Today’s GMAT challenge question comes from our friends at ManhattanGMAT. To help you with your GMAT studying, try to solve the problem on your own, and then read on for the explanation of its solution:
Problem
At the beginning of year 1, an investor puts p dollars into an investment whose value increases at a variable rate of xn% per year, where n is an integer ranging from 1 to 3 indicating the year. If 85 < xn < 110 for all n between 1 and 3, inclusive, then at the end of 3 years, the value of the investment must be between
(A) $p and $2p
(B) $2p and $5p
(C) $5p and $10p
(D) $10p and $25p
(E) $25p and $75p
Solution
The quick solution to this problem is to pick a convenient number in the allowed range of growth rates. If xn is always between 85 and 110, then the most convenient growth rate to pick is 100. An annual growth rate of 100% is exactly equivalent to doubling one’s money. Doubling one’s money three times is equivalent to multiplying one’s investment by a factor of 8 (= 23). The only range that includes $8p is the third range ($5p to $10p).
Computing the exact outer limits of the allowed range is much more cumbersome. We would have to cube 1.85 for the lower limit and 2.1 for the upper limit. The cube of 1.85 is 6.331625, and the cube of 2.1 is 9.261. These values fall between 5 and 10.
The correct answer is (C).
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Read the full article: GMAT Tip: Mo’ Money Mo’ Problems