The number of different sets of coins that she can have is; Two sets which are; (0, 7, 4, 2) or (1, 3, 7, 2)
Let the numbers of coins be;
We have:
Q + D + N + P = 13
We know that;
25 quarters make 1 cent
10 dimes makes one cent
5 nickels make one cent
1 penny makes one cent
Thus;
25Q + 10D + 5N + P = 92
P will have to be 2 because the others will give a value with the unit digit as 0. Thus;
Q + D + N = 11 ----(1)
25Q + 10D + 5N + 2 = 92
25Q + 10D + 5N = 90
Divide through by 5 to get;
5Q + 2D + N = 18 (2)
Subtract eq (1) from eq (2) to get;
4Q + D = 7
Let us put D = 7 and Q = 0, and so from eq (1), N = 4
The set here is now; (Q, D, N, P) = (0, 7, 4, 2)
Let us put D = 3, Q = 1, and from eq (1), N = 7
The set here is now; (Q, D, N, P) = (1, 3, 7, 2)
If D = 6, 5, 4, 2, 1, 0, then Q is not an integer, these can be ignored.
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