2017 AMC 10A Problem 14

Call the ticket cost $m$ and the soda cost $s$.

1. Turn each sentence into an equation. $20\%$ is one fifth and $5\%$ is one twentieth:
$$m = \tfrac15 (A - s), \qquad s = \tfrac{1}{20} (A - m).$$
2. Clear the fractions by multiplying by $5$ and by $20$:
$$5m = A - s, \qquad 20s = A - m.$$
Move terms so both equations read "something $= A$":
$$5m + s = A, \qquad m + 20s = A.$$
3. Pick a handy value for $A$. Only the fraction of $A$ matters, so $A$ can be any number. Solve with a letter first: both left sides equal $A$, so
$$5m + s = m + 20s \quad\Longrightarrow\quad 4m = 19s.$$
The smallest whole numbers that fit are $m = 19$, $s = 4$. Then $A = 5 \cdot 19 + 4 = 99$. Check the other equation: $19 + 20 \cdot 4 = 99$. Both work.
4. Find the fraction. Roger spends $m + s = 19 + 4 = 23$ out of $99$:
$$\frac{23}{99} \approx 0.232 = 23.2\%.$$
5. To the nearest whole percent, that is $23\%$.

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

Each cost is described using the other, so neither can be found alone; write both as equations and solve them together. Since the question asks for a fraction of $A$, you are free to pick $A$ so that everything is a whole number. Keep the exact fraction until the end, because $22\%$ and $23\%$ are close choices.

Estimate first. The soda is small, so the ticket is a bit under $20\%$ of $A$, say about $19\%$. Then the soda is $5\%$ of about $81\%$ of $A$, roughly $4\%$. Together about $23\%$. The exact solution above confirms it.

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