Bunuel
The numbers of nuts stored by two squirrels, Burl and Shirl, at the beginning of September were in the ratio A : B. At the end of September, the numbers of nuts stored by Burl and Shirl were in the ratio C : D. If Burl had n times as many nuts at the end of September as at the beginning of September, then by what percent did the number of nuts stored by Shirl increase during September?
(A) 100nAD/(BC)
(B) 100n(AD - B)/(BC)
(C) 100(D/(BC)*(n + A) - 1)
(D) 100D(n + A)/(BC)
(E) 100(nAD/(BC) - 1)
Kudos for a correct solution.Because no actual quantities are given, the problem technically requires at least one unknown multiplier: for instance, using the unknown multiplier x, you could define the numbers of nuts at the beginning of September as Ax and Bx. However, the answer choices contain only the quantities defined in the problem statement, so you must be able to answer the question using only those quantities. Therefore, without loss of generality, you can let Burl’s and Shirl’s nut collections be just A and B, respectively, at the beginning of September. Because Burl had n times as many nuts at the end of September as at the beginning, Burl must have had nA nuts at the end of September.
Make a table:

The ratio of the two quantities in the second row is C : D, so set up a proportion to solve for the unknown quantity:
nA/[unknown] = C/D
Solve for the unknown by cross-multiplication:
C × [unknown] = nAD
[unknown] = nAD/C
Finally, find the desired percentage change, which is the percent change from an initial value of B to a final value of nAD/C:
Percent change = \(100(\frac{\frac{nAD}{C}-B}{B})\)
Since this is a complex fraction, multiply through by the common denominator of the inside terms (which, in this case, is C, the only denominator appearing on the inside at all):
Percent change = \(100(\frac{nAD-BC}{BC})\)
\(= 100(\frac{nAD}{BC}-1)\)
The correct answer is E.Attachment:
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