gmat prep -- donuts Mmmmmmm....

Using either equaion we can get to know the price of each Cake and Doughnut.

But we cant find the quantity purchased. Tricky question.. seemd to be (D) is the firt instance.

Original equation can be wrtten as [(PriceC * QtyofC)+(PriceOfD*Qty of D) =6]

mav800rick wrote: Original equation can be wrtten as [(PriceC * QtyofC)+(PriceOfD*Qty of D) =6]

Let's rely on our old DS buddy, "# of equations vs # of unknowns".

From the original, we have 1 equation and 4 unknowns. To solve the entire system, we need 3 more equations.

We're only asked to solve for D, so it's possible we can get away with fewer equations if they eliminate multiple variables in one shot. For example, "the revenue from cupcakes is $4" would eliminate two variables and we'd be left with PriceD * D = 2.

(1) one equation involving 2 of our variables, but not in a way that allows us to eliminate them both. Insufficient.

(2) one equation involving 2 of our variables, but not in a way that allows us to eliminate them both. Insufficient.

Together: 3 distinct linear equations for 4 unknowns, without anything special to allow us to sneakily solve for D: insufficient.

You're correct.

In this particular question, we can ignore positive/negative solutions (since real world things like items and prices will never be negative), but we do need to be careful.

Hello,

Edit: here's my work. Could anyone tell me what I'm missing?:

Stem:

C=cost of cupcake
D=cost of doughnut

1C + 1D = 6. What is qty of D?

Statement 1:

2D = 3C - 0.10

multiply both sides by 10 to get rid of the fraction:

20D = 30C - 1

Divide both sides by 20:

D=3/2C - 1/20

Multiply 3/2 by 10 to get a common denominator:

D=30/20C - 1/20
D=29/20C

This give me the cost of D in terms of C. I have no info about the quantity. Not Sufficient. Scratch out A and D.

Statement 2:

(C + D) / 2 = 0.35

I'm not sure what to do here. It just doesn't seem sufficient. Scratch out B.

Together:

It seems that if I plug the results from statement 1 into the equation for statement 2, I still don't get a quantity for D. Could someone please explain a more efficient approach for this question?

Thanks so much

Poisson wrote:Hello,

Edit: here's my work. Could anyone tell me what I'm missing?:

Stem:

C=cost of cupcake
D=cost of doughnut

1C + 1D = 6. What is qty of D?

That equation is problematic. If you make C the cost of each cupcake, you'd need to multiply that by the number of cupcakes you have to find the total cost of the cupcakes.

For instance, if I buy 3 cupcakes and 2 donuts, my cost is 3c + 2d. But if I buy 10 cupcakes and 12 donuts, my cost in 10c + 12d.

Since we don't know how many cupcakes or donuts Lew bought, we'd have to assign variables to each (say x and y), then make the total cost of a function of them, something like xc + yd = $6.

From there, we need to puzzle out four variables (x, y, c, and d).