Second radio station plays the song in every = 3 days.

So, both of the radio station will play the song per every = 2 × 3 = 6 days.

So, the radio stations plays the same song on the same day in 30 days = 30/6 = 5 times.

[Shortcut Trick of This Question is Given Below]

Exercise 1.2 Class 10 Maths

Real Numbers

Class 10 Math Ex 1.2 Question 1

Express each number as a product of its prime factors:

1. 140
2. 156
3. 3825
4. 5005
5. 7429

Solution: (i) 140 = 2 × 2 × 5 × 7 = 2² × 5 × 7

(ii) 156 = 2 × 2 × 3 × 13 = 2² × 3 × 13

(iii) 3825 = 3 × 3 × 5 × 5 × 17 = 3² × 5² × 17

(iv) 5005 = 5 × 7 × 11 × 13

(v) 7429 = 17 × 19 × 23

Class 10 Maths Exercise 1.2 Question 2

Find the LCM and HCF of the following pairs of integers and verify that

LCM × HCF = Product of the two numbers.

1. 26 and 91
2. 510 and 92
3. 336 and 54

Solution: (i) Here, 26 = 2 × 13

and 91 = 7 × 13

So, HCF(26, 91) = 13

LCM(26, 91) = 2 × 7 × 13 = 182

Now, LCM × HCF = Product of the two numbers

⇒ 13 × 182 = 26 × 91

⇒ 2366 = 2366

Which is true, so, the given condition has verified.

(ii) 510 = 2 × 3 × 5 × 17

and 92 = 2 × 2 × 23

So, HCF(510, 92) = 2

LCM(510, 92) = 2 × 2 × 3 × 5 × 17 × 23 = 23460

Now, LCM × HCF = Product of the two numbers

⇒ 2 × 23460 = 510 × 92

⇒ 46920 = 46920

Which is true, so, the given condition has verified.

(iii) 336 = 2⁴ × 3 × 7

and 54 = 2 × 3³

So, LCM(336, 54) = 2 × 3 = 6

HCF(336, 54) = 2⁴ × 3³ × 7 = 3024

Now, LCM × HCF = Product of the two numbers

⇒ 6 × 3024 = 336 × 54

⇒ 18144 = 18144

Class 10 Maths Exercise 1.2 Question 3

Find the LCM & HCF of the following integers by applying the prime factorisation method;

(i) 12, 15 & 21

(ii) 17, 23 & 29

(iii) 8, 9 & 25

Solutions: (i) 12 = 2² × 3

15 = 3 × 5

21 = 3 × 7

So, LCM = 2² × 3 × 5 × 7 = 420

HCF = 3

(ii) 17, 23 & 29 – all are prime numbers.

So, LCM = 17 × 23 × 29 = 11339

HCF = 1

(iii) 8 = 2³

9 = 3²

25 = 5²

LCM = 8 × 9 × 25 = 1800

HCF = 1

Class 10 Maths Exercise 1.2 Question 4

Given that HCF(306, 657) = 9, Find LCM(306, 657).

Solution: Here, HCF(306, 657) = 9

LCM(306, 657) = ?

We know that, LCM × HCF = Product of two number

LCM = \cfrac {Product  of  two  numbers} {HCF}

⇒ LCM = \cfrac {306 × 657} {9} = 22338

Class 10 Maths Exercise 1.2 Question 5

Check whether 6ⁿ can with the digit 0 for any natural number n.

Solution: We know that, any number can end with the digit 0, if 5 is a factor of 6ⁿ

Now,  6ⁿ  = (2 × 3)ⁿ

We see that, 5 is not a factor of 6ⁿ

That means , 6ⁿ can’t end with “0”

Class 10 Maths Exercise 1.2 Question 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)

= 13 × 78

That means, the given number can be expressed as a product of two numbers other than 1 & itself. So, 7 × 11 × 13 + 13 is composite number.

Second Number = 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5

= 5 × ( 7 × 6 × 4 × 3 × 2 × 1 + 1 )

= 5 × 1009

That means, the given number can be expressed as a product of two numbers other than 1 & itself. So, 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 is composite number.

Class 10 Maths Exercise 1.2 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: Here, we need to find the LCM of 18 & 12, which will be the required time they meet again.

So, 18 = 2 × 3 × 3 and 12 = 2 × 2 × 3

LCM(18, 12) = 2² × 3² = 36

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

Class 10 Maths Exercise 1.2 Question 8(i)

The soldiers of a regiment can stand in 15, 20 or 25 rows. Find the least no of soldiers in the regiment.

Solution: We need to find the LCM of 15, 20 and 25, which is the least number of soldiers.

15 = 3 × 5, 20 = 2 × 2 × 5

And 25 = 5 × 5

So, LCM = 4 × 25 × 3 = 300

Hence, Least Number of Soldiers in the regiment is 300.

Class 10 Maths Exercise 1.2 Question 8(ii)

A bell rings every 18 seconds and another every 60 seconds. If suppose they ring at a same time, at what time will the bell ring again at the same time?

Solution: We need to find the LCM of 18 & 60 to get the next moment.

18 = 2 × 3² and 60 = 2² × 3 × 5

So, LCM = 2² × 3² × 5 = 180

Hence, after 180 seconds, both of the bells will ring again at the same moment.

Class 10 Maths Exercise 1.2 Question 8(iii)

A radio station plays assam sangeet once every two days another radio station plays the same song once every three days. how many times in 30 days will both the radio stations plays the same song on the same day.

Solution: First radio station plays the song in every = 2 Days.

Second radio station plays the song in every = 3 days.

So, both of the radio station will play the song per every = 2 × 3 = 6 days.

So, the radio stations plays the same song on the same day in 30 days = 30/6 = 5 times.