Quantitative GRE

Z is the set of all fractions of the form n/n−1, where n is a positive integer and 1 < n < 20. a) The product of all the fractions that are in set Z b) 19

Since n is a positive integer that lies between 1 and 20, then n can have integer values from 2 through 19, inclusive. When we evaluate Quantity A, we get the following: 2/1 × 3/2 × 4/3 × 5/4 ×...× 16/15 × 17/16 × 18/17 × 19/18 Notice that the numerator of the first term cancels with the denominator of the second term, the numerator of the second term cancels with the denominator of the third term, the numerator of the third term cancels with the denominator of the fourth term, and so on. Therefore we get that 2/1 × 3/2 × 4/3 × 5/4 ×...× 16/15 × 17/16 × 18/17 × 19/18 = 19. Choice (C) is correct. Alternatively, use factorials. Even though 1 will not be multiplied in the numerator, we know that whatever one is multiplied by will be its identity (19 * 1 = 19). 19!/18! = (19 * 18!)/18! = 19

a) The number of distinct prime factors of 100,000. b) The number of distinct prime factors of 99,000.

Since only the number of distinct prime factors matter, not what they are or how many times they are present, it is possible to tell on sight that Quantity A has two distinct prime factors, because 100,000 is a power of 10. (Any prime tree fr 10, 100, or 1,000, etc. will contain only the prime factors 2 and 5, occurring in pairs.) In Quantity B, 99,000 breaks down as 99 * 1,000. Since 1,000 also contains even more factors (specifically 3, 3, and 11), Quantity B is greater. It is not necessary to make prime factor trees for each number. But if you did, you would find that 100,000 only has 2 and 5 as its distinct prime factors and 2, 3, 5, and 11 distinct prime factors for 99,000, making B greater than A.

a and b are both greater than 4 but less than 10. c and d are both greater than 2 but less than 5. a) a + b b) c + d

Since we are dealing with ranges for the variables, the sums of these variables will also be ranges. Since 4 < a < 10 and 4 < b < 10, 4 + 4 < a + b < 10 + 10, and then 8 < a + b < 20. Since 2 < c < 5 and 2 < d < 5, 2 + 2 < c + d < 5 + 5, and then 4 < c + d < 10. Thus, 8 < a + b < 20 and 4 < c + d < 10. Given that these ranges overlap (don't forget that we could pick 4.9 for a and b), either quantity could be greater, or the quantities could be equal. Therefore, the relationship between the quantities cannot be determined. Choice (D) is correct. Use smart numbers if needed, specifically their endpoints (do not forget the values could be decimals). If both an and b are 3 then the left endpoint, excluding decimals, will be 3 and the right endpoint, if using 9 and 9, will be 18. For c and d, test endpoints, excluding decimals, 1 and 1 and 4 and 4 to get a range of 2 and 8. Notice that the range for an and b, 6 to 18, and the range for c and d, 2 to 8, overlap. This means the quantities can be equal, greater than, or less than each other.

*****If n is an integer and (n^3) is divisible by 24, what is the largest number that must be a factor of n? - 1 - 2 - 6 - 8 - 12

Start by considering the relationship between n and (n^3). Because n is an integer, for every prime factor n has, (n^3) must have three of them. Thus, (n^3) must have prime numbers in multiples of 3. If (n^3) has one prime factor of 3, it must actually have two more, because (n^3)'sprite factors can only come in triples. The question says that (n^3) is divisible by 24, so (n^3)'s prime factors must include at least three 2's and a 3 (24: 2, 2, 2, and 3). But since (n^3) is a cube, it must contain at least three 3's. Therefore, n must contain at least one 2 and one 3, or 2 * 3 = 6

If p is divisible by 7 and q is divisible by 6, pq must have at least how many factors greater than 1? - One - Three - Six - Seven - Eight

The best way to do this problem is by testing the lowest common multiple and seeing how many factors it has. 7 * 6 = 42 (Note, 7 and 6 share no prime numbers and as a result the lowest common multiple will be the numbers multiplied by each other). 42: 1 * 42, 2 * 21, 3 * 14, and 6 * 7) => 7 (Remember that the question tells you to exclude 1). For extra practice do Question 20 in Chapter 13.

Rectangular solid A is 8 centimeters wide, 8 centimeters long, and 12 centimeters high. Right cylinder B is 10 centimeters in diameter and 16 centimeters high. a) The percent increase in the surface area of rectangular solid A when its height is increased by 50%. b) The percent increase in the surface area of right cylinder B when its height is increased by 50%.

The centered information describes the dimensions of a rectangular solid and a right cylinder. Quantity A is the percent increase in the surface area of the rectangular solid when its height is increased by 50%; Quantity B is is the percent increase in the surface area of the right cylinder when its height is increased by 50%. Note that, because you have all of the dimensions for each shape and the amounts of the increases, you have enough information to compare the quantities, so (D) cannot be correct. Approach strategically The surface area of a rectangular solid is 2(l × w) + 2(l × h) + 2(w × h), so rectangular solid A is 2(8 × 8) + 2(8 × 12) + 2(8 × 12) = 512. The surface area of a right cylinder is 2πr2+(2πr×height), so right cylinder B is 2π×25+(10×π×16)=50π+160π=210π. Since the rectangular solid is increasing in height by 50%, or 6 centimeters, the increase in surface area will be 2(8 × 6) + 2(8 × 6) = 192. Use the percent change formula (change/original) to determine the percent increase in rectangular solid A's surface area: 192/512=0.375=37.5%. Since the right cylinder is increasing in height by 8 centimeters, the increase in the surface area of the cylinder is 10π×8=80π. Therefore, the percent increase in the surface area of right cylinder B is 80π/210π≈0.38=38%. Thus, Quantity B is greater and (B) is correct. TAKEAWAY: Knowing how each part of a geometry formula relates to the whole is often the key to answering high-difficulty questions.

*****Question 3 (Test 2/Quant 1) (4-hour exam)

The centered information gives us a triangle, with an interior angle at the top measuring b degrees and two exterior angles measuring adegrees 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. result: a+c = 180 + b 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? a) 50cd b) (50d)/c c) d/(50c) d) (50c)/d e) c/(50d)

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. Or figure out the formula by using dummy variables. Say that 10 cherries cost $5 ($5 = 10 cherries). This means that 10 cherries/$5 = $2.00 per cherry. Which ultimately means 50 cherries would cost 50 * 2 = $100.

Wall-to-wall carpeting is installed in a certain hallway. The carpeting costs $4.25 per square foot. If the perimeter of the hallway (in feet) is equal to 44% of the area of the hallway (in square feet) and the hallway is 50 feet long, how much did the carpeting cost? a) $182.50 b) $212,50 c) $505.25 d) 1,062.50 e) $1,100.00

The equation for the perimeter of a space is 2W + 2L = P, where W is width and L is length. The equation for the area is A = W * L. Thus: 0.44(W * L) = 2W + 2L 0.44(50W) = 2W + 2(50) 22W = 2W + 100 20W = 100 W = 5 If W = 5 and L = 50, then the area of the hallway is 250 sq. ft., and the total cost is: $4.25 * 250 = $1,062.50.

*****Question 12: Number line with 0---a---b---c---d a) a * c b) b * d

The exact values of a, b, c, and d are union, as is whether they are evenly spaced (do not assume that they are, just because the figure looks that way). However, it is known that all of the variables are positive such that 0 < a < b < c < d. Because a < b and c < d and all the variables are positive, a * c < b * d. In words, the product of the two smaller numbers is less than the product of the two greater numbers. Quantity B is greater. You could also try this with real numbers. You could try a = 1, b = 2, c = 3, and d = 4, or you could mix up the spacing, as in a = 0.5, b = 7, c = 11, and d = 45. For any scenario that matches the conditions of the problem, Quantity B is greater.

w, x, y, and z are consecutive odd integers such that w < x < y < z. Which of the following statements must be true? Indicate all such statements. * wxyz is odd * w + x + y + z is odd * w + z = x + y

The first choice is true, as multiplying only odd integers together (and no evens) always yields an odd answer. However, when adding, the rule "an odd number of odds makes an odd". Summing an even number of odds produces an even, so the second choice is false. The third choice is true. Since w, x, y, and z are consecutive odd integers, all can be defined in terms of w: w = w x = w + 2 y = w + 4 z = w + 6 Thus, w + z + = w + (w + 6) => 2w + 6, and x + y => (w + 2) + (w + 4) = 2w + 6. Therefore, w + z = x + y. Alternatively, try real numbers, such as 1, 3, 5, and 7. It is true that 1 + 7 = 3 + 5. This would hold true for any set of four consecutive, ordered odd numbers tested.

