# Real Numbers Exercise – 1.2 NCRT Download pdf || UP Board

# __Real Numbers__

__Exercise – 1.2__

__Real Numbers__

__Exercise – 1.2__

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

**(****i****) 140**

**Solution****:**

^{2}× 5 × 7

**(ii)
156**

**Solution****:**

(Solve as first question)

The prime factor of 156 = 2 × 2 × 3 × 13

= 2^{2} × 3 × 7

**(iii)
3825**

**Solution****:**

The prime factor of 3825 = 3 × 3 × 5 × 5 × 17

= 3^{2} × 5^{2} × 17

**(iv)
5005**

**Solution****:**

The prime factor of 5005 = 5 × 7 × 11 × 13

**(v)
7429**

**Solution****:**

The prime factor of 7429 = 17 × 19 × 23

**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**

**Solution****:**

The prime factor of 26 = 2 × 13

The prime factor of 91 = 7 × 13

HCF = 13

LCM = 2×7×13 = 182

product of the two numbers = 26×91 = 2366

HCF×LCM = 13×182 = 2366

product of the two numbers = LCM
× HCF

**(****ii)
510 and**** ****92**

**Solution****:**

510 = 2×3×5×17

92 = 2×2×23

HCF = 2

LCM = 2×2×3×5×17×23 = 23460

product of the two numbers = 510×92 = 46920

HCF×LCM = 2×23460 = 46920

product of the two numbers = LCM × HCF

**(****iii)
336 and**** ****54**

**हल****:**

336 = 2×2×2×2×3×7

54 = 2×3×3×3

LCM = 2^{4}×3^{3}×7 = 3024

HCM = 2×3 = 6

product of the two numbers = 336×54 = 18144

HCF×LCM = 6×3024 = 18144

product of the two numbers = LCM × HCF

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

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

**Solution****:**

12 = 2×2×3

15 = 3×5

21 = 3×7

HCF = 3

LCM = 2^{2}×3×5×7 = 420

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

**Solution****:**

17 = 17×1

23 = 23×1

29 = 29×1

HCF = 1

LCM = 17×23×29 = 11339

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

**Solution****:**

8 = 2×2×2×1

9 = 3×3×1

25= 5×5×1

HCF = 1

LCM = 2×2×2×3×3×5×5 = 1800

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

**Solution****:**

HCF(306, 657) = 9

We know that,

LCM × HCF = product of the two numbers

LCM(306, 657)
= 22338

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

**Solution****:**

If a number ends on 0, it is divisible by 2 and 5 (eg- 10)

The prime factor of 6^{n} = (2×3)^{n}

There is no 5 in the prime factorization of 6^{n}. So 6^{n} will not be divisible by 5.

Hence, for any natural number n, the number 6^{n} cannot end with the digit 0.

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

**Solution****:**

First Number = 7×11×13+13

= 13×(7×11×1+1)

= 13×(77+1)

= 13×78

= 13×13×6

This
number has more than two factors (1, 6, 13)
so it is a composite
number, because composite numbers have more than two factors.

Second
Number =
7×6×5×4×3×2×1+5

= 5×(7×6×4×3×2×1+1)

=
5×(1008+1)

=
5×1009

This
number has more than two factors (1, 5, 1009)
so it is a composite
number.

**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****:**

Sonia takes 18 minutes while Ravi takes 12 minutes, after a time both will be at their starting
positions that time will be the LCM of 12 and 18.

18 = 2×3×3

12 = 2×2×3

LCM(18,
12) = 2×2×3×3 = 36

Hence,
Ravi and Sonia will meet again at the starting place after 36
minutes.

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