I get an answer of E.

Let x = price of cupcake, y = price of donut, C = # of cupcakes, D = # of donuts.

You are given that xC + yD = 6, and want to find D.

(1) 2y = 3x - 0.1
(2) (x + y) / 2 = 0.35, or x + y = 0.70

Either statement by itself is insufficient to solve for D in the first equation given.

(1 + 2): combining the statements will allow you to solve for x and y, where x = 0.4 and y = 0.3

Your original equation becomes 0.4C + 0.3D = 6

There are multiple combinations that can satisfy this equation (0 cupcakes and 20 donuts, 5 cupcakes and 10 donuts, etc), so we do not have a definite answer.

Choice E.

mp2437 wrote:I get an answer of E.

Let x = price of cupcake, y = price of donut, C = # of cupcakes, D = # of donuts.

You are given that xC + yD = 6, and want to find D.

(1) 2y = 3x - 0.1
(2) (x + y) / 2 = 0.35, or x + y = 0.70

Either statement by itself is insufficient to solve for D in the first equation given.

(1 + 2): combining the statements will allow you to solve for x and y, where x = 0.4 and y = 0.3

Your original equation becomes 0.4C + 0.3D = 6

There are multiple combinations that can satisfy this equation (0 cupcakes and 20 donuts, 5 cupcakes and 10 donuts, etc), so we do not have a definite answer.

Choice E.

Yes tks, but is there any way to know that the equation gives multiple answers? If not, so every time we have some problem like this we should develop it to the end. DS problems are supposed to be solved without so many steps.

El Cucu wrote:

mp2437 wrote:I get an answer of E.

Let x = price of cupcake, y = price of donut, C = # of cupcakes, D = # of donuts.

You are given that xC + yD = 6, and want to find D.

(1) 2y = 3x - 0.1
(2) (x + y) / 2 = 0.35, or x + y = 0.70

Either statement by itself is insufficient to solve for D in the first equation given.

(1 + 2): combining the statements will allow you to solve for x and y, where x = 0.4 and y = 0.3

Your original equation becomes 0.4C + 0.3D = 6

There are multiple combinations that can satisfy this equation (0 cupcakes and 20 donuts, 5 cupcakes and 10 donuts, etc), so we do not have a definite answer.

Choice E.

Yes tks, but is there any way to know that the equation gives multiple answers? If not, so every time we have some problem like this we should develop it to the end. DS problems are supposed to be solved without so many steps.

when we have two variables and one eqn we can't solve it in abscence of any supporting info...

Well, a quick look at the statements they give only gives you info to solve for price, but nowhere does it mention anything about quantity of either donuts or cupcakes, which should raise an eyebrow. Sometimes you could stop there, but other times, I have seen questions where just solving for the 2 variables on price will lead to a single answer (because there is only one combination for plugging in the variables).

I usually go through a problem to the end; when you have an idea on how to setup the equation, the rest of the work doesn't take too long.