New cars leave a factory in a repeating pattern of red, blue, black, and grey cars. If the first car to exit the factory was red, what color is the 463rd car to exit the factory? - Red - Blue - Black - Gray - It cannot be determined from the information given.

The key to this problem is the remainder. If we divide 463/4 (the number of colors possible/options possible) we get 115.75. This means we can do 115 complete cycles (of produce 460 cars) before we do not complete a full cycle. As a result car 461 will be red, 462 will be blue, and 463 will be black. The correct answer is Black.

*****A fair coin will be tossed twice and a fair die with sides numbered 1, 2, 3, 4, 5, and 6 will been rolled twice. What is the probability that at least one head will be tossed and at least one number greater than 1 will be rolled? a) 35/144 b) 1/2 c) 25/48 d) 9/16 e) 35/48

The probability that an event occurs is equal to 1 minus the probability that the event does not occur. The probability that at least one head is tossed is equal to 1 minus the probability that no head is tossed. It is easier to find the probability that no head is tossed, which is the probability that both tosses of the coin are tails. The probability that one coin toss is tails is 1/2. The coin tosses are independent. The probability that two independent events both occur is equal to the probability that one event occurs multiplied by the probability that the other event occurs. So the probability that both tosses are tails is 1/2 × 1/2 = 14. Then the probability that at least 1 head is tossed is 1 − 1/4 = 4/4 − 1/4 = (4 − 1)/4 = 3/4. The probability that at least one number greater than 1 shows when the dice are rolled is equal to 1 minus the probability that both numbers on the dice are 1. The probability that when one die is rolled, a 1 shows, is 16. The rolls of the dice are independent. So the probability that both dice show a 1 is 16×16=136. Then the probability that at least one number greater than 1 shows is 1 − 1/36 = 36/36 − 1/36 = 36−1/36=35/36. Thus, the probability that at least one head is tossed is 3/4, and the probability that at least one number greater than 1 is rolled is 35/36. The tosses of the coin and the rolls of the dice are all independent of one another. So the probability that at least one head is tossed and at least one number greater than 1 shows on the dice is 3/4×35/36 = 1/4 × 35/12 = 1×35/4×12 = 35/48. Choice (E) is correct.

*****|x + 5| < 3 On the number line, the distance between x and 0 is 6. What is the distance between -|x| and |x - 10| ? a) 2 b) 6 c) 8 d) 10 e) 22

The question asks about the distance between the values of two expressions on a number line. To determine their values, we need to know more about x. Because the distance between x and 0 on the number line is 6, x is either 6 or -6. We are given |x + 5| < 3. If x = 6, then |6 + 5| = 11, which does not satisfy the inequality |x + 5| < 3. If x = -6, then |-6 + 5| = 1, which satisfies the inequality |x+ 5| < 3. Therefore, x = -6. To find the distance between -|x| and |x - 10|, let's determine their values. -|x| = -|(-6)| = -(6) = -6 |x - 10| = |(-6) - 10| = |(-16)| = 16 The distance between the two values on the number line is the absolute value of their difference: |(-6) - 16| = |(-22)| = 22. The correct answer is choice (E). Note, that the distance of the points is non-inclusive. Also, one could have also determined the value of x by solving for the absolute value. Set up you two equations: (x + 5) < 3 = x < -2 -(x + 5) < 3 = x > 8 Seeing that x is 6 away from 0, we know that x < -2 must be the correct inequality to look at. Meaning that x = -6, since anything above 8 would have a greater than 6 distance from 0.

Question 5/Test 2/Quant 3 => try doing it without using a calculator

The question displays a table listing the vehicles in the inventories of two car dealers and asks for a comparison between the percentage of pickups in each dealer's inventory. Although Dealer K has more pickups, that dealer also has more vehicles overall. The total number of vehicles at Dealer J is 20 + 17 + 31 + 9 + 3 = 80, so the percentage that are pickups is 31/80×100%. Rather than activating the calculator to determine that this is 38.75%, just note that this amount is a bit more than 3/8. The total number of vehicles at Dealer K is 25 + 33 + 44 + 11 + 7 = 120. One-eighth of 120 is 15, so 3/8×120=45. Since Dealer K has only 44 pickups, that is a bit less than 3/8, so the correct answer is (A). (For the record, the calculator would have shown the percentage for Dealer K to be 36.67%.) To confirm your answer, check your calculations and be certain that you extracted the correct data from the table.

*****Question 7 (Test 2/Quantitative Reasoning 2) (Test 2/Quantitative 2/Question 6)

The trap here is that these two angles appear to form a right angle. Remember that we cannot assume that figures are drawn to scale unless we are told so. Likewise, we cannot assume this is a coordinate geometry problem (coordinate geometry is drawn to scale) because there is no indication that the lines represent the x— or y-axis. Therefore, it is possible that m + n is less than 90, it is possible that m + n is equal to 90, and it is possible that m + n is greater than 90. There is more than one possible relationship between the quantities, and choice (D) is correct.

*****A rectangular public park has an area of 3,600 square feet. It is surrounded on three sides by a chain link fence. If the entire length of the fence measures 180 feet, how many feet long could the unfenced side of the rectangular park be? Indicate all such lengths. * 30 * 40 * 60 * 90 * 120

The two values given are the area of the park ad three out of the four sides of the perimeter of the park. If the side without fencing is a length, the equation for the overall lentgth of the existing fence is 180 = 2W + L, so L = 180 - 2W. The equation for the area of the park is LW = 3,600. With two variables and two equations, it is now possible to solve for the passive values of L: L * W = 3,600 L = 180 - 2W (180-2W)W = 3,600 180W - 2W^2 = 3,600 90W - W^2 = 1,800 0 = (W^2) -90W + 1,800 0 = (W - 60)(W - 30) So W = 30 or 60. Plug each value back into either of the original two equations to solve for the corresponding length, which is 120 or 60, respectively.

*****The yoga company Yoga for Life offers 45-minute classes at $12 per class. If the number of minutes Randolph spent doing yoga this month was 132 greater than the number of dollars he paid, how many classes did he attend? a) 3 b) 4 c) 5 d) 6 e) 8

The typical way to so this problem would be to assign variables and set up equations, using x to represent the number of classes Randolph took, 12x to represent the amount he paid, and 45x to represent the number of minutes he spent. A quicker way might be to notice that with every class Randolph takes, the difference between the number of minutes he spends and the amount he pays increases by 22. If Randolph takes 1 class, then the number of minutes he spends is 33 greater than the number of dollars he pays. If he takes 2 classes, the number of minutes is 66 greater than the number of dollars, and so on. Since 132 = 4 * 33, Randolph must have taken 4 classes.

a) (3^(19)) + (3^(19)) + (3^(19)) b) (3^(20))

The values are a bit too difficult to compute, so try making the quantities look alike by factoring out a 319 in Quantity A: 3^19 + 3^19 + 3^19 = 3^19(1 + 1 + 1) = 3^19(3) = 3^19(31) = 3^19 + 1 = 3^20. This is the same as Quantity B, so the correct answer is choice (C). We used the law of exponents that says that babc = ba + c, when we said above that 3^19(31) = 3^19 + 1.

*****If k is a multiple of 24 but not a multiple of 16, which of the following cannot be an integer? - k/8 - k/9 - k/32 - k/36 - k/81

There are two ways of doing this problem. First, and quickest is by determining the prime factors of 24 and 16. 24: 2, 2, 2, and 3 16: 2, 2, 2, and 2 A multiple of 24 can only have three 2s and no more, otherwise it could be a multiple of 16. Note the number could have less than three 2s (even none). The number could also have as many threes as it wants. Note it does not need a three. 8: 2, 2, and 2 (Note that even though there are no threes this would result in integer values for k/8. Or in other words, 8 shares multiples with 24). 9: 3 and 3 (Note even though this number does not have any 2s it still shares multiples with 24) 32: 2, 2, 2, 2, and 2 (More than three 2s, this number will result in non integer values for multiples of 24 that are not also multiples of 16. 36: 2, 2, 3, and 3 (Multiple of 32 that is not a multiple of 16) 81: 3, 3, 3, and 3 (You can have as many 3s as you want, just no more 2s. Another way to do this problem is by listing out the multiples of both 24 and 16 and crossing out the multiples that are shared by both numbers. What you will notice is that only the odd numbers multiplied by 24 are not shared between 24 and 64, so k = (an odd integer) * 24. k/8 => k = 24 (Incorrect) k/9 => k = 72 (Incorrect) k/32 => is correct by process of elimination (continue on from this point). k/36 => k = 72 (Incorrect) k/81 => 81 * 24 will result in a multiple of 24, that is not a multiple of 16 ((81 * 24)/16 /= an integer) - Best method is to use the prime factors.

