As promised, here is the answer to yesterday’s Manhattan GMAT Challenge Question!

The key to this problem is to make the common term in both ratios equal. We should set up a table to display both ratios:

Now, we can double any of the ratios. In fact, we can multiply any row by any positive-integer factor (x2, x3, x10, etc.). We are constrained to positive-integer factors, though, because the actual number of any boat must be a positive integer.

The number of canoes in each row should be the same, so that we can merge the ratios. The least common multiple of 7 and 5 is 35, so we multiply the top row by 5 and the bottom row by 7:

This means that the “3-way” ratio of catamarans, canoes, and kayaks is 20:35:63. That is, for every 20 catamarans on the lake, there are 35 canoes and 63 kayaks. This ratio is already reduced to lowest integers, because there are no prime factors in common among all 3 integers.

The smallest number of boats that can be on the lake is 20 + 35 + 63 = 118. Since the total number of boats must be an integer, any possible number of boats must be a multiple of 118. To check which number is a multiple of 118, we factor 118 into its primes: 2 and 59. This allows us to spot 590, which is 59 x 10 and therefore 118 x 5.

The answer is D.

Below is the answer to yesterday’s GMAT Challenge Question!

(A) 15
(B) 24
(C) 36
(D) 54
(E) 90

Answer
Let’s start by finding the cost of the lobster, per bowl, in terms of the variables given (d, v, and b).

(d dollars/6 pounds) x (1 pound/v vats) x (1 vat/b bowls) = (d/6vb).

The problem states that this value, the cost of the lobster per bowl, or (d/6vb), is an integer. In other words, d is divisible by 6vb. To make d as small as possible, we need to make 6vb as small as possible. Since v and b are different prime integers, the smallest value of 6vb is 36 (using the two smallest prime integers, v = 2 and b = 3, or v = 3 and b = 2).

In order to make the cost of the lobster per bowl an integer, d must be divisible by 36. In other words, d must be a multiple of 36. What’s the smallest possible multiple of 36? The smallest multiple of 36 is 36.

The correct answer is C, 36.

Question

A certain club has exactly 5 new members at the end of its first week. Every subsequent week, each of the previous week’s new members (and only these members) brings exactly x new members into the club. If y is the number of new members brought into the club during the twelfth week, which of the following could be y?

(A)
(B)
(C)
(D)
(E)

Answer

At the end of the first week, there are 5 members. During the second week, 5x new members are brought in (x new members for every existing member). During the third week, the previous week’s new members (5x) each bring in x new members:new members. If we continue this pattern to the twelfth week, we will see that new members join the club that week. Since y is the number of new members joining during week 12, .

If , we can set each of the answer choices equal to and see which one yields an integer value (since y is a specific number of people, it must be an integer value). The only choice to yield an integer value is (D):

Therefore x = 15.

Since choice (D) is the only one to yield an integer value, it is the correct answer.

As promised, below is the answer to our weekly Challenge Problem. Enjoy!

Question

A gambler began playing blackjack with $110 in chips. After exactly 12 hands, he left the table with $320 in chips, having won some hands and lost others. Each win earned $100 and each loss cost $10. How many possible outcomes were there for the first 5 hands he played? (For example, won the first hand, lost the second, etc.)

(A) 10
(B) 18
(C) 26
(D) 32
(E) 64

Answer

Let W be the number of wins and L be the number of losses. Since the total number of hands equals 12 and the net winnings equal $210, we can construct and solve the following simultaneous equations:

So we know that the gambler won 3 hands and lost 9. We do not know where in the sequence of 12 hands the 3 wins appear. So when counting the possible outcomes for the first 5 hands, we must consider these possible scenarios:

1) Three wins and two losses
2) Two wins and three losses
3) One win and four losses
4) No wins and five losses

In the first scenario, we have WWWLL. We need to know in how many different ways we can arrange these five letters:

So there are 10 possible arrangements of 3 wins and 2 losses.

The second scenario — WWLLL — will yield the same result: 10.

The third scenario — WLLLL — will yield 5 possible arrangements, since the one win has only 5 possible positions in the sequence.

