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07 Jan 2022, 08:15
Kudos
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C
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Difficulty:
45%
(medium)
Question Stats:
71% (02:00) correct
29%
(02:07)
wrong
based on 282
sessions
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A nation’s population triples every 100 years. If the population was 100,000 in the year 1200, in what year was the population 72,900,000?
(A) 600
(B) 1700
(C) 1800
(D) 1900
(E) 2000
(A) 600
(B) 1700
(C) 1800
(D) 1900
(E) 2000
07 Jan 2022, 09:24
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Given
A nation’s population triples every 100 years.
Thus the percentage at which the population grows = 200%
Let us assume that every t represents one block of 100 years.
Thus -
\(72,900,000 = 100,000 * (1 + \frac{200}{100})^ t \)
Dividing the equation by 100,000 on both the sides and reducing the fractions we get -
\(729 = (1 + 2)^ t \)
\(27^2 = 3^ t \)
\(3^{2*3} = 3^ t \)
Therefore t = 6
We had assumed 't' to be a block of 100 years, so 6 represents 600 years.
Year in which the population becomes \(729*10^5 \)
= 1200 + 600 = 1800
A nation’s population triples every 100 years.
Thus the percentage at which the population grows = 200%
Let us assume that every t represents one block of 100 years.
Thus -
\(72,900,000 = 100,000 * (1 + \frac{200}{100})^ t \)
Dividing the equation by 100,000 on both the sides and reducing the fractions we get -
\(729 = (1 + 2)^ t \)
\(27^2 = 3^ t \)
\(3^{2*3} = 3^ t \)
Therefore t = 6
We had assumed 't' to be a block of 100 years, so 6 represents 600 years.
Year in which the population becomes \(729*10^5 \)
= 1200 + 600 = 1800
20 Jan 2026, 16:12
Kudos
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