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10 Jul 2019, 04:59
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Difficulty:
45%
(medium)
Question Stats:
49% (00:56) correct
51%
(00:45)
wrong
based on 53
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Does the population of rats in a zoo double every day?
(1) The population increased by 12 on a particular day.
(2) The population increased by 18 on a particular day.
Source: Nova's
(1) The population increased by 12 on a particular day.
(2) The population increased by 18 on a particular day.
Source: Nova's
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10 Jul 2019, 06:57
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I don't think this question makes much sense. If the population increased by 12 one day, then if it were true that the population were doubling each day, the population would need to be increasing from 12 to 24 that day. But then what happened three days ago? The population cannot have increased from 1.5 rats to 3 rats. Since the question doesn't specify a time period or starting point for the doubling, the answer to the question is "no" regardless of what the statements say. That said, it's certainly impossible that the population was doubling each day if it increased by 12 one day and 18 another, because that would mean one day the population was exactly 12, and another day it was exactly 18, and you can't get from 12 to 18 by doubling.
10 Jul 2019, 13:57
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I translated it this way:
we have a geometric sequence \((a , ar , ar^2 , ... , ar^x, ... , ar^y , ... )\) and we are asked whether \(r = 2\)?
statement (1) says that \(ar^x = 12\) --> clearly insufficient
statement (2) says that \(ar^y = 18\) --> clearly insufficient
combining (1) and (2),
if \(r = 2\), then dividing any value in the sequence by any lower value in the sequence will give a power of 2 (\(2^n\) where n is an integer) (such as 2,4,8,16,.....),
but dividing \(\frac{ar^y}{ar^x} = r^{y-x} = \frac{3}{2} = 1.5\), which can't be rewritten into the \(2^n\) formula. Thus the population can't be doubling every day --> sufficient
we have a geometric sequence \((a , ar , ar^2 , ... , ar^x, ... , ar^y , ... )\) and we are asked whether \(r = 2\)?
statement (1) says that \(ar^x = 12\) --> clearly insufficient
statement (2) says that \(ar^y = 18\) --> clearly insufficient
combining (1) and (2),
if \(r = 2\), then dividing any value in the sequence by any lower value in the sequence will give a power of 2 (\(2^n\) where n is an integer) (such as 2,4,8,16,.....),
but dividing \(\frac{ar^y}{ar^x} = r^{y-x} = \frac{3}{2} = 1.5\), which can't be rewritten into the \(2^n\) formula. Thus the population can't be doubling every day --> sufficient
