GRE Quan. Practice Problems

How many seconds does it take to travel 180 miles at 60 miles per hour?

Use the speed formula and apply it to this rate with the three numbers to make the following equation based on Speed = Distance/Time. (Recall that speed = rate and distance = quantity). 60 miles/1 hour = 180 miles/X hours.

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In a class, 1/3 of the students play a sport and 1/5 of the remaining students play a musical instrument. What could be the number of students in the class? a. 10 b. 15 c. 18 d. 20 e. 25 f. 30 g. 33

Yet another case of translating the words into variables to make a ratio. N is the total number of students, S is the sport players, M is the instrument players. S equals N/3 and M would equal (N-S)/5. M can be simplified to M = (2N)/15. Since S equals N/3, that means all even answers or non-divisible by 2 numbers are eliminated. Since M equals (N-S)/5, 2N has to be divisible by 15, further eliminating answer choices. Only ones left should be b and f.

Manjeet is putting up a fence around his rectangular yard. The length of his yard is 24 feet. If the yard has an area of 768 square feet, how many feet of fencing does he need?

You are given the length and the area of the rectangular yard, and you are asked for the perimeter. Begin by calculating the area's width: width = (area/length) = 768/24 = 32 feet Thus, the perimeter is twice the length of 24 feet plus twice the width of 32 feet: 112 feet.

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Answer is C

A circular pie is placed in a square box whose sides are 12 inches long. The pie touches the box on all four sides. What is the area, in square inches, of the space in the box not covered by the pie?

Begin by finding the area of the square and the area of the circle individually, and then subtract the area of the circle from the area of the square. Notice that the diameter of the circle equals the side of the square (12 inches). Therefore, the radius of the circle is 6 inches. The area of the square equals 122, that is, 144. The area of the circle equals π62, that is, 36π Thus, the area of the uncovered space equals 144 - 36π.

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Answer is C. Check full screenshot for more info.

Quantity A: The number of hours in a week Quantity B: The number of days in six consecutive months

Calculating column A is straightforward: there are 24 hours in a day and 7 days in a week, so there are 24 times 7, or 168, hours per week. Now, not all months have the same number of days, but you can at least calculate a minimum value for quantity B. Assume that each month has 28 days. Then, six months have 28 times 6, or 168, days. However, in reality every month except one has more than 28 days, so the number of days in 6 consecutive months must be greater than 168. Quantity B is greater.

A teacher graded six tests, averaging 80. If fives of the tests received grades of 64+76+81+87+94, what was the sixth grade.

X is the sixth test grade. Set up an equation with the known test values on one side equaling 80 (on the other side). (64+76+81+87+94+x)/6 = 80 x = 78

For all numbers a and b, the operation # is defined by a # b = 3a(b - 2). If x # (-4) = 36, then what is (-4) # x?

Solve x # (-4) = -36 for x: x # (-4) = -36 3x(-4-2) = -36 3x(-6) = -36 -18x = -36 x = 2 Next, solve (-4) # x, substituting 2 for x: (-4)#x = -4#2 =3(-4)(2-2) =0

j = 4k If J is equal to 4k, then 50% of j is 2k. 100% of k is k.

No values defined for k and j. Therefore, cannot be determined.

Quantity A: k/(10^22) Quantity B: k/(10^21)

Relationship cannot be determined because k has NO set value or range of values.

In the school band, 40 students play a stringed, brass, or percussion instrument. Two-fifths of the students play a stringed instrument. Three more than half the number of students who play a brass instrument play a percussion instrument. Quantity A: The number of students who play a percussion instrument Quantity B: 14

1.) Determine how many students play stringed instruments, and subtract that number from the total. Two-fifths of 40 is 16, and 40 minus 16 equals 24. Thus, the number of brass and percussion players is 24. Next, let b be the number of brass players and p be the number of percussion players. You can now set up an equation relating b to p: p = 1/2b + 3. Since the total number of these two groups is 24, you have a second equation: b + 1/2b + 3 = 24. Solving for b, you find that b equals 14. In other words, there are 14 brass players, which means that there are 10 percussion players, Quantity B is greater.

(15³-15²)/15² = ?

14/1. Simplify the fraction. Begin by factoring 15² out of each term in the numerator. (15²(15-1))/15² = 15 - 1 = 14

Each card in a certain deck of cards is 164 of an inch thick. If there are 122 cards in the deck, how thick is the deck, in inches?

Simply multiply the thickness of each card, in inches, by the number of cards by the to get the total thickness, in inches. (1/64)×122=122/64=61/32=1(29/32)

Quantity A: The number of faces of a cube that are perpendicular to its top face. Quantity B: 2

A cube has six faces. Two of them- the top and bottom ones- are parallel to each other. The other four are perpendicular to the first two. Thus, quantity A equals 4, so its greater than quantity B. Therefore, the answer is A.

In a certain movie theater, the center rows are 7 seats across, with aisles on either side. A group of 6 friends goes to the movies and finds an empty center row to sit in. One of the friends needs to sit in a seat along the aisle. Which of the following represents the number of different possible seating arrangements for the 6 friends?

