# Class 10 NCERT Solutions- Chapter 1 Real Numbers – Exercise 1.2

**Question 1: Express each number as a product of its prime factors:**

**(i) 140**

**(ii) 156**

**(iii) 3825**

**(iv) 5005**

**(v) 7429**

**Solutions:**

(i) 140Taking the LCM of 140,

140 = 2 × 2 × 5 × 7 × 1 = 2

^{2 }× 5 × 7Therefore, the product of the prime factors is 2

^{2 }× 5 × 7

(ii)156Taking the LCM of 156,

156 = 2 × 2 × 3 × 13 × 1 = 2

^{2 }× 3 × 13Therefore, the product of the prime factors is 2

^{2}× 3 × 13

(iii) 3825Taking the LCM of 3825,

3825 = 3 × 3 × 5 × 5 × 17 × 1 = 3

^{2}× 5^{2}× 17Therefore, the product of the prime factors is 3

^{2 }× 5^{2 }× 17

(iv)5005Taking the LCM of 5005,

5005 = 5 × 7 × 11 × 13 × 1 = 5 × 7 × 11 × 13

Therefore, the product of the prime factors is 5 × 7 × 11 × 13

(v) 7429Taking the LCM of 7429,

7429 = 17 × 19 × 23 × 1 = 17 × 19 × 23

Therefore, the product of the prime factors is 17 × 19 × 23

**Question 2: Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = product of the two numbers.**

**(i) 26 and 91**

**(ii) 510 and 92**

**(iii) 336 and 54**

**Solutions:**

(i)26 and 91Taking the LCM of 26,

26 = 2 × 13 × 1

Taking the LCM of 91,

91 = 7 × 13 × 1

Therefore, LCM of 26 and 91 together = 2 × 7 × 13 × 1 = 182

HCF of 26 and 91 = 13

Now, the product of 26 and 91 = 26 × 91 = 2366

And the product of LCM and HCF = 182 × 13 = 2366

Therefore, LCM × HCF = product of the 26 and 91.

(ii)510 and 92Taking the LCM of 510,

510 = 2 × 3 × 17 × 5 × 1

Taking the LCM of 92,

92 = 2 × 2 × 23 × 1

Therefore, LCM of 510 and 92 = 2 × 2 × 3 × 5 × 17 × 23 = 23460

HCF of 510 and 92 = 2

Now, the product of 510 and 92 = 510 × 92 = 46920

And the product of LCM and HCF = 23460 × 2 = 46920

Therefore, LCM × HCF = product of the 510 and 92.

(iii)336 and 54Taking the LCM of 336,

336 = 2 × 2 × 2 × 2 × 7 × 3 × 1

Taking the LCM of 54,

54 = 2 × 3 × 3 × 3 × 1

Therefore, LCM of 336 and 54 = 3024

HCF of 336 and 54 = 2×3 = 6

Now, the product of 336 and 54 = 336 × 54 = 18144

And the product of LCM and HCF = 3024 × 6 = 18144

Therefore, LCM × HCF = product of the 336 and 54.

**Question 3: Find the LCM and HCF of the following integers by applying the prime factorisation method.**

**(i) 12, 15 and 21**

**(ii) 17, 23 and 29**

**(iii) 8, 9 and 25**

**Solutions:**

(i) 12, 15 and 21Taking the LCM of 12,

12=2×2×3

Taking the LCM of 15,

15=5×3

Taking the LCM of 21,

21=7×3

Therefore,

HCF of 12, 15 and 21 = 3

LCM of 12, 15 and 21 = 2 × 2 × 3 × 5 × 7 = 420

(ii)17, 23 and 29Taking the LCM of 17,

17=17×1

Taking the LCM of 23,

23=23×1

Taking the LCM of 29,

29=29×1

Therefore,

HCF of 17, 23 and 29 = 1

LCM of 17, 23 and 29 = 17 × 23 × 29 = 11339

(iii) 8, 9 and 25Taking the LCM of 8,

8=2×2×2×1

Taking the LCM of 9,

9=3×3×1

Taking the LCM of 25,

25=5×5×1

Therefore,

HCF of 8, 9 and 25 =1

LCM of 8, 9 and 25 = 2×2×2×3×3×5×5 = 1800

**Question 4: Given that HCF (306, 657) = 9, find LCM (306, 657).**

**Solution:**

Given: HCF (306, 657) = 9

We know that,

HCF×LCM=Product of the two given numbers

Therefore, by substituting the value we get,

9 × LCM = 306 × 657

LCM = (306×657)/9 = 22338

LCM (306,657) = 22338

Therefore, the LCM (306,657) = 22338

**Question 5: Check whether 6**^{n} can end with the digit 0 for any natural number n.

^{n}can end with the digit 0 for any natural number n.

**Solution:**

If the given number 6

^{n}ends with the digit 0, then it should be divisible by 5.We know that if any number with the unit place as 0 or 5 is divisible by 5.

Therefore,

By prime factorization of 6

^{n}= (2×3)^{n}Since the prime factorization of 6

^{n}doesn’t contain prime number 5.Therefore, 6

^{n}cannot end with the digit 0 for any natural number.

**Question 6: Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.**

**Solution:**

We know by the definition of a composite number, that, the composite number has factors other than 1 and itself.

Therefore, in the given expression

7 × 11 × 13 + 13

By taking 13 as a common factor, we get,

=13(7×11×1+1)

= 13(77+1)

= 13×78 [taking prime factors of 78]

= 13×3×2×13

Therefore, 7 × 11 × 13 + 13 is a composite number.

Now, for the 2

^{nd}number7 × 6 × 5 × 4 × 3 × 2 × 1 + 5

By taking 5 as a common factor, we get,

=5(7×6×4×3×2×1+1)

= 5(1008+1)

= 5×1009

Therefore, 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 is a composite number.

**Question 7: There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, while Ravi takes 12 minutes for the same. Suppose they both start at the same point and at the same time and go in the same direction. After how many minutes will they meet again at the starting point?**

**Solution:**

Given: Both Sonia and Ravi move in the same direction and at the same time.

Now, the time when they will be meeting again at the starting point can be calculated by finding the LCM of 18 and 12

Therefore,

LCM (18, 12) = 2×3×3×2×1 = 36

Finally, they both will meet again at the starting point after 36 minutes.

## Please

Loginto comment...