The **Chapter 1** of **Class 10 Mathematics** based on **NCERT** syllabus is about the **Real Numbers**. This page contains the **Solution** of **Exercise 1.**2, **Chapter 1** of **Class 10 Mathematics**. The **Exercise 1.**2 of **Class 10 Mathematics** consists of 7 questions which are based on **The Fundamental Theorem of Arithmetic**.

Description | NCERT Maths Solutions |

Subject | Mathematics |

Class | 10 (X) |

Chapter | 1 |

Exercise | 1.2 |

Topic | Fundamental Theorem of Arithmetic |

Syllabus | NCERT |

The solution of all the 7 questions of **Exercise 1.2** of **Class 10**, **Chapter 1** is given below.

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

**i. **140

**Solution:**

The prime factors of 140 are

140=2\times 2\times 5\times 7 \Rightarrow 140=2^2\times 5\times 7**ii. **156

**Solution:**

The prime factors of 156 are

156=2\times 2\times 3\times 13 \Rightarrow 156=2^2\times 3\times 13**iii.** 3825

**Solution:**

The prime factors of 3825 are

3825=3\times 3\times 5\times 5\times 17 \Rightarrow 3825=3^2\times 5^2\times 17**iv.** 5005

**Solution:**

The prime factors of 5005 are

5005=5\times 7\times 11\times 13**v.** 7429

**Solution:**

The prime factors of 7429 are

7429=17\times 19\times 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

**Solution:**

The prime factors of 26 are

26=2\times 13The prime factors of 91 are

91=7\times 13Therefore,

HCF=13 LCM=2\times 7\times 13=182**Verification:**

We know that

LCM\times HCF=a\times b \therefore LHS=LCM\times HCF \Rightarrow LHS=183\times 13 \Rightarrow LHS=2366Now

RHS=a\times b \Rightarrow RHS=26\times 91 \Rightarrow RHS=2366Therefore, LHS=RHS

Hence verified.

**ii.** 510 and 92

**Solution:**

The prime factors of 510 are

510=2\times 3\times 5\times 17The prime factors of 92 are

92=2\times 2\times 23Therefore,

HCF=2 LCM=2\times 2\times 2\times 3\times 3\times 5\times 17=23460**Verification:**

We know that

LCM\times HCF=a\times b \therefore LHS=LCM\times HCF \Rightarrow LHS=23460\times 2 \Rightarrow LHS=46920Now

RHS=a\times b \Rightarrow RHS=510\times 92 \Rightarrow RHS=46920Therefore, LHS=RHS

Hence verified.

**iii.** 336 and 54

**Solution:**

The prime factors of 336 are

336=2\times 2\times 2\times 2\times 3\times 7The prime factors of 54 are

54=2\times 3\times 3\times 3Therefore,

HCF=2\times 3=6 LCM=2\times 2\times 2\times 2\times 3\times 3\times 3\times 7=3024**Verification:**

We know that

LCM\times HCF=a\times b \therefore LHS=LCM\times HCF \Rightarrow LHS=3024\times 6 \Rightarrow LHS=18144Now

RHS=a\times b \Rightarrow RHS=336\times 54 \Rightarrow RHS=18144Therefore, LHS=RHS

Hence verified.

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

**i. **12, 15 and 21

**Solution:**

The prime factorisation of 12 is

12=2\times 2\times 3The prime factorisation of 15 is

15=3\times 5And the prime factorisation of 21 is

21=3\times 7Therefore,

HCF=3 LCM=2\times 2\times 3\times 5\times 7=420**ii.** 17, 23 and 29

**Solution:**

All the three numbers 17, 23 and 29 are prime numbers.

Therefore,

HCF=1 and

LCM=17\times 23\times 29=11339**iii.** 8, 9 and 25

**Solution:**

The prime factorisation of 8 is

8=2\times 2\times 2The prime factorisation of 9 is

9=3\times 3And the prime factorisation of 25 is

25=5\times 5Therefore,

HCF=1 LCM=8\times 9\times 25=1800

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

**Solution:**

Given,

a=306 b=657And HCF=9

We know that,

LCM\times HCF =a\times b \Rightarrow LCM=\frac{a\times b}{HCF} \Rightarrow LCM=\frac{306\times 657}{9} \Rightarrow LCM=22338Therefore, LCM (306,\, 657) is 22338.

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

**Solution:**

6^n can be written as

6^n=(2\times 3)^n 6^n=2^n\times 3^nWe know that, any number that ends with 0 is divisible by 5. Therefore, the prime factors of such number must have 5.

Here, the factors of 6^n does not have 5. Therefore, the number 6^n cannot end with 0.

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

**Solution:**

The first number

7\times 11\times 13+13 =(7\times 11\times 13)+13 =13\times (7\times 11+1) =13\times 78 =2\times 3\times 13\times 13The second number

7\times 6\times 5\times 4\times 3\times 2\times 1+5 =5\times (7\times 6\times 4\times 3\times 2\times 1+1) =5\times (1008+1) =5\times 1009In both the above numbers, all the factors are prime numbers. Therefore, the given numbers are composite numbers.

**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:**

Time taken by Sonia=18 minutes

Time taken by Ravi=12 minutes

The time after which Sonia and Ravi will meet at the starting point is the LCM of 18 and 12.

The prime factorisation of 18 is

18=2\times 3\times 3The prime factorisation of 12 is

12=2\times 2\times 3 \therefore LCM=2\times 2\times 3\times 3=36Therefore, Sonia and Ravi will meet at the starting point after 36 minutes.

I hope the solutions of Class 10, Exercise 1.2, Chapter 1 of Mathematics have helped you. If you have any issues/queries related to the NCERT Solutions of Class 10 Mathematics, Exercise 1.2, feel free to contact me at [email protected] or [email protected] or fill the form here.

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