A permutation problem. One friend has to sit in an aisle seat, labeled as A and G in this case, so they have two choices. The remaining letters between A and G are six options for the second person to choose a seat. The second seat taken over leaves 5 seats open and so on. Therefore, the total possible arrangements is 2 (coming from the aisle seat choices for person 1), then 6 times 5 times 4 times 3 times 2, which is 1,440.

If x and y are different integers and xy = (y²/x), which of the following MUST be true? a. y = -x b. y ≥ 0 c. y > 0 d. x = 1 e. y = 0

A problem where the variables needs clear values to reduce the possible answers. Solve the equation so it equals 0 on the other side. The given equation can be simplified to x²y - y² = 0. Or, after factoring, y(x² - y) = 0. x² is always positive, y will either be 0 or a positive number to remain true. So, choice B has to be true. Important note: C and E only give some possible values for y, but doesn't define x, so it must be eliminated. A and D can be eliminated for the same reason.

P (a, b), which lies in quadrant I, is a point on the graph of equation 2x - y + 4 = 0. Quantity A: -a Quantity B: -b

A slope based question. Rewrite the equation in slope intercept form, m being the equations slope. The slope, after the equation is rewritten as y = 2x + 4, is 2. Since it is quadrant I, a and b, its x- and y- coordinates are both positive. (Ask Ihab about P in relation to the equation, since QP2S2L3Q7's explanation states it's somehow relevant). If values are being tested, it means that as X increases, y increases. This means -a is greater than -b, thus Quantity A is bigger.

Working alone, Tina, Sam, and Nick can type a long document in 2, 3, and 5 hours respectively. What is the ratio of the time it takes Tina, working alone at her rate, to type the document, to the time it takes Sam and Nick, working together at their individual rates, to type the document?

A work formula based problem. This means that the problem can take Sam and Nick's work per hour and can be rewritten as 1/3 and 1/5 respectively. Adding both, it is 8/15 per hour. (Ask Ihab how QP2S2L3Q8 got the working together number.) Working together, it takes them 15/8 to finish that job. The ratio of time for Sam and Nick typing it together is: 2/(15/8) = 2 X 8/15 = 16/15

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Answer is C

The area of a square having sides of length x meters is z square meters. Quantity A: z Quantity B: x

Because x is less than 1, its square is less than x. Thus, z, which equals x², is less than x, and quantity B is greater

The area of a square having sides of length x meters is z square meters. x < 1 Quantity A: z Quantity B: x

Because x is less than 1, its square is less than x. Thus, z, which equals x2, is less than x, and quantity B is greater.

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Quantity A: (1/2x²+x) Quantity B: (1/2x³-x)

Choose different values for x in order to see the relationship between the quantities. If x = 0, then both quantities equal 0. If x = 1, quantity A equals 3/2 and quantity B equals -1/2. So, the relationship cannot be determined.

Which of the following are equal to (1/4)4 ? Indicate all correct answer choices. (1/2)^2 (1/8)^2 (1/16)^2 (8)^-2 (4)^-4 (1/2)^8

Convert each choice to compare: (A) (12)2=14≠(14)4. Answer choice (A) is not equivalent. Eliminate it. (B) There is no easy way to make the bases equivalent, so we need to calculate the answers and then compare them. Now (18)2=18×8 and (14)4=14×4×4×4. Since 8 × 8 ≠ 4 × 4 × 4 × 4, answer choice (B) is not equivalent. Eliminate it. (C) (116)2=((14)2)2=(14)4. Answer choice (C) is equivalent. (D) 8−2=182=(18)2. This is equivalent to answer choice (B), which we have eliminated. (E) 4−4=144=(14)4. Answer choice (E) is equivalent. (F) (12)8=(12)2(4)=[(12)2]4=[1222]4=(14)4. Answer choice (F) is equivalent. Answer choices (C), (E), and (F) are correct.

Five hamburgers and three sodas cost a total of $6.50. Quantity A: The cost of a hamburger Quantity B: The cost of a soda

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Quantity A: The sum of all even integers from 33 to 91, inclusive Quantity B: The sum of all odd integers from 34 to 92, inclusive

Do not waste time performing the additions. Rather, list the integers in the two quantities. The evens in quantity A are: 34, 36, 38, ... 90. The odds in quantity B are: 35, 37, 39, ... 91. Each quantity features the same number of integers, and each integer in quantity B is one greater than its corresponding integer in quantity A. Thus, quantity B is greater.

(c+3)(d-2) = 0 Quantity A: c Quantity B: d

Even though c has to equal -3 and/or d has to equal 2, it DOES NOT mean both need to be that number. C and D can be anything else as long as one () equals 0. So, it is indeterminate.

The average score for a class on a certain test was 76. Among the female students, the average score was 84. Among the male students, the average score was 64. What is the ratio of female to male students in the class?

F = number of female students in the class. M = number of male students. Since M and F are the only students in the class, M + F = ? of students. Setting up the average equation, T/(M+F) = 76. T would need to be total score. For the groups; female scores Y/F = 84. For male scores, Z/M = 64. T would need to equal, after getting the unknown scores for each group to one side, Y = 84F, Z = 64M. Plugging it all back in the average equation is (84F + 64M)/F + M = 76. This new equation ensures the scores can be removed using algebra to just get the F:M ratio. Solving until the fractional ratio, F/M = 3/2.

xy ≠ 0 Quantity A: (x+y)² Quantity B: x² + y²

FOIL quantity A to get x² + 2xy + y. Regardless, it is immediately unknown whether x and y are positive or negative so the relationship cannot be determined.

