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CATEGORY - WORKBOOK WEDNESDAYS

Thursday, October 30, 2008

Workbook Wednesdays: Where-4 Art Thou Divisible? Answer

Dost thou seek the answer to yesterday’s Challenge Problem?  As promised, find it below!

Question
If p, x, and y are positive integers, y is odd, and p = x2 + y2, is x divisible by 4?

(1) When p is divided by 8, the remainder is 5.

(2) x – y = 3

(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.

Solution
Statement (1): SUFFICIENT. We know that p is odd. We know from the problem stem that y is odd, which means that y² is odd. Therefore, x² must be even (because O = E + O), so x must be even.

However from all of this, we can infer something else—specifically, that x is NOT a multiple of 4. Here’s why:

y² = (y² — 1) + 1, which using the quadratic property, yields: y = (y+1)(y—1) +1. Because y is odd, then (y+1)(y—1) is even times even, which is a multiple of 4. However we ALSO know that either y+1 or y—1 is a multiple of 4, because they are consecutive multiples of 2. Therefore, y² is 1 greater than a multiple of 8. (You can confirm this by thinking about all squared odd numbers: 1² = 1, 3² = 9, 5² = 25, 7² = 49, etc.)

Since y² divided by 8 yields a remainder of 1, from statement (1) we need x² divided by 8 to yield a remainder of 4. This means that x must be even, but NOT be a multiple of 4, because if it were a multiple of 4, then x² would be a multiple of 16, and therefore a multiple of 8.

Statement (2): INSUFFICIENT. We know that x is even. However, we can come up with two different examples which lead to different answers. For example, if x = 8 and y = 5, then x is a multiple of 4, but if x = 6 and y = 3, then x is not a multiple of 4.

The answer is A: Statement (1) is sufficient to answer the question, but statement (2) is insufficient.

Wednesday, October 29, 2008

Workbook Wednesdays: Where-4 Art Thou Divisible?

Welcome to this week’s edition of Workbook Wednesdays, featuring the type of problem one would see if scoring above a 700 on the GMAT. As always, thanks to ManhattanGMAT for providing this week’s challenge problem! Be sure to check back tomorrow for the answer!

Question
If p, x, and y are positive integers, y is odd, and p = x² + y², is x divisible by 4?

(1) When p is divided by 8, the remainder is 5.

(2) x – y = 3

(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.

Thursday, October 23, 2008

Workbook Wednesdays: X to the Nth Power Answer

Check out the answer to yesterday’s Challenge Problem!

Question
If n is a positive integer and x does not equal zero, is x^n > x^(n+1)?

1) x < 1

2) n is even.

(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.

Solution
The answer to the question depends on the values of both x and n. Specifically, we care about the value of x^n, since this will determine how we can rephrase the question.

If n is even, then x^n > 0, no matter what the value of x is (remember that x is nonzero).
Likewise, if x > 0, then x^n > 0, no matter what the value of n is.

The reason that we care about the value of x^n is that we can simplify the question by dividing by x^n:

After we divide both sides of the inequality by x^n, the question “Is x^n > x^(n+1)?” becomes “Is 1 > x?” ONLY IF x^n > 0, which is true if x > 0 OR if n is even. (Recall that x is nonzero; thus, we are allowed to divide by x^n.) On the other hand, if x^n < 0, then the question rephrases to “Is 1 < x?”

Statement 1: INSUFFICIENT. We know that x < 1, but x could be positive or negative. Moreover, we do not know whether n is even or odd. As a result, we do not know the sign of x^n, and thus we do not know the answer to either the rephrased question or to the original question.

Alternatively, you can choose positive and negative values of x and an odd n, in order to test the question. If n = 1 and x is positive (but less than 1), then x^n > x^(n+1). But if n = 1 and x is negative, then x^n > x^(n+1).

Statement 2: INSUFFICIENT. We know that n is even, so we know that x^n > 0, and therefore we can rephrase the question as “Is 1 > x?” However, we do not know the answer to that question.

Statements 1 & 2 TOGETHER: SUFFICIENT. Using Statement (2), we can rephrase the question as “Is 1 > x?”, to which Statement (1) gives us a definitive answer.

The answer is C: BOTH statements TOGETHER are sufficient to answer the question, but neither statement alone is sufficient.

Wednesday, October 22, 2008

Workbook Wednesdays: X to the Nth Power

Welcome to this week’s edition of Workbook Wednesdays! If you’re ready to sharpen your quantitative skills, check out the Challenge Problem below, courtesy of our friends at ManhattanGMAT.

Question
If n is a positive integer and x does not equal zero, is x^n > x^(n+1)?

1) x < 1

2) n is even.

(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.

Thursday, October 16, 2008

Workbook Wednesdays: The Cost of Fuel Answer

Check below to see what was paid at the pump in yesterday’s GMAT question!

Question
A certain military vehicle can run on pure Fuel X, pure Fuel Y, or any mixture of X and Y. Fuel X costs $3 per gallon; the vehicle can go 20 miles on a gallon of Fuel X. In contrast, Fuel Y costs $5 per gallon, but the vehicle can go 40 miles on a gallon of Fuel Y. What is the cost per gallon of the fuel mixture currently in the vehicle’s tank?

1) Using fuel currently in its tank, the vehicle burned 8 gallons to cover 200 miles.

2) The vehicle can cover 7 and 1/7 miles for every dollar of fuel currently in its tank.

(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.

Solution
This problem has to do with weighted averages. To find the cost per gallon of the fuel mixture currently in the vehicle’s tank, we need to know the ratio (by gallons) of Fuel X to Fuel Y in that mixture.

Statement 1: SUFFICIENT. The statement tells us that the vehicle covered 200 miles on 8 gallons of the fuel mixture; that is, the fuel mixture delivers 25 miles per gallon. As a result, we can find the ratio of Fuel X (20 mpg) to Fuel Y (40 mpg) in the mixture (this ratio turns out to be 3:1, but we do not need to find the exact ratio; we simply need to know that there will be one unique ratio). Finally, we can use this ratio to find the weighted average cost per gallon of the fuel mixture (this weighted average turns out to be $3.50 per gallon, but again, we do not need the exact number).

Statement 2: SUFFICIENT. The statement tells us that $1 of fuel “buys” 7 and 1/7 miles. Multiplying through by 7, we can rephrase the ratio as $7 for every 50 miles. Let’s now express the weighted average cost per gallon as $3w + $5(1-w) = 5 – 2w, where w represents the percent of Fuel X in the mixture (and 1-w represents the percent of Fuel Y in the mixture). The number of gallons bought by the $7 is then $7 divided by the average cost per gallon, or 7/(5 – 2w). Likewise, we can express the weighted average fuel efficiency (miles per gallon) as 20w + 40(1-w) = 40 – 20w. The number of gallons burned to cover 50 miles is then 50 divided by the average miles per gallon, or 50/(40 – 20w). We can now set these numbers of gallons equal to each other:

7/(5 – 2w) = 50/(40 – 20w)
280 – 140w = 250 – 100w
30 = 40w
3/4 = w

Knowing that the mixture is 75% Fuel X and 25% Fuel Y, we can now in theory calculate the average cost per gallon of the mixture. (In fact, we should have stopped before calculating w – simply knowing that we could calculate that percentage is sufficient.)

The answer is D: EACH statement ALONE is sufficient to answer the question.

Wednesday, October 15, 2008

Workbook Wednesdays: The Cost of Fuel

Welcome to this week’s edition of Workbook Wednesdays, featuring the type of problem one would see if scoring above a 700 on the GMAT. As always, thanks to ManhattanGMAT for providing this week’s challenge problem!  Be sure to check back tomorrow for the answer!

A certain military vehicle can run on pure Fuel X, pure Fuel Y, or any mixture of X and Y. Fuel X costs $3 per gallon; the vehicle can go 20 miles on a gallon of Fuel X. In contrast, Fuel Y costs $5 per gallon, but the vehicle can go 40 miles on a gallon of Fuel Y. What is the cost per gallon of the fuel mixture currently in the vehicle’s tank?

1) Using fuel currently in its tank, the vehicle burned 8 gallons to cover 200 miles.

2) The vehicle can cover 7 and 1/7 miles for every dollar of fuel currently in its tank.

(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.

Thursday, October 09, 2008

Workbook Wednesdays: Medical Test Answer

The results from yesterday’s GMAT question are in!  See below for a detailed answer:

Question
A medical test for a certain liver enzyme can be given in the morning, in the afternoon, or in the evening; moreover, the result of the test can be low, average, or high. At least three-quarters of low and medium readings are not given in the evening. Sixty percent of exams are given in the morning or in the afternoon, and 20% of exams result in a high reading. What percent of exams given in the evening result in low or medium readings?

(A) 20%
(B) 30%
(C) 40%
(D) 50%
(E) 60%

Answer
The key to this problem is to realize that you can collapse certain categories together. The distinction between low and medium readings does not matter, because we are never given data about just low or just medium readings. Likewise, morning and afternoon tests can be combined, because the given information never distinguishes those times of day. Thus, we only have two categories for each dimension: time of day is either “morning+afternoon” or evening, and result is either “low+medium” or high.
We can now set up a 2×2 table, plus a total row and column (and labels):

We choose 100 for the total of all tests, since we are only dealing with percents. Filling in the table with the given information and completing the total row and column, we get the following:

Since the number of high morning+afternoon exams cannot be less than zero, it must actually be zero. This means that the table fills in this way:

Thus, the percentage of evening exams that do NOT result in a high reading is 20/40 x 100%, or 50%.

The correct answer is (D).

Wednesday, October 08, 2008

Workbook Wednesdays: Medical Tests

Get ready for your quantitative check-up! As always, we’d like to thank ManhattanGMAT for providing this week’s Challenge Question and - more importantly - tomorrow’s answer!

Question
A medical test for a certain liver enzyme can be given in the morning, in the afternoon, or in the evening; moreover, the result of the test can be low, average, or high. At least three-quarters of low and medium readings are not given in the evening. Sixty percent of exams are given in the morning or in the afternoon, and 20% of exams result in a high reading. What percent of exams given in the evening result in low or medium readings?

(A) 20%
(B) 30%
(C) 40%
(D) 50%
(E) 60%

Thursday, October 02, 2008

Workbook Wednesdays: A Slippery Slope Answer

Check out the answer to yesterday’s Challenge Problem below!

Question
Line k lies in a coordinate plane. Is the slope of line k positive?

(1) Line k and the graph of the function f(x) = x^2 – bx, where b is positive, intersect on the x-axis.

(2) Line k and the graph of the function g(x) = –x^2 – c, where c is positive, intersect on the y-axis.

(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
The given information simply sets the stage: a coordinate plane.

Statement (1) states that the graph of the function f(x) = x^2 – bx, where b is positive, and line k intersect on the x-axis. Since the function is a quadratic, we know that the graph of the function is a parabola. Let’s find the places that this function intersects the x-axis. This is equivalent to finding the values of x for which the function equals zero.
f(x) = x^2 – bx = x(x – b) = 0. Thus, x = 0 or x = b, a positive number. Therefore, line k touches the x-axis at one (or possibly both) of the points (0,0) and (b, 0). However, we do not know another point guaranteed to be on the line. Thus, we do not know the slope of line k. INSUFFICIENT.

Statement (2) states that the graph of the function g(x) and line k intersect on the y-axis. Again, we know that the graph of the function is a parabola, since the function itself is quadratic. To find where the parabola intersects the y-axis, we find the value of the function at x = 0.
g(x) = –x^2 – c
g(0) = –0^2 – c = –c
Since c is positive, –c is negative. Thus, the line k touches the y-axis at the point (0, –c). However, we do not know another point guaranteed to be on the line. Thus, we do not know the slope of line k. INSUFFICIENT.

Statements (1) and (2) together: Line k goes through (0, –c) AND either (0,0) or (b, 0). If line k goes through (0, –c) and (b, 0), where both b and c are positive, then we know that the slope of line k is positive (both the rise and the run are positive). However, line k could go through (0, –c) and (0, 0), which would mean that line k is vertical (it actually would coincide with the y-axis). A vertical line has an undefined slope. Since the slope could either be positive or undefined, the two statements together are INSUFFICIENT.

The correct answer is E: Statements (1) and (2) TOGETHER are INSUFFICIENT to answer the question.

Wednesday, October 01, 2008

Workbook Wednesdays: A Slippery Slope

Welcome to this week’s edition of Workbook Wednesdays, courtesy of our friends at ManhattanGMAT! Get ready to wrap your brain around an advanced quantitative problem - the kind you would see if you are scoring a 700 or higher on the GMAT - then be sure to check back tomorrow for the answer.

Question
Line k lies in a coordinate plane. Is the slope of line k positive?

(1) Line k and the graph of the function f(x) = x^2 – bx, where b is positive, intersect on the x-axis.

(2) Line k and the graph of the function g(x) = –x^2 – c, where c is positive, intersect on the y-axis.

(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.

Thursday, September 25, 2008

Workbook Wednesdays: Solve for N Answer

Find the answer to yesterday’s Challenge Problem below!

Question
If n is a positive integer, what is n?

(1) 3^n – 1 has three prime factors, not necessarily distinct.
(2) n^2 = 2^n.

(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
The given information simply guarantees that n is a positive integer.

Statement (1) indicates that 3^n – 1 has three prime factors, not necessarily distinct. There is no way to intuit or derive a general solution for n; we must simply test values of n and see how many prime factors the expression has.

If n = 1, then 3^n – 1 = 2, which has just one prime factor. N cannot be 1.
If n = 2, then 3^n – 1 = 8, which has three non-distinct prime factors (8 = 2^3). N could be 2. However, we should not stop here. We need to see whether other values of n generate expressions that have three prime factors.
If n = 3, then 3^n – 1 = 26, which has two prime factors: 2 and 13. N cannot be 3.
If n = 4, then 3^n – 1 = 80, which has six prime factors: 2 (counted five times) and 5, since 80 = (5)(2^5). N cannot be 4. There does not seem to be a simple pattern in the number of prime factors, so we should suspect that for some other value of n, the expression has three prime factors, but we need to prove it. Let’s try one more.
If n = 5, then 3^n – 1 = 242. 242 = (2)(121) = (2)(11^2). 242 has three prime factors, 2 and 11 (counted twice). N could be indeed be 5. Since n could be 2 or 5 (at least), we do not know what n is. INSUFFICIENT.

Statement (2) indicates that n^2 = 2^n. By testing small numbers, we see that n could only be 2 or 4. For values of n above 4, the right side of the equation grows faster than the left side; for instance, if n = 5, then n^2 = 25, but 2^n = 32. The powers of 2 grow faster than the squares. So n cannot be any larger value. However, we do not know whether n is 2 or 4. INSUFFICIENT.

Statements (1) and (2) together: n can only be 2. SUFFICIENT.

The correct answer is C: Statements (1) and (2) TOGETHER are sufficient, but neither statement ALONE is sufficient.

Wednesday, September 24, 2008

Workbook Wednesdays: Solving for N

Welcome to this week’s edition of Workbook Wednesdays! As always, we’d like to thank our friends at ManhattanGMAT for providing this week’s Challenge Problem. Check back tomorrow for an in-depth look at the answer!

Question
If n is a positive integer, what is n?

(1) 3^n – 1 has three prime factors, not necessarily distinct.
(2) n^2 = 2^n.

(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.

Thursday, September 18, 2008

Workbook Wednesdays: Learn Your ABC’s Answer

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

Question

If a, b, and c are integers, and the product abc is even, is b even?
(1) (ab)/c is an even integer.
(2) (ac)/b is an odd integer.

(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

The given information simply guarantees that at least one of the integers a, b, or c is even.

Statement (1) indicates that ab = (c)(some even integer). This means that the left side of the equation must be even. However, we could have this result either with an even b or with an odd b (if a is even). INSUFFICIENT.

Statement (2) indicates that ac = (b)(some odd integer). If b is even, then we also know that a or c (or both) is even; this scenario fits the constraint that at least one of the variables is even. However, if b is odd, then the right side of the equation is odd. Therefore, BOTH a and c are odd, since ac is odd. This scenario contradicts the constraint that at least one of the variables is even. Thus, we know that b must be even. SUFFICIENT.

The correct answer is B: Statement (2) ALONE is sufficient, but statement (1) alone is insufficient.

Wednesday, September 17, 2008

Workbook Wednesdays: Learn Your ABC’s

Welcome to this week’s edition of Workbook Wednesdays, courtesy of our friends at ManhattanGMAT!  Just like last week, this problem mimics the most advanced quantitative problems on the exam, the type of problem you will see if you are scoring around 700 or higher.  Check back tomorrow for the answer!

If a, b, and c are integers, and the product abc is even, is b even?

(1) (ab)/c is an even integer.
(2) (ac)/b is an odd integer.

(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.

Thursday, September 11, 2008

Workbook Wednesdays: Recursive Curse Answer

As promised, here is the answer to yesterday’s Challenge Problem!

Question
The function f(x) is defined as f(x) = 1 – 1/(1-x) for all x not equal to 1. The sequence A(n) for all integers n > 1 is defined as A(n) = f(A(n-1)).

What values of A(1) create a sequence such that A(n) = A(n-2) for all n > 2?

(I) x < 0
(II) x = 0
(III) 0 < x < 1
(IV) x > 1

(A) I only
(B) II only
(C) II and III only
(D) II, III, and IV only
(E) I, II, III, and IV

Answer
The definition of the sequence means that you apply the function f to A(1) to get A(2); then you apply the function again to A(2) to get A(3), and so on.

Now, you could try various values of A(1) and see whether you get the same value for A(3). A more general approach is to rephrase the question: you are looking for values of x such that f(f(x)) = x. (In other words, when you apply the function to the value twice, you get the same value back.)

If f(x) = 1 – 1/(1-x), then f(f(x)) = 1 – 1/(1-[1 – 1/(1-x)], where brackets [] indicate the insertion of f(x) in place of x. This expression looks ugly, but let’s try to simplify it.

In general, 1 – [1 – Z] = Z. So 1 – 1/(1-[1 – 1/(1-x)] = 1 – 1/(1/(1-x)).

Next, 1/(1/Z) = Z. So 1 – 1/(1/(1-x)) = 1 – (1-x) = x.

Amazingly, that ugly expression 1 – 1/(1-[1 – 1/(1-x)] = x, for all legal values of x.

Thus, for any value of x besides 1, if you apply the function to it twice, you get the original value back. Try it!

The answer is E: I, II, III, and IV.

Wednesday, September 10, 2008

Workbook Wednesdays: Recursive Curse

Welcome to this week’s edition of Workbook Wednesdays, brought to you by our friends at ManhattanGMAT. Check back tomorrow for an in-depth look at the answer!

Question
The function f(x) is defined as f(x) = 1 – 1/(1-x) for all x not equal to 1. The sequence A(n) for all integers n > 1 is defined as A(n) = f(A(n-1)).

What values of A(1) create a sequence such that A(n) = A(n-2) for all n > 2?

(I) x < 0
(II) x = 0
(III) 0 < x < 1
(IV) x > 1

(A) I only
(B) II only
(C) II and III only
(D) II, III, and IV only
(E) I, II, III, and IV

Thursday, September 04, 2008

Workbook Wednesdays: Consecutive Divisibility Answer

Pencil’s up! Here is the answer to yesterday’s Challenge Question.

Question
If x and n are positive integers, is n = 1?

(1) The sum of n consecutive integers, starting at x, is divisible by xn.
(2) The product of n consecutive integers, starting at x, is divisible by x^n.

(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
The problem asks a “Yes or No” question about n: is it equal to 1? Note that we do not need to know the value of n in order to answer this question definitively. For instance, knowing that n is even would be sufficient to answer the question “No” (which would be a sufficient answer).

Statement (1) asserts that the sum of n consecutive integers, starting at x, is divisible by xn. First, we can connect this with other facts about consecutive integers. It turns out that the sum of n consecutive integers is divisible by n if and only if n is odd. (The reason is that the average number in a set of consecutive integers is actually the middle integer if you have 3, 5, or some other odd number of integers. But if you have an even number of integers, there is no middle integer, and the average number is not an integer. This matters because the sum of n consecutive integers divided by n IS the average number in a set of consecutive integers.) So we rule out even values of n. However, if we simply let x be 1, then the condition is satisfied by n = 1, 3, 5, or any other positive odd integer. We do not know whether n is equal to 1. INSUFFICIENT.

Statement (2) seems even more complicated, but it can be defeated again by a judicious choice of x as 1. If x = 1, then x^n = 1, no matter what n is. Since all positive integers are divisible by 1, then there is no restriction on the value of n, which could be equal to 1 OR to any other positive integer. INSUFFICIENT.

Statements (1) and (2) together: Putting what we have learned together, we know that if we let x = 1, then n might equal 1, but it could also equal any other positive odd number. We cannot answer the question definitively. INSUFFICIENT. The correct answer is (E): Statements (1) and (2) TAKEN TOGETHER are NOT sufficient to answer the question.

Wednesday, September 03, 2008

Workbook Wednesdays: Consecutive Divisibility

Welcome to the another installment of Workbook Wednesdays. Thanks again to Manhattan GMAT for supplying today’s question and (more importantly) tomorrow’s answer! Just like last week, this problem mimics the most advanced quantitative problems on the exam, the type of problem you will see if you are scoring around 700 or higher.

Question
If x and n are positive integers, is n = 1?

(1) The sum of n consecutive integers, starting at x, is divisible by xn.
(2) The product of n consecutive integers, starting at x, is divisible by x^n.

(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.

Get out your pencils and scratch paper, and be sure to check back tomorrow for an in depth look at the answer!

Thursday, August 28, 2008

Workbook Wednesdays: Boats on the Lake Answer

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.

Wednesday, August 27, 2008

Workbook Wednesdays: Boats on the Lake

Welcome back to another edition of Workbook Wednesdays, brought to you by our friends at Manhattan GMAT. Take a look at the problem below, and be sure to check back in with us tomorrow for an explanation of the answer!

Question:

On Lake Coheeries, there are only three kinds of boats: catamarans, canoes, and kayaks. The ratio of catamarans to canoes is 4:7, and the ratio of canoes to kayaks is 5:9. Which of the following could be the total number of boats on the lake?

A) 575
B) 580
C) 585
D) 590
E) 595

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