Advanced Quant
The following questions are extremely advanced for the GRE. We have included them by popular demand—students who are aiming for perfect math GRE scores often wish to practice on problems that may be even harder than any seen on the real GRE. We estimate that a GRE test-taker who does well on the first math section and therefore is given a difficult second section might see one or two problems, at most, of this level of difficulty.
If you are not aiming for a perfect math score, we absolutely recommend that you skip these problems!
If you are taking the GRE for business school or another quantitative program, you may wish to attempt some of these problems. For instance, you might do one or two of these problems—think of them as “brain teasers”—to cap off a study session from elsewhere in the book. (For reference, getting 50% of these problems correct would be a pretty incredible performance!)
Even if you are aiming for a perfect math score, though, please make sure you are flawless at the types of math problems in the rest of this book before you work on these. You will gain far more points by reducing minor mistakes (through practice, steady pacing, and good organization) on easy and medium questions than you would by focusing on ultra-hard questions.
For more such problems, visit the Manhattan Prep GRE blog for our monthly Challenge Problem. (Access to the archive of over fifteen dozen Challenge Problems is available to our course and Guided Self-Study students for free and to anyone else for a small fee.)
That said, attempt these Advanced Quant problems—if you dare!
1.21 people per minute enter a previously empty train station beginning at 7:00:00 pm (7 o’clock and zero seconds). Every 9 minutes beginning at 7:04:00 pm, a train departs and everyone who has entered the station in the last 9 minutes gets on the departing train. If the last train departs at 8:25:00 pm, what is the average number of people who get on each of the trains leaving from 7:00:00 pm to 8:25:00 pm, inclusive?
(A)84
(B)136.5
(C)178.5
(D)189
(E)198.5
2.The random variable x has the following continuous probability distribution in the range 0 ≤ x ≤ 
, as shown in the coordinate plane with x on the horizontal axis:
The probability that x < 0 = the probability that x > = 0.
What is the median of x?
(A)
(B)
(C) – 1
(D)
(E)
x < 0
| 3. | Quantity A x2 – 5x + 6 |
Quantity B x2 – 9x + 20 |
4.If x is a positive integer, what is the units digit of (24)5 + 2x(36)6(17)3?
(A)2
(B)3
(C)4
(D)6
(E)8
5.In the figure above, the circumference of the circle is 20π. Which of the following is the maximum possible area of the rectangle?
(A)80
(B)200
(C)300
(D)
(E)
6.The length of each edge of a cube equals 6. What is the distance between the center of the cube to one of its vertices?
(A)
(B)
(C)
(D)
(E)
7.If c is randomly chosen from the integers 20 to 99, inclusive, what is the probability that c3 – c is divisible by 12?
Give your answer as a fraction.
8.The remainder when 120 is divided by single-digit integer m is positive, as is the remainder when 120 is divided by single-digit integer n. If m > n, what is the remainder when 120 is divided by m – n?
9.A circular microchip with a radius of 2 centimeters is manufactured following a blueprint scaled such that a measurement of 1 centimeter on the blueprint corresponds to a measurement of 0.8 millimeters on the microchip. What is the diameter of the blueprint representation of the microchip, in centimeters? (1 centimeter = 10 millimeters)
centimeters
For a certain quantity of a gas, pressure P, volume V, and temperature T are related according to the formula PV = kT, where k is a constant.
| 10. | Quantity A The value of P if V = 20 and T = 32 |
Quantity B The value of T if V = 10 and P = 78 |
11.The circle with center O has a circumference of 
. If AC is a diameter of the circle, what is the length of line segment AB?
(A)
(B)6
(C)
(D)18
(E)
12.A batch of widgets costs p + 15 dollars for a company to produce and each batch sells for p(9 – p) dollars. For which of the following values of p does the company make a profit?
(A)3
(B)4
(C)5
(D)6
(E)7
13.If k is the sum of the reciprocals of the consecutive integers from 41 to 60 inclusive, which of the following are less than k?
Indicate all such statements.
14.Half an hour after car A started traveling from Newtown to Oldtown, a distance of 62 miles, car B started traveling along the same road from Oldtown to Newtown. The cars met each other on the road 15 minutes after car B started its trip. If car A traveled at a constant rate that was 8 miles per hour greater than car B’s constant rate, how many miles had car B driven when they met?
(A)14
(B)12
(C)10
(D)9
(E)8
| 15. |
x and y are positive integers such that x25y = 10,125
| Quantity A x2 |
Quantity B 5y |
16.Which of the following is equal to 
for all values of n > 1?
(A)–1
(B)1
(C)
(D)
(E)
17.Bank account A contains exactly x dollars, an amount that will decrease by 10% each month for the next two months. Bank account B contains exactly y dollars, an amount that will increase by 20% each month for the next two months. If A and B contain the same amount at the end of two months, what is the ratio of 
to 
?
(A)4 : 3
(B)3 : 2
(C)16 : 9
(D)2 : 1
(E)9 : 4
Body Mass Index (BMI) is calculated by the formula 
, where w is weight in pounds and h is height in inches.
| 18. | Quantity A The number of pounds gained by a 74-inch-tall person whose BMI increased by 1.0. |
Quantity B The number of pounds lost by a 65-inch-tall person whose BMI decreased by 1.2. |
19.How many times does the digit grouping “57” (in that order) appear in all of the five-digit positive integers? For instance, “57” appears once in 12,357, twice in 57,057, and does not appear in 24,675.