Quantity A: The greatest odd factor of 120. Quantity B: The greatest even factor of 120 that is less than 24.

Factor 120 using the 'branching' technique. i.e. make pairs starting with 1 and 120, then from 120 go down to 60 and 2, etc. The greatest odd factor is 15. Greatest even factor less than 24 is 20.

x² + 2x - 24 = 0 y² - 2y - 24 = 0 Quantity A: The sum of the possible values of x. Quantity B: The sum of the possible values of y.

Factor both equations. First one: (x+6)(x-4) = 0. Second one: (y-6)(y+4) = 0. Branching method for 24 can work in this case. Inputting the possible values that are true for both equations, quantity B is greater.

Kelly's job pays $10 per hour plus $0.40 per mile traveled. Which of the following could be Kelly's salary for one day's work? a. $62.50 b. $74.90 c. $83.60 d. $86.00

Find out what is consistent in Kelly's salary. Or, even what CANNOT be consistent. So, select C and D because they are the only even multiples.

A computer password must be 5 characters long. The first character must be a capital letter. The second character must be a digit from 0 through 9, inclusive. The third character must be one of eight specified symbols. Each of the fourth and fifth characters can be any combination of capital or lowercase letters, digits, or symbols, where the symbols are from the 8 specified symbols. How many different passwords can be made using these rules?

First, we need to determine how many choices are available for each of the 5 characters. The first character must be a capital letter, so there are 26 letters to choose from. The second character must be a digit from 0 through 9, inclusive, so there are 10 digits to choose from. The third character must be one of 8 symbols. The fourth and fifth characters must be a capital or lowercase letter (52 total), a digit from 0 to 9 (10 total), or a symbol (8 total), for a total of 52 + 10 + 8 = 70 choices each. This results in a total of 26 × 10 × 8 × 70 × 70 = 10,192,000 possible passwords. Choice (D) is correct.

M, N, and O are positive integers. M + N is odd. M + O is odd. N × O is even. Which of the following MUST be true? a. M is even b. N is even c. O is odd d. M x O is odd e. M - N is positive

In order to answer this question, you may have to test several possibilities, as none of the information you are given allows you to figure out immediately whether the various integers are even or odd. Consider the first piece of information: M + N is odd. This means that either M is even and N is odd or vice versa. Suppose first that M is even. Then O is odd, since M + O is odd. That would make N × O odd, since N and O are odd. However, you are told that N × O is even, so the assumption that M is even and N odd was incorrect. Thus, the only alternative—that M is odd and N is even—is correct. This means that answer choice B is correct, and choice A is incorrect. Incidentally, because M is odd and M + O is odd, O is even. Thus, choice C is incorrect. Next, because O is even, M × O must be even, so choice D is incorrect. Finally, you don't know whether M is greater than N or not, so choice E need not be correct.

t > 0 Quantity A: The area of a square with a side length of t√5 Quantity B: The area of a right triangle with sides of lengths 3t, 4t, and 5t

Let's begin with Quantity A. The area of a square is its side squared. Quantity A, which is the area of a square with a side of length t5‾√ is (t5‾√)2=t5‾√×t5‾√=t×t×5‾√×5‾√=t2×5=5t2. Now let's look at Quantity B. Notice that this right triangle is a 3:4:5 right triangle. The area of a right triangle is equal to 12 the product of the legs. The hypotenuse of a right triangle is the longest side, so the hypotenuse of this right triangle has length 5t and the legs of this right triangle have lengths 3t and 4t. Quantity B, which is the area of a right triangle with sides of lengths 3t, 4t, and 5t, is 12×3t×4t=6t2. We're comparing Quantity A, which is 5t2, with Quantity B, which is 6t2. Since t > 0, t ≠ 0. The square of any nonzero number is positive. Here, t2 is positive. So 5t2 < 6t2 and Quantity B is greater. Choice (B) is correct.

After a class action lawsuit, 10% of the settlement money is given to the lawyers. IF 45 claimants split the rest of the money evenly, and each receives $3,000, what was the total amount of the settlement?

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Amanda is choosing photos to display in 2 frames. Each frame holds 4 photos. She is choosing from a number of family photos to arrange in the first frame and a number of vacation photos to arrange in the second frame. Which numbers of family photos and vacation photos would result in more than 500,000 ways to arrange the photos in the frames? Indicate all that apply. a. 5 family photos and 9 vacation photos b. 6 family photos and 8 vacation photos c. 7 family photos and 7 vacation photos d. 10 family photos and 4 vacation photos

Only (B) and (C) have more than 500,000 possibilities, so the correct answers are choices (B) and (C).

A math professor predicted a 20% increase in the number of points scored by his students on the second test over that of the first test. However, the number of points scored on the second test decreased by 5%. Approximately what percent of the predicted number of points scored was the actual number of points scored on the second test? Give your answer to the nearest integer.

P = number of predicted first test points. For second test, 1.20P. Actual number of points on 2nd test is 0.95. Input both numbers in the percent decrease formula to acquire (0.95P/1.20P) times 100%. Which equals 79%.

