Question

Mr. Hayden's Bible class collected $11

in coins for a needy family. They col-
lected twice as many quarters as dimes
and 12 more nickels than dimes. How
many of each coin did the class collect?

Answer 1

This is the only way I can think of this mathematically. This took a good 20 minutes for me. You need to guess and check. Like try 17 quarters, 50 nickels, and 37 dimes etc. until you get 11.00. My numbers are mathematically correct but we both know you cant have 18.909 quarters so. lol

Answer:

Quarters: 18.909

Nickels: 49.818

Dimes: 37.818

Somewhere around those numbers.

Explanation:

$11.00 total in coins

Nickel = $0.05

Dime = $0.10

Quarter = $0.25

(n*0.05) + (d*0.10) + (q*0.25)

Solve the equation for q.

`q=(1)/(2) d`

(n*0.05) + (d*0.10) +(
`(1)/(2) d` *0.25)

Simplify the expression.

9d + 2n = 440

d - n = -12

Multiply both sides of the equation by -9.

-9d + 9n = 108

9d + 2n = 440

Sum the equations vertically to eliminate at least one variable.

11n = 548

`n=(548)/(11)`

Substitute the given value of n into the equation d - n = -12

`d-(548)/(11) =-12`

`d=(416)/(11)`

Substitute the given value of n into the equation
`q=(1)/(2) d`

`q=(1)/(2) *(416)/(11)`

`q=(208)/(11)`

Finally

`d=(416)/(11)` d = 37.818

`n=(548)/(11)` n = 49.818

`q=(208)/(11)` q = 18.909