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Difference between revisions of "1996 AJHSME Problems/Problem 23"

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Revision as of 23:25, 4 July 2013

Problem

The manager of a company planned to distribute a <dollar/>50 bonus to each employee from the company fund, but the fund contained <dollar/>5 less than what was needed. Instead the manager gave each employee a <dollar/>45 bonus and kept the remaining <dollar/>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 (Problems • Answer Key • Resources)
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

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