# EX.2.2 Q15 Linear Equations in One Variable Solutions - NCERT Maths Class 8

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## Question

I have a total of $$\rm{Rs}. \,300$$ in coins of denomination $$\rm{Rs}.\,1$$, $$\rm{Rs}.\,2$$ and $$\rm{Rs}.\,5$$. The number of $$\rm{Rs}.\,2$$ coins is $$3$$ times the number of $$\rm{Rs}.\,5$$ coins. The total number of coins is $$160$$. How many coins of each denomination are with me?

Video Solution
Linear Equations
Ex 2.2 | Question 15

## Text Solution

What is known?

(i) Total amount $$= \rm{Rs}.\, 300$$

(ii) Coin denominations are $$\rm{Rs.} 1$$, $$\rm{Rs}.\,2$$ and $$\rm{Rs}. 5$$

(iii) Number of $$\rm{Rs}.\,2$$ coins are $$3$$ times $$\rm{Rs.} \,5$$ coins.

(iv) Total coins ➔$$160$$

What is unknown?

How many coins of each denominations are there.

Reasoning:

Assume the number of $$\rm{Rs}.\,5$$ coin to be a variable. Use the second condition to obtain the number of coins for $$\rm{Rs}.\,2$$ in terms of the variable. Use the third condition to find the number of coin for $$\rm{Rs}.1$$. Finally, use the first condition to form a linear equation.

Steps:

Let the number of $$\rm{Rs} \,5$$ coins be $$x.$$

Then the number of $$\rm{Rs}.2$$ coins is $$3x.$$

Number of $$\rm{Rs.}1$$ coin is $$= 160 - (x + 3x)$$ [$$\therefore$$as total coins are 160]

$= 160 - 4x$

 Denomination Number of coins Amount $$\rm{Rs}.1$$ $$160-4x$$ $$160-4x$$ $$\rm{Rs}.2$$ $$3x$$ $$6x$$ $$\rm{Rs}.5$$ $$x$$ $$5x$$

$\to 160 - 4x + 6x + 5x = 300$

\begin{align}160 + 7x &= 300 \\7x &= 300 - 160 \\7x &= 140 \\x &= \frac{{140}}{7} \\x &= 20 \\\end{align}

 Denomination Number of coins $$\rm{Rs}.1$$ \begin{align} 160 - 4x &= 160 - 80 \\&= 80 \end{align} $$\rm{Rs}.2$$ $$3x = 3 \times 20 = 60$$ $$\rm{Rs}.5$$ $$x = 20$$

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