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

\(\rm{Rs}.1\)

\(\begin{align} 160 - 4x &= 160 - 80 \\&= 80 \end{align} \)

\(\rm{Rs}.2\)

\(3x = 3 \times 20 = 60\)

\(\rm{Rs}.5\)

\(x = 20\)