# 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.