Question

a bag contains 50 Paisa 25 paisa and 10 Paisa coin in the ratio 5:9:4 amounting to rupees find the number of coins of each type respectively

Answer 1

$\huge\bold\blue{Solution : }$

$\mathsf{Let \: \: the \: \: number \: \: of \: \: 50 \: paisa \: \: 25}$
$\mathsf{paisa \: \: and \: \: 10 \: paisa \: \: be \: \: 5x \: 9x \: and \: 4x}$
$\mathsf{respectively.}$

$\mathsf{Total \: \: value \: \: of \: \: these \: \: coins}$

$\mathsf{ = Rs \: \: (5x \times \frac{50}{100} + 9x \times \frac{25}{100} + 4x \times \frac{10}{100} ) }\\ \\ \mathsf{= Rs \: \: ( \frac{5x}{2} + \frac{9x}{4} + \frac{2x}{5} )}$

$\mathsf{ = Rs \: \: (\frac{50x +45x + 8x }{20} )} \\ \\ \mathsf{= Rs \: \: \frac{103x}{20}}$

$\mathsf{But, \: \: total \: \: value \: \: of \: \: coins \: \: = Rs \: \: 206}$

$\mathsf{ Therefore,\: \: \: \: \: \ \frac{103x}{20} = 206} \\ \\ \mathsf{= > 103x = 206 \times 20} \\ \\ \mathsf{= > x = \frac{206 \times 20}{103}} \\ \\ \mathsf{= > x = 40}$

$\mathsf{Number \: \: of \: \: 50p \: \: coins = 5x = (5 \times 40) = 200.}$

$\mathsf{Number \: \: of \: \: 25p \: \: coins = 9x = (9 \times 20) = 360.}$

$\mathsf{Number \: \: of \: \: 10p \: \: coins = 4x = (4 \times 20) = 160.}$

Answer 2

$<b>Answer:</b>$

105, 126, 672

$<b>Step-by-step explanation:</b>$

Given, Ratio of values = 5 : 6 : 8

                                     = 5 : (6/2) : (8/4)

                                     = 5 : 3 : 2

Given,Total = 210.

Divide 210 in the ratio of 5 : 3 : 2.

(i)

1st part = (210 * 5/10)

            = 105

2nd part = (210 * 3/10)

              = 63.

3rd part = (210 * 8/10)

             = 168

Therefore,

$\textsf{\large{Number of 1-rupee coins = 105.}}$

$\textsf{\large{Number of 50-paise coins = 63 * 2 = 126.}}$

$\textsf{\large{Number of 25-paise coins = 168 * 4 = 672.}}$

Hope it helps!