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Parallel lines k and m are cut by a transversal, so the angles marked x° and y° are equal. So x = y. Because x = 30, Quantity A is 180 - x = 180 - 30 = 150, and Quantity B is 30 because y = x = 30. Therefore, since Quantity A is 150 and Quantity B is 30, answer choice (A) is correct

The price of a sweater is increased by x percent. Then the sweater is put on sale for x percent off. Quantity A: The original price of the sweater Quantity B: The price of the sweater after it is put on sale

Quantity A is greater. Pick numbers, with the cost being a static value and x being a percentage. Then, it'll be found the original price is higher.

In a taste test matching soft drink P against soft drink Q, each participant chose either P or Q, and two fifths more people chose P than Q. Quantity A: The fraction of test participants who chose P. Quantity B: 7/12

Set up a rate. Create a proportion that can represent the participants that chose P. P = participants that chose P. T = total # of participants. In this case, P/T. T equals P + Q in this case. Can be rewritten as P/(P+Q). With the other info, it's known that P = 7Q/5. Substituting this value in after Q is on the other side, the original equation is now... P/(P + 5P/7) = 1/ (1 + 5/7) = 1/(12/7) = 7/12. So, both quantities are equal. TAKEAWAY: Break down each part of the problem into an equation before finding any sort of relation.

Quantity A: 0.9 Quantity B: √0.9

Since both quantities are positive, you can square them both in order to compare them. Quantity A becomes 0.81, while quantity B becomes 0.9. B is greater.

The probability is img/img1268.png that a red marble will be selected from a bag. If a marble is selected and replaced three times, what is the probability that a red marble will never be selected? a. 1/27 b. 2/27 c. 8/27 d. 1/3 e. 2/3

The correct answer is C. On any one selection, the probability that red will not be selected is 1 minus 1/3 or 2/3. The probability that red will not be selected three times equals 2/3 to the third power, or 8/27.

The operation x ∩ y is defined for all values of x and y by the equation x ∩ y = (3x²+y)/(4xy). What is the value of 5 ∩ (-2 ∩ 3)?

This rule largely follows PEMDAS. Start with the equation in the parenthesis first, which should equal -5/8. The whole value of the formula can be calculated using 5 as x and -5/8 as y. This should equal -5.95.

In 2007, the value of a house was 3/4 of the original purchase price. In 2011, the price was 2/3 the original purchase price. By approximately what percent did the value of the house decrease from 2007 to 2011?

V = original price of the house. In relation to 2007 and 2011, that means 2007 = 3V/4 and 2011 = 2V/3. Applying the decrease formula, this means that the decrease amount can be found by subtracting 2007 from 2011. The original is still 2007. Therefore, ((2007-2011)/(2007))(100%). After using algebra, this equals 11.1%

The average (arithmetic mean) score on a test of a class consisting of 20 boys and 30 girls was 76. The average score of the boys on the test was 70. Quantity A The average score of the girls on the test Quantity B 79

We know that there are 30 girls in the class, so we could compute their average if we knew how many total points they scored. The total number of points for the girls must be the total number of points for the class as a whole minus the total number of points for the boys. So let's start by computing the total number of points for the class and the total number of points for the boys. The average formula is Average=Sum of the termsNumber of terms. We can use this formula in the rearranged form Sum of the terms = Average × Number of terms. For the class as a whole, we have 20 + 30 = 50 students, and a 76 average. This gives us a total number of points for the class of 76 × 50 = 3,800 points. The 20 boys had an average of 70, so the total the total number of points for the boys was 70 × 20 = 1,400. Therefore, the total number points for the girls = total number of points for the class - total number of points for the boys, which is 3,800 - 1,400 = 2,400. The 30 girls scored a total of 2,400 points, so their average was 2,40030=80. Quantity A is 80. Quantity B is 79. Quantity A is greater, and choice (A) is correct. We could also have approached this as a weighted average problem. The overall average will be a weighted average of the average of the boys and the average of the girls, where the weights are the percents of boys and girls in the class. Since 20 out of 50 students are boys, the class consists of 40% boys and 60% girls. Using x to represent the girls' average, we have the following equation for the overall average: (% boys)(boys' average) + (% girls)(girls' average) = overall average, or (0.40)(70) + (0.60)(x) = 76. Then 28 + 0.60x = 76, or 0.60x = 48. Dividing both sides by 0.60, we have x = 80. Once again, we see that Quantity A is greater, and choice (A) is correct.

If 2a is an odd integer, which of the following must also be an integer? a a/2 a² 3a 4a

With variables in the question stem and answer choices, let's try Picking Numbers. When using the method of Picking Numbers, all 4 incorrect answer choices must be eliminated because sometimes one or more incorrect answer choices will work for the particular value that we select. If 2a = 3, then a = 1.5. Eliminate choice (A). With a = 1.5, choices (B), (C), and (D) are equal to 0.75, 2.25, and 4.5, respectively. These are non-integers, so eliminate those answer choices. Now that all 4 incorrect answer choices have been eliminated, we know that choice (E) must be correct. Let's check to confirm that choice (E) is an integer: 4(1.5) = 6, so answer choice (E) is indeed an integer. Here is why choice (E) must be an integer. Choice (E) is 4a. Now 4a = 2(2a). Since 2a is an odd integer, 2a is an integer. Then 4a, which equals 2(2a), is 2 times an integer. Thus, 4a is an integer times an integer, so 4a must be an integer. Furthermore, since 4a equals 2(2a), 4a is twice an integer, so 4a must be an even integer.