(A)279
(B)3,091
(C)3,519
(D)3,671
(E)4,077
| 20. | Quantity A The average of all the multiples of 5 between 199 and 706 |
Quantity B The average of all the multiples of 10 between 199 and 706 |
AC || FD
| 21. | Quantity A a + d – c – 90 |
Quantity B 90 – e – b – f |
22.A man travels to his home from his current location on the rectangular grid shown above. If he may choose to travel north or east at any corner, but may never travel south or west, how many different paths can the man take to get home?
(A)12
(B)24
(C)32
(D)35
(E)64
23.A bag contains 3 white, 4 black, and 2 red marbles. Two marbles are drawn from the bag. If replacement is not allowed, what is the probability that the second marble drawn will be red?
(A)
(B)
(C)
(D)
(E)
x < 0
| 24. | Quantity A ((25x)–2)3 |
Quantity B ((5–3)2)–x |
| 25. | Quantity A The sum of all the multiples of 6 between –126 and 342, inclusive |
Quantity B 8,502 |
x is an integer.
| 26. | Quantity A (–1)x2 + (–1)x3 + (–1)x4 |
Quantity B (–1)x + (–1)2x + (–1)3x + (–1)4x |
d > c
| 27. | Quantity A a |
Quantity B d – b |
The circumference of a circle is 
the perimeter of a square.
| 28. | Quantity A The area of the square |
Quantity B The area of the circle |
| 29. | Quantity A The ratio of the area of the larger square to the area of the smaller square |
Quantity B Twice the ratio of the area of the smaller circle to the area of the larger circle |
m = 216317418519
n = 219318417516
| 30. | Quantity A The number of zeros at the end of m when written in integer form |
Quantity B The number of zeros at the end of n when written in integer form |
The sequence of numbers a1, a2, a3, …, an, … is defined by 
for each integer n ≥ 1.
| 31. | Quantity A The sum of the first 32 terms of this sequence |
Quantity B The sum of the first 31 terms of this sequence |
32.Each of 100 balls has an integer value from 1 to 8, inclusive, painted on the surface. The number nx of balls representing integer x is defined by the formula nx = 18 – (x – 4)2. What is the interquartile range of the 100 integers?
(A)1.5
(B)2.0
(C)2.5
(D)3.0
(E)3.5
The operator ! is defined such that a!b = ab × b–a.
| 33. | Quantity A |
Quantity B |
34.What is the ratio of the sum of the odd positive integers between 1 and 100, inclusive, and the sum of the even positive integers between 100 and 150, inclusive?
(A)2 to 3
(B)5 to 7
(C)10 to 13
(D)53 to 60
(E)202 to 251
35.For integer n ≥ 3, a sequence is defined as an = (an – 1)2 – (an – 2)2 and an > 0 for all positive integers n. The first term a1 is 2, and the fourth term is equal to the first term multiplied by the sum of the second and third terms. What is the third term, a3?
(A) 0
(B) 3
(C) 5
(D)10
(E)16
36.In a certain sequence, each term beyond the second term is equal to the average of the previous two terms. If a1 and a3 are positive integers, which of the following is not a possible value of a5?
(A)–
(B)0
(C)
(D)
(E)
37.The operator @ is defined by the following expression: a@b = 
where ab ≠ 0. What is the sum of the solutions to the equation x@2 = 
?
(A)–1
(B)–0.75
(C)–0.25
(D) 0.25
(E) 0.75
x is a non-negative number and the square root of (10 – 3x) is greater than x.
| 38. | Quantity A |x| |
Quantity B 2 |
The area of an equilateral triangle is greater than 
but less than 
.
| 39. | Quantity A The length of one of the sides of the triangle |
Quantity B 9 |
40.The inequality |8 – 2x| < 3y – 9 is equivalent to which of the following?
(A)
(B)3y + 2x > 1
(C)6y – 2 < 2x
(D)1 – y < 2x < 17 + y
(E)3y – 1 > 2x > 17 – 3y
In the sport of mixed martial arts, more than 30% of all fighters are skilled in both the Muy Thai and Brazilian Jiu Jitsu styles of fighting. 20% of the fighters who are not skilled in Brazilian Jiu Jitsu are skilled in Muy Thai. 60% of all fighters are skilled in Brazilian Jiu Jitsu.
| 41. | Quantity A The percent of fighters who are skilled in Muy Thai |
Quantity B 37% |
The rate of data transfer, r, over a particular network is directly proportional to the bandwidth, b, and inversely proportional to the square of the number of networked computers, n.
| 42. | Quantity A The resulting rate of data transfer if the bandwidth is quadrupled and the number of networked computers is more than tripled |
Quantity B |
Advanced Quant Answers
1. (C). From 7 pm to 7:04, 84 people enter the station (21 per minute). These 84 people will get on the 7:04 train.
After that, for each 9-minute period, 9(21) = 189 people will enter the station and then get on a train. These trains will leave at 7:13, 7:22, 7:31, 7:40, 7:49, 7:58, 8:07, 8:16, and 8:25.
Since 9 trains each have 189 people and the first train has 84 people, the average is:
Note that the strange time format (minutes and seconds) doesn’t make the problem any harder—the problem is actually clearer because the train comes at 7:04 and 0 seconds, rather than 7:04 and 30 seconds, at which point more people would have entered the station.
2. (C). A continuous probability distribution has a total area of 100%, or 1, underneath the entire curve. The median of such a distribution splits the area into two equal halves, with 50% of the area to the left of the median and the other 50% to the right of the median:
In simpler terms, the random variable x has a 50% chance of being above the median and a 50% chance of being below the median. You can ignore the regions to the right or the left of this triangle, since the probability that x could fall in either of those regions is zero. So the question becomes this: what point on the x-axis will divide the large right triangle into two equal areas?
