Question

$\ \textless \ br /\ \textgreater \ \huge\mathtt\pink{I've \: total \: of \: 300rs \: in \: coins \: of \: denomination \: 1rs , 2rs \: and \: 5rs. the \: number \: of \: 2rs \: coins \: is \: 3 \: times \: the \: number \: of \: 5rs \: coins. \: the \: total \: number \: of \: coins \: is \: 160. \: how \: many \: coins \: of \: each \: denomination \: are \: with \: me \: ? }\ \textless \ br /\ \textgreater$
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Answer 1

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Answer 2

Answer:

Let the number of coin 5 Rs.x so the number of coin 2 = 3x so the number of coin 1 = 160 - 4x

so amount of rs.5 = 5x so the amount of rs.2 6x

a/q

5x + 6x + 160 - 4x = 300 7x = 300 - 160

7x = 140

X= 20

so the number of coin of 5 = 20 coins

so the number of coin of 2 = 60 coins

so the number of coin of 1 = 80 coins

$\mathfrak\purple {Hope\: this\: helps\: you....}$