(2/x) + 1/5 = 1/3x Quantity A: x Quantity B: - 8

Solve for x: 2/x + 1/5 = 1/3x 2/x - 1/3x = -1/5 (6-1)/3x = -1/5 5/3x = -1/5 x = -25/3 So x is less than -8, and column B is greater.

x - y = 9 11 - y = 2 Quantity A: x Quantity B: 0

Solve the second equation. y = 9. Put that y value in the first equation. x needs to be 18 for this to be true. Therefore, x would be greater. So, answer A.

The vertices of equilateral triangle XYZ lie on a circle. Quantity A: Length of minor arc XY Quantity B: Length of minor arc YZ

After drawing a diagram, with the vertices touching the circles edges, inscribed angle XZY intercepts minor arc XY. Inscribed angle ZXY intercepts minor arc YZ. An inscribed angle is equal to half of the arc it intercepts, in degrees. In the case of equilateral triangle XYZ, the two angles are equal to each other. Thus, the two arcs are equal to each other, as well.

x ≠ 0 Quantity A: |((-2x)³y⁴)/x^y)| Quantity B: -|(y^x)/9x³|

An absolute value is NEVER negative so B will always be a negative value and A will always be positive. Key note though, is that it is possible for y to equal 0. Since y is in the numerator of both, it is possible for both to equal 0. Therefore, the relationship cannot be determined. TAKEAWAY: Find something that both may have in common before finding the major differences.

A certain type of tree has leaves and branches. Each individual branch must end with a single leaf or produce two more branches. A tree always starts with a single branch (the trunk) at the bottom. The figure shows two sample trees: the first with one branch and one leaf, the second with 5 branches and 3 leaves. What CANNOT be a possible combination of branches and leaves for this type of tree.

Another case of finding a common multiple. Only an even number of branches can be added to the original trunk. It can end there or add branches in original pairs. It is also possible to create an equation knowing the pattern that the number of branches always goes up by 2, while the number of leaves goes up by 1. So, a tree with n branches will have (n+1)/2. Then, choices that follow that pattern can be found.

In 2010, a company sold a line of clothing to retailers for a fixed price per item. In 2011, the number of items that the company sold to retailers decreased by 30% while the price per item increased by 30%. If the company's revenue from the sale of clothing in 2011 was $5.2 million, what was the approximate revenue from the sale of clothing in 2010?

As with other ratio problems, the values of interest need to become variables. In this case, P equals the price per item in 2010 and N is the number of items sold in 2010. The revenue from the sale of clothing in 2010 is P, the items solid in 2010 is N. Revenue can be found by multiplying N and P. The 2011 price being 30% higher than 2010 means it is equal to 1.3P. Similarly, the number of items decreasing by 30% equals 0.7N. The revenue, once both variables are multiplied, is equal to the given 5.2 million. Solve for NP to find the total revenue. PN ≈ 5.7 million.

Weather Station L recorded precipitation for every day in July. x is the number of days for which L recorded more than 0.0 inches, but fewer than 0.15 inches of precipitation, and y is the number of days for which L recorded 0.15 inches or more of precipitation. Quantity A: 31 - (x+y) Quantity B: Number of days in July for which L recorded no precipitation

Figure out what the Quantity A represents. In this case, x and y are all the days in July for which any precipitation greater than 0 inches was recorded, 31 - (x+y) represents the number of days for which no precipitation was recorded. Take note that there is NO variable that represents days with exactly zero participation. So, the two quantities are equal.

a, b, and c are integers. 0 < a < b < c Quantity A a! + b! Quantity B c! - a!

Given that a, b, and c are positive integers, we can use Picking Numbers to compare the quantities. Let a = 1, b = 2, and c = 3. Quantity A is 1! + 2! = (1) + (2 × 1) = 3. Quantity B is 3! − 1! = (3 × 2 × 1) − (1) = 5. For this example, Quantity B is greater than Quantity A. Let's now consider how the behavior of the quantities changes as the variables change. If we increase c without changing b, Quantity B will increase and will still be greater than Quantity A. However, if we also increase b and a, it is not obvious whether Quantity B will still be greater than Quantity A. So, let's look more closely. If we increase a to 2, b to 3, and c to 4, Quantity A becomes 2! + 3! = (2 × 1) + (3 × 2 × 1) = 2 + 6 = 8 and Quantity B becomes 4! - 2! = (4 × 3 × 2 × 1) - (2 × 1) = 24 - 2 = 22. We can see that with larger numbers for all three variables, the difference between the two quantities only becomes larger, and this trend will increase no matter what values a, b, and c take on. Therefore, Quantity B will always be greater and the correct answer is choice (B). Here is an algebraic proof that Quantity B is greater. Quantity A is a! + b! and Quantity B is c! - a!. We can add the same amount to both quantities and the relationship between the quantities will remain the same. Adding a! to Quantity A, we have a! + b! + a! = 2a! + b! for Quantity A. Adding a! to Quantity B, we have c! - a! + a! = c! for Quantity B. We now have 2a! + b! for Quantity A and c! for Quantity B. Since 0 < a < b < c, and a, b, and c are integers, a ≥ 1, b ≥ 2, and c ≥ 3. Now a < b, so a! < b!. Then 2a! + b! < 2b! + b! = 3b!. Thus, a! + 2b! < 3b!. Now 3 ≤ c. Since b < c, b ≤ c - 1. Then 3b! ≤ c(b!) ≤ c(c - 1)! = c!. Thus, 3b! ≤ c!. Since a! + 2b! < 3b! and 3b! ≤ c!, a! + 2b! < c!. Since 2a! + b! for Quantity A is less than c! for Quantity B, Quantity A is less than Quantity B. Quantity B is greater than Quantity A, and choice (B) is correct.

