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27 Sep 2021, 07:15
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
25%
(medium)
Question Stats:
76% (02:07) correct
24%
(02:38)
wrong
based on 34
sessions
History
Date
Time
Result
Not Attempted Yet
If we divide 5000 into \((k^2-16)\) equal parts where \((\frac{k+4}{25})\)% of each part is equal to 1 what is k ?
A. 2
B. 3
C. 4
D. 5
E. 6
A. 2
B. 3
C. 4
D. 5
E. 6
27 Sep 2021, 07:25
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Bunuel
I'd normally do a question like this algebraically, but there's only really one plausible answer among the choices, and it's easiest just to confirm that answer is right. We're dividing 5000 into "k^2 - 16" equal parts. If k were 2, 3 or 4, we'd be dividing 5000 into either a negative number of parts, or into zero parts, and it's nonsensical to do either of those things. So k must be 5 or 6 from the answer choices. If k were 5, we'd be dividing 5000 into 9 equal parts, and 5000 isn't divisible by 9. That fact alone is not dispositive, but it makes k = 5 unlikely to be the right answer to a divisibility-related question. So I'd just check the most likely answer, k = 6: then we're dividing 5000 into 36-16 = 20 equal parts, so each part is 5000/20 = 250, and we want to confirm that (k+4)/25 % of each part, or (10/25) %, or 10/2500 of each part is 1. Since (10/2500)(250) = 1, the answer is 6.
I wouldn't expect a similar strategy to be optimal on an official question, however.
27 Sep 2021, 07:35
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Bunuel
Value of each part: \(\frac{5000}{(k^2-16)}\)
Given:
\((\frac{k+4}{25})*\frac{1}{100}*\frac{5000}{(k^2-16)}=1\)
\((\frac{k+4}{25})*\frac{1}{100}*\frac{5000}{(k+4)(k-4)}=1\)
2 = k - 4
k = 6
IMO, (E)!
