Chapter 4.10, Problem 10P

To determine

To Find: the number of coins Janet has of each category..

Answer to Problem 10P

(1,30,22),(8,34,19),(15,38,16),(22,42,13),(29,46,10),(36,50,7),(43,54,4) and (50,58,1).

Explanation of Solution

Given: Janet has coins of nikels,dimes and quarters worth $8.55$ $.

She has $7$ more dimes than nickels and quarters combined.

Calculation:

Let number of nickels be $n$ ,dimes be $d$ and quarters be $q$ .

Equation for their value can be written as.

  $\begin{array}{l}n\times 5+d\times 10+q\times 25=855\\ 5n+10d+25q=855\\ n+2d+5q=171\end{array}$

Since, Janet has $7$ more dimes than nickels and quarters combined therefore.

  $\begin{array}{l}d-\left(n+q\right)=7\\ d-n-q=7\\ d=n+q+7\end{array}$

Substitute second equation into first.

  $\begin{array}{l}n+2d+5q=171\\ n+2\left(n+q+7\right)+5q=171\\ 3n+7q+14=171\\ 3n+7q=157\\ 7q=157-3n\\ q=\frac{157-3n}{7}\end{array}$

This equation allows multiple set of answer which are integers only therefore by trial error, following answers have been found.

n=1,d=30,q=22

n=8,d=34,q=19

n=15,d=38,q=16

n=22,d=42,q=13

n=29,d=46,q=10

n=36,d=50,q=7

n=43,d=54,q=4

n=50,d=58,q=1

So the answer is (1,30,22),(8,34,19),(15,38,16),(22,42,13),(29,46,10),(36,50,7),(43,54,4) and (50,58,1).