If x² = -a, which of the following is NOT a possible value of a? -9 -4 -1 0 2

If x² = -a, then -a is either positive or zero, which means that a is either negative or zero. Choice E is the only one that is not less than or equal to zero.

x² + 3x - 70 = 0 Quantity A x Quantity B 4

Let's try to use reverse FOIL to factor the left side of the equation x2 + 3x - 70 = 0, which is x2 + 3x - 70. We know that x2 + 3x - 70 can be factored into (x + a)(x + b), where a and b are constants. Testing some numbers, we find that x2 + 3x - 70 = (x - 7)(x + 10). So the equation x2 + 3x - 70 = 0 becomes (x - 7)(x + 10) = 0. When the product of a group of numbers is 0, at least one of the numbers must be 0. Since (x - 7)(x + 10) = 0, x - 7 = 0 or x + 10 = 0. If x - 7 = 0, then x = 7. If x + 10 = 0, then x = -10. The possible values of x are 7 and -10. If x = 7, then Quantity A is 7 and Quantity B is 4. In this case, Quantity A is greater. If x = −10, then Quantity A is -10 and Quantity B is 4. In this case, Quantity B is greater. Since different relationships between the quantities are possible, the relationship between the quantities cannot be determined. Choice (D) is correct.

Eighteen small black 1' x 1' x 1' cubes are combined to form one 2' x 3' x 3' rectangular solid. The solid is then painted white. Quantity A: The number of smaller cubes that have exactly two white sides. Quantity B: The number of smaller cubes that have exactly three white sides.

Make the entire figure first according the labeled L, W, and H. Mark each one according to the amount of white sides they may have. The corners have 3 white, the "middle" of the sides of have the 8 white cubes. The center cubes have only 1 white side.

5x ≤ 2y y < 0 Quantity A: The ratio of x to y Quantity B: The ratio of y to x

Simplify the top inequality. It will become 2x≤y. Therefore, the full inequality would be 0<y≥2x. Pick numbers for x and y that fulfill the inequality. As with all ratio and comparison problems with variables, pick numbers that would allow the variables to equal each other and if any others present a difference from that. In this case, if x equals -2 and y equals -2, both quantities equal 1. Any other presents a difference in comparison.

There are 4 queens in a standard deck of 52 playing cards. Quantity A: The probability that a queen will be picked at random from a standard deck of 52 cards Quantity B: 0.1

Probability=Number of desired outcomesNumber of possible outcomes. There are 52 cards in a standard deck of cards, 4 of which are queens. The number of desired outcomes is 4 and the number of possible outcomes is 52. Quantity A, which is the probability of picking a queen, is therefore 452=113. Quantity B is 0.1 or 110. Both fractions have a numerator of 1, but Quantity A has a larger denominator, which makes that fraction smaller. Therefore, Quantity B is greater than Quantity A. Choice (B) is the correct answer.

x, y, and z are integers. If x = y² and z = x + y, which of the following statements INDIVIDUALLY provide(s) sufficient additional information to determine whether z is odd? 3y = 2k + 1, where k is an integer x is even z is a perfect cube (that is, a number that equals the cube of an integer)

Read QP3Q19S1 reasoning

An automatic searchlight makes 19 sweeps of a given area in t minutes. 0 < t ≤ 1.5 Quantity A: The number of sweeps of the area the searchlight makes in 6 minutes. Quantity B: The number of minutes in which the searchlight makes 760 sweeps.

Read QP3Q8S1 screenshot

If 2 - 3x ≥ 10 + 2x, then what describes all values of x?

Recall that the inequality sign is flipped when multiplying or dividing by negative numbers. 2-3x ≥ 10+2x -3x ≥ 8 + 2x -5x ≥ 8 -5x/-5 ≥ 8/-5 x ≤ -8/5

On a community baseball team consisting of 25 members, some of whom are men and some of whom are women, exactly 3/5 of the men and exactly 2/5 of the women attended the season's opening ceremony. What is the greatest number of members on the team who could have attended the ceremony?

Set up equation with two variables equal to 25. Ex: X + Y = 25. X equals males. Y equals females. A second equation for the ceremony value would be 2/5(x) + 3/5(y) = ?. Then, use multiples of 5 up to 20 (since 25 is the original number and denominator is in fifths) and plug combinations of the multiples adding up to 25 into the variables until the lowest answer is a possibility. In this case, 14.

x > 1 y > 1 Quantity A: x/(1+y) Quantity B: (1/x)/(1+y)

Since y > 1, 1 + y must be positive, so you can multiply both quantities by 1 + y to eliminate the denominators. This leaves x for quantity A and 1/x for quantity B. Since x > 1, x is greater than 1/x.

