A box contains $6.80 in nickels, dimes, and quarters. There are 40 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) 7 nickels, 12 dimes, and 21 quarters
B) 11 nickels, 11 dimes, and 18 quarters
C) 6 nickels, 13 dimes, and 21 quarters
D) 6 nickels, 10 dimes, and 25 quarters

Question

Quizplus · Accepted Answer

Let's denote the number of nickels, dimes, and quarters as N, D, and Q, respectively. We are given three pieces of information:1. The total value of the coins is $6.80, which translates to the equation: $0.05N + 0.10D + 0.25Q = 6.80$.2. The total number of coins is 40, leading to the equation: $N + D + Q = 40$.3. The sum of the numbers of nickels and dimes is two less than the number of quarters, which gives us: $N + D = Q - 2$.Rearranging the third equation gives us $N + D + 2 = Q$. Substituting this into the second equation, we get $N + D + (N + D + 2) = 40$, simplifying to $2(N + D) + 2 = 40$, and further to $N + D = 19$.Given these equations, we need to find a set of values for N, D, and Q that satisfy all the conditions. Among the options provided, only option A (7 nickels, 12 dimes, and 21 quarters) satisfies all the given conditions:- The total number of coins is $7 + 12 + 21 = 40$.- The sum of the numbers of nickels and dimes is $7 + 12 = 19$, which is indeed two less than the number of quarters (21).- The total value is $0.05*7 + 0.10*12 + 0.25*21 = 0.35 + 1.20 + 5.25 = 6.80$.Therefore, option A is correct.

A Box Contains $6