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29 Apr 2015, 04:09
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
53% (03:33) correct
47%
(03:32)
wrong
based on 357
sessions
History
Date
Time
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Not Attempted Yet
On January 1, 2010, Dave invests 70% of his retirement savings in Antarctic largecap stocks, 20% in Antarctic midcaps, and 10% in Antarctic smallcaps. In 2010, largecaps rise 5%, midcaps rise 10%, and smallcaps rise 15% in the Antarctic stock market; however, in 2011, largecaps fall 10% and midcaps fall 20%, while smallcaps rise x% in Antarctica. If, on January 1, 2012, Dave has the same total amount of retirement savings as he did two years before, then x is between
A. 10 and 20
B. 20 and 30
C. 30 and 40
D. 40 and 50
E. 50 and 60
A. 10 and 20
B. 20 and 30
C. 30 and 40
D. 40 and 50
E. 50 and 60
04 May 2015, 04:15
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Bunuel
MANHATTAN GMAT OFFICIAL SOLUTION:
The difficulty of this “percent change” problem is not so much conceptual as it is “executional”—you want to be able to solve it quickly, easily, and of course accurately. Let’s get to the workout!
You’ve got three investments (at various percent allocations) changing by various other percents over two time periods. The numbers don’t look too ugly, but you might suspect that the result will be hard to compute exactly, because the problem only asks for a range. Thus, you should be ready to switch to estimation at some point.
Pick a smart number for the total retirement savings Dave starts with—say, $10,000. (If you pick $100, you’ll wind up needing to track decimals, so give yourself a couple more zeros to start with.)
Here are the starting values:
L = $7,000
M = $2,000
S = $1,000
Apply the first year’s changes, so that you have these numbers on 1/1/2011:
Newer L = $7,000 + 5% = $7,350
Newer M = $2,000 + 10% = $2,200
Newer S = $1,000 + 15% = $1,150
Now apply the second year’s changes to L and M:
Newest L = $7,350 – 10% = $7,350 – $735 = $6,615
Newest M = $2,200 – 20% = $2,200 – $440 = $1,760
Add these to get $8,375. So the newest S must be $10,000 (the target final total of Dave’s retirement savings) minus $8,375, or $1,625.
The dollar change in S from 1/1/11 to 1/1/12 is $1,625 – $1,150 = $475. So the question is this: what percent change does $475 represent, from a starting point of $1,150? Since $1,150 is a nasty divisor, switch to benchmarks:
10% of $1,150 = $115.
So 20% is just double that, or $230.
And 40% is double that, or $460.
Since $475 is just slightly larger than $460 (but not enough to get you to 50%, which would be $460 + $115), x must be between 40 and 50.
Intuitively, it should make sense that you’d need a much bigger positive percent change in the smallest investment (S) to make up for even a moderate downturn in the larger investments (L and M), so if you were completely stuck for time and needed to guess in a hurry, you should favor C/D/E over A or B.
The correct answer is D.
29 Apr 2015, 08:00
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Bunuel
Let's \(s\) be common savings, than
\(large = 0.7s*1.05*0.9 = 0.66s\)
\(middle = 0.2s*1.1*0.8=0.17s\)
\(small = 0.1s*1.15*x = 0.12sx\)
and we can make equation
\(large + middle + small = s\)
\(0.66s + 0.17s + 0.12sx = s\)
\(0.12sx = 0.17s\)
\(x = 1.41\)
So x = 41%
Answer is D
