# NCERT Solutions For Class 10 Maths Chapter 1 Exercise 1.2

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## Question 1

Express each number as a product of its prime factors:

(i) $$140$$

(ii) $$156$$

(iii) $$3825$$

(iv) $$5005$$

(v) $$7429$$

### Solution

What is known?

A number.

What is unknown?

The expression of the given number as a product of its prime factors.

Reasoning:

Find the prime factors of the given numbers by prime factorisation method and then multiply the obtained prime numbers to get the product of the prime numbers.

Steps:

(i) $$140$$

Prime factors of $$140$$

\begin{align}&=2\times 2 \times 5 \times 7 \\&={{2}^{2}}\times 5\times 7\end{align}

(ii) $$156$$

Prime factors of $$156$$

\begin{align}&=2\times 2\times 3\times 13 \\ &={{2}^{2}}\times 3\times 13 \end{align}

(iii) $$3825$$

Prime factors of $$3825$$

\begin{align}&=3\times 3\times 5\times 5\times 17 \\ &={{3}^{2}}\times {{5}^{2}}\times 17\end{align}

(iv) $$5005$$

Prime factors of $$5005$$

$=5\times7\times11\times 13$

(v) $$7429$$

Prime factors of $$7429$$

$=17\times 19\times 23$

## Question 2

Find the LCM and HCF of the following pairs of integers and verify that LCM $$\times$$ HCF $$=$$ Product of the two numbers.

(i) $$26$$ and $$91$$

(ii) $$510$$ and $$92$$

(iii) $$336$$ and $$54$$

### Solution

What is known?

Pairs of numbers.

What is unknown?

The LCM and HCF of the following pairs of integers and verify that LCM $$\times$$ HCF $$=$$ Product of the two numbers

Reasoning:

• To find the LCM and HCF of the given pairs of the integers, first find the prime factors of the given pairs of integers.
• Then, find the product of smallest power of each common factor in the numbers. This will be the HCF.
• Then find the product of greatest power of each prime factor in the number. This would be the LCM.
• Now, you have to verify LCM $$\times$$ HCF $$=$$ product of the two numbers, find the product of LCM and HCF and also the two given numbers. If LHS is equal to the RHS then it will be verified.

Steps:

(i) $$26$$ and $$91$$

Prime factors of $$26=2\times 13$$

Prime factors of $$91=7\times 13$$

HCF of $$26$$ and $$91=13$$

LCM of $$26$$ and $$91$$\begin{align}&=2 \times7 \times 13\\&=14 \times 13 \\&=\text{ }182 \\ \end{align}

Product of two numbers \begin{align}&=26\times 91\\ &=2366 \end{align}

LCM $$\times$$ HCF

\begin{align}&= 182\,\times\,13\\ &= 2366\end{align}

So,

Product of two numbers $$=$$ LCM $$\times$$HCF

(ii) $$510$$ and $$92$$

Prime factors of $$510=2\times 3\times 5\times 17$$

Prime factors of $$92=2\,\times \,2\times 23$$

HCF of two numbers $$=2$$

LCM of two numbers

\begin{align}&=2\times 2\times 3\times 5\times 17\times 23 \\ &=23460\end{align}

Product of two numbers

\begin{align}&=510\,\times\,92 \\ &= 46920\end{align}

LCM $$\times$$ HCF \begin{align}&=2\times 23460 \\&=46920\end{align}

Product of two numbers $$=$$ LCM $$\times$$ HCF

(iii) $$336$$ and $$54$$

Prime factors of $$336=2\times 2\times 2\times 2\times 3\times 7$$

Prime factors of $$54=2 \times 3 \times 3\times 3$$

HCF of two numbers $$=6$$

LCM of two numbers

\begin{align} &=2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 7 \\ &=2^{4} \times 3^{3} \times 7 \\ &=3024 \end{align}

Product of two numbers

\begin{align} &=336 \times 54 \\ &=18144 \end{align}

LCM $$\times$$ HCF

\begin{align}&=3024 \times 6 \\ &=18144\end{align}

Product of two numbers $$=$$ LCM $$\times$$ HCF

## 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

What is known?

(i)  $$12, 15$$ and $$21$$

(ii)  $$17, 23$$ and $$29$$

(iii)  $$8, 9$$ and $$25$$

What is unknown?

The LCM and HCF of the given integers by applying the prime factorisation method.

Reasoning:

To solve this question, follow the following steps-

• First find the prime factors of the given integers.
• Find the HCF of the given pair of integers i.e. product of smallest power of each prime factor, involved in the number.
• Lastly, Find the LCM of the given pair of integers i.e. product of greatest power of each prime factor, involved in the number.

