1996 AJHSME Problems/Problem 23


The manager of a company planned to distribute a $$50$ bonus to each employee from the company fund, but the fund contained $$5$ less than what was needed. Instead the manager gave each employee a $$45$ bonus and kept the remaining $$95$ in the company fund. The amount of money in the company fund before any bonuses were paid was

$\text{(A)}\ 945\text{ dollars} \qquad \text{(B)}\ 950\text{ dollars} \qquad \text{(C)}\ 955\text{ dollars} \qquad \text{(D)}\ 990\text{ dollars} \qquad \text{(E)}\ 995\text{ dollars}$

Solution 1

Let $p$ be the number of people in the company, and $f$ be the amount of money in the fund.

The first sentence states that $50p = f + 5$

The second sentence states that $45p = f - 95$

Subtracing the second equation from the first, we get $5p = 100$, leading to $p = 20$

Plugging that number into the first equation gives $50\cdot 20 = f + 5$, leading to $f = 995$, which is answer $\boxed{E}$.

Solution 2

Since the company must employ a whole number of employees, the amount of money in the fund must be $5$ dollars less than a multiple of $50$. Only options $A$ and $E$ satisfy that requirement.

Additionally, the number must be $95$ more than a multiple of $45$. Since $45 \cdot 20 = 900$, the only number that is $95$ more than a multiple of $45$ out of options $A$ and $E$ is option $\boxed{E}$.

(Option $B$ is also $95$ more than a multiple of $45$, but it was eliminated previously.)

See Also

1996 AJHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 22
Followed by
Problem 24
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS