Smita04 wrote:

At the bakery Lew spent a total of 6$ for one kind of cupcake and one kind of doughnut. How many donuts did he buy?

(1) Price of 2 doughty was $.10 less than 3 cupcakes

(2) Average price of 1 doughnut and 1 cupcake was $.035

There is an error in this question.

"Average price of 1 doughnut and 1 cupcake was $

0.35"

It doesn't make sense if the average price is 3.5 cents. The numbers don't work. It must be 35 cents. Convert everything to cents.

Using both statements together, you get

3C - 2D = 10 (C is the price of cupcakes and D is the price of doughnuts)

C + D = 35*2

Solving them simultaneously, you get C = 30, D = 40.

If Nc is number of cupcakes and Nd is number of doughnuts,

30*Nc + 40*Nd = 600

3*Nc + 4*Nd = 60

The first solution I get is Nd = 3, Nc = 16

We will get many more solutions such as Nd = 6, Nc = 12. Also, Nd = 9, Nc = 8 etc

Hence both statements together are not sufficient.

Answer (E)

Note that in this equation: 3*Nc + 4*Nd = 60, if instead of 60, the total price were say 18, there would have been only one solution.

3*Nc + 4*Nd = 18

Nc = 2, Nd = 3

Other possible solutions will be Nc = 6, Nd = 0; Or Nc = -2, Nd = 6 etc

Since we cannot buy negative or 0 number of items (she spends money on doughnuts AND cupcakes so she must have bought at least one of each), no other solution works.

Refer to this post for further details on this method:

http://www.veritasprep.com/blog/2011/06 ... -of-thumb/
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