Cashews cost $4.75 per pound and hazelnuts cost $6.50 per pound. a) The number of pounds of cashews in a mixture of cashews and hazelnuts that costs $5.50 per pound b) 1.25

There is no way to calculate the number of pounds of either nut in the mixture. We can calculate the ratio of the number of pounds of cashews to the number of pounds of hazelnuts required for the mixture to cost $5.50 per pound, but without knowing how many total pounds the mixture is, we cannot calculate the number of pounds of either component. For instance, if there is only one pound of the $5.50 mixture, then the number of pounds of cashews would be approximately .57 of a pound, which is less than Quantity B. However, if there are 10 pounds of the $5.50 mixture, then there are approximately 5.7 pounds of cashews, which is far more than Quantity B. Choice (D) is correct. Note we have four variables and only two equations: 4.75(weight of cashews) + 6.50(weight of hazelnuts) = total cost (weight of cashews) + (weight of hazelnuts) = total weight We can not determine the correct value with only the cost per pound of the final mixture. Alternatively, look at it by setting up your equations: 4.75(c) + 6.50(h) = 5.50 c + h = unknown With three unknowns and only two equations, we cannot solve this problem, hence answer D).

If (12^x) is odd and x is an integer, what is the value of (x^12)?

This is a bit of a trick question. (12^x) is odd? How strange! (12^1) is 12, (12^2) is 144, 123 is 1,728 ... every "normal" power of 12 is even. (An even number such as 12 multiplied by itself any number of times will yield an even answer.) These normal powers are 12 raised to a positive integer. What about negative integer exponents? They are all fractions of this form: (1/(12^positive integer)). The only way for 12^x to be odd is for x to equal 0. Any non-zero number to the 0th power is equal to 1. Since x = 0 and the question asks for x^12, the answer is 0. Note, that an even number multiplied by an even number always yields and even number (12 * 12 = 144). As a result the only way for there to be an answer is for x = 0 since 12^0 = 1 (an odd number). Remember that 0 is an integer.

Jason deposits money at a bank on Thursday and returns to the bank 100 days later to withdraw the money. On what day of the week did Jason withdraw the money from the bank? - Monday - Tuesday - Wednesday - Thursday - Friday

This is a pattern problem. An efficient method is to recognize that the 7th day after the initial deposit would be Tuesday, as would the 14th day, the 21st day, etc. Divide 100 by 7 to get 14 full weeks comprising 98 days, plus 2 days left over For the 2 leftover days, think about when they would fall. The first day after the deposit would be a Wednesday, as would the first day after waiting 98 days. The second day after the deposit would be a Thursday, and so would the 100th day. The answer is Thursday. Note that we count days as the first day being tomorrow, not today.

*****In 8 years, Polly's age, which is currently p, will be twice Quan's age, which is currently q. a) p - 8 b) 2q

This is an algebraic translation question, so stat by translating the given information into equations. Remember to add 8 to both Polly and Quan's ages, because they will both be 8 years older in 8 years! p + 8 = 2(q + 8) p + 8 = 2q + 16 p = 2q + 8 p - 8 = 2q The two quantities are equal. (Look at question 7, 12, and especially 24 in Two-Variable Word Problems.)

What is the perimeter of a square inscribed in a circle with a circumference of 8π((2)^1/2)?

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, 8π((2)^1/2)=πd. Solving for the diameter, we get d=8((2)^1/2). 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)^1/2, so we can find the leg length by setting up a proportion: (s/(8*((2)^1/2)) = 1/((2)^1/2) s((s)^1/2) = 8((2)^1/2) s = 8 Be careful not to stop too soon! The square has a side of 8, so its perimeter is 8 × 4 = 32.

*****In triangle ABC, the measure of angle ABC is 90 degrees, and the lengths of two sides of triangle ABC are shown. Triangle DEF is similar to ABC and has integer side lengths. Which of the following could be the area of triangle DEF (Question 18/Test 2/Quant 2)? Indicate all possible areas. * 6 * 15 * 24 * 48 * 54 * 96 * 120 * 150

Triangle ABC is a right triangle with legs 6 and 8, so it must be a 3:4:5 right triangle with a hypotenuse of 10. The problem states that DEFis similar to ABC with integer side lengths, so DEF must also be a 3:4:5 right triangle. As we do not know the actual dimensions of DEF, the best way to solve this problem is to work out the areas of the first few 3:4:5 right triangles: Area: 1/2 (3 * 1)(4* 1) = 6 Area: 1/2 (3 * 2)(4 * 2) = 24 Area: 1/2 (3 *3)(4 * 3) = 54 Area: 1/2 (3 * 4)(4 * 4) = 96 Area: 1/2 (3 * 5)(4 * 5) = 150 The correct answer choices are (A), (C), (E), (F), and (H).

T/F: The range is always greater than the standard deviation?

True

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. a) The average score of the girls on the test 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 terms/Number 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 2,400/30=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.

A car traveled 24 miles, at an average speed of 40 miles per hour. What is the average speed, in miles per hour, at which the car must travel another 48 miles if the total travel time for both distances is to be 116 minutes? a) 26 b) 28 c) 30 d) 36 e) 48

We will be using the distance formula, which is Distance = Rate × Time. The car traveled the first 24 miles at an average speed of 40 miles per hour. So using the formula in the rearranged form Time=DistanceRate, the time the car took to travel the first 24 miles was 24 miles/40 miles-hour = 24/40 hours = 35 hours = 0.6hours. There are 60 minutes in an hour, so in 0.6 hours there are 0.6(60) = 36 minutes. The car took 36 minutes to travel the first 24 miles. Since the total time of travel was 116 minutes, the amount of time in which the car is to travel the remaining 48 miles is 116 - 36 = 80 minutes. Thus, the car is to travel the remaining 48 miles in 80 minutes. Since we want to find the rate in miles per hour, let's convert 80 minutes to hours. There are 60 minutes in an hour, so in 80 minutes, there are 80/60 = 8/6 = 4/3 hours. Let's keep the fraction 4/3 in this form, since we will be dividing by 4/3 hours. Now let's use the distance formula in the rearranged form Rate = Distance/Time to find the car's average speed over the remaining 48 miles. The car's average speed is to be 48 miles(4/3hours) = 48 (4/3) miles per hour. Now let's find the value of 48/(4/3). We have 48/(4/3)=48×3/4 = 12 × 3 = 36. So, the car's average speed will have to be 36 miles per hour for the remainder of the trip. Choice (D) is correct. Note, you could have also done this problem working in minutes and then converting the final answer to hours (specifically: going from miles per minute to miles per hour).

If (3^x)(5^2) is divided by (3^5)(5^3), the quotient terminates with one decimal digit. If x > 0, which of the following statements must be true? - x is even - x is odd - x < 5 - x >= 5 - x = 5

When a non-multiple of 3 is divided by 3, the quotient does not terminate (for instance, 1/3 = 0.333...). Since ((3^x)(5^2)/(3^5)(5^3)) does not repeat forever, x must be large enough to cancel out the (3^5) in the denominator. Thus, x must be at least 5. Note that the question asks what must be true. Choice D) must be true. Choice E), x=5, represents one value that would work, but this does not have to be true.

When the positive integer x is divided by 6, the remainder is 4. Each of the following could also be an integer Except: - x/2 - x/3 - x/7 - x/11 - x/17

When dealing with remainder questions on the GRE, the best thing to do is test a few real numbers: Multiples of 6 are 0, 6, 12, 18, 24, 30, 36, etc. Numbers with a remainder of 4 when divided by 6 are those 4 greater than the multiples of 6: x could be 4, 10, 16, 22, 28, 34, 40, etc. You could keep listing numbers, but this is probably enough to establish a pattern. x/2 => All of the listed x values are divisible by 2. Eliminate (A) x/3 => None of the listed x values are divisible by 3, but continue checking. x/7 => 28 is divisible by 7. x/11 => 22 is divisible by 11. x/17 => 34 divisible by 17. The question is "Each of the following could also be an integer EXCEPT." Since four of the choices could be integers, (B) must be the answer.

On a number line, the distance from A to B is 4 and the distance from B to C is 5. a) The distance from A to C b) 9

Whenever a question looks this straightforward (4 + 5 = 9, so the quantities initially appear equal), be suspicious. Draw the number line described. If the points, A, B, and C are alphabetical order from left to right, then the distance from A to C will be 9. However, the alphabetical order is not required. If the points are in order C, A, and B from left to right, then the distance from A to C is 5 - 4 = 1. Therefore, the relationship cannot be determined.

