Let's call p_d price of donut and p_c price of cupcake.

We know from statements 1 and 2 that: 2p_d = 3p_c -.1 and (p_d + p_c)/2 = 0.35. This is 2 equations and 2 unknowns as far as the price goes.

I can solve for p_d and p_c algebraically, and I get p_c = 0.3 and p_d = 0.4.

In the case presented in the original problem, that Lew spent $6... we could have that he bought 20 cupcakes and no donuts, or 15 donuts and no cupcakes, or some other possible combinations.

It is actually a little dangerous to assume that because we only have 3 equations and 4 unknowns that we automatically have insufficient information. For example, it we knew Lew spent $0.70, he would have HAD to have bought 1 of each item. The additional info we have beyond the equations is the assumption that he is buying whole quantities (ie not half a donut).

I didn't use your example of $5.76 because that is not a possible amount to spent given the prices of donuts and cupcakes, but yes... your point is valid that given certain amounts spent, there may be 1 and only one way for Lew to have spent that amount, in which case the answer would be C.

Hope that helped,

Tatiana

Thanks a lot Tatiana,
That really helped me a lot!

Shilo