1974 AHSME Problems/Problem 7

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Problem

A town's population increased by $1,200$ people, and then this new population decreased by $11\%$. The town now had $32$ less people than it did before the $1,200$ increase. What is the original population?

$\mathrm{(A)\ } 1,200 \qquad \mathrm{(B) \ }11,200 \qquad \mathrm{(C) \  } 9,968 \qquad \mathrm{(D) \  } 10,000 \qquad \mathrm{(E) \  }\text{none of these}$

Solution

Let $n$ be the original population. When $1,200$ people came, the new population was $n+1200$. Then, since the new population decreased by $11\%$, we must multiply this new population by $.89$ to get the final population. This is equal to $(0.89)(n+1200)=0.89n+1068$. We're given that this is $32$ less than the original popluation, $n$. Therefore, $0.89n+1068=n-32\implies 0.11n=1100\implies n=\frac{1100}{0.11}=10000, \boxed{\text{D}}$.

See Also

1974 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 6
Followed by
Problem 8
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