Chapter 4.10, Problem 15P

To determine

To calculate: Types ofcoinsJane had.

Answer to Problem 15P

This is unsolvable problem because we cannot have a fractional number of nickels.

Explanation of Solution

Given information:

Total money Jane have in nickels, dimes and quarters = $8.60.

Jane have 6 quarters less than nickels and 3 more dimes than twice the number of nickels.

Calculation:

Let us consider,

Number of nickels=n

Number of dimes=d

Number of quarters=q

Value of the nickels in cents=5n

Value of the dimes in cents=10d

Value of the quarters in cents=25q

The problem is in dollars so we will convert it to cents,

Thus,

Jane has $\$8.60=860cents$

According to question,

  $5n+10d+25q=860cents$

She has 6 fewer quarters than nickels $\therefore q=n-6$

3 more dimes than twice the number of nickels $\therefore d=2n+3$

Now substituting the value of q and d in the above equation we get-

  $\begin{array}{l}5n+10d+25q=860\\ 5n+10\left(2n+3\right)+25\left(n-6\right) =860\\ 5n+20n+30+25n-150=860 \\ 50n–120 =860\\ 50n=860+120\\ 50n=980\\ n=\frac{980}{50}\\ n=19.6\end{array}$

Since, n = 19.6 that means Number of nickels is 19.6.

And we know that number of nickels cannot be in fractions or point so it has no proper solution.

Thus, it has no proper solution.