At a certain office supply store, the price of a pencil is half the price of a pen and the price of a notebook is twice the price of a pen. For which of the following combinations is the total price equal to the price of 5 notebooks? Indicate all that apply. a) 2 pencils, 1 pen, 4 notebooks b) 3 pencils, 2 pens, 3 notebooks c) 4 pencils, 4 pens, 2 notebooks d) 8 pencils, 3 pens, 2 notebooks e) 8 pencils, 4 pens, 1 notebook f) 4 pencils, 6 pens, 1 notebook g) 6 pencils, 5 pens, 1 notebook h) 10 pencils, 4 pens, 1 notebook

With only a relative relationship in the question stem among prices and no actual numbers, try Picking Numbers. If a pencil costs $1, then a pen costs $1 × 2 = $2 and a notebook costs $2 × 2 = $4. The price of 5 notebooks would then be $4 × 5 = $20. Plug these values into the choices and look for the ones that come out to $20: Choice (A): ($1 × 2) + ($2 × 1) + ($4 × 4) = $2 + $2 + $16 = $20 This is the correct amount. Choice (B): ($1 × 3) + ($2 × 2) + ($4 × 3) = $3 + $4 + $12 = $19 This is not the correct amount. Discard choice (B). Choice (C): ($1 × 4) + ($2 × 4) + ($4 × 2) = $4 + $8 + $8 = $20 This is the correct amount. Choice (D): ($1 × 8) + ($2 × 3) + ($4 × 2) = $8 + $6 + $8 = $22 This is not the correct amount. Discard choice (D). Choice (E): ($1 × 8) + ($2 × 4) + ($4 × 1) = $8 + $8 + $4 = $20 This is the correct amount. Choice (F): ($1 × 4) + ($2 × 6) + ($4 × 1) = $4 + $12 + $4 = $20 This is the correct amount. Choice (G): ($1 × 6) + ($2 × 5) + ($4 × 1) = $6 + $10 + $4 = $20 This is the correct amount. Choice (H): ($1 × 10) + ($2 × 4) + ($4 × 1) = $10 + $8 + $4 = $22 This is not the correct amount. Discard choice (H). The correct choices are (A), (C), (E), (F), and (G).

The sum of four consecutive integers is -2. a) The smallest of the four integers b) -2

Write an equation x + (x + 1) + (x + 2) + (x + 3) = -2. Now solve: 4x + 6 = -2 4x = -8 x = -2 Thus, the integers are -2, -1, 0, and 1. The smallest of the four integers equals -2, so the quantities are equal.

*****What is the greatest prime factor of (2^99) - (2^96)?

You cannot subtract (2^99) - (2^96) to get (2^3)! You cannot directly combine numbers raised to powers when adding or subtracting. Instead, factor out the greater common factor of (2^99) and (2^96): (2^99) - (2^96) = (2^96)((2^3) - 1) = (2^96)(7) Since (2^99) - (2^96) is equal to (2^96)(7^1), its greatest prime factor is 7.

Eva meditates for 20 minutes at a time, with a 5-minute break in between sessions. If she begins meditating at 10:10, what time will it be when she completes her third 20-minute meditation session? a) 11:20 b) 11:25 c) 11:50 d) 11:55 e) 12:25

You could list out all the times (make sure to stop before the third 5 minute break). Or you could add 20(3) + 5(2) to get 70 minutes, and 70 minutes after 10:10 is 11:20.

*****A washing machine takes 35 minutes to wash one load of laundry, and in between washing different loads of laundry it takes Derek 2 minutes to unload and another 4 minutes to reload the machine. If the washing machine begins washing one load of laundry at 12:30pm, how many loads of laundry can Derek wash and unload before 6:35pm? a) 8 b) 9 c) 10 d) 14 e) 15

You could list out the times and calculate to 6:35 but this is not an optimum use of time. A better approach to determine how many minutes are available for Derek to do laundry. From 12:30 to 6:35 is 6 hours and 5 minutes or 365 minutes. Divide 365 minutes by 41 to get 8.9... So, Derek can definitely do 8 total loads of laundry plus switching time. What about the extra 0.9...? You need to figure out whether Derek can fit in one more laundry load. Importantly, for the last load he needs only 2 extra minutes to unload, since he will not be reloading the machine. Multiply 8 (the total number of loads Derek can definitely do) by 41 minutes to get 328 minutes.. Subtract 328 from the 365 available minutes to get 37 minutes. That is exactly how much time it takes, Derek ti do one load of laundry (35 minutes) and then unload it (2 minutes). So, Derek can wash and unload 9 total loads of laundry.

If p is the sum of all integers from 1 to 150, inclusive, and q is the sum of all the integers fro m1 to 148, inclusive, what is the value of p - q?

p is the large number, but it consists entirely of q + 149 + 150. Thus, p - q is what's left of p once the common terms are subtracted: 149 + 150 = 299.

As the radius of a right cylinder increases, the volume increases by the __________ of the multiplicative factor of the radius; for example, when the radius triples, the volume is multiplied by 9 (Question 20/Quant 3/Test 2). Also, when all other elements remain fixed, increasing the height of a right cylinder by a given percentage will increase its volume by the ___________ percentage.

square, same

(2x^2) - 7x - 4 (factor)

(2x + 1)(x - 4)

If (3^11) = (9^x), what is the value of x?

(3^11) = ((3 * 3)^x) or (3^11) = (3^2)^x or (3^11) = (3^2x) 11 = 2x => x = 11/2

If x is a positive integer, which one of the following could be the remainder when (73^x) is divided by 10? Indicate all such remainders. * 0 * 1 * 2 * 3 * 4 * 5 * 6 * 7 * 8 * 9

1, 3, 7, and 9 only. As with multiplication, when an integer is raised to a power, the units digit is determined solely by the product of the units digits. These products will form a repeating pattern. Here, (3^1) = 3, (3^2) = 9, (3^3) = 27, (3^4) = 81, and (3^5) = 243. Here the pattern returns to its original value of 3 and any larger power of 3 will follow this same pattern`; 3, 9, 7, and then 1. Thus, the units digit of (73^x) must be 1, 3, 7, or 9. When dividing by 10, the remainder is the units digits, so those same values are the complete list of possible remainders.

If set S consists of all positive integers that are multiples of both 2 and 7, how many numbers in set S are between 140 and 240?

A positive integer that is a multiple of both 2 and 7 is a multiple of 14. Since 140 is a multiple of 14, start listing there and count the terms in the range: 140, 154, 168, 182, 196, 210, 224, and 238. Alternatively, note that 140 is the 10th multiple of 14, and 240/14 = (approximately) 17.143 (use the calculator). Therefore, the 10th through the 17th multiples of 14, inclusive, are in this range. The number of terms is 17 - 10 + 1 = 8 ("add one before you are done" for an inclusive list).

|x-1| = |y| a) x^2 b) y^2

Analyze the centered information and quantities The centered information shows that the absolute value of x − 1 is equal to the absolute value of y. Quantity A is x2 and Quantity B is y2. Approach strategically Trying to determine which value is larger using algebra would get complicated quickly. Thankfully, there is a much simpler, and more strategic approach: pick a number for x and determine the resulting values for y. Then try again with a different value for x. Start with x = 3. In that case, |x − 1| would equal |2|, or simply 2. So 2 = |y| and therefore y = +2 or −2. Plugging those values into Quantity A and Quantity B, then, would produce x2=9 for Quantity A and y2=4 for Quantity B. Using those numbers, Quantity A is greater. Now try a different number for x. When x = 0, y would equal +1 or −1. In this case, x2=0 and y2=1, which means that Quantity B is greater. You don't need to check any other numbers. (D) is correct. TAKEAWAY: Because absolute values do not lend themselves well to being stated in equations, picking numbers is often an excellent strategy for solving these questions.