In the correctly worked multiplication problems above, the symbols Δ and $ represent distinct digits. 2Δ6 X 5 _____________ 1,280 1,280 + 7,680 _____________ 8,$60 Quantity A Δ Quantity B $

The best way to approach a problem like this is to decide which symbol is the easiest to determine and start there. It seems that the $ is easier to solve for. Considering only the addition, we can see that $ = 9. The digits are distinct, so $ must be bigger, since there is no digit larger than 9, which means Quantity B must be larger than Quantity A. But solving further, use division to reverse the multiplication. 1,280 divided by 5 is 256, so Δ = 5.

KP2Q3

The centered information gives us a triangle, with an interior angle at the top measuring b degrees and two exterior angles measuring a degrees and c degrees, respectively. The angle marked a° and the interior angle of the triangle adjacent to the angle marked a° make up a straight line. So the interior angle of the triangle adjacent to the angle marked a° has a measure of (180 - a)°. The angle marked c° and the interior angle of the triangle adjacent to the angle marked c° make up a straight line. So the interior angle of the triangle adjacent to the angle marked c° has a measure of (180 - c)°. The sum of the interior angles of any triangle is 180°. The interior angles of this triangle have measures of (180 - a)°, b°, and (180 - c)°. Therefore, (180 - a) + b + (180 - c) = 180. Let's work with the equation (180 - a) + b + (180 - c) = 180. (180−a)+b+(180−c)180−a+b+180−c360−a+b−c180−a+b−c180+b=====1801801800a+c Thus, a + c = 180 + b. Now the centered information says that b < 47. Since a + c = 180 + b, a + c < 180 + 47, and then a + c < 227. Quantity A, which is a + c, is less than 227, while Quantity B is 230. Quantity B is greater and choice (B) is correct.

KP2Q5

The centered information gives us a triangle, with an interior angle at the top measuring b degrees and two exterior angles measuring a degrees and c degrees, respectively. The angle marked a° and the interior angle of the triangle adjacent to the angle marked a° make up a straight line. So the interior angle of the triangle adjacent to the angle marked a° has a measure of (180 - a)°. The angle marked c° and the interior angle of the triangle adjacent to the angle marked c° make up a straight line. So the interior angle of the triangle adjacent to the angle marked c° has a measure of (180 - c)°. The sum of the interior angles of any triangle is 180°. The interior angles of this triangle have measures of (180 - a)°, b°, and (180 - c)°. Therefore, (180 - a) + b + (180 - c) = 180. Let's work with the equation (180 - a) + b + (180 - c) = 180. Thus, a + c = 180 + b. Now the centered information says that b < 47. Since a + c = 180 + b, a + c < 180 + 47, and then a + c < 227. Quantity A, which is a + c, is less than 227, while Quantity B is 230. Quantity B is greater and choice (B) is correct.

If c cherries cost d dollars, then how many dollars would 50 cherries cost?

The easiest way to handle this question if the math isn't apparent to you right away is to Pick Numbers. When using the method of Picking Numbers, all 4 incorrect answer choices must be eliminated because sometimes one or more incorrect answer choices will work for the particular values that we select. Let's pick numbers that are convenient to work with. Let's say that c = 7 and d = 14. Then 7 cherries cost $14.00, making the price $2.00 per cherry. So 50 cherries would cost $100. Now let's substitute 7 for c and 14 for d into all 5 answer choices. Any answer choice that does not result in a value of 100 when c = 7 and d = 14 can be eliminated. We find that only choice (B) gives us a value of 100. Therefore, choice (B) must be correct.

What is the circumference of a circle with diameter 20?

The formula for the circumference of a circle is 2pir, where r is the radius of the circle. If the diameter of the circle is 20, then the radius is 10. Insert 10 for r in the formula and you find that the circumference equals 2pi10, that is 20pi.

In an increasing sequence of 10 consecutive integers, the sum of the first five integers is 435. What is the sum of all 10 integers in the sequence?

The single digit has to remain consistent. Therefore, the integers increase by 5 more than the previous one. To solve for the other values, sum the sixth through the tenth integer equaling the sum of the first five integers plus 5 times 5. Therefore, 435 + 5 x 5 which equals 460. 435 + 460 = 895. Bottom line: In problems like these, find what's consistent.

The post office is 7 miles from the bank. The bank is 5 miles from the school. If the school is x miles from the post office, what is the range of possible values for x?

There are three possible arrangements for the three buildings. All three may be in a line, with the bank in the middle; or all three may be in a line, with the school in the middle; or maybe the three are not collinear, but rather define a triangle. In the first scenario, add the distance between the post office and the bank to the distance between the bank and the school in order to find the distance between the post office and the school. That distance is 12 miles. In the second scenario, the school is between the bank and post office, so its distance from the post office is the distance between the bank and the post office minus the distance between the bank and the school. That distance is 2 miles. In the third scenario, the distance you're looking for follows the Triangle Inequality, which states that the sum of the lengths of any two sides of a triangle is greater than the length of the third side, and the difference of the lengths of any two sides of a triangle is less than the length of the third side. In this scenario, then, the distance between the post office and the school can take any value between 2 and 12 miles.

