EX.2.2 Q15 Linear Equations in One Variable Solutions - NCERT Maths Class 8
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?
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 |
\(\rm{Rs}.1\) |
\(\begin{align} 160 - 4x &= 160 - 80 \\&= 80 \end{align} \) |
\(\rm{Rs}.2\) |
\(3x = 3 \times 20 = 60\) |
\(\rm{Rs}.5\) |
\(x = 20\) |