*****A manufacturer produced n number of devices. Each device cost c dollars to make. The manufacturer then sold (n − 10) such devices at r dollars each. The revenue generated from the sale of the devices exceeded the cost of the production of the devices. a) (r-c)/10 b) r/n

Analyze the centered information and quantities The quantities are algebraic expressions that both contain r as well as one other variable (c or n). The centered information provides information about each variable as it relates to cost, revenue, or profit. The cost of producing n devices would be nc, and the revenue generated from selling (n - 10) devices would be (n - 10)r. Since the revenue generated from the sale of (n - 10) devices was greater than the cost of producing n devices, set up an inequality: (n − 10)r > nc. Approach strategically Manipulate your inequality to try to solve for either Quantity. Since Quantity A contains r - c in the numerator, attempt to isolate r - c on one side of the inequality. (n−10)r > nc nr − 10r > nc nr − nc > 10r n(r−c) > 10r r − c > 10r/n After isolating r - c, keep going. Quantity A contains r - c in the numerator and 10 in the denominator. Divide both sides by 10. r − c/10 > r/n The inequality now consists of the two values in Quantity A and Quantity B. Clearly, Quantity A is larger, so (A) is correct. If you hadn't thought to rearrange the inequality in the manner that is shown above, you could have picked numbers that keep the previously determined inequality r(n − 10) > cn true. Start by picking a number for n that is an integer (since n represents the number of devices made) and that keeps the given statement n − 10 > 0 true; let n equal, say, 11; thus, n − 10 = 11 − 10 = 1, and the inequality simplifies to r(1) > 11c, or r > 11c. Now pick a permissible number for each of r and c (since each represents a cost, both should be positive values); for example, let c = 1 and r = 12, such that 12 > 11. Next, substitute your picked numbers into the Quantities to compare the latter: Quantity A becomes 12 − 1/10 = 11/10 = 1 1/10, and Quantity B becomes 12/11=1 1/11. The fraction 1/10 is greater than the fraction 1/11, and adding a larger fraction to 1 results in a larger overall quantity than does adding a smaller fraction to 1. Thus, Quantity A is larger given your initial group of picked numbers. To check whether Quantity A could ever not be greater than Quantity B, test a different group of numbers. To try to make Quantity B larger than Quantity A or at least eq

*****If ((x^2)^3 * y^−3)/(x^(1/2))^6 = 8, which of the following could be the values of x and y? a) x = 1, y = 2 b) x = 1, y = 8 c) x = 2, y = -1 d) x = 4, y = 2 e) x = 8, y = 1

Analyze the question The question presents an two-variable equation with a fraction, exponents, and a radical. Identify the task Determine which of the choices could be the values of x and y. Approach strategically Use the rules of exponents to solve. To simplify a term with an exponential term raised to another exponent, multiply the exponents. So, (x^2)^3 = x^6. A negative exponent is equivalent to the same value but with the exponent expressed as a positive value in the denominator of a fraction, so y^(−3) = 1/y^3. A square root is the same as the exponent 1/2, so (x^1/2)^6 = (x^1/2)^6 = x^3. And since only the square roots of positive numbers are real numbers, this fact limits the allowable values of x to positive numbers. Next, divide the x term in the numerator by the x term in the denominator by subtracting the exponents to get x^(6−3) = x^3. So, x^3/y^3 = 8. Multiply both sides of the equation by y3 to yield x^3 = 8y^3 . Take the cube root of both sides to see that x = 2y. Thus, the correct choice will have a value for x that is twice the value of y. Only (D) has a value of x that is 2y. Confirm your answer Check that you've properly simplified the given equation. You could also quickly verify that (2)^3 = 8(1)^3 = 8. TAKEAWAY: Memorize the rules for exponents so you can apply them efficiently on Test Day.

*****How many three-digit numbers with a hundreds digit of 1, 2, or 6 are divisible by 7 ? a) 14 b) 21 c) 42 d) 43 e) 63

Analyze the question You're asked to consider the properties of three-digit numbers that contain either a 1, 2, or 6 in the hundreds place. If this seems a bit abstract, it might help to think of some examples. Examples of such numbers are 100, 250, and 699. Identify the task Use your understanding of divisibility to find how many of these three-digit numbers are divisible by 7. Approach strategically Here is one approach. Start with the numbers that have a 1 in the hundreds place. There are 100 such numbers: 100-199. One way to determine how many of these numbers are divisible by 7 is to find the smallest of the numbers that is a multiple of 7, then count up from there. You can find the smallest such number by (either in your head or using the calculator) dividing 7 into numbers starting with 100 until you find a number that 7 goes into evenly. The smallest number that's divisible by 7 is 105 (which is 7 × 15). Now start adding 7 to find that the numbers in this range that are divisible by 7 are 105, 112, 119, 126, 133, 140, 147, 154, 161, 168, 175, 182, 189, 196. Count them up: that's 14 numbers. Follow the same process with the numbers in the range 200-299 and 600-699. That approach is very methodical and will get you to the right answer. However, it involves a lot of calculation and counting. Here is a potentially more efficient way. First, divide 100 by 7, getting ≈ 14.29. Since there can't be "partial" numbers that are divisible by 7, then there must be either 14 or 15 numbers divisible by 7 in every group of 100 consecutive numbers. To ensure that you find all of the numbers that are divisible by 7, multiply 7 × 14 = 98. This means that in every consecutive group of 98 numbers, there are 14 factors of 7. So, for 100-197, there are 14 numbers that are divisible by 7. Now check 198 and 199 to see if either of these numbers is also divisible by 7; they are not, so you know that there are definitely 14 numbers divisible by 7 in the range of numbers 100-199. Follow the same procedure for 200-299 and 600-699. You know that there are 14 factors of 7 in the range 200-297, so now check 298 and 299. They are not divisible by 7. There are 14 factors of 7 in the range 600-697, so now check 698 and 699. They are no

When x is divided by 10, the quotient is y with a remainder of 4. If x and y are both positive integers, what is the remainder when x is divided by 5? - 0 - 1 - 2 - 3 - 4

Any number that yields remainder 4 when divided by 10 will also yield remainder 4 when divided by 5. This is because the remainder 4 is less than both divisors, and all multiples of 10 are also multiples of 5. For example, 14 yields remainder 4 when divided either by 10 or by 5. This also works for 24, 34, 44, 54, etc.

a < b < c < d < 0 a) a - d b) bc

Anytime a number is subtracted from a smaller one, the result will be negative. Thus, a - d < 0. Conversely, since the product of two negatives is positive, bc > 0. Because any positive value is greater than all negative values, Quantity B must be greater. Remember that all the variables are negative and that larger value negative numbers are smaller than small value numbers.

*****Which of the following values times 12 is not a multiple of 64? Indicate all such values. * (6^6) * (12^2) * (18^3) * (30^3) * 222

Because 64 = (2^6), multiples of 64 would have at least six 2's among their prime factors. Since 12 (which is 2 * 2 * 3) has two 2's already, a number that could be multiplied by 12 to generate a multiple of 64 would need to have, at minimum, the other four 2's needed to generate a multiple of 64. Since you want the choices that don't multiply with 12 to generate a multiple of 64, select the choices that have fewer than four 2's within their prime factors. (6^6) = (2 * 3)^6 => incorrect, this value has more that enough 2s. (12^2) => (2 * 2 * 3)^2 => Incorrect, it has four 2s. (18^3) => (2 * 3 * 3)^3 => Only has three 2s, and as a result will not be a multiple of 64 when multiplied by 12; correct answer. (30^3) => (2 * 3 * 5)^3 => Only three 2s, so this will also be a correct answer. 222 => 2 * 3 * 37 => only one 2, so this is the correct answer.

Company H distributed $4000 and 180 pencils evenly among its employees, with each employee getting an equal integer number of dollars and an equal integer number of pencils. What is the greatest number of employees that could work for company H? - 9 - 10 - 20 - 40 - 180

Because the values have to be distributed evenly and we are looking for the greatest number of employees you must divide the 4000 and 180 evenly among the number of employees. Start from the highest answer to lowest. 180 does not go evenly into 4000 and 40 does not go evenly into 180 so the correct answer is 20 since it goes evenly into both 4000 and 180. Alternatively, find the greatest common factor: 4000: 2, 2, 2, 2, 2, 2, 5, 5, and 5. 180: 2, 2, 3, 3, and 5. As a result the greatest common factor of the two numbers is 2 * 2 * 5 = 20.

Jamal got three monthly electric bills over the course of three months. If his average monthly bill over these months was $44 more than the median bill, and the sum of the largest and the smallest bills was $412, what was the total amount of the three electric bills Jamal got over the course of the three months? a) $456 b) $552 c) $600 d) $824 e) $1,000

Call the smallest bill S, the middle bill M, and the largest bill L. M is the same as the median, since there are only three values. The equation for the average is: (sum of bills)/(number of bills) = Average Incorporate the equation for average into the following equation: (S + M + L)/3 = M + 44 S + M + L = 3M + 132 S+ L = 2M + 132 While the individual values of S and L are not given, their sum is: 412 - 2M + 132 280 = 2M 140 = M Finally, add M to the sum of S and L: 140 + 412 = 552

*****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. 5 family photos and 9 vacation photos 6 family photos and 8 vacation photos 7 family photos and 7 vacation photos 10 family photos and 4 vacation photos

Consider choice (A) first. There are 5 different photos Amanda could put into the first picture slot of the first frame. Once she commits a photo to the first slot, there are 4 photos left, any of which could go into the second slot. That leaves 3 photos for the third slot and 2 for the last. Thus, the number of arrangements for the first frame of choice (A) is: 5 × 4 × 3 × 2 = 120 By a similar logic, the number of arrangements in the second frame is 9 × 8 × 7 × 6 = 3,024. For each of the 120 possible arrangements in the first frame, there are 3,024 arrangements for the second. Because "for each" is a multiplication keyword, multiply these two results to get the total number of possibilities: 120 × 3,024 = 362,880. This is less than 500,000, so eliminate choice (A). Now apply the same logic to the other choices: (You could also use the permutation formula.) Choice (B), First Frame: 6 × 5 × 4 × 3 = 360 Choice (B), Second Frame: 8 × 7 × 6 × 5 = 1,680 Choice (B), Total: 360 × 1,680 = 604,800 Choice (C), First Frame: 7 × 6 × 5 × 4 = 840 Choice (C), Second Frame: 7 × 6 × 5 × 4 = 840 Choice (C), Total: 840 × 840 = 705,600 Choice (D), First Frame: 10 × 9 × 8 × 7 = 5,040 Choice (D), Second Frame: 4 × 3 × 2 × 1 = 24 Choice (D), Total: 5,040 × 24 = 120,960 Only (B) and (C) have more than 500,000 possibilities, so the correct answers are choices (B) and (C).