One shortcut is to note that the area of the large isosceles right triangle must be 1, which equals the total area under any probability distribution curve. Confirm by means of the area formula for this right triangle: 
.
The quickest way to find the median is to consider the small isosceles right triangle, ABC, as shown:
Triangle ABC must have an area of 
. So what must be the length of each of its legs, AB and BC? From the formula bh = , and noting that the base BC equals the height AB, the base BC must be 1 (the same as the height). Since the coordinates of point C are (, 0), the coordinates of point B must be ( – 1, 0). That is, the median is – 1.
3. (B). One way to solve is to set up an implied equation or inequality, then make the same changes to both quantities, and finally compare after simplifying:
Because x is negative, –4x + 14 = –4(neg) + 14 = pos + 14, which is greater than 0.
Another way to solve is to factor and then compare based on number properties. Quantity A factors to (x – 2)(x – 3). Quantity B factors to (x – 4)(x – 5). Because x is negative, “x minus a positive number” is also negative. Each quantity is the product of two negative numbers, which is positive:
Quantity A: (x – 2)(x – 3) = (neg)(neg) = pos
Quantity B: (x – 4)(x – 5) = (more neg)(more neg) = more pos
Quantity B is greater.
4. (A). 24 to any power ends in the same units digit as 4 to the same power (if considering only the last digit of the product, consider only the last digits of the numbers being multiplied).
4 to any power ends in either 4 or 6 (41 = 4, 42 = 16, 43 = 64, etc.). If the power is odd, the answer ends in 4; if the power is even, the answer ends in 6. Since the exponent 5 + 2x is odd for any integer x, 245 + 2x ends in 4.
36 to any power ends in the same units digit as 6 to the same power. Powers of 6 always end in 6, so 366 ends in 6.
17 to any power ends in the same units digit as 7 to the same power. While the units digits of the powers of 7 do indeed create a pattern, 73 is just 343, which ends in 3. Thus:
245 + 2x ends in 4
366 ends in 6
73 ends in 3
Multiplying three numbers that end in 4, 6, and 3 yields answer that ends in 2, because (4)(6)(3) = 72, which ends in 2.
5. (B). What are the possibilities for the inscribed rectangle?
The inscribed rectangle can be stretched and pulled to extremes: extremely long and thin, extremely tall and narrow, and somewhere in between:
The “long and thin” and “tall and narrow” rectangles have a very small area, and the “in between” rectangle has the largest possible area. In fact, the largest possible rectangle inscribed inside a circle is a square:
In this problem, the circumference is equal to 20π = 2πr. Thus r = 10. The square then has a side length of 
and an area of ()2 = 200.
6. (C).
The length of any side of the cube is 6, and the question asks for the distance between the center of the cube and any of its vertices (corners). Chopping up the cube into 8 smaller cubes, the distance from the center of the 6 × 6 × 6 cube to any corner is the diagonal of a 3 × 3 × 3 cube.
There are several ways to find the diagonal of a cube. Probably the fastest is to use the Super Pythagorean theorem, which extends to three dimensions:
a2 + b2 + c2 = d2
In the special case when the three sides of the box are equal, as they are in a cube, then this is the equation, letting s represent any side of the cube:
| s2 + s2 + s2 | = | d2 |
| 3s2 | = | d2 |
s |
= | d |
Since s = 3, d = 
.
7. 
(or any equivalent fraction). Probability is 
. There are 99 – 20 + 1 = 80 possible values for c, so the unknown is how many of these c values yield a c3 – c that is divisible by 12.
The prime factorization of 12 is 2 × 2 × 3. There are several ways of thinking about this: numbers are divisible by 12 if they are divisible by 3 and by 2 twice, or if they are multiples of both 4 and 3, or if half of the number is an even multiple of 3, etc.
The expression involving c can be factored:
c3 – c = c(c2 – 1) = c(c – 1)(c + 1)
These are consecutive integers. It may help to put them in increasing order: (c – 1)c(c + 1). Thus, this question has a lot to do with consecutive integers, and not only because the integers 20 to 99 themselves are consecutive.
In any set of three consecutive integers, a multiple of 3 will be included. Thus, (c – 1)c(c + 1) is always divisible by 3 for any integer c. This takes care of part of the 12. So the question becomes “How many of the possible (c – 1)c(c + 1) values are divisible by 4?” Since the prime factors of 4 are 2’s, it makes sense to think in terms of odds and evens.
(c – 1)c(c + 1) could be (E)(O)(E), which is definitely divisible by 4, because the two evens would each provide at least one separate factor of 2. Thus, c3 – c is divisible by 12 whenever c is odd, which are the cases c = 21, 23, 25, …, 95, 97, 99. That’s 
possibilities.
Alternatively, (c – 1)c(c + 1) could be (O)(E)(O), which will only be divisible by 4 when the even term itself is a multiple of 4. Thus, c3 – c is also divisible by 12 whenever c is a multiple of 4, which are the cases c = 20, 24, 28, …, 92, 96. That’s 
possibilities.
The probability is thus 
.
8. 0. Since the remainder is defined as what is left over after one number is divided by another, it makes sense that the leftover amount would be positive. So why is this information provided, if the remainder is “automatically” positive? Because there is a third possibility: that the remainder is 0. If the remainder when 120 is divided by m is positive, the salient point is that 
does not have a remainder of 0. In other words, 120 is not divisible by m, or m is not a factor of 120. Similarly, n is not a factor of 120.
