Question

a bag contains some Re 1,Rs 2 and Rs 5 coins in the ratio 8:5:3.If the total value of the coins was Rs 330,find the number of Rs 2 coins?

Answer 1

50
let x be common term in ratio
then no of 1rs coin =8x
2rs coins=5x
and 5rs coins=3x
then
total cost = 1*8x+2*5x+5*3x=330
33x=330
X= 10
no of 5rs coins=5x=50

Answer 2

Answer:

Required numbers of 2 rupees coin are 50.

Step-by-step explanation:

Given,a bag contains some Rs 1,Rs 2 and Rs 5 coins in the ratio 8:5:3.

Let number of 1 rupee coin be 8x

and number of 2 rupee coin be 5x

and number of 5 rupee coin be 3x.

Therefore now total amount by 1 rupee coin is

$(8x \times 1) = 8x \: rupees$

and total amount by 2 rupee coin

$(5x \times 2) = 10x \: rupees$

and total amount by 5 rupee coin

$(3x \times 5) = 15x \: rupees$

Therefore total amount in the bag

$= 8x + 10x + 15x \\ = 33x \: rupees$

According to question,

$33x = 330 \\ x = \frac{330}{33} \\ x = 10$

So,number of 1 rupee coin

$= (8 \times 10) = 80$

and number of 2 rupee coin

$=( 5 \times 10) = 50$

and number of 5 rupee coin

$= (3 \times 10) = 30$

This is a problem of ratio.

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