*****The length of a rectangle is 2 more than twice its width, and the area of the rectangle is 40. What is the rectangle's perimeter?

Convert the word problem into two equations with two variables. "The length is two more than twice the width" can be written as: L = 2W + 2 Since the area is 40 and area is equal to length * width: LW = 40 Since the first equation is already solved for L, plug (2W + 2) in for L into the second equation: (2W + 2)W = 40 (2W^2) + 2W = 40 Since this is now a quadratic (there are both a W^2 and a W term), get all terms on one side to set the expression equal to zero: (2W^2) + 2W - 40 = 0 Simplify as much as possible-in this case, divide the entire equation by 2 - before trying to factor: (W^2) + W -20 = 0 (W - 4)(W + 5) = 0 W = 4 or -5 Since a width cannot be negative, the width equal to 4. Since LW is equal to 40, the length must be 10. Now use the equation for perimeter to solve: Perimeter = 2L + 2W Perimeter = 2(10 + 2(4) Perimeter = 28 Note that guessing would have been a bad idea since the question does not state the values have to be integers.

Ramon wants to cut a rectangular board into identical square pieces. If the board is 18 inches by 30 inches, what is the least number of square pieces he can cut without wasting any of the board? - 4 - 6 - 9 - 12 - 15

Cutting a rectangular board into square pieces means Ramon needs to cut pieces that are equal in length and width. "without wasting any of the board" means that he needs to choose a side length that divides evenly into both 18 and 30. "The least number of square pieces" means that he needs to choose the largest possible squares. With these three stipulations, choose the largest integer that divides evenly into 18 and 30, or the greatest common factor, which is 6. This would give Ramon 3 pieces going one way and 5 pieces going the other. He would cut 3 * 5 = 15 squares of dimension 6 inches x 6 inches. Note that this solution ignored squares with non-integer side length for the sake of convenience, a potentially dangerous thing to do. (After all, identical squares of 1.5 inches by 1.5 inches could be cut without wasting any of the board.) However, to cut squares any larger than 6 inches by 6 inches could only cut 2 squares of 9 inches or 1 square of 18 inches from the 18 inches dimension of the rectangle. The computed answer is correct.

80 is divisible by (2)^x a) x b) 3

D) Construct a prime factor tree for 80; it has four factors of 2 and one factor of 5. That doesn't mean x is 4, however! The problem does not say "80 is equal to 2^x". Rather it says "divisible by." 80 is divisible by 2^4, and therefore also divisible by 2^3, 2^2, 2^1, and 2^0 (any non-zero number to the 0th power equals 1.) Thus x could be 0, 1, 2, 3, or 4, and could therefore be less than, equal to, or greater than 3. Thus, the relationship cannot be determined.

*****If (125^14)(48^8) were expressed as an integer, how many consecutive zeros would that integer have immediately to the left of its decimal point? A) 22 B) 32 C) 42 D) 50 E) 112

Exponent questions usually involve prime factorization, because you always want to find common bases, and the fundamental common bases are prime numbers, Test some values to see what leads to zeros at the end of an integer. 10 = 5 * 2 40 = 8 * 5 * 2 100 = 10 * 10 = 2 * 5 * 2 * 5 1,000 = 10 * 10 * 10 * 10 = 2 * 5 * 2 * 5 * 2 * 5 Ending zeros are created by 10's, each of which is the product of one 2 and one 5. So, to answer this question, determine how many pairs of 2's and 5's are in the expression: (125^14)(48^8) = ((5^3)^14) * (((2)^4) * 3)^8 = (5^42) * (2^32) * (3^8) Even though there are 42 powers of 5, there are only 32 powers of 2, so you can only form 32 pairs of one 5 and one 2.

If (2^k) - (2^(k+1)) + (2^(k-1)) = (2^k)m, what is the value of m?

Factor out a 2^k, divide by 2^k, and then solve for m. m = -1/2

How many factors greater than 1 do 120, 210, and 270 have in common? - One - Three - Six - Seven - Thirty

First determine the factors of 120: 120: 1 * 120, 2 * 60, 3 * 40, 4 * 30, 5 * 24, 6 * 20, 8 * 15, and 10 * 12. Cross off the numbers that are not divisible by 210 and keep the ones that are divisible by 210: 1, 2, 3, 30, 5, 6, 15, and 10. Then keep only the ones that are also divisible by 270: 1, 2, 3, 30, 5, 6, 15, and 10. Ignore 1 and you have seven factors.

If x^y = 64 and x and y are positive integers, which of the following could be the value of x + y? Indicate all such values. * 2 * 6 * 7 * 8 * 10 * 12

First note that the values (x and y) must be all positive integers. If a product is even we know that what ever numbers are being multiplied must be both even to produce the even number. As a result go up the number line and see which even numbers when brought to a power can equal 64. 2 * 2 * 2 * 2 * 2 * 2 or 2^6 = 64 (best found by using a prime factor tree). As a result 2 + 6 = 8 is an answer. 4 * 4 * 4 or 4^3 = 64, so 4 + 3 = 7 is an answer. 6 does not go into 64 when brought to a power (6 * 6 = 36 and 6 * 6 * 6 = 216) 8 * 8 or 8^2 = 64, so 8 + 2 = 10 is an answer. 10: 10 * 10 = 100, too big. 12: we already know this one will be too big.

***** 10! is divisible by (3^x)(5^y), where x and y are positive integers. a) The greatest possible value for x. b) Twice the greatest possible value for y.

First, expand 10! as 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 (Note do not multiply it out.) Note that 10! is divisible by (3^x)(5^y), and the question asks for the greatest possible values of x and y, which is equivalent to asking "What is the maximum number of times you can divide 3 and 5, respectively, out of 10! while still getting an integer answer?" In the product 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1, only the multiples of 3 have 3 in their prime factors, and the multiples of 5 have 5 in their prime factors. Here are all the primes contained in 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 and therefore in 10!: 10: 5 * 2 9: 3 * 3 8: 2 * 2 * 2 7: 7 6: 2 * 3 5: 5 4: 2 * 2 3: 3 2: 2 1: no primes There are four 3's and two 5's total. The maximum values are x = 4 and y = 2. Therefore, the two quantities are equal.

x > 0 > y a) x - y b) (x + y)^2

From the constraint, x is positive and y is negative. So quantity A is definitely positive: x - y = positive - negative = positive. Quantity B is the square of a number, which cannot be negative. Quantity B could be zero, if, for example, x = 2 and y = -2. Note, that when x is so much larger, then b) becomes greater (and sometimes equal to) to value of a).

a, b, and c are integers. 0 < a < b < c a) a! + b! 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 ctake 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.

(x^2) > (y^2) and x > -|y| a) x b) y

If (x^2) > (y^2), then we know the magnitude of x must be greater than y. But, that doesn't help us with the fact that the magnitude of negative numbers increase as they get smaller (-1 > -5), while positive numbers magnitudes get larger as one goes further to the right of the number line (5 > 1). The best way to determine this is by making a chart and testing each example to see if the meet the second condition. Possible options based on the first condition alone: x = 3 and y = 2 x = 3 and y = -2 x = -3 and y = 2 x = -3 and y = -2 Test the second statement now: x > -|y| 3 > -2 (True) 3 > -2 (True) -3 > -2 (False) -3 > -2 (False) Only the first two choices fit the first and the second criteria, and in both situations x is greater than y. As a result a) is correct.

