HCF Questions for Class 7 with Answers | CBSE Maths Practice

Understanding HCF (Highest Common Factor) in Class 7 lays a strong foundation for future success in exams like GATE, NEET, and other competitive exams. HCF helps students master number sense, divisibility, and problem-solving—core skills used in higher-level mathematics. When Class 7 students practice HCF questions, they not only score well in school but also prepare their minds for logical thinking and fast calculations needed in Olympiads, NTSE, and entrance exams later on. 

Starting early with such concepts builds confidence and sharpens their ability to solve real-world math problems with ease, making HCF an essential step in every student’s learning journey.

What is HCF? Formulas to Find HCF

HCF (Highest Common Factor) is the greatest number that can exactly divide two or more given numbers without leaving a remainder. Also Named as the Greatest Common Divisor (GCD). Finding the HCF is useful in simplifying fractions, solving word problems, and understanding the structure of numbers. 

There are a few common methods to find the HCF: one is the prime factorization method, where we break each number into its prime factors and take the common ones. Another method is the division method, where you repeatedly divide the bigger number by the smaller one, then use the remainder as the new divisor until the remainder becomes zero. 

The final number that divides exactly is the HCF. These methods are not only simple to learn in Class 7 but are also the building blocks for complex problem-solving in higher classes and competitive exams.

Also Read: CBSE Class 7 English Informal Letter Writing | Story Writing for Class 7

HCF Questions for Class 7 with Answers

Question 1: Three numbers are in the ratio 2:3:4. Their highest common factor is 5. Find the numbers.

Answer:
10, 15, and 20

Explanation:
Since the numbers are in the ratio 2:3:4, we can represent them as 2k, 3k, and 4k, where k is the common factor.
Given that the HCF is 5, it means k = 5.
So, the actual numbers are:
2 × 5 = 10, 3 × 5 = 15, and 4 × 5 = 20.

Question 2: The HCF of two numbers is 11 and their sum is 132. If both numbers are greater than 42, what is the difference between the two numbers?
Answer: 22
Explanation:
If the HCF is 11, the numbers can be written as 11a and 11b, where a and b are coprime (have no common factors). We know:
11a + 11b = 132
a + b = 12
Since both numbers are greater than 42, we need 11a > 42 and 11b > 42, which means a ≥ 4 and b ≥ 4.
With a + b = 12, the possible pairs are (4,8), (5,7), and (6,6).
Only (5,7) makes a and b coprime.
 Numbers - 11×5 = 55 and 11×7 = 77.
Their difference is 77 - 55 = 22.

Question 3: Find the HCF of 153, 117, and 405.
Answer: 9
Explanation:
Using the Euclidean algorithm:
First, let's find the HCF of 153 and 117:
153 = 117 × 1 + 36
117 = 36 × 3 + 9
36 = 9 × 4 + 0
HCF of 153 and 117 is 9.
Now, find the HCF of this result and 405:
405 = 9 × 45 + 0
So the HCF of 9 and 405 is 9.
Therefore, the HCF of 153, 117, and 405 is 9.

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Question 4: Find the HCF of 1/2, 2/3, and 4/5.
Answer: 1/30
Explanation:
The fractions are 1/2, 2/3, and 4/5.
HCF of numerators (1, 2, 4) = 1
LCM of denominators (2, 3, 5) = 30
Therefore, HCF = 1/30

Question 5: If the HCF of two numbers is 18 and their LCM is 720, what is their product?
Answer: 12,960
Explanation:
Product of two numbers = HCF × LCM
Therefore, product = 18 × 720 = 12,960

Question 6: The HCF of two consecutive odd numbers is:
Answer: 1
Explanation:
Any two consecutive odd numbers are always coprime, meaning they have no common factor other than 1. This is because if a number divides two consecutive odd numbers, it must also divide their difference, which is 2. But since both numbers are odd, no number greater than 1 can divide both.

Question 7: Two numbers have a Highest Common Factor (HCF) of 6 and a Least Common Multiple (LCM) of 198. If one of the numbers is 18, what is their total sum?

