A box contains $7.15 in nickels, dimes, and quarters. There are 44 coins in all, and the sum of the numbers of nickels and dimes is two less than the number of quarters. How many coins of each kind are there?
A) 17 nickels, 6 dimes, and 21 quarters
B) 16 nickels, 5 dimes, and 23 quarters
C) 16 nickels, 10 dimes, and 27 quarters
D) 14 nickels, 7 dimes, and 23 quarters
E) 18 nickels, 9 dimes, and 17 quarters

Question

Quizplus · Accepted Answer

Let's use the following variables to represent the number of each coin:
- N: number of nickels
- D: number of dimes
- Q: number of quarters

From the problem, we know that:
- N + D + Q = 44 (the total number of coins is 44)
- 0.05N + 0.1D + 0.25Q = 7.15 (the total value of the coins is $7.15)
- N + D = Q - 2 (the sum of nickels and dimes is 2 less than the number of quarters)

Note that we can rewrite the third equation to be N + D + 2 = Q. Then we can substitute this into the first equation to get N + D + (N + D + 2) = 44, which simplifies to 2N + 2D + 2 = 44 or N + D = 21.

Now we have two equations involving N and D: N + D = 21 and 0.05N + 0.1D + 0.25Q = 7.15. We can use the first equation to solve for one variable in terms of the other, for example N = 21 - D. Substituting this into the second equation gives:

0.05(21 - D) + 0.1D + 0.25Q = 7.15
Simplifying and rearranging gives:
0.2D + 0.25Q = 6.9
4D + 5Q = 138

At this point, we can try different values of D and Q that satisfy this equation and check if they also satisfy the other conditions. One approach is to use trial and error, but we can also use some reasoning to narrow down our options. Note that D and Q must be integers, and D cannot be greater than Q (because a quarter is worth more than a dime).

Some possible pairs of D and Q that sum to 44 are: (2, 22), (3, 21), (4, 20), (5, 19), (6, 18), (7, 17), (8, 16), (9, 15), (10, 14), (11, 13), and (12, 12).

Out of these pairs, only (7, 17) and (14, 10) satisfy the equation 4D + 5Q = 138 (you can check this by plugging in the values). However, if we use (7, 17), we get N = 21 - 7 = 14, which doesn't work because 14 + 7 + 17 = 38, not 44. So the only valid pair is (14, 10), which gives N = 21 - 14 = 7. We can double-check that this works by verifying that 7 + 14 + 10 = 31 coins and 0.05(7) + 0.1(14) + 0.25(10) = 0.35 + 1.4 + 2.5 = 4.25 dollars.

Therefore, the answer is choice D: 14 nickels, 7 dimes, and 23 quarters.

A Box Contains $7