Manhattan GMAT Challenge Problem of the Week - 8 Jan 10

by , Jan 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. In past weeks, the Challenge Problem was a 750+ level question, but in order to mix things up a bit, Manhattan GMAT has slightly expanded the difficulty range of the Challenge Problem to 700+ level questions. If you are up for the challenge, set your timer for 2 minutes and go!

Question

Several friends in a dinner group decide to contribute equally to the purchase of a $36 gift. How many people are in the group?

(1) The number of people in the group is equal to the size of each persons contribution, in dollars.

(2) If three more people joined the group, each persons individual contribution would fall by $2.

(A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.

(B) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.

(C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

(D) EACH statement ALONE is sufficient.

(E) Statements (1) and (2) TOGETHER are NOT sufficient to answer the question asked, and additional data are needed.

Solution

Lets call the number of people in the group n, and lets call each contribution $x. Then we know from the stem that 36 = nx. We are asked for n, which is equivalent to asking for x (because of the equation we are given).

(1) SUFFICIENT. This statement tells us that n = x. We can substitute into the given equation:

36 = [pmath]n^2[/pmath]

Since n represents a number of people, only the positive root makes sense, and n must be equal to 6.

(2) SUFFICIENT. This statement requires more work. We are told that if the number of people increased by 3, the contribution would fall by $2.

In other words, the new number of people is n + 3, and the new contribution is x 2. The product will still be $36.

Thus, we know that 36 = (n + 3)(x 2). We also still know that 36 = nx, or 36/n = x. Lets expand the new equation and swap out x.

36 = (n + 3)(x 2) = nx + 3x 2n 6

Since nx equals 36, we can substitute in 36 for nx as follows:

36 = 36 + 3x 2n 6

6 = 3x 2n

Now substitute in 36/n for x:

6 = 3(36/n) 2n

6 = 108/n 2n

[pmath]2n^2[/pmath] + 6n 108 = 0

Now, we need to try to factor this quadratic equation. At first glance, we can set up the following template:

(2n ...)(n ...) = 0

Since the 108 in the distributed quadratic equation has a minus sign in front of it, we know that one of the missing numbers is negative and that the other one is positive. This means that one of the solutions for n will be positive, while the other one will be negative. We could stop here, since only one positive solution exists.

If we wanted to keep going with the factoring, we could observe that 108 = 336 = [pmath]3^3[/pmath][pmath]2^2[/pmath]. We need a pair of factors relatively close in value to each other, since their difference (after multiplying one of the factors by 2) is only 6. The pair of factors that works is {9, 12}, as we can see by trial and error.

(2n 12)(n + 9) = 0

n = 6 or n = 9

The negative solution is impossible, so we know that n is 6.

The correct answer is (D). EACH statement ALONE is sufficient.

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