Answer:
Total = 90

Explanation:
Using the relation:
HCF × LCM = First Number × Second Number
⇒ 6 × 198 = 1188
Now, divide 1188 by 18 to get the other number:
Second number = 1188 ÷ 18 = 66
Adding both numbers: 18 + 66 = 90

Question 8: What is the HCF of (3^4 × 5^2 × 7) and (3^2 × 5^3 × 11)?
Answer: 3^2 × 5^2 = 9 × 25 = 225
Explanation:
HCF takes the common factors with the smallest powers:
Common factors: 3 with power min(4,2) = 2, and 5 with power min(2,3) = 2
HCF = 3^2 × 5^2 = 9 × 25 = 225

Question 9: Find the largest number that divides 285 and 1239 leaving the same remainder in each case.
Answer: 318
Explanation:
We need to find the HCF of (1239-285) and 285.
HCF of 954 and 285:
954 = 285 × 3 + 99
285 = 99 × 2 + 87
99 = 87 × 1 + 12
87 = 12 × 7 + 3
12 = 3 × 4 + 0
HCF = 3
The largest number = 285 - 3 = 282

Question 10: The smallest number which when divided by 16, 18, and 20 leaves remainder 4 in each case is:
Answer: 76
Explanation:
We need to find LCM of 16, 18, and 20, then subtract the common remainder and add 4.
First, find the HCF of each pair:
HCF of 16 and 18 = 2
HCF of 18 and 20 = 2
Now, find the LCM: LCM = (16 × 18 × 20) / (HCF)
= 5760 / 2 = 2880
The smallest such number = 4 + 72 = 76

Question 11: If the HCF of 408 and 1032 is expressible in the form 1032 - 408x, where x is a positive integer, find the value of x.
Answer: 2
Explanation:
Using the Euclidean algorithm:
1032 = 408 × 2 + 216
408 = 216 × 1 + 192
216 = 192 × 1 + 24
192 = 24 × 8 + 0
HCF = 24
Now, 1032 - 408x = 24
408x = 1008
x = 1008/408 = 2.47
Since x must be a positive integer, x = 2

Question 12: The greatest possible length of a scale that can measure exactly both 63 cm and 147 cm is:
Answer: 21 cm
Explanation:
We need to find the HCF of 63 and 147.
Using Euclidean algorithm:
147 = 63 × 2 + 21
63 = 21 × 3 + 0
HCF = 21
Therefore, the greatest possible length is 21 cm.

Question 13: Find the maximum number of students among whom 1001 pens and 910 pencils can be distributed in such a way that each student gets the same number of pens and the same number of pencils.
Answer: 91
Explanation:
The maximum number of students = HCF of 1001 and 910
Using Euclidean algorithm:
1001 = 910 × 1 + 91
910 = 91 × 10 + 0
HCF = 91
Therefore, 91 students.

Question 14: If the HCF of 65 and 117 is expressible in the form 65m - 117n, where m and n are non-negative integers, then the minimum value of m + n is:
Answer: 4
Explanation:
HCF of 65 and 117 = 13
We need to find m and n such that 65m - 117n = 13
One solution is m = 2, n = 1 (65×2 - 117×1 = 130 - 117 = 13)
Therefore, m + n = 2 + 1 = 3

Question 15: The product of two numbers is 936 and their HCF is 12. Find the number of possible pairs.
Answer: 4
Explanation:
If two numbers are a and b, then a×b = 936 and HCF(a,b) = 12
So a = 12p and b = 12q where p and q are coprime
This means p×q = 936/144 = 6.5, which isn't an integer
Let's check again: p×q = 936/12² = 936/144 = 6.5
Since we need integers, let's factor 936 = 12 × 78 = 12 × 6 × 13 = 72 × 13
So we can have the pairs: (12, 78), (24, 39), (36, 26), (72, 13)
That's 4 possible pairs.

Question 16: What is the largest number which divides 1305, 4665, and 6905 to give the same remainder in each case?
Answer: 1320
Explanation:
We need the HCF of (4665-1305) and (6905-1305)
HCF of 3360 and 5600
Using Euclidean algorithm:
5600 = 3360 × 1 + 2240
3360 = 2240 × 1 + 1120
2240 = 1120 × 2 + 0
HCF = 1120
The largest number is = 1120 + (1305 % 1120) = 1120 + 185 = 1305

Question 17: Given that the HCF of two numbers is 12 and their LCM is 900, how many distinct pairs of numbers can be formed?

Answer: 4 pairs

Explanation:
Let the two numbers be a and b.
From the formula a × b = HCF × LCM, we get:
a × b = 12 × 900 = 10800
Now, we need to find pairs of numbers whose product is 10800 and the HCF is 12.
Let a = 12p and b = 12q, where p and q are coprime (i.e., their HCF is 1). Then:
12p × 12q = 10800
This simplifies to:
p × q = 10800 ÷ 144 = 75
Now, we find the coprime pairs of 75, which are:
(1, 75), (3, 25), and (5, 15).
Thus, the number pairs are:
(12 × 1, 12 × 75), (12 × 3, 12 × 25), and (12 × 5, 12 × 15).
So, the possible pairs are: (12, 900), (36, 300), and (60, 180).
That gives us a total of 3 pairs.

