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31 Jul 2019, 01:30
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E
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Difficulty:
35%
(medium)
Question Stats:
62% (00:38) correct
38%
(00:52)
wrong
based on 78
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From 1984 to 1987, what's the percent change in number of the insured worker in ABC program?
(1) 65% workers are insured in 1984
(2) 78% are insured in 1987
(1) 65% workers are insured in 1984
(2) 78% are insured in 1987
31 Jul 2019, 08:53
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Bunuel
From 1984 to 1987, what's the percent change in number of the insured worker in ABC program?
(1) 65% workers are insured in 1984
(2) 78% are insured in 1987
(1) 65% workers are insured in 1984
(2) 78% are insured in 1987
IMO C.
There is a 20% increase from 1984 to 1987 and the base doesn't matter in this case.
01 Aug 2019, 12:26
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For those of you wondering why the answer is E and not C.
If the population was the same then C is SUFFICIENT.
Let us assume that the population is indeed the same. 100 people in 1984 and 1987.
Then, in 1984, 65% of workers translates to 65 workers out of 100.
Similarly, in 1987, 78% of workers implies 78 workers out of 100.
Therefore, the number of insured workers rose from 65 to 78 which translates to \(\frac{(78-65)}{65}\)
However, the problem is we don't know the population in 1984 and 1987.
Let us assume the population is 1000 in 1984 but only 100 in 1987.
Thus, in 1984, 650 (65% from 1000) workers are insured but only 78 (78% from 100) in 1978.
Thus, the answer is E.
If the population was the same then C is SUFFICIENT.
Let us assume that the population is indeed the same. 100 people in 1984 and 1987.
Then, in 1984, 65% of workers translates to 65 workers out of 100.
Similarly, in 1987, 78% of workers implies 78 workers out of 100.
Therefore, the number of insured workers rose from 65 to 78 which translates to \(\frac{(78-65)}{65}\)
However, the problem is we don't know the population in 1984 and 1987.
Let us assume the population is 1000 in 1984 but only 100 in 1987.
Thus, in 1984, 650 (65% from 1000) workers are insured but only 78 (78% from 100) in 1978.
Thus, the answer is E.
