exercise 1.2 class 10 maths

exercise 1.2 class 10 maths involve complete  answers for each question in the exercise 1.2. The solutions provide students a strategic methods  to prepare for their exam. exercise 1.2 class 10 maths questions and answers helps students to perform better in exam and it will  clear doubts definitely. Students will find it extremely easy to understand the questions and learn solving the problems.exercise 1.2 class 10 maths prepared by www.mathematicsandinformationtechnology.com team in very delicate, easy and creative way. 

Question 1. 

Express each number as a product of its prime factors:

(i) 140 

(ii) 156

(iii) 3825 

(iv) 5005 

(v) 7429 

Solution 1:

(i)\(140 = 2 × 2 × 5 × 7 = 2^2 × 5 × 7\)

(ii)\(156 = 2 × 2 × 3 × 13 = 2^2 × 3 × 13\)

(iii)\(3825 = 3 × 3 × 5 × 5 × 17 = 3^2 × 52 × 17\)

(iv)\(5005 = 5 ×7× 11 × 13\)

(v)\(7429 = 17 × 19 × 23\)
 

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 

(ii) 510 and 92 

(iii) 336 and 54 

(i)\(26\) and \(91\)

\(26 = 2 × 13\)

\(91 = 7 × 13\)

\(\text{HCF} = 13\)

\(\text{LCM} = 2 × 7 × 13 = 182\)

\(\text{Product of the two numbers} = 26 × 91 = 2366\)

\(\text{HCF × LCM} = 13 × 182 = 2366\)

\(\text{Hence, product of two numbers = HCF × LCM}\)

(ii)\(510\) and \(92\)

\(510 = 2 × 3 × 5 × 17\)

\(92 = 2 × 2 × 23\)

\(\text{HCF} = 2\)

\(\text{LCM} = 2 × 2 × 3 × 5 × 17 × 23 = 23460\)

\(\text{Product of the two numbers }= 510 × 92 = 46920\)

\(\text{HCF × LCM }= 2 × 23460= 46920\)

\(\text{Hence, product of two numbers = HCF × LCM}\)

(iii)\(336\) and \(54\)

\(336 = 2 × 2 × 2 × 2 × 3 × 7\)

\(336 = 24× 3× 7\)

\(54 = 2 × 3 × 3 × 3\)

\(54 = 2 × 33\)

\(\text{HCF} = 2 × 3 = 6\)

\(\text{LCM} = 24 × 33 × 7 = 3024\)

\(\text{Product of the two numbers }= 336 × 54 = 18144\)

\(\text{HCF × LCM} = 6 × 3024 = 18144\)

\(\text{Hence, product of two numbers  = HCF × LCM}\).

Question 3. 

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

(i) 12, 15 and 21 

(ii) 17, 23 and 29 

(iii) 8, 9 and 25 

Solution 3:

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

\(12 = 22× 3\)

\(15 = 3 × 5\)

\(21 = 3 × 7\)

\(\text{HCF} = 3\)

\(\text{LCM }= 22× 3 × 5 × 7 = 420\)

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

\(17 = 1 × 17\)

\(23 = 1 × 23\)

\(29 = 1 × 29\)

\(\text{HCF} = 1\)

\(\text{LCM} = 17 × 23 × 29 = 11339\)

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

\(8 = 2 × 2 × 2\)

\(9 = 3 × 3\)

\(25 = 5 × 5\)

\(\text{HCF} = 1\)

\(\text{LCM } = 2 × 2 × 2 × 3 × 3 × 5 × 5 = 1800\)

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

\(\text{HCF} (306, 657) = 9\)

\(\text{Product of two numbers  = HCF × LCM}\).

\(\therefore \text{HCF × LCM}=306 × 657 \)

\(\text{LCM}= {306 × 657 \over \text{HCF}} = {306 × 657 \over 9}\)

\(\text{LCM}= {22338}\)

Question 5:

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

Solution 5:

If any number ends with the digit \(0\), it should be divisible by \(10\) or in other words, it will also be divisible by \(2\) and \(5\) as \(10 = 2 × 5\)

Prime factorization of \(6^n = (2 ×3)^n\)

It can be observed that \(5\) is not in the prime factorization of \(6^n\).

Hence, for any value of \(n, 6^n\) will not be divisible by \(5\).

Therefore, \(6^n\) cannot end with the digit \(0\) for any natural number \( n\).

Question 6:

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

Solution 6:

Numbers are of two types - prime and composite. A prime number can be divided by \(1\) and only itself, whereas a composite number have factors other than \(1\) and itself. It can be observed that

\(7 × 11 × 13 + 13 \)

\(= 13 × (7 × 11 + 1) \)

\(= 13 × (77 + 1) \)

\(= 13 × 78\)

\(= 13 ×13 × 6\)

The given expression has \(6\) and \(13\) as its factors.

Therefore, it is a composite number.

\(7 × 6 × 5 × 4 × 3 × 2 × 1 + 5\)

\(= 5 ×(7 × 6 × 4 × 3 × 2 × 1 + 1)\)

\(= 5 × (1008 + 1)\)

\(= 5 ×1009\)

\(1009\) cannot be factorized further.

Therefore, the given expression has \(5\) and \(1009\) as its factors.

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?

Solution 7:

It can be observed that Ravi takes lesser time than Sonia for completing \(1\) round of the circular path. As they are going in the same direction, they will meet again at the same time when Ravi will have completed \(1\) round of that circular path with respect to Sonia. And the total time taken for completing this \(1\) round of circular path will be the LCM of time taken by Sonia and Ravi for completing \(1\) round of circular path respectively i.e.,

LCM of \(18\) minutes and \(12\) minutes.

\(18 = 2 ×3 ×3\)

And, \(12 = 2 ×2 ×3\)

LCM of \(12\) and \(18 = 2 × 2 × 3 × 3 = 36\)

Therefore, Ravi and Sonia will meet together at the starting point after \(36\) minutes.