Tuesday, January 10, 2012

this week's MGRE Math Beast Challenge

This week's challenge:

My answer will eventually appear in the comments, but you're free to take a stab at the problem and provide your own answer. Show your work! Note to SAT students: this might be a GRE problem, but it isn't so different from the sort of problem you might see on the SAT I. Feel free to join in.


1 comment:

Kevin Kim said...

This looks like a "systems of equations" problem, so let's set up the two equations:

Calories for both types of snacks have to total up to 3250, so:

Let A = # of servings of Snack A
Let B = # of servings of Snack B

200A + 350B = 3250, or

20A + 35B = 325, or

4A + 7B = 65

We also need to create an equation that reckons the cost which, for the purchasing of both types of snack, totals $11.

1.7A + .6B = 11, or

17A + 6B = 110

Let's pair these equations up:

4A + 7B = 65
17A + 6B = 110

Let's get rid of the B by multiplying each equation by a factor that gives us 42B in each case:

6(4A + 7B = 65)
7(17A + 6B = 110)

which gives us:

24A + 42B = 390
119A + 42B = 770

Switch and subtract:

[119A + 42B = 770]
-[24A + 42B = 390]

95A = 380

A = 4

We can stop here. The correct answer would appear to be (C): the two quantities are equal.

QED. (Or is it?)