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

A classic linear system problem. This is a perfect time to work with your answers. We’ll start with (D)

(D) If there are 9 quarters, there must be 7 dimes (16 total coins). The total value would be:

\(9(25) + 7(10) = 225 + 70 = 295\)

That’s too big. In other words, too many quarters. We can eliminate (D) and (E). Let’s try (B)

(B) If there are 5 quarters, there must be 11 dimes. The total value would be:

\(5(25) + 11(10) = 125 + 110 = 235\)

Shazam! It works.

**The correct answer is (B).**

For all you algebra nerds, let \(x\) be the number of quarters, and let \(y\) be the number of dimes.

The total number of coins is 16, so we know:

\(x + y = 16\)

And the total value of the coins is 235 (just ignore the decimals – you’re going to be a CEO, not a CFO), so we know:

\(25x + 10y = 235\)

Now we solve using the elimination method (generally more reliable and faster than substitution unless one variable is already isolated). Since we’re solving for the number of quarters, it’s preferable to eliminate the variable that represents the number of dimes. This might seem like a technicality, but the clock is ticking.

Let’s multiply our first equation by 10:

\(10x + 10y = 160\)

Now, we have the following system of equations:

\(\begin{cases} 25x + 10y = 235 \\ 10x + 10y = 160\end{cases}\)

When we subtract, the \(10y\) terms cancel and we get:

\(15x = 75\)

Thus, \(x = 5\)