A school supply store sells only one kind of desk and one

Hi All,

We're told that a school supply store sells only one kind of desk and one kind of chair, at a uniform cost per desk or per chair. The total cost of 3 desks and 1 chair is TWICE that of 1 desk and 3 chairs. We're asked - the total cost of 4 desks and 1 chair is how many times that of 1 desk and 4 chairs. This question is ultimately about ratios - and you have to do a little Algebra to work through it.

The information that we're given can be translated into the following equation:
3D + C = 2(D + 3C)

We can then do a couple Algebra steps to simplify this equation:
3D + C = 2D + 6C
D = 5C

We now know that 1 desk is the same price as 5 chairs. With this information, we can substitute in to the given question:
Cost of 4 desks and 1 chair = 4D + 1C = 4(5C) + 1C = 21C

Cost of 1 desk and 4 chairs = 1D + 4C = 1(5C) + 4C = 9C

21C/9C = 7/3

Final Answer: E

GMAT assassins aren't born, they're made,
Rich

BTGmoderatorDC wrote:A school supply store sells only one kind of desk and one kind of chair, at a uniform cost per desk or per chair. If the total cost of 3 desks and 1 chair is twice that of 1 desk and 3 chairs, then the total cost of 4 desks and 1 chair is how many times that of 1 desk and 4 chairs?

A. 5
B. 3
C. 8/3
D. 5/2
E. 7/3

OA E

Source: Official Guide

Let’s let D = the cost of a desk and C = the cost of a chair. We can create the equation:

3D + C = 2(D + 3C)

3D + C = 2D + 6C

D = 5C

Let’s let K = the number of times greater that one desk and 4 chairs costs, compared to the cost of 4 desks and 1 chair. So, we have:

4D + C = K(D + 4C) ?

Substituting 5C for D, we have:

20C + C = K(5C + 4C) ?

21C= K(9C) ?

K = 21C/9C = 7/3

Answer: E