Another constraint on both m and n is that they are single-digit positive integers. So m and n are integers between 1 and 9, inclusive, that are not factors of 120. Only two such possibilities exist: 7 and 9.
Since m > n, m = 9 and n = 7. Thus, m – n = 2, and the remainder when 120 is divided by 2 is 0.
9. 50 centimeters. Microchip radius = (2 cm) 
= 20 mm. Microchip diameter = 40 mm.
Blueprint diameter = 1 cm on blueprint per every 0.8 mm on the microchip:
| = |  (40 mm diameter on microchip) |
|
| = | (1 cm on blueprint) |
|
| = | (1 cm on blueprint) |
|
| = | 50 cm on blueprint |
10. (D). If PV = kT, then 
. Quantity A is 
.
If PV = kT, then 
. Quantity B is 
.
Don’t rush to judgment, thinking that 780 > 
means that Quantity B is greater. Notice that the k term is in the numerator of one quantity (so Quantity A increases with k) and the denominator of the other (so the larger k is, the smaller Quantity B is).
If k = 1, then Quantity B is greater 
. But if k = 100, Quantity A is greater (160 > 7.8). The relationship cannot be determined from the information given.
11. (C). Since AC is a diameter of the circle, triangle ABC is a right triangle and angle ABC is a right angle. This means that angle CAB is 60°, and the ratios of a 30–60–90 triangle can be used to solve the problem.
The circumference of the circle, πd = 
, so the diameter, which is also AC, is 
.
Now use ratios (specifically via the unknown multiplier x to find AB):
Line segment AB has a length of 
.
Alternatively, you could use estimation here if you forgot the ratios for 30–60–90 triangles. Since the longest side is always opposite the largest angle (and the shortest opposite the smallest), the sequence of sides must be AC > BC > AB. The diameter AC = ≈ 12(1.7) ≈ 20.4. From the looks of it, AB is about of AC’s length (definitely not close to the length of AC), so answer choices (D) and (E) seem too long. Choices (A) and (B) are each less than 
of AC, which doesn’t seem long enough.
12. (B). Profit equals revenue minus cost. The company’s profit is:
| p(9 – p) – (p + 15) | = | 9p – p2 – p – 15 |
| = | –p2 + 8p – 15 | |
| = | –(p2 – 8p + 15) | |
| = | –(p – 5)(p – 3) |
Profit will be zero if p = 5 or p = 3, which eliminates answers (A) and (C). For p > 5, both (p – 5) and (p – 3) are positive. In that case, the profit is negative (i.e., the company loses money). The profit is only positive if (p – 5) and (p – 3) have opposite signs, which occurs when 3 < p < 5.
Alternatively, plug in the answer choices to see which value corresponds to a revenue that is higher than cost.
(A) Cost = 18, Revenue = 18, Profit = 0 Incorrect
(B) Cost = 19, Revenue = 20, Profit = 1 Correct
(C) Cost = 20, Revenue = 20, Profit = 0 Incorrect
(D) Cost = 21, Revenue = 18, Profit < 0 Incorrect
(E) Cost = 22, Revenue = 14, Profit < 0 Incorrect
13. 
and only. The sum 
has 20 fractional terms. It would be nearly impossible to compute if you had to find a common denominator and solve without a calculator and a lot of time. Instead, look at the maximum and minimum possible values for the sum.
Maximum: The largest fraction in the sum is 
; k is definitely smaller than 20 × , which is itself smaller than 20 × 
= .
Minimum: The smallest fraction in the sum is 
; k is definitely larger than 20 × = .
Therefore, < k < .
| I. |  < < k |
YES |
| II. | < k |
YES |
| III. | > k |
NO |
14. (A). Draw a diagram to illustrate the moment at which car A and car B pass each other moving in opposite directions:
You could test the answer choices:
Or solve algebraically, using an RTD chart. Convert 15 minutes to (or 0.25) hours:
Set up and solve an equation for the total distance:
| (0.75)(r + 8) + (0.25r) | = | 62 |
| 0.75r + 6 + 0.25r | = | 62 |
| r | = | 56 |
Therefore, car B traveled a distance of 0.25r = (0.25)(56) = 14 miles.
15. (D). Factor 10,125 to its prime factors: 10,125 = 3453.
So, x25y = 3453.
In order to have 53 on the right side, there have to be three factors of 5 on the left side. All three could be in the 5y term (i.e., y could equal 3). Or, one of the 5’s could be in the 5y term, and two of the 5’s in the x2 term (i.e., y could equal 1 and x could have a single factor of 5).
In order to have 34 on the right side, x2 must have 34 = (32)2 as a factor. In other words, x must have 32 as a factor, because 32 is certainly not a factor of 5. Thus, x is a multiple of 9.
The possibilities:
In one case, Quantity A is greater. In the other, Quantity B is greater. The relationship cannot be determined from the information given.
16. (D). Since there are variables in the answer choices, pick a number and test the choices. If n = 2, then 
, which is greater than 2 (around 2.86). Now test the answer choices to see which one matches the target:
| (A) –1 | Too low |
| (B) 1 | Too low |
(C)  |
Too high |
(D)  |
Correct |
(E)  |
Too low |
Alternatively, solve this problem algebraically. The expression is in the form of 
, where a = 
and b = 
.