Steps:

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

Prime factors of $$12$$

\begin{align}&= {2 \times 2 \times 3}\\&={{2^2} \times 3}\end{align}

Prime factors of $$15= 3 \times 5$$

Prime factors of $$21=2 \times 2 \times 3$$

HCF of $$12,15$$ and $$21= {3}$$

LCM of $$12,15$$ and $$21$$  \begin{align}&= {{2^2} \times 3 \times 5 \times 7}\\ &={ 420}\end{align}

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

Prime factors of $$17= 17\,\times\,1$$

Prime factors of $$23=23\,\times\,1$$

Prime factors of $$29 = 29 \times 1$$

HCF of $$7,\,23$$ and $$29={1}$$

LCM of $$17,\,23$$ and $$29$$

\begin{align}&= 17 \times 23 \times 29\\&= 11339\end{align}

(iii) $$8, 9$$ and $$25$$

Prime factors of $$8$$

\begin{align}&= 2 \times 2 \times 2\\ &= {2^3}\end{align}

Prime factors of $$9$$

\begin{align}&= 3 \times 3\\ &= {3^2}\end{align}

Prime factors of $$25$$

\begin{align}&= 5 \times 5\\ &= {\text{ }}{5^2}\end{align}

HCF of $$8, 9$$ and $$25 = 1$$

LCM of $$8,\,9$$ and $$25$$

\begin{align} &=2 \times 2 \times 2 \times 3 \times 3 \times 5 \times 5 \\ &=1800 \end{align}

## Question 4

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

### Solution

What is known?

HCF of two numbers $$(306, 657) = 9$$

What is unknown?

LCM of the given numbers

Reasoning:

We know that LCM $$\times$$ HCF $$=$$ product of two given integers

We have the given numbers as $$306$$ and $$657$$ and we can find the product of $$306$$ and $$657$$. The HCF of this two numbers is $$9$$. Put the values in the above property and find the value of unknown i.e. HCF.

Steps:

Given, HCF $$(306, 657) = 9.$$

We have to find,

LCM $$(306, 657) = ?$$

We know that

LCM$$\times$$ HCF $$=$$ Product of two numbers

\begin{align}\rm{LCM}\times\, 9 &= 306\, \times \,657\\\rm{LCM}&= {\frac{{306\, \times \,657}}{9}}\\\rm{LCM}&= 34\, \times \,657\\\rm{LCM}&= 22338 \end{align}

## Question 5

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

### Solution

What is unknown?

Whether $$6^n$$ can end with the digit $$0$$ for any natural number $$n$$.

Reasoning:

If any number ends with the digit $$0$$ that means it should be divisible by $$5$$. That is, if $$6^n$$ ends with the digit $$0$$, then the prime factorization of $$6^n$$ would contain the prime $$5$$.

Steps:

Prime factors of $${6^{{n}}} = {\left( {2 \times 3} \right)^{{n}}} = {\left( 2 \right)^{{n}}}{\left( 3 \right)^{{n}}}$$

You can observe clearly, $$5$$ is not in the prime factors of $$6^n$$.

That means $$6^n$$ 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

What is unknown?

Whether $$7{{ }} \times {{ }}11{{ }} \times {{ }}13{{ }} + {{ }}13$$ and $$7{{ }} \times {{ }}6{{ }} \times {{ }}5{{ }} \times {{ }}4{{ }} \times {{ }}3{{ }} \times {{ }}2{{ }} \times {{ }}1{{ }} + {{ }}5$$ are composite numbers?

Reasoning:

To solve this question, recall that:

• Prime numbers are whole numbers whose only factors are $$1$$ and itself.
• Composite number are the positive integers which has factors other than $$1$$ and itself.

Now, simplify $$7{{ }} \times {{ }}11{{ }} \times {{ }}13{{ }} + {{ }}13$$ and $$7{{ }} \times {{ }}6{{ }} \times {{ }}5{{ }} \times {{ }}4{{ }} \times {{ }}3{{ }} \times {{ }}2{{ }} \times {{ }}1{{ }} + {{ }}5.$$ On simplifying them, you will find that both the numbers have more than two factors. So, if the number has more than two factors, it will be composite.

Steps:

It can be observed that,

\begin{align} 7\times 11\times &13+13\,\\&=13\left( 7\times 11+1 \right) \\ \,\,&=13\left( 77+1 \right) \\ &=13\,\,\times \,78 \\ &=13\times 13\times 6\times 1 \\&=13\times13\times2\times3\times1\end{align}

The given number has $$2,3,13$$ and $$1$$ as its factors. Therefore, it is a composite number.

\begin{align}7\times &6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5 \\&= {5 \times \left( {7\!\times\! 6 \!\times\! 4 \!\times\! 3 \!\times\! 2 \!\times\! 1 \!+\!1} \right)}\\&= {5 \times \left( {1008 + 1} \right)}\\ &={ 5 \times 1009 \times 1}\end{align}

$$1009$$ cannot be factorised further. Therefore, the given expression has $$5,1009$$ and $$1$$ as its factors. Hence, it is a composite number.

## Question 7

There is a circular path around a sport 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

What is known?

• Sonia takes $$18$$ minutes to drive one round of the field.
• Ravi takes $$12$$ minutes for the same.
• They both start at the same point and at the same time and go in the same direction.

What is unknown?

After how many minutes will they meet again at the same point.

Reasoning:

Time taken by Sonia is more than Ravi to complete one round. Now, you have to find after how many minutes will they meet again at the same point. For this, there will be a number which is a divisible by both $$18$$ and $$12$$ and that will be the time when both meet again at the starting point. To find this you have to take LCM of both the numbers.

Steps:

LCM of $$18$$ and $$12$$,

\begin{align}18 &= 2 \times 3 \times 3\\12 &= 2 \times 2 \times 3\end{align}

LCM of $$12$$ and $$18$$

\begin{align}&= 2 \times 2 \times 3 \times 3\\ &= 36\end{align}

Therefore, Ravi and Sonia will meet together at starting point after $$36$$ minutes.