word problem?

Question

the numbers of the grades D,C,B and A in a class were consecutive integers. the number of D's was the same as the sum of one-third the number of A's and half the number of C's. how many students received a B

Answer 1

D = 1/3A + 1/2C

Multiply both sides by 6:
6D = 2A + 3C

Try out different values for D (6D will be even, so C must be even too, but it can never be more than 2 away from D).
D = 1, A = 3, C = 0
Thus B = 2 (to be consecutive, but I don't think zero is allowed).
D = 1, A = 0, C=2
Thus B = 3 (to be consecutive, but I don't think zero is allowed).

D = 2, A = 6, C = 0 (non-consecutive)
D = 2, A = 3, C = 2 (non-consecutive)

D = 3, A = 6, C = 2 (non-consecutive)
D = 3, A = 3, C = 4 (non-consecutive)

D = 4, A = 9, C = 2 (non-consecutive)
D = 4, A = 3, C = 6
B = 5 (to be consecutive)

The answer is:
3 As
5 Bs <--
6 Cs
4 Ds