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

Answer:

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

Step-by-step 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