A collection of 16 coins, each with a face value of either

BTGmoderatorDC wrote:A collection of 16 coins, each with a face value of either 10 cents or 25 cents, has a total face value of $2.35. How many of the coins have a face value of 25 cents?

A) 3
B) 5
C) 7
D) 9
E) 11

OA B

Source: Official Guide

Let D = the NUMBER of 10-cent coins
Let Q = the NUMBER of 25-cent coins

Notice that the VALUE of Q 25-cent coins = ($0.25)Q
For example, the VALUE of six 25-cent coins = ($0.25)6 = $1.50
And the VALUE of ten 25-cent coins = ($0.25)10 = $2.50
etc
Likewise, the VALUE of D 10-cent coins = ($0.10)D

The collection has 16 coins
We can write: D + Q = 16

The collection has a total value of $2.35
So, (0.10)D + (0.25)Q = 2.35

So, we have the following system:
(0.10)D + (0.25)Q = 2.35
D + Q = 16

Take the top equation and multiply both sides by 10 to get:
D + 2.5Q = 23.5
D + Q = 16

Subtract the bottom equation from the top to get:
1.5Q = 7.5
Solve: Q = 5

Answer: B

Cheers,
Brent

BTGmoderatorDC wrote:A collection of 16 coins, each with a face value of either 10 cents or 25 cents, has a total face value of $2.35. How many of the coins have a face value of 25 cents?

A) 3
B) 5
C) 7
D) 9
E) 11

OA B

Source: Official Guide

We can let the number of 10-cent coins = d and the number of 25-cent coins = q. We are given that there are 16 total coins; thus, d + q = 16.

We are also given that the total face value is $2.35, thus:

0.1d + 0.25q = 2.35

10d + 25q = 235

Isolating d in our first equation, we have d = 16 - q. We can substitute 16 - q for d in our second equation, and we have:

10(16 - q) + 25q = 235

160 - 10q + 25q = 235

15q = 75

q = 5

There are 5 coins with a face value of 25 cents.

Answer: B