# Exercise 1.2 class 10 maths – Real Numbers Complete Solutions

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163 # Exercise 1.2 Class 10 Maths

## class 10 maths chapter 1 exercise 1.2

Class 10 Maths Chapter 1 Exercise 1.2 : We are providing Class 10 Maths solutions  for 1st chapter “Real Numbers” textbook provided by NCERT for SEBA, CBSE etc complete exercise solutions with Additional Most Important Questions in our website – online, absolutely for FREE for everyone. No need to do any sign in process, just browse and enjoy all of the Class 10 Maths Chapter 1 exercise 1.2 solutions from us. From the First exercise “Real Numbers”, there are total 4 exercise given in bellow table. Click on the exercise number to browse the complete solutions from ex 1.2 class 10 ..

Sub-Contents of “Chapter 1 Real Numbers” : There are  total 6 sub contents exists in Class 10 maths chapter 1 Real Numbers. We need to study all of the sub-contents first. The Sub Contents are-

1. Introduction
2. Euclid’s Division algorithm
3. The Fundamental Theorem of Arithmetic
4. Revisiting Irrational Numbers
5. Revisiting Rational Numbers & Their Decimal Expansion
6. Summary
 FROM LATEST UPDATED NCERT SYLLABUS

## Exercise 1.2 Class 10 Maths

Real Numbers

#### 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,  6= (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.

## Ex 1.2 Class 10 Question 7 Concepts :

In that ex 1.2 class 10 maths question 7 , We are given to find the next exact same moments so that the different moments together occur again. This type of question asked in SSC GD, ARMY GD and various state jobs. That swhy these types of questions are some of the most important problems that we must have to remember for our whole life. So, what is the main technique to solve this type of questions, do you know? Okay, the reason is bellow:

If any two or more sources starts from a same point, to same direction in the same instances, and we are suggested to find the next moment, when they will meet again – what we need to do is simply just find the LCM of the given numbers. Thats it. Just remember this rule and you can solve any problems related to this rule.

## Ex 1.2 Class 10 Question 8(i) Concepts :

These questions are also considered as some of the most important problems from Class 10 Mathematics Exercise 1.2 ncert. Specially, for anyone, who is aiming to appear in any entrance exams and any primary level job interviews. So, we have enclosed the tricks to solve such problems bellow:

If we need to find the least number of some special cases which are well established by some special conditions like the question, you should find the Least Common Multiplier or LCM of the given numbers. Conversely, If we need to find the greatest or the biggest number, for which the well ordered conditions will not violet, we need to find the Greatest Common Divisor or the GCD means HCF of the given numbers.

## NCERT SOLUTIONS FOR CLASS 10 MATHS :

We are providing complete Class 10 Maths Solution . You can get complete and updated solution from our website. Please click on your prefered chapter bellow from NCERT CLASS 10 MATHS and see the solutions. You can get complete ncert solutions for class 10 maths from here.

 2 Chapter 2  Polynomials 3 Chapter 3  Pair of Linear Equations in Two Variables 4 Chapter 4  Quadratic Equations 5 Chapter 5  Arithmetic Progressions 6 Chapter 6  Triangles 7 Chapter 7  Co-ordinate Geometry 9 Chapter 9  Application of Trigonometry 10 Chapter 10  Circle 11 Chapter 11  Constructions 12 Chapter 12  Area Related to Circles 13 Chapter 13  Surface area and volumes 14 Chapter 14 Statistics 15 Chapter 15  Probability