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A historian asserts that at the beginning of 1852, the popu
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17 Dec 2017, 14:08
17 Dec 2017, 14:08
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38% (01:28) correct
61% (01:53) wrong based on 13 sessions
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A historian asserts that at the beginning of 1852, the population of a certain mining town was 16,000. The historian also asserts that for each of the years from 1849 through 1853, thetown’s population at the beginning of the year was twice that of the preceding year. According to the historian, what was the range of the town’s populations at the beginning of each year from 1848 through 1853?
A. 14,000
B. 15,000
C. 28,000
D. 30,000
E. 31,000
A. 14,000
B. 15,000
C. 28,000
D. 30,000
E. 31,000
Re: A historian asserts that at the beginning of 1852, the popu
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18 Dec 2017, 02:38
18 Dec 2017, 02:38
1
We know that in 1852 the population was the double that of 1851.
So in 1851 = 16000/2 = 8000
1850 = 8000/2 = 4000
1949 = 2000
1948 = 1000
But the question asks for year 1853. Since we know that it doubkes, 16 000 becomes 32000.
So the range from Year 1853 - Year 1948 = 31 000
Answer E
So in 1851 = 16000/2 = 8000
1850 = 8000/2 = 4000
1949 = 2000
1948 = 1000
But the question asks for year 1853. Since we know that it doubkes, 16 000 becomes 32000.
So the range from Year 1853 - Year 1948 = 31 000
Answer E
Re: A historian asserts that at the beginning of 1852, the popu
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08 Mar 2018, 01:18
08 Mar 2018, 01:18
1
A faster method could be to realize that doubling of population actually means raising the power by \(2\) hence if in 1852 the population was \(16000\) than in 1853 the population would be \(32000\)
Now we can do away with calculating the population for intermediaries between 1948 and 1953 because we know the lowest population will be that for the year 1848.
\(1953-1948 = 5\) hence population of \(1948 = \frac{32000}{2^5} = 1000\)
Therefore Range = \(32000-1000 = 31000\)
option E
Now we can do away with calculating the population for intermediaries between 1948 and 1953 because we know the lowest population will be that for the year 1848.
\(1953-1948 = 5\) hence population of \(1948 = \frac{32000}{2^5} = 1000\)
Therefore Range = \(32000-1000 = 31000\)
option E
