Manhattan GMAT Challenge Problem of the Week - 8 Apr 10

by , Apr 8, 2010

Welcome back to this week's Challenge Problem! As always, the problem and solution below were written by one of our fantastic instructors. Each challenge problem represents a 700+ level question. If you are up for the challenge, however, set your timer for 2 mins and go!

Question

At the beginning of year 1, an investor puts p dollars into an investment whose value increases at a variable rate of [pmath]x_n[/pmath]% per year, where n is an integer ranging from 1 to 3 indicating the year. If 85 < [pmath]x_n[/pmath] < 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 [pmath]x_n[/pmath] 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 (= [pmath]2^3[/pmath]). 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) $5p and $10p

Special Announcement: Manhattan GMAT is now offering you a chance to win prep materials by solving the Challenge Problem. On our website, we will post a new question (without the answer) every week. Submit a solution to the problem, and if we pick your name out of those who answer correctly, you could win free prep material from Manhattan GMAT. To view the current question, simply visit our Challenge Problem Showdown.