2017 AMC 10A Problem 14

Let the ticket cost $m$ and the soda cost $s$. The two statements say
$$m = \tfrac{1}{5}(A - s), \qquad s = \tfrac{1}{20}(A - m).$$
Clear the fractions: $5m = A - s$ and $20s = A - m$.

From the first equation, $A = 5m + s$. Substitute into the second:
$$20s = 5m + s - m \quad\Longrightarrow\quad 19s = 4m \quad\Longrightarrow\quad m = \tfrac{19}{4}s.$$
Then $A = 5 \cdot \tfrac{19}{4}s + s = \tfrac{99}{4}s$, so $s = \tfrac{4}{99}A$ and $m = \tfrac{19}{99}A$.

Together the two purchases cost
$$m + s = \frac{23}{99}A \approx 0.2323A,$$
which is $23\%$ of $A$ to the nearest percent.

The answer is $\boxed{\textbf{(D)}\ 23\%}$.

Each cost is defined in terms of the other, so neither can be found alone; the pair of statements is a linear system, and substitution resolves it in two lines. Working with the exact fraction $\tfrac{23}{99}$ before rounding avoids accumulated approximation error, and the answer choices $22\%$ and $23\%$ are close enough that rounding early could mislead.

Add the two cleared equations in a helpful form. From $5m + s = A$ and $m + 20s = A$, multiply the second by $5$ and subtract to eliminate $m$: $99s = 4A$. The rest is the same. Alternatively set $A = 99$ so the numbers are integers: $s = 4$, $m = 19$, total $23$ out of $99$.

• Approximating each cost against the whole allowance (20% + 5% = 25%) instead of against the stated differences.
• Reporting only the ticket's share, $19\%$ (choice B), instead of ticket plus soda.