1960 AHSME Problems/Problem 29

Problem

Five times $A$'s money added to $B$'s money is more than $51.00$. Three times $A$'s money minus $B$'s money is $21.00$. If $a$ represents $A$'s money in dollars and $b$ represents $B$'s money in dollars, then:

$\textbf{(A)}\ a>9, b>6 \qquad \textbf{(B)}\ a>9, b<6 \qquad \textbf{(C)}\ a>9, b=6\qquad \textbf{(D)}\ a>9, \text{but we can put no bounds on} \text{ } b\qquad \textbf{(E)}\ 2a=3b$

Solution

Use math symbols to write an equation and an inequality based on the conditions in the problem. $$5a+b>51$$ $$3a-b=21$$ From the second equation, $b = 3a-21$ and $a = 7 + b/3$.

Substituting $b$ in the inequality results in $$5a + 3a - 21 > 51$$ $$8a > 72$$ $$a > 9$$

Substituting $a$ in the inequality results in $$35 + \frac{5b}{3} + b > 51$$ $$35 + \frac{8b}{3} > 51$$ $$\frac{8b}{3} > 16$$ $$b > 6$$

The answer is $\boxed{\textbf{(A)}}$.