Problem Solving

BTGModeratorVI wrote: ↑

Thu Jul 23, 2020 6:40 am

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

Answer: E
Source: Official guide

Let C = the cost of ONE chair
Let D = the cost of ONE deck

The total cost of 3 desks and 1 chair is twice that of 1 desk and 3 chairs
We can write: 3D + C = 2(D + 3C)
Expand to get: 3D + C = 2D + 6C
Subtract 2D from both sides: D + C = 6C
Subtract C from both sides: D = 5C


The total cost of 4 desks and 1 chair is how many times that of 1 desk and 4 chairs?
This is asking us to determine how many times the price of 1 desk and 4 chairs DIVIDES INTO the price of 4 desks and 1 chair

In other words, we want to find the value of (4D + C)/(D + 4C)

Since we now know that D = 5C, we can replace D with 5C to get:
(4D + C)/(D + 4C) = [4(5C) + C]/[5C + 4C]
= [20C + C]/9C
= 21C/9C
= 21/9
= 7/3

Answer: E

Cheers,
Brent

BTGModeratorVI wrote: ↑

Thu Jul 23, 2020 6:40 am

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

Answer: E
Source: Official guide

Solution:

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