****** (xy)^(1/2) is a prime number, xy is even, and x > 4y > 0. a) y b) 1

If (xy)^(1/2) is a prime number, (xy)^(1/2) could be 2, 3, 5, 7, 11, 13, etc. Square these possibilities to get a list of possibilities for xy: 4, 9, 25, 49, 121, 169, etc. (note that an odd times an odd (squared) is an odd => only two would work in this situation). However, xy is even, so xy must equal 4. Finally, x > 4y > 0, which implies that both x and y are positive. Solve xy = 4, then substitute to eliminate the variable x and solve for y: If xy = 4, then x = 4/y If x > 4y, then (4/y) > 4y. Because y is positive, you can manipulate both sides of the inequality by y and you don't have to flip the sign of the inequality: 4 > 4(y^2). Finally, divide both sides of the inequality by 4: 1 > y^2. Thus, y is a positive fraction less than 1 (it was already given that y > 0). Quantity B is greater. Try Question 29 in Number Properties for more practice.

a) ab > 0 b) bc < 0

If ab > 0, then an and b have the same sign. If bc < 0, then b and c have opposite signs. Therefore, an and must have opposite signs. Therefor, ac is negative, so Quantity B is greater.

During a sale, the local outlet of Chasm sold three times as many jeans as chinos. If they made twice as much profit for a pair of chinos as for a pair of jeans, and sold no other items, what percent of their profits during the sale came from chinos? a) 16(2/3)% b) 20% c) 40% d) 60% e) 83(1/3)%

If all the values given in a problem and its answers are percents, ratios, or fractions of some unknown, then the problem will probably be easiest to solve by stipulating values for the unknowns. In this problem, the two ratios given are 3:1 (jeans sold:chinos sold) and 2:1 (profits per pair of chinos:profits per pair of jeans). The easiest numbers so stipulated are: 3 pairs of jeans sold 1 pair of chinos sold $2 profit/pair of chinos $1 profit/pair of jeans This yields $2 profit from the chinos out of a total $5 in profit: 2/5 = 40%.

n is divisible by 14 and 3. Which of the following statements must be true? Indicate all such statements. * 12 is a factor of n * 21 is a factor of n * n is a multiple of 42

If n is divisible by both 14 and 3, then n must be divisible by the prime factors of both 14 (2 and 7) and 3 (3). As a result any combination of these three number s will create a factor for n. One cannot make twelve from 2, 3, and 7, but they can make 21 (3 * 7). Lastly n must be at least 42 since it is its least common multiple (2 * 3 * 7 = 42).

*****If t is divisible by 12, what is the least possible integer value of a for which (t^2)/(2^a) might not be an integer? - 2 - 3 - 4 - 5 - 6

If t is divisible by 12 then t^2 must be divisible by 144 or 2 * 2 * 2 * 2 * 3 * 3. Therefore, t^2 can be divided evenly by 2 at least four times, so a must be at least 5 before (t^2)/(2^a) might not be an integer. Alternatively, test values. If t = 12, (t^2)/(2^a) = 144/(2^a). Plug in the choices as possible a values, starting with the smallest choice and working up: 144/(2^2) = 36, eliminate 144/(2^3) = 18, eliminate 144/(2^4) = 9, eliminate 144/(2^5) = 4.5, the first choice for which (t^2)/(2^a) might not be an integer is D).

a, b, and c are consecutive integers such that a < b < c < 4. a) The range of a, b, and c b) The average of a, b, and c

If the variables were also contained to be positive, they would have to be 1, 2, and 3, making the quantities both equal to 2. However, the variables could be negative, for example, a = -10, b = -9, c = -8. The range of a, b, and c will always be 2 because the integers are consecutive, but the average can vary depending on the specific values. There is not enough information to determine the relationship.

x > 0 a) (5^x)/(5^(x-1)) b) (5^(x-1))/(5^(x-2))

If you know the rule for dividing powers with the same base, this question becomes quite simple. The law of exponents for dividing powers with the same base says that bcbd=bc−d. Quantity A is 5x/5^(x−1)=5^(x−(x−1))=5^(x−x+1)=5^1=5. Quantity B is 5^((x - 1) - (x - 2)) = 5^(x - 1 - x + 2) = 5^1 = 5. So Quantity A is equal to Quantity B. Choice (C) is correct.

*****Set X is a set of n distinct integers, where n > 1. p is the mean of Set X but not a member of Set X. Which of the following must be true? Indicate all such statements. The range of Set X is greater than the standard deviation of Set X. If Set Y contains all the members of Set X and p(and has no other members), the standard deviation of Set X is greater than the standard deviation of Set Y. The standard deviation of Set X is not a member of Set X. * The range of Set X is greater than the standard deviation of Set X. * If Set Y contains all the members of Set X and p(and has no other members), the standard deviation of Set X is greater than the standard deviation of Set Y. * The standard deviation of Set X is not a member of Set X.

In this All-That-Apply question, we need to keep track of a lot of information and test different scenarios to determine whether each statement must be true. Starting with the information in the question stem, we know that Set X is a set of n distinct integers, where n > 1. This means that Set X has at least 2 terms (up to an infinite number of terms), but they are all different. We also know that p is the mean of Set X, and that it is NOT a member of Set X. We should keep the stipulation that p is not a member of Set X in mind to ensure we use permissible scenarios when we test the answer choices. Choice (A) asks us to compare the range of Set X to the set's standard deviation. We could test this by calculating the range and standard deviation for many different sets, but it is more efficient to think critically: The range represents the difference between the largest and smallest elements of a set. Roughly speaking, the standard deviation measures the "standard" difference between any element and the mean. As such, the range represents the span from top to bottom, whereas the standard deviation can be no bigger than the difference between the top or bottom number and the mean, which is towards the middle of the set. Therefore, the range is larger than the standard deviation with one exception—when all the numbers in the set are equal, as in {7, 7, 7}, which results in both the range and standard deviation being zero. However, the question stem stipulates that the integers in Set X are distinct, and that the mean p is not a member of the set. Therefore, this special case is not permissible according to the rules of the problem. Thus, the range is always greater than the standard deviation, and Choice (A) is a correct choice. Choice (B) asks us to understand what happens to the standard deviation when we add a new element to the set. Remember that standard deviation is a measure of how tightly packed or loosely spread a set of data is. The choice stipulates that Set Y is the same as Set X, except that it also includes p, the mean of Set X. By inserting p into the set to form Set Y, we are making the data more compact and the standard deviation smaller. After all, standard deviation is a measure of "standard"

a) The remainder when (10^11) is divided by 2. b) The remainder when (3^13) is divided by 3.

It is not necessary to calculate (10^11) or (3^13). Because 10 is an even number, so is (10^11), and 0 is the remainder when any even is divided by 2. Similarly, (3^13) is a multiple of 3 (it has 3 among its prime factors), and 0 is the remainder when any multiple of 3 is divided by 3. Therefore, the quantities are equal.

*****Three friends ate 14 slices of pizza. If two of the friends ate the same number of slices, and the third ate 2 more slices than each of the other two, how many slices were eaten by the third friend? a) 3 b) 4 c) 5 d) 6 e) 7

Let P = the number of slices of pizza eaten by each of the two friends who eat the same amount. Let T = the number of slices of pizza eaten by the third friend. T = P + 2 P + P + T = 14 Substitute (P + 2) for T in the second equation: P + P + (P + 2) = 14 3P + 2 = 14 3P = 12 p = 4 Solve for T: T = P + 2 = 4 + 2 = 6

t > 0 a) The area of a square with a side length of t((5)^1/2) 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 t((5)^1/2) is (t((5)^1/2))^2=t((5)^1/2‾× t((5)^1/2) = 5t^2. 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 5t^2, with Quantity B, which is 6t2. Since t > 0, t ≠ 0. The square of any nonzero number is positive. Here, t^2 is positive. So 5t^2 < 6t^2 and Quantity B is greater. Choice (B) is correct.

How many positive integers that are less than 741 are NOT multiples of both 4 and 7 ? a) 483 b) 635 c) 684 d) 714 e) 728

Let's first find the number of positive integers less than 741 that are multiples of both 4 and 7, and then let's subtract that number from the number of positive integers that are less than 741. Since the number of positive integers less than 741 is 740, we will subtract from 740 the number of positive integers less than 741 that are both multiples of 4 and 7. Now 4 and 7 have no common factors greater than 1. So the least common multiple of 4 and 7 is 4 × 7 = 28. Now let's divide 740 by 28 to find how many positive multiples of 28 there are that are less than or equal to 740. Now 740 divided by 28 is 26 with a remainder of 12. The positive multiples of 28 that are less than 741 are 1 × 28, 2 × 28, 3 × 28, 4 × 28, ..., 26 × 28. There are 26 (740/28 [without remainder]) positive integers less than 741 that are both multiples of 4 and 7. Let's note that 26 × 28 = 728 is the greatest multiple of 28 that is less than 741. Since the number of positive integers that are less than 741 and are multiples of 4 and 7 is 26, and the number of positive integers that are less than 741 is 740, the number of positive integers that are less than 741 and are not multiples of 4 and 7 is 740 − 26 = 714. Choice (D) is correct.

