Real Numbers Class 10th Important Questions with Answers Mathematics
Updated on June 1, 2025 | By Learnzy Academy
Q1. If the product of two positive integers is equal to the product of their HCF and LCM is true then, the HCF (32 , 36) is
Q2. The LCM of 60, 84 and 108 is
Q3. The product of HCF and LCM of 60,84 and 108 is
Q4. Find the smallest number that is divisible by 6, 10, and 15, and leaves a remainder of 5 in each case.
Q5. Find the smallest number that when divided by 12, 16, and 24 leaves a remainder of 4 in each case.
Q6. If the HCF of two numbers is 16 and their product is 3072, find their LCM.
Q7. Prove that 7×11×13 +13 is a composite number.
7× 11 × 13 + 13 can be wriiten as 13 (7× 11 + 1).
=> 13 x (77 + 1)
=> 13 x 78
=> 13 x 2 x 39
Clearly, this has more than 2 factors.
So, 7× 11 × 13 + 13 is a composite number.
Q8. Three tankers contain 403 litres, 434 litres and 465 litres of diesel respectively. Find the maximum capacity of a container that can measure the diesel of the three containers exact number of times.
To solve this problem, we need to find the greatest common divisor (GCD) of the three given quantities: 403 litres, 434 litres, and 465 litres.
The GCD will be the maximum capacity of a container that can measure all three amounts an exact number of times.
Find the GCD of 403, 434, and 465
Step 1: GCD of 403 and 434
Use the Euclidean algorithm:
434 ÷ 403 = 1 remainder 31
403 ÷ 31 = 13 remainder 0
So, GCD(403, 434) = 31
Step2: Now, GCD 31 and 465
465 ÷ 31 = 15 remainder 0
So, GCD(31, 465) = 31
Hence the maximum capacity of the container that can measure all three diesel quantities exactly is 31 litres.
Q9. Find the least number which when divided by 6, 15 and 18 leave remainder 5 in each case.
To solve this, we need to find the least number which, when divided by 6, 15, and 18, leaves a remainder of 5 in each case.
Step 1: Find LCM of 6, 15, and 18
- 6 = 2 × 3
- 15 = 3 × 5
- 18 = 2 × 32
Take the highest powers of all primes:
LCM =21 × 32 × 51 = 90
So, the least number is: 90 + 5 = 95
Q10. If HCF(6, a) = 2 and LCM(6, a) = 60 then find the value of a.
It is given -
- HCF of 6 & a is 2
- LCM of 6 & a is 60
Now we use the identity
HCF x LCM = Product of both Number
2 x 60 = 6 x a
=> a = 20
Final Answer: 20
Q11. Determine the number nearest 110000 but greater than 100000 which is exactly divisible by each of 8, 15 and 21.
Step 1: Find the Least Common Multiple (LCM) of 8, 15, and 21.
Prime factorization:
8 = 2³
15 = 3 × 5
21 = 3 × 7
LCM = 2³ × 3 × 5 × 7 = 8 × 3 × 5 × 7 = 840
So, any number divisible by 8, 15, and 21 must be a multiple of 840.
Step 2: Find the multiple of 840 that is nearest to 110000 and greater than 100000.
Divide 110000 by 840:
110000 ÷ 840 ≈ 130.95
Now check nearby multiples:
130 × 840 = 109200
131 × 840 = 110040
132 × 840 = 110880
Among these, 110040 is the nearest to 110000 and is greater than 100000.
Hence answer is 110040
Q12. In a school there are two sections – section A and section B of class X. There are 32 students in section A and 36 students in section B. Determine the minimum number of books required for their class library so that they can be distributed equally among students of section A or section B.
To find the minimum number of books that can be equally distributed among:
32 students in section A
36 students in section B
This means we are looking for the Least Common Multiple (LCM) of 32 and 36.
Step 1: Prime Factorization
32 = 2⁵
36 = 2² × 3²
Step 2: Find LCM
LCM takes the highest powers of all prime factors involved:
LCM = 2⁵ × 3² = 32 × 9 = 288
Hence answer is 288 books