Either simplify or cancel the denominator, as none of the answer choices have the same denominator as the original, and most of the choices have no denominator at all. To be able to manipulate a denominator with radical signs, first try to eliminate the radical signs entirely, leaving only a2 and b2 in the denominator. To do so, multiply by a fraction that is a convenient form of 1:
Notice the “difference of two squares” special product created in the denominator with the choice of (a + b).
Substitute for a and b:
The correct answer is (D).
17. (A). First, note the answer pairs (A) and (C), and (B) and (E), in which one ratio is the square of the other. This represents a likely trap in a problem that asks for the ratio of 
to 
rather than the more typical ratio of x to y. It is fairly safe to eliminate (D) as it is not paired with a trap answer and therefore probably not the correct answer. You should also suspect that the correct answer is (A) or (B), the “square root” answer choice in their respective pairs.
For problems involving successive changes in amounts—such as population growth problems, or compound interest problems—it is helpful to make a table:
| Account A | Account B | |
| Now | x | y |
| After 1 month |  |
 |
| After 2 months |  |
 |
If the accounts have the same amount of money after two months, then:
 |
= |  |
| 81x | = | 144y |
This can be solved for 
:
18. (A). First, note that the height of each person in question is fixed (no one grew taller or shorter); only weights changed. Second, note that BMI is always positive and is proportional to w; as weight increases, BMI increases, and vice versa. So the language of the quantities—“pounds gained … BMI increased” and “pounds lost … BMI decreased”—is aligned with this proportionality. Both quantities are a positive number of pounds.
Since BMI = 
, change in BMI = 
.
To simplify things, you can write this in terms of ΔBMI and Δw, the positive change in BMI and weight, respectively:
(The triangle symbol indicating positive change in a quantity does not appear on the GRE—it is used here for convenience in notating an explanation.)
Since the quantities both refer to Δw, rewrite the relationship as 
. Both ΔBMI and h are given in each quantity, so Δw can be calculated and the relationship between the two quantities determined. (The answer is definitely not (D).)
Quantity A |
Quantity B |
Since the 703 in the denominator is common to both quantities, the comparison is really between 742 = 5,476 and 652(1.2) = 4,225(1.2) = 5,070. Quantity A is greater.
19. (D). There are four different cases that you must count: 5 7 _ _ _ ; _ 5 7 _ _ ; _ _ 5 7 _; and _ _ _ 5 7. In the case of 5 7 _ _ _, all three of the empty spaces can have any digit from 0–9, which is 10 possibilities, for a total of 10 × 10 × 10 = 1,000 possible numbers. In the case of _ 5 7 _ _, there are only 9 choices for the first digit since you cannot put a zero there if it is to be a five-digit positive integer. For the last two digits any number from 0–9 is still allowed, for a total of 9 × 10 × 10 = 900 possible numbers for the second case. The third and fourth cases are similar to the second; both include 900 possible numbers. Summing up the four cases, there are 1,000 + 900(3) = 3,700 such integers. This result double-counts, however, the cases with two 57’s in them: 5757X, 57X57, and X5757. The first two cases yield 10 numbers each, while the last case yields only 9 (because 05757 doesn't count as a 5-digit number, by convention). Subtracting these 29 numbers, you get 3,671.
20. (A). To find the average of any evenly spaced set, take the average of the first and last values. For the case of Quantity A, the first multiple of 5 is 200 and the last multiple of 5 is 705. The average then is 
. For Quantity B, the first multiple of 10 is also 200, however, the last multiple is 700, thus the average is 
. Quantity A is greater.
21. (C). Set up an implied inequality and perform identical operations on each quantity, grouping variables:
| Quantity A | Quantity B | |
| a + d – c – 90 | ? | 90 – e – b – f |
| a + d – c | ? | 180 – e – b – f |
| a + d – c + e + b + f | ? | 180 |
| (a + d + e + f) + (b – c) | ? | 180 |
In the last step above, only the order of the variables was changed and parentheses added to group certain terms. Notice that the angle at point C and point D is the same, as AC and FD are parallel lines intersected by the transversal CE. So, the first set of parentheses holds the sum of the interior angles of the biggest triangle ACE, which is 180. Also because AC and FD are parallel lines intersected by transversal BE, b = c, so b – c = 0 in the second set of parentheses.
| Quantity A | Quantity B | |
| (a + d + e + f) + (b – c) | ? | 180 |
| (180) + (0) | = | 180 |
The two quantities are equal.
22. (D). Given that the man can only move north and east, he must advance exactly 7 blocks from his current location to get home regardless of which path he takes. Of these 7 blocks, 4 must be moving east and 3 must be moving north. An example path is shown below:
The problem can then be rephrased as follows: “Of the 7 steps, when does the man choose to go east and when does he choose to go north?” Labeling each step as N for north and E for east, you can see the problem as the number of unique rearrangements of NNNEEEE (e.g., this arrangement corresponds to going north 3 times and then east 4 times straight to home). This is determined by 
.
23. (D). There are 2 red marbles and 7 not-red marbles. (It is irrelevant whether white or black is drawn; the question is about red or not.) There are two ways in which the second marble drawn is red: either not-red first, then red second OR red first, and then red again. Using P(R) to indicate the probability of drawing red and P(not R) to indicate the probability of drawing not-red, we have:
Since a red on the second draw could happen one way or the other, sum these probabilities:
.
24. (A). Simplify both quantities, remembering that a power to a power means multiply the exponents. Also, 25 is 5 squared, so you can substitute, putting both quantities in terms of a base of 5:
Quantity A: ((25x)–2)3 = 25–6x = (52)–6x = 5–12x
Quantity B: ((5–3)2)–x = 56x
Typically, when comparing exponents with the same base, the one with the larger exponent is greater. It might be tempting to conclude that 6x > –12x, but be careful with negative variables.