*****Each new car produced in a certain factory includes at least one of the following three options: Bluetooth, navigation system, and sunroof. Of the cars that came off the production line yesterday, 25 had Bluetooth, 13 had a navigation system, and 30 had a sunroof. Yesterday, 50 cars were completed. Which of the following could be true about the cars that came off the production line yesterday? Indicate all possible answers. * There were 2 cars that had exactly two of the options and there were 8 cars that had all three options. * There were 3 cars that had exactly two of the options and there were 7 cars that had all three options. * There were 4 cars that had exactly two of the options and there were 7 cars that had all three options. * There were 6 cars that had exactly two of the options and there were 6 cars that had all three options. * There were 7 cars that had exactly two of the options and there were 5 cars that had all three options. * There were 8 cars that had exactly two of the options and there were 4 cars that had all three options.

Let's see how we would take apart this complicated question stem on Test Day. If we add up 25 cars with Bluetooth, 13 with the navigation system, and 30 with the sunroof, we get 68 cars. We know that the number 68 is too high, however, not only because we were told that exactly 50 cars were manufactured but also because cars with exactly two features would be counted twice and cars with all three features would be counted three times. The correct answer choice(s) will give us a way to account for the discrepancy between 68 and 50. Choice (A) tells us that 2 cars had exactly two options and that 8 cars had all three options. The 2 cars with two options each would have been double-counted in the total of 68, and the 8 cars with three options each would have been triple-counted in the total of 68. If we subtract the number of double-counted cars and twice the number of triple-counted cars, we'll have the actual number of cars manufactured: 68 - 2 - (2)(8) = 68 - 18 = 50. Since we are looking for an answer choice that gives us exactly 50 cars, choice (A) is one of the correct answer choices. Here's how this same math and logic apply to the remaining choices: Choice (B): 68 - 3 - (2)(7) = 68 - 17 = 51. Discard. Choice (C): 68 - 4 - (2)(7) = 68 - 18 = 50. This is correct. Choice (D): 68 - 6 - (2)(6) = 68 - 18 = 50. This is correct. Choice (E): 68 - 7 - (2)(5) = 68 - 17 = 51. Discard. Choice (F): 68 - 8 - (2)(4) = 68 - 16 = 52. Discard. Choices (A), (C), and (D) are correct.

x^2 + 3x - 70 = 0 a) x b) 4

Let's try to use reverse FOIL to factor the left side of the equation x^2 + 3x - 70 = 0, which is x^2 + 3x - 70. We know that x^2 + 3x - 70 can be factored into (x + a)(x + b), where an and b are constants. Testing some numbers, we find that x^2 + 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.

A: The least common multiple of 22 and 6. B: The greatest common factor of 66 and 99.

List out the prime factors of 22 (2 and 11) and 6 (2 and 3). Then multiple the highest repeated values of each: 2 * 3 * 11 = 66. List out the prime factors of 66 (2, 3, and 11) and 99 (3, 3, and 11). Then multiply the shared prime factors including repeats if they have a partner. So we have 3 * 11 = 33.

A movie theatre charges $6 per ticket, and pays $1,750 of expenses each time a movie is shown. How many tickets must be sold each time a movie is shown for the theater to make $1 of profit per ticket. a) 300 b) 325 c) 350 d) 375 e) 400

Make sure you know Profit = Revenue - Cost (of Profit = Revenue - Expenses). The cost each time a movie is shown is $1,750. If the theatre charges $6 per ticket and t is the number of tickets, the revenue is 6t. In order for the profit to be $1 per ticket, the profit must be $1t dollars. Plug these values into the equation Profit = Revenue - Cost: 1t = 6t - 1,750 -5t = -1,750 t = 350

a) 100/0.25 b) 100 * (1/4)

Make the quantities look more alike. Replace the decimal 0.25 that appears in Quantity A with its fractional equivalent of 1/4. Then Quantity A is 100(1/4). To divide by a fraction, invert the fraction and then multiply. Then Quantity A is 100(4)1. Quantity A, which is equal to 100(4), is greater than Quantity B, which is equal to 100/4. Choice (A) is correct.

a) The sum of all integers from 1 to 100, inclusive. b) The sum of all the even integers from 1 to 100, inclusive.

No math is required to solve this problem. Note that the number from 1 to 100 include 50 even integers and 50 odd integers. The first few odds are 1, 3, 5, etc. The first few evens are 2, 4, 6, etc. Every even is 1 greater than its counterpart ( 2 is 1 greater than 1, 4 is 1 greater than 3, 6 is 1 greater than 5, etc.) Not only is Quantity B greater, it's greater by precisely 50.

a, b, and c are integers such that a < b < c. a) (a + b + c)/3 b) b

Note that (a + b + c)/3 is just another way to express "the average of a, b, and c." The average of a, b, and c would equal b if the numbers were evenly spaced (such as 1, 2, and 3 or 5,7, and 9), but that is not specified. For instance, the integers could be 1, 2, 57 and still satisfy that a < b < c constraint. In that case, the average is 20, which is greater than b = 2. The relationship cannot be determined from the information given.

What is the value of (44^(5/2)/(11^(3/2))) (Exponents and Roots: Question 20)?

Note that 44 can be broken into 11 * 4. ((11 * 4)^5/2) / (11^(3/2)) = (11^(5/2))/(11^(3/2))) * 4^(5/2)/1 (11^1) * (4^1/2)^5 = (11) * (2)^5 = (11) * (32) = 352

If a and b are integers such that a > b > 1, which of the following cannot be a multiple of either a or b? - a - 1 - b + 1 - b - 1 - a + b - ab

Note that it can be a multiple of a OR b. Before trying to test numbers, think of this problem logically. a multiple of a number must be greater or equal to that number. If b is the smallest number and you subtract 1, or any number for it, you get a number that is smaller that b and as a result would give a number that is not a multiple of a or b. As a result, c is the correct answer.

If a = 16b and b is a prime number greater than 2, how many positive distinct factors does a have?

Note that they are asking for distinct factors not distinct prime factors. Since this is a fill in the blank question we know that what ever prime factor create than two we put in for b, the number of distinct factors for a must be the same. For example: b = 3: 16 * 3 = 48 (1 * 48, 2 * 24, 3 * 16, 4 * 12, and 6 * 8) => 10 distinct factors If we were to put in 5 for be, we would still get 10 distinct factors for 16 * 5 = 80. (1 * 80, 2 * 40, 4 * 80, 5 * 16, and 8 * 10)

Question 13/Quant 3/Test 2

Note the easier way is to determine the quantity unwanted and then subtract that percentage from 100% (3 + 14)/(11 + 5 + 6 + 16 + 2) = approximately 30% 100 - 30 = 70%

If x is a number such that 0 < x <= 20, for how many values of x is 20/x an integer? - Four - Six - Eight - Ten - More than ten

Notice that the problem did not say that x had to be an integer, Therefore, the factors of 20 will work (1,2,3,4,5,10, and 20), but so will 0.5, 0.1, 0.25, 2.5, etc. It is possible to divide 20 into fractional parts-for instance, something 20 inches long could be divided evenly into quarter inches (there would be 80 of them, as 20/0.25 = 80). There are an infinite number of x values that would work (it is possible to divide 20 into thousandths, millionths, etc.), so the answer (E). It is very important on the GRE to notice whether there is an integer constraint on a variable or not! Any answer like "More than 10" should be a clue that this problem may be less straightforward than it seems.

*****If (0.000027 * (10^x))/(900 * (10^-4)) = 0.03 * 10^11, what is the value of x?

One good approach is to convert 0..000027, 900, and 0.03 to powers of 10: (27 * (10^-6) * 10^x) /(9 * (10^2) * (10^-4)) = 3 * (10^-2) * 10^11 Now combine the exponents from the tens with base 10: (27 * (10^(-6 + x)) * 10^x) / (9 * (10^-2)) = 3 * 10^9 Since 27/9 = 3, cancel the 3 from both sides, then combine powers of 10: (10^(-6 + x))/(10^-2) = 10^9 10^(-6 + x - (-2)) = 10^9 10^(-4 + x) = 10^9 Thus, -4 + x = 9, and x =13.

The "aspect ratio" of a computer monitor is the ratio of the monitors width to its height. If a particular monitor has an aspect ratio of 16:9, and a perimeter of 100 inches, how many inches wide is the monitor? a) 18 b) 25 c) 32 d) 36 e) 64

Rather than assigning separate variables to width and height, define them both un terms id the same unknown multiplier, based on the ratio given: Width = 16m Height = 9m Remember that the question asks for the width, so answer for 16m, not for m! The perimeter of a rectangle is equal to 2(length + width), or in this case 2(width + height): 100 = 2 * (16m + 9m) 100 = 50m m = 2 16m = 32 An alternative method depends on the same underlying logic, but forgoes the algebra. Suppose the dