Question

katha has coins in the denominations of rs.2, rs.5, and rs.10. The number of rs.2 coins is three times the number of rs.5 coins and total number of coins is 160.If the total amount of the coins is rs.730, howmany coins of each denomination does Katha have?

Answer 1

Solutions :-

Let the number of Rs 5 coins be x
Number of Rs 2 coins = 3x
Total number of coins = 160
So,
Number of Rs 10 coins = 160 - 4x
The total amount of the coins = Rs 730

A/q

=> 5(x) + 2(3x) + 10(160 - 4x) = 730
=> 5x + 6x + 1600 - 40x = 730
=> 11x - 40x = 730 - 1600
=> -29x = -870
=> x = -870/-29 = 30

Hence,
Number of Rs 5 coins = x = 30
Number of Rs 2 coins = 3x = 30 × 3 = 90
Number of Rs 10 coins = (160 - 4x) = 160 - 4×30 = 160 - 120 = 40

Answer 2

GIVEN :-

• Katha have denominations of 2rs,5rs and 10rs.

• 2 coins is three times the number of 5

• total no of coins 160

• total amount of coins is rs 730

TO FIND :-

• no of coins of each denomination does Katha have

SOLUTION :-

let the number of 5rs coins be x

HENCE ,

$\rm{no \: of \: 2 \: rs \: coins \: =3x }$

$\rm{total \: no \: of \: coins \: =160 }$

so,

$\rm{no \: of \: 10 \: rs \: coins =160 - 3x - x }$

$\rm{no \: of \: 10 \: rs \: coins =160 - 4x }$

now according to question :-

$\implies \rm{5(x) + 2(3x) + 10(160 - 4x) = 730}$

$\implies \rm{5x + 6x+ 1600 - 40x = 730}$

$\implies \rm{11x + 1600 - 40x = 730}$

$\implies \rm{11x - 40x = 730 - 1600}$

$\implies \rm{- 29x = - 870}$

$\implies \rm{x = \dfrac{ - 870}{ - 29} }$

$\implies \rm{ \bf \: x = 30}$

now ,

$\implies \rm{no \:of \: 5 \: rs \: coins = x}$

$\implies \boxed{\rm{no \:of \: 5 \: rs \: coins = 30}}$

$\implies \rm{no \:of \: 2 \: rs \: coins = 3x}$

$\implies \boxed {\rm{no \:of \: 2 \: rs \: coins = 90}}$

$\implies \rm{no \:of \: 10 \: rs \: coins = 160 - 4x}$

$\implies \rm{no \:of \: 10 \: rs \: coins = 160 - 120}$

$\implies \boxed{ \rm{no \:of \: 10 \: rs \: coins = 40}}$