Question

A newspaper carrier has $3.65 in change he has four more quarters then dimes but two times as many nickels as quarters how many coins of each type does he have

Answer 1

The number of quarter coins is 9, the number of dime coins is 5 and the number of nickel coins is 18

Explanation:

Let

x ----> the number of quarter coins

y ---> the number of dime coins

z ---> the number of nickel coins

Remember that

`1\ quarter=\$0.25`

`1\ dime=\$0.10`

`1\ nickel=\$0.05`

we know that

`0.25x+0.10y+0.05z=3.65` ----> equation A

`x=y+4`

`y=x-4` ----> equation B

`z=2x` ----> equation C

Solve the system by substitution

substitute equation B and equation C in equation A

`0.25x+0.10(x-4)+0.05(2x)=3.65`

solve for x

`0.25x+0.10x-0.40+0.10x=3.65`

`0.45x=3.65+0.40`

`0.45x=4.05\\x=9`

Find the value of y

`y=x-4`

substitute the value of x

`y=9-4=5`

Find the value of z

`z=2x`

`z=2(9)=18`

therefore

The number of quarter coins is 9, the number of dime coins is 5 and the number of nickel coins is 18