A certain computer has five identical ports into which accessories can be plugged. If a user has two different accessories, and there is only one correct port for him to plug each accessory into, how many INCORRECT ways are there to plug in the accessories?

This problem asks you to count how many incorrect ways there are to plug the accessories into the computer. Since it states there is only one correct way to plug them in, you should first count the number of total possible ways to plug them in, then subtract one to get the answer you are looking for. So think about the first accessory. Since there are five ports, there are five different places to plug it in. Now, when each of the five possible ports is used for the first accessory, four empty ports remain. Thus, for each of the five ways the first accessory can be plugged in, there are four ways the second accessory can be plugged in. That is, there are 5 times 4 possible different ways to plug the accessories into the computer. If only one of those 20 combinations is the right way, there are 19 incorrect ways to do it. The correct answer is B.

Quantity A: The time required to travel 2k kilometers at t kilometers per hour Quantity B: The time required to travel k kilometers at t/2 kilometers per hour

Time equals distance divided by rate, so quantity A equals 2k/T and quantity B equals k/2/t which also equals 2k/T. So, they are equal.

S = {2, 4, 8, x, 13, 16, 20} x is the median of the numbers in set S. Quantity A: The average of the numbers in Set S. Quantity B: x

To find the median of a set of numbers, arrange the numbers in increasing order. If the number of numbers is odd, the median is the middle number, and if the number of numbers is even, the median is the average of the two middle numbers. Because x is the median of the numbers in set S, x can be any value greater than or equal to 8 and less than or equal to 13. We can calculate the average by using the average formula. The average formula is Average =Sum of the termsNumber of terms. If we pick 9 for x, the average is equal to 2+4+8+9+13+16+207=727=1027, which is greater than 9, so Quantity A is greater. Now let's pick 12 for x. If x = 12, the average is equal to 2+4+8+12+13+16+207=757=1057, which is less than 12, so Quantity B is greater. Since different relationships between the quantities are possible, the correct answer is choice (D).

What is the perimeter of a square inscribed in a circle with a circumference of 8π√2?

To find the perimeter of a square, we need its side length. For a square inscribed in a circle, the diagonal of the square is equal to the diameter of the circle. Because we know the circumference, we can calculate the diameter. The circumference of a circle is C = πd. In this case, 8π2‾√=πd. Solving for the diameter, we get d=82‾√. This is also the diagonal of the square as well as the hypotenuse of a 45-45-90 triangle. In this special triangle, the ratio of leg length to leg length to hypotenuse length is 1 to 1 to 2‾√, so we can find the leg length by setting up a proportion: s/(8√2)=1/√2 s√2= 8√2 s=8 Be careful not to stop too soon! The square has a side of 8, so its perimeter is 8 × 4 = 32.

4x - 8y = 0 x and y are positive integers Quantity A: x^y Quantity B: y^x

Use algebra to find what x equals in relation to y. In this case x = 2y. Use Picking Values, choosing ones that will remain true for the new equation and find when A will be greater than B, or vice-versa, or even equal to each other. Doing so, different relations will be found with different values so it cannot be determined.

In a bag full of marbles, 1/4 of the marbles are green, 1/6 of the marbles are red, 1/8 of the marbles are blue, and the remaining 22 marbles are black. How many marbles are red?

Use the work formula. If there are x marbles in the bag, then the number of green, red, and blue marbles equals: (1/4 + 1/6 + 1/8)x = (13/24)x This means the remaining (11/24x) marbles are black. Solve for x, which is 48. Since 1/6 of the marbles are red, 48 x 1/6. So, 8 red marbles.

2/5 of the land on a farm is planted with corn, 1/4 is planted with wheat, 3/10 is set aside for livestock, and the remaining 4 acres are set aside for buildings. What is the total farm, in acres?

Use the work formula. f can be the total size of the farm, in acres. (2/5)f + (1/4)f + (3/10)f + 4 = f f - (2/5)f - (1/4)f - (3/10)f = 4 (20/20-8/20-5/20-6/20)f = 4 (1/20)f = 4 f = 80

KP2Q15

With variables in the question stem and answer choices, let's try Picking Numbers. When using the method of Picking Numbers, all 4 incorrect answer choices must be eliminated because sometimes one or more incorrect answer choices will work for the particular value that we select. If 2a = 3, then a = 1.5. Eliminate choice (A). With a = 1.5, choices (B), (C), and (D) are equal to 0.75, 2.25, and 4.5, respectively. These are non-integers, so eliminate those answer choices. Now that all 4 incorrect answer choices have been eliminated, we know that choice (E) must be correct. Let's check to confirm that choice (E) is an integer: 4(1.5) = 6, so answer choice (E) is indeed an integer. Here is why choice (E) must be an integer. Choice (E) is 4a. Now 4a = 2(2a). Since 2a is an odd integer, 2a is an integer. Then 4a, which equals 2(2a), is 2 times an integer. Thus, 4a is an integer times an integer, so 4a must be an integer. Furthermore, since 4a equals 2(2a), 4a is twice an integer, so 4a must be an even integer.