Question 18: The HCF of two numbers is 8 and their LCM is 240. If one numbers is 48, then find the other number.
Answer: 40
Explanation:
We know: a × b = HCF × LCM
Therefore, 48 × x = 8 × 240
x = (8 × 240)/48 = 40

Question 19: Find the least number which when divided by 12, 15, 20, and 54 leaves remainder 7 in each case.
Answer: 547
Explanation:
The least such number = LCM(12, 15, 20, 54) + 7
First, find the LCM:
LCM(12, 15) = 60
LCM(60, 20) = 60
LCM(60, 54) = 540
Therefore, the required number = 540 + 7 = 547

Question 20: Three numbers are in the ratio 3:4:5 and their HCF is 6. Find their LCM.
Answer: 360
Explanation:
Since the ratio is 3:4:5 and HCF is 6, the numbers are:
3 × 6 = 18, 4 × 6 = 24, and 5 × 6 = 30
To find LCM, we find the prime factorizations:
18 = 2 × 3²
24 = 2³ × 3
30 = 2 × 3 × 5
LCM = 2³ × 3² × 5 = 8 × 9 × 5 = 360

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Question 21: If the HCF of 3703 and 1813 is expressible in the form 3703x + 1813y, where x and y are integers, find the value of (x + y).
Answer: -1
Explanation:
Using the Euclidean algorithm:
3703 = 1813 × 2 + 77
1813 = 77 × 23 + 42
77 = 42 × 1 + 35
42 = 35 × 1 + 7
35 = 7 × 5 + 0
So HCF = 7
Now we need to find x and y such that 3703x + 1813y = 7
Working backwards:
7 = 42 - 35 = 42 - (77 - 42) = 2×42 - 77
= 2×(1813 - 23×77) - 77 = 2×1813 - (2×23 + 1)×77
= 2×1813 - (2×23 + 1)×(3703 - 2×1813)
= 2×1813 - (2×23 + 1)×3703 + (2×23 + 1)×2×1813
= (2 + (2×23 + 1)×2)×1813 - (2×23 + 1)×3703
= (2 + 46 + 2)×1813 - 47×3703
= 50×1813 - 47×3703
So x = -47 and y = 50, which means x + y = 3

Question 22: What is the greatest possible length of rods that can be used to measure exactly 42 meters and 56 meters?
Answer: 14 meters
Explanation:
We need to find the HCF of 42 and 56
Using the Euclidean algorithm:
56 = 42 × 1 + 14
42 = 14 × 3 + 0
HCF = 14
Therefore, the greatest possible length is 14 meters.

Question 23: Find the least 4-digit number that is divisible by both 18 and 45.
Answer: 1080
Explanation:
First, find the LCM of 18 and 45:
18 = 2 × 3²
45 = 3² × 5
LCM = 2 × 3² × 5 = 90
The smallest 4-digit number divisible by 90 is 1080 (= 90 × 12)

Question 25: If the HCF of two numbers is 16 and their difference is 80, what are the possible values of the two numbers?
Explanation:
If HCF is 16, then both numbers must be multiples of 16
Let the numbers be 16a and 16b where a and b are coprime
Their difference is 80, so:
|16a - 16b| = 80
16|a - b| = 80
|a - b| = 5
Since a and b are positive integers, possible values for (a,b) are: (1,6), (2,7), (3,8), (4,9), (5,10), (6,11)
Therefore, the pairs are: (16, 96), (32, 112), (48, 128), (64, 144), (80, 160), (96, 176)

NCERT Solutions for Class 7 Chapter Wise:

Why HCF Questions Are Important for Class 7 Students?

Solving HCF questions for Class 7 helps students strengthen their understanding of number properties and improves their problem-solving skills

1. Strengthens Math Basics: HCF questions in Class 7 help students understand number properties and divisibility rules, which are the base of advanced math concepts.
2. Prepares for Competitive Exams: A strong grasp of HCF supports logical thinking and fast calculations—skills needed for exams like GATE, NEET, NTSE, and Olympiads.
3. Improves Problem-Solving Skills: Solving HCF problems boosts analytical thinking and helps in tackling real-world mathematical situations.
4. Useful in Higher Classes: Topics like LCM, algebra, fractions, and ratios in higher classes are easier to learn with a solid understanding of HCF.
5. Foundation for Logical Reasoning: Many competitive exams include HCF-type questions in their reasoning or aptitude sections, making early practice important.
6. Builds Confidence in Numbers: Regular HCF practice helps Class 7 students feel more confident with numbers, patterns, and step-by-step calculations.
7. Enhances Exam Performance: With clear HCF concepts, students can solve questions faster and more accurately in school and entrance exams.