If x = –1, Quantity A = 512 and Quantity B = 5–6, or 
. In this case, Quantity A is much greater.
If x = –, Quantity A = 56 and Quantity B = 5–3, or 
. Again, Quantity A is much greater.
If x = –10, Quantity A = 5120 and Quantity B = 5–60. The more negative x gets, the greater the difference between Quantity A and Quantity B becomes. Quantity A is always greater.
Another way to look at it:
Quantity A: 5–12x = 5–12×negative = 5positive
Quantity B: 56x = 56×negative = 5negative
Even if |x| is a tiny fraction, that is, the expression is some high order root of 5 such as 
or 
, these quantities would approach 1 such that Quantity A is 5positive > 1 and Quantity B is 5negative < 1.
Since Quantity A is greater than 1 and Quantity B is less than 1, Quantity A is greater.
25. (A). First, note that some of the positive multiples of 6 in Quantity A are canceled out by the negative multiples of 6 in Quantity A (e.g., –126 + 126 = 0, –120 + 120 = 0, etc.).
To find the sum of the remaining multiples of 6 (i.e., 132 through 342, inclusive), find both the number of terms and the average of those terms: sum = (average) × (number of terms).
To find the number of terms, take the last multiple of 6 minus the first multiple of 6, divide by 6 and then add 1:
| Number of terms (multiples of n) | = |  |
| = |  |
To find the average of any evenly spaced set, take the average of the first and last values:
Therefore, the sum is equal to (Average) × (Number of terms) = (237)(36) = 8,532. Quantity A is greater.
26. (B). When a negative base is raised to an integer power, the question is about positives and negatives and odds and evens: (–1)odd = –1 and (–1)even = +1.
If x is even, all of the exponents in this question are even:
Quantity A: (–1)even2 + (–1)even3 + (–1)even4 = (–1)even + (–1)even + (–1)even = 1 + 1 + 1 = 3
Quantity B: (–1)even + (–1)2 × even + (–1)3 × even + (–1)4 × even = 1 + 1 + 1 + 1 = 4
If x is odd, some of the exponents in this question are odd:
Quantity A: (–1)odd2 + (–1)odd3 + (–1)odd4 = (–1)odd + (–1)odd + (–1)odd = (–1) + (–1) + (–1) = –3
Quantity B: (–1)odd + (–1)2 × odd + (–1)3 × odd + (–1)4 × odd = (–1)odd + (–1)even + (–1)odd + (–1)even
= (–1) + 1 + (–1) + 1
= 0
In both cases, Quantity B is greater than Quantity A.
27. (B). Because an exterior angle of a triangle is equal to the sum of the two opposite interior angles of the triangle (in this case, the top small triangle), c = a + b.
Therefore, d > c and a + b = c taken together imply that d > a + b.
Subtract b from both sides: d – b > a.
Quantity B is greater.
28. (A). This problem introduces a square and a circle, and states that the circumference of the circle is 
the perimeter of the square.
This is license to plug in. Both a square and a circle are regular figures—that is, all squares are in the same proportion as all other squares and all circles are in the same proportion as all other circles—plugging in only one set of values yields the same result as would plugging in any set of values. Because the figures are regular and related in a known way (circumference is equal to × square perimeter), there is no need to repeatedly try different values as is often necessary on Quantitative Comparisons.
Because the circumference of a circle depends on π(C = πd), it is best to pick values for the square. If the side of the square is 2, the perimeter is 4(2) = 8 and the area is (2)(2) = 4. Then, circumference of the circle is ()(8) = 7. Since circumference is 2πr = 7, the radius of the circle is r = 
.
Using these numbers:
Quantity A: The area of the square = 4.
Quantity B: The area of the circle = πr2 = 
.
(Use the calculator and the approximation 3.14 for π to determine that Quantity A is greater.)
29. (A). One good way to work through this problem is to pick a number, ideally starting with the innermost shape, the small circle. Say this circle has radius 1 and diameter 2, which would also make the side of the smaller square equal to 2.
If the small square has side 2, its diagonal would be 
(based on the 45–45–90 triangle ratios, or you could do the Pythagorean theorem using the legs of 2 and 2). If the diagonal is , then the diameter of the larger circle is also (and the radius of the larger circle is one-half of that, or ), making the side of the larger square also equal to . Therefore:
Small circle: radius = 1, area = π
Large circle: radius = , area = 2π
Small square: side = 2, area = 4
Large square: side = , area = 8
Thus, the large circle has twice the area of the small circle, and the large square has twice the area of the small square. This will work for any numbers you choose. In fact, you may wish to memorize this as a shortcut: if a circle is inscribed in a square that is inscribed in a circle, the large circle has twice the area of the small circle; similarly, if a square is inscribed in a circle that is inscribed in a square, the large square has twice the area of the small square.
In Quantity A, the ratio of the area of the larger square to the smaller square is 
= 2.
In Quantity B, twice the ratio of the area of the smaller circle to the area of the larger circle is equal to 2
= 1.
30. (A). This problem is not as bad as it looks! Of course, the integers are much too large to fit in your calculator. However, all you need to know is that a pair consisting of one 2 and one 5 has a product of 10 and therefore adds a zero to the end of a number. For instance, a number with two 2’s and two 5’s in its prime factors ends with two zeros, because the number is a multiple of 100.
