and 108 m² are to be divided into identical rectangular flower beds, each having a length of 6 m. Find the breadth of each flower bed.
5. Find the HCF of fractions 108/18, 36/15, and 60/17.