We use the fact that if 3 quantities are in the ratio
of a:b:c, then the fraction of the total which the first,
second and third quantities are, are respectively,
..
There were 5c 10c and 20c coins in Joel's wallet in the ratio
of 2:6:7 respectively. If he takes out half the number of 5c
and half the number of10c coins they would make up $4.20.
How many 20c coins in his wallet?let x = number of 5c coins.
let y = number of 10c coins.
let z = number of 20c coins.
The fraction of 5c coins is .
The fraction of 10c coins is .
the fraction of 20c coins is .
Since he had x 5c coins, then the value of them would be 5x cents.
When he takes out half of them, only 2.5x cents.
Suppose he had y 10c coins, then the value of them would be 10y cents.
When he takes out half of them, that's only 5y cents.
2.5x + 5y = 420
Multiply through by 2
5x + 10y = 840
Divide through by 5
x + 2y = 168
x = 168 - 2y
Substitute for x
Divide the first through by 2 and the second through by 3
Solve the first for z
z = 1092-14y
Substitute in the second equation
-336+7y-2(1092-14y) = 0
-336+7y-2184+28y = 0
35y-2520 = 0
35y = 2520
y = 72
Substitute in
z = 1092-14y
z = 1092-1008
z = 84
x = 168 - 2y
x = 168 - 2(72)
x = 168 - 144
x = 24
Substitute in
So there are 24 5c coins, 72 10c coins, 84 20c coins
Edwin