Quantity A has nineteen 5’s and many more 2’s (since 216 and 418 together is more than nineteen 2’s—if you really want to know, it’s 216 and (22)18, or 216 and 236, or 252, or fifty-two 2’s). Considering pairs made up of one 2 and one 5, exactly nineteen pairs can be made (the leftover 2’s don’t matter), and the number ends in 19 zeros.
Quantity B has sixteen 5’s and many more 2’s (specifically, there are fifty-three 2’s, but there’s no need to calculate this). Considering pairs made up of one 2 and one 5, exactly sixteen pairs can be made (the leftover 2’s don’t matter), and the number ends in 16 zeros. Thus, Quantity A is greater.
31. (A). Calculate several terms of the sequence defined by an = 2n – = 
and look for a pattern:
| a1 | = | 21 –  = 21 – 232 |
| a2 | = | 22 –  = 22 – 231 |
| … | ||
| a16 | = | 216 –  = 216 – 217 |
| a17 | = | 217 –  = 217 – 216 |
| … | ||
| a31 | = | 231 –  = 231 – 22 |
| a32 | = | 232 –  = 232 – 21 |
Notice that the 16th and 17th terms (the two middle terms in a set of 32 terms) are arithmetic inverses, that is, their sum is zero. Likewise, the 1st and 32nd terms sum to zero, as do the 2nd and 31st terms. In the first 32 terms of the sequence, there are 16 pairs that each sum to zero. Thus, Quantity A is zero.
For the sum of the first 31 terms, you could either:
- Subtract a32 from the sum of the first 32 terms: 0 – (232 – 21) = 21 – 232 = 2 – (a very large number) = negative, or
- realize that in the first 31 terms, all terms except a1 can be paired such that the pair sums to zero, so the sum of the first 31 terms = a1 = 21 – 232 = 2 – (a very large number) = negative.
Thus, Quantity B is negative, which is less than zero. Quantity A is greater.
32. (C). The “interquartile range” of a group of 100 integers is found by splitting the 100 integers into two groups, a lower 50 and an upper 50. Then find the median of each of those groups. The median of the lower group is the first quartile (Q1), while the median of the upper group is the third quartile (Q3). Finally, Q3 – Q1 is the interquartile range.
The median of a group of 50 integers is the average (arithmetic mean) of the 25th and the 26th integers when ordered from smallest to largest. Order the list of 100 integers from smallest to largest, then, find #25 and #26 and average them to get the first quartile. Likewise, find #75 and #76 and average them to get the third quartile. Then perform the subtraction.
| x = the integer label on the ball | nx = 18–(x – 4)2 = the number of balls with this label |
Cumulative number of balls |
| 1 | 18 – (1 – 4)2 = 18 – (– 3)2 = 18 – 9 = 9 | 9 |
| 2 | 18 – (2 – 4)2 = 18 – (– 2)2 = 18 – 4 = 14 | 9 + 14 = 23 |
| 3 | 18 – (3 – 4)2 = 18 – (– 1)2 = 18 – 1 = 17 | 23 + 17 = 40 |
| 4 | 18 – (4 – 4)2 = 18 – (0)2 = 18 | 40 + 18 = 58 |
| 5 | 18 – (5 – 4)2 = 18 – (1)2 = 18 – 1 = 17 | 58 + 17 = 75 |
Stop here. Ball #75 has a 5 on it (in fact, the last 5), while ball #76 must have a 6 on it (since 6 is the next integer in the list). Thus, the third quartile Q3 is the average of 5 and 6, or 5.5. Count carefully—if you are off by even just one either way, you’ll get a different number for the third quartile. Balls #25 and #26 both have a 3 on them. So the first quartile Q1 is the average of 3 and 3, namely 3.
Finally, Q3 – Q1 = 5.5 – 3 = 2.5.
33. (C). Compute the expressions for each of the terms:
x!4 = x4 × 4–x and 4!x = 4x × x–4
Dividing the first by the second yields:
There are a number of ways to write x84–2x:
The two quantities are equal.
34. (C). First find each of the sums. To find a sum of an evenly spaced set, use the formula:
sum = (average) × (number of terms)
For the odd positive integers between 1 and 100, inclusive (use “2” as the multiple; while odds are not multiples of two per se, they are evenly spaced every two numbers):
| Number of terms (odds) | = |  |
| = |  |
|
| = | 50 |
To find the average of any evenly spaced set, take the average of the first and last values:
Average (odds) = 
Sum of the odd integers between 1 and 100, inclusive is equal to (average) × (number of terms) = 50 × 50 = 2,500.
For the even positive integers between 100 and 150, inclusive:
| Number of terms (evens) | = |  |
| = |  |
|
| = | 26 | |
| Average (evens) | = |  |
Therefore, the sum of the even integers between 100 and 150, inclusive is equal to (average) × (number of terms) = 125 × 26 = 3,250.
The ratio of the sum of the odd positive integers between 1 and 100, inclusive, to the sum of the even positive integers between 100 and 150, inclusive is equal to 
35. (C). The problem gives two ways to calculate the fourth term: (1) the definition of the sequence tells you that a4 = a32 – a22 and (2) the words indicate that a4 = a1(a2 + a3) = 2(a2 + a3). Setting these two equal gives a32 – a22 = 2(a2 + a3). Factor the left side: (a3 + a2)(a3 – a2) = 2(a2 + a3). Since an > 0 for all possible n’s, (a3 + a2) does not equal 0 and you can divide both sides by it: a3 – a2 = 2 and a3 = a2 + 2. Using the definition of a3, you know a3 = a22 – a12 = a22 – 4. Substituting for a3 yields: a2 + 2 = a22 – 4 and a22 – a2 – 6 = 0. Factor and solve: (a2 – 3)(a2 + 2) = 0; a2 = 3 or –2. an must be positive, so a2 = 3 and a3 = a2 + 2 = 3 + 2 = 5.
