Exercise 1.1 (NCERT) – Real Numbers, Class 10 – All Solutions

02 Mar
Class 10, Class 10 - Chapter 1, Class 10 - Real Numbers, , , ,
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Ex: 8.1, Class 10

Exercise 1.1 (NCERT) - Real Numbers, Class 10

Chapter 1 Exercise 1.1 Class 10

Below are the Quick links for all questions of Exercise 1.1, Real Numbers, Class 10.

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Below are the solutions of every question of exercise 1.1 of chapter 1, Real Numbers, NCERT of class 10. YouTube video tutorial of every question is also embedded along with the written solution.  

Question 1

Express each number as a product of its prime factors:

You must learn the Ladder method to solve this question. Watch the video embedded below.

(i) 140

140 = 2 x 2 x 5 x 7 = 22 x 5 x 7

(ii) 156

156 = 2 x 2 x 3 x 13 = 22 x 3 x 13

(iii) 3825

3825 = 3 x 3 x 5 x 5 x 7 = 32 x 52 x 17

(iv) 5005

5005 = 5 x 7 x 11 x 13

(v) 7429

7429 = 17 x 19 x 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

26=2×1391=7×13𝐿𝐶𝑀(26,91)𝐻𝐶𝐹(26,91)=21×70×131=20×71×131=21×71×131=182=20×70×131=13

Verification: LCM × HCF = product of the two numbers

LCM (26, 91) x HCF (26, 91) = 182 x 13 = 2366 … (1)

26 x 91 = 2,366 … (2)

From (1) and (2) 

LCM (26, 91) x HCF (26, 91) = 26 x 91

(ii) 510 and 92

510=2×3×5×1792=2×2×23𝐿𝐶𝑀(510,92)𝐻𝐶𝐹(510,92)=21×31×51×171×230=22×30×50×170×231=22×31×51×171×231=23460=21×30×50×170×230=2

Verification: LCM × HCF = product of the two numbers

LCM (510, 92) x HCF (510, 92) = 23460 x 2 = 46920 … (1)

510 x 92 = 42960 … (2)

From (1) and (2) 

LCM (510, 92) x HCF (510, 92) = 510 x 92

(iii) 336 and 54

336=24×31×7154=21×33𝐿𝐶𝑀(336,54)𝐻𝐶𝐹(336,54)=24×31×71=21×33×70=24×33×71=3024=21×31×70=6

Verification: LCM × HCF = product of the two numbers

LCM (336, 54) x HCF (336, 54) = 3024 x 6 = 18144 … (1)

336 x 54 = 18144 … (2)

From (1) and (2) 

LCM (336, 54) x HCF (336, 54) = 336 x 54

Watch the video embedded below in case you have any doubts about finding LCM and HCF.

Question 3

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

(i) 12, 15 and 21

12=22×315=3×521=3×7𝐿𝐶𝑀(12,15,21)𝐻𝐶𝐹(12,15,21)=22×31×50×70=20×31×51×70=20×31×50×71=22×31×51×71=420=20×31×50×70=3

(ii) 17, 23 and 29

17=1723=2329=29𝐿𝐶𝑀(17,23,29)𝐻𝐶𝐹(17,23,29)=171×230×290=170×231×290=170×230×291=171×231×291=11339=170×230×290=1

(iii) 8, 9 and 25

8=239=3225=52𝐿𝐶𝑀(8,9,25)𝐻𝐶𝐹(8,9,25)=23×30×50=20×32×50=20×30×52=23×32×52=1800=20×30×50=1

Question 4

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

Let a = 306, b = 657

Now, we know that 

𝐿𝐶𝑀(𝑎,𝑏)×𝐻𝐶𝐹(𝑎,𝑏)⇒𝐿𝐶𝑀(306,657)×𝐻𝐶𝐹(306,657)⇒𝐿𝐶𝑀(306,657)×9⇒𝐿𝐶𝑀(306,657)=𝑎×𝑏=306×657=306×657=306×6579=34×657=22338

Question 5

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

Any number that ends with 0, is divisible by 10. 

For example, 20, 30, 50 etc. 

This means the number ending with 0 is always divisible by 2 and 5. 

Let us now check if both 2 and 5 are factors of 6n for some natural n or not.

6𝑛=(2×3)𝑛=2𝑛×3𝑛

We can see that for any natural n, only 2 and 3 are the prime factors of 6n.

So, it can never end with 0 because 5 is not a factor.

Question 6

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

A number is a composite number if it can be expressed as product of at least 2 prime numbers. 

7 × 11 × 13 + 13

7×11×13+13=13(7×11+1)=13(77+1)=13×78=13×2×3×13

We can see that 7 × 11 × 13 + 13 can be written as product of 4 prime numbers. 

Hence it is a composite number. 

7 × 6 × 5 × 4 × 3 × 2 × 1 + 5

7×6×5×4×3×2×1+5=5(7×6×4×3×2×1+1)=5(1008+1)=5×1009

We can see that 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 can be written as product of 2 prime numbers. 

Hence it 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?

Time taken by Sonia to complete 1 round = 18 minutes

Time taken by Ravi to complete 1 round = 12 minutes

This means after every 18 minutes, Sonia reaches the Starting Point and after every 12 minutes, Ravi reaches the Starting Point.

Or we can say that:

Sonia reaches the Starting Point at multiple of 18

Ravi reaches the Starting Point at multiple of 12

So we have to find the LCM (because both will need at starting point at multiple of 18 and 12 respectively)

12 = 2 x 2 x 3

18 = 2 x 3 x 3

LCM (12, 18) = 2 x 2 x 3 x 3 = 36

Both will meet at starting point after 36 minutes.

Ex: 8.2, Class 10
Ex: 8.1, Class 10
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