The fourth scenario — LLLLL — will yield only 1 possible arrangement, since rearranging these letters always yields the same sequence.

Altogether, then, there are 10 + 10 + 5 + 1 = 26 possible outcomes for the gambler’s first five hands.

The correct answer is C.

Here is the answer to Wednesday’s Challenge Question from Manhattan GMAT. Check back next week for another Workbook Wednesday question!

Question

If n is an integer and n⁴ is divisible by 32, which of the following could be the remainder when n is divided by 32?

(A) 2
(B) 4
(C) 5
(D) 6
(E) 10

Answer

The prime factors of n⁴ are really four sets of the prime factors of the integer n.

Since n⁴ is divisible by 32 (or 2⁵), n⁴ must be divisible by 2 at least 5 times. What does this tell us about the integer n?

If n is divisible by only one 2, then n⁴ would be divisible by exactly four 2’s (since the prime factors of n⁴ have no source other than the integer n).

But we know that n⁴ is divisible by at least five 2’s! This means that n must be divisible by at least two 2’s (which means that n⁴ must be divisible by eight 2’s). Thus, we know that the integer n must be divisible by 4.

Now that we know that n is divisible by 4, we can consider what happens when we divide n by 32.

If we divide n by 32 we can represent this mathematically as follows:

n = 32b + c (where b is the number of times 32 goes into n and c is the integer remainder)

We know that n is divisible by 4 so we can rewrite this as:

4x = 32b + c(where x is an integer)

This equation can be simplified, by dividing both sides by 4 as follows:

x = 8b + c/4

Since we know that x is an integer, the sum of 8b and c/4 must yield an integer. We know that 8b is an integer so c/4 must be also be an integer. Therefore, c, the remainder, must be divisible by 4.

Only answer choice B qualifies. The remainder when n is divided by 32 could be 4. It could not be any of the other answer choices. The correct answer is B.

Here is the answer for yesterday’s Workbook Wednesday challenge problem!

Question

Two missiles are launched simultaneously. Missile 1 launches at a speed of x miles per hour, increasing its speed by a factor of every 10 minutes (so that after 10 minutes its speed is , after 20 minutes its speed is , and so forth). Missile 2 launches at a speed of y miles per hour, doubling its speed every 10 minutes. After 1 hour, is the speed of Missile 1 greater than that of Missile 2?

1)
2)

(A) Statement (1) ALONE is sufficient to answer the question, but statement (2) alone is not.
(B) Statement (2) ALONE is sufficient to answer the question, but statement (1) alone is not.
(C) Statements (1) and (2) TAKEN TOGETHER are sufficient to answer the question, but NEITHER statement ALONE is sufficient.
(D) EACH statement ALONE is sufficient to answer the question.
(E) Statements (1) and (2) TAKEN TOGETHER are NOT sufficient to answer the question.

Answer

Since Missile 1’s rate increases by a factor of every 10 minutes, Missile 1 will be traveling at a speed of miles per hour after 60 minutes:

minutes	0-10	10-20	20-30	30-40	40-50	50-60	60+
speed

And since Missile 2’s rate doubles every 10 minutes, Missile 2 will be traveling at a speed of after 60 minutes:

minutes	0-10	10-20	20-30	30-40	40-50	50-60	60+
speed

The question then becomes: Is ?

Statement (1) tells us that . Squaring both sides yields . We can substitute for y: Is ? If we divide both sides by , we get: Is ? We can further simplify by taking the square root of both sides: Is ? We still cannot answer this, so statement (1) alone is NOT sufficient to answer the question.

Statement (2) tells us that , which tells us nothing about the relationship between x and y. Statement (2) alone is NOT sufficient to answer the question.

Taking the statements together, we know from statement (1) that the question can be rephrased: Is ? From statement (2) we know certainly that , which is another way of expressing . So using the information from both statements, we can answer definitively that after 1 hour, Missile 1 is traveling faster than Missile 2.

The correct answer is C: Statements (1) and (2) taken together are sufficient to answer the question, but neither statement alone is sufficient.