Real Numbers – Class 10

Important Questions for CBSE Class 10 Mathematics
Real Numbers – Chapter 1
Questions Based on the topics – Euclid’s Division Algorithm and The Fundamental Theorem of Arithmetic.

1. Use Euclid’s division algorithm to find the HCF of 135 and 225?
Solution:
By Euclid’s division algorithm, we can write
225 = 135 x 1 + 90
135 = 90 x 1 + 45
90 = 45 x 2 + 0
The remainder is now zero.
Therefore, HCF (135, 225) = 45.

2. Show that the numbers 143 and 187 are not co prime.
Solution:
By the Prime factorisation method, we can write 143 = 11 x 13
and 187 = 11 x 17.
Here common factor is 11.
Therefore these numbers are not co prime.

3. Express 156 as a product of its prime factors.
Solution:
By the prime factorisation of 156, we will get 156 = 2 x 2 x 3 x 13 x 1.

4. Given that HCF (306, 657) = 9, find the LCM (306, 657)?
Solution:
We know that HCF x LCM = Product of the two given numbers.
Therefore, HCF x LCM = 306 x 657.
Given HCF = 9.
So LCM = 306 x 657/9 = 22338.

5. Explain why 7 x 11 x 13 + 13 is a composite number?
Solution:
7 x 11 x 13 + 13 = 13(7 x 11 + 1)
= 13(77 + 1)
= 13 x 78
= 13 x 2 x 3 x 13, which is a product of primes.
Since 7 x 11 x 13 + 13 can be expressed as product of primes, it is a composite number.

6. Find HCF of 336 and 54 by prime factorisation method. Hence find their LCM also.
Solution:
By prime factorisation method, 336 = 2 x 2 x 2 x 2 x 3 x 7 and 54 = 2 x 3 x 3 x 3.
Common factors are 2 and 3. Therefore, HCF = 2 x 3 = 6.
We know that HCF x LCM = Product of two numbers.
6 x LCM = 336 x 54
LCM = 336 x 54 /6 = 3024.

7. Three bells toll at intervals of 18, 24, 30 minutes respectively. If they start tolling together, after what time will they next toll together?
Solution:
Here we need to find the LCM of 18, 24 and 30.
LCM (18, 24, 30) = 2 x 2 x 3 x 3 x 2 x 5 = 360 minutes.
i.e., after 6 hours.

8. Given that HCF (75, 69) = 3. Find LCM (75, 69)?
Solution:
We know that LCM x HCF = 75 x 69
LCM x 3 = 75 x 69
LCM = 75 x 69/3 = 1725.

9. Use Euclid’s Division Algorithm to find the largest number which divides 872 and 260 leaving remainder 5 in each case.
Solution:
Since these numbers leaving remainder 5 in each case,
we can consider 872 – 5 = 867 and 260 -5 = 255.
The largest number is the HCF of 867 and 255.
By Euclid’s Division Algorithm, 867 = 255 x 3 + 102
255 = 102 x 2 + 51
102 = 51 x 2 + 0
Now the remainder is zero, so HCF (867, 255) = 51.

10. Find the HCF and LCM of 6, 72 and 120, using the prime factorisation method.
Solution:
We have 6 = 2 x 3
72 = 2 x 2 x 2 x 3 x 3
120 = 2 x 2 x 2 x 3 x 5
Here the common factors are 2 and 3.
So HCF (6, 72, 120) = 2 x 3 = 6.
LCM (6, 72, 120) = 2 x 2 x 2 x 3 x 3 x 5 = 360.