1974 AHSME Problems/Problem 7

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, $0.89n+1068=n-32\implies 0.11n=1100\implies n=\frac{1100}{0.11}=10000, \boxed{\text{D}}$ .

1974 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
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 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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

Page

Toolbox

Search

1974 AHSME Problems/Problem 7

Problem

Solution

See Also