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.

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.

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 . . . → Continue Reading

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):

  Low+Avg High Total Morning +Afternoon       Evening       Total      

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:

  Low+Avg High Total Morning +Afternoon At least 3/4 of 80 = 60   60 Evening     40 Total 80 20 100

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:

  Low+Avg High Total Morning +Afternoon 60 0 60 Evening 20 20 40 Total 80 20 100

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).

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: . . . → Continue Reading

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.

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.