36. (D). Since a1 and a3 are integers, a2 must also be an integer: a3 = 
or INT = 
so 2(INT) = INT + a2 and a2 = 2(INT) – INT, which is itself an integer. a4 is thus the average of two integers. If a2 + a3 is even, a4 will be an integer. If a2 + a3 is odd, a4 will be a decimal ending in 0.5. If a4 is an integer, a5 can be an integer or can be a decimal ending in 0.5. If a4 is a decimal ending in 0.5, a5 must be a decimal ending in 0.25 or 0.75. a5 cannot be a decimal ending in 0.375 such as 
= 9.375. Note that a5 can be negative: even if a1 and a3 are positive, that does not rule out the possibility that a2 (and subsequent terms) could be negative.
37. (D). Use the definition of @ to rewrite the equation: 
. Simplifying yields: 
. Let z = 
. Substitute z into the equation: z – 
= ()z or z = 3. To solve = 3, take two cases:
> 0, so = 3 or x = 0.5.- < 0, so = –3 or x = –0.25.
The sum of the solutions is 0.5 + (–0.25) = 0.25.
38. (B). Expressed algebraically, 
> x. Because both sides of this inequality are non-negative, you can square both sides to result in the following:
| 10 – 3x | > | x2 |
| 0 | > | x2 + 3x – 10 |
| 0 | > | (x + 5)(x – 2) |
Now, because the product of (x + 5) and (x – 2) is negative, you can deduce that the larger of the two expressions, (x + 5), must be positive, and the smaller expression, (x – 2), must be negative. Therefore, x > –5 and x < 2. Combining these yields –5 < x < 2.
However, because the question indicates that x is non-negative, x must be 0 or greater. Therefore, 0 ≤ x < 2. The absolute value sign in Quantity A doesn’t change anything—x is still greater than or equal to zero and less than 2, and Quantity B is larger.
Alternatively, plug the value from Quantity B into > x:
 |
> | 2 |
 |
> | 2 |
| 2 | > | 2 |
This is FALSE, so x cannot be 2.
Now, plug in a smaller or larger value to determine whether x needs to be greater than or less than 2. If x = 1:
> 1
> 1
is between 2 and 3, so this is true.
Trying values will show that only values greater than or equal to zero and less than 2 make the statement true, so Quantity A must be less than 2.
39. (A). The area of an equilateral triangle is 
where b is the length of one side. Since this area is between 25 
and 
, substitute to get 
< < . Dividing all sides by yields 25 < 
< 36. Multiplying all sides by 4 yields 100 < b2 < 144, and taking the square root of all sides yields 10 < b < 12. Since every possibility for b is greater than 9, Quantity A is greater.
40. (E). When dealing with absolute values, consider two outcomes. First determine the outcome if the expression within the absolute value sign is positive. So, if 8 – 2x > 0, then |8 – 2x| = 8 – 2x, and therefore 8 – 2x < 3y – 9 or 2x > 17 – 3y.
You also must determine the outcome if the expression within the absolute value sign is negative. So if 8 – 2x < 0, then |8 – 2x| = 2x – 8, and therefore 2x – 8 < 3y – 9 or 2x < 3y – 1. Combining these two inequalities yields 3y – 1 > 2x > 17 – 3y.
Now a quick sanity check to make sure the inequality makes sense: 3y – 9 must be greater than 0 or the absolute value could not be less than 3y – 9. So y > 3. This means 17 – 3y < 8, and 3y – 1 > 8, so there is definitely room for 2x to fit between those values. If the potential values of 17 – 3y and 3y – 1 had overlapped, this would be an indication either that a mistake had been made or that the problem required further investigation to refine the result.
41. (A). This is an overlapping set problem. Matrix 1 shows an initial setup for a double-set matrix. The columns are headed “Skilled in BJJ” and “Not Skilled in BJJ.” The rows are headed “Skilled in Muy Thai” and “Not Skilled in Muy Thai.” There is also a total row and a total column.
When dealing with overlapping sets, consider whether the question is giving information regarding the population as a whole or regarding a subset of the population. While the first statement (“30% of all fighters”) refers to the whole population, the second statement (“20% of the fighters who are not skilled in Brazilian Jiu Jitsu”) refers to a subset of the population, in this case the 40% who are not skilled in Brazilian Jiu Jitsu. Thus, 8% are skilled in Muy Thai but not in Brazilian Jiu Jitsu, as seen in Matrix 1:
Matrix 2 shows how to fill out additional cells. Notably, there are some ranges of values that are possible for the cells in the first column. These ranges are limited by 0 on the low end and 60 on the high end:
Matrix 3 sums across the rows to get subtotals. Particularly, the percent of fighters who are skilled in Muy Thai is greater than 38 but less than or equal to 68. Thus, Quantity A is greater:
42. (B). Express this situation with the equation 
, where k is a constant. Quadrupling b and more than tripling n yields the following equation: r1 = 
, where r1 represents the new rate of data transfer.
Squaring a value that is greater than 3 produces a value that is greater than 9, allowing the equation to be rewritten as r1 = 
. Rearranging this equation yields r1 = 
= less than 
. Thus, Quantity B is greater.