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20 Sep 2010, 10:10
Kudos
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C
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Difficulty:
35%
(medium)
Question Stats:
72% (02:08) correct
28%
(02:19)
wrong
based on 502
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The population of a certain town increases by 50 percent every 50 years. If the population in 1950 was 810, in what year was the population 160?
A. 1650
B. 1700
C. 1750
D. 1800
E. 1850
A. 1650
B. 1700
C. 1750
D. 1800
E. 1850
20 Sep 2010, 10:20
Kudos
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cano
Increase by 50% = times 1.5, so if the population at a certain year were 10 then in 50 years it would be 10*1.5 and in 50*2=100 years it would be 10*1.5*1.5=10*1.5^2.
So, we are given that \(160*1.5^n=810\), where \(n\) is the number of 50-year periods;
\(160*(\frac{3}{2})^n=810\);
\((\frac{3}{2})^n=\frac{810}{160}\);
\((\frac{3}{2})^n=\frac{81}{16}\);
\((\frac{3}{2})^n=(\frac{3}{2})^4\);
\(n=4\), so there were 4 50-year periods;
Thus the population was 160 at year \(1950-4*50=1750\).
Answer: C.
20 Sep 2010, 13:42
Kudos
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I just worked this one backwards until I reached 160. If the population increases by 50% then you're multiplying the previous number by 3/2. So to work backwards, divide by 3/2 (which is the same as multiplying by 2/3).
So we have:
1950: 810
1900: 810*(2/3) = 540
1850: 540*(2/3) = 360
1800: 360*(2/3) = 240
1750: 240*(2/3) = 160
Answer is C
So we have:
1950: 810
1900: 810*(2/3) = 540
1850: 540*(2/3) = 360
1800: 360*(2/3) = 240
1750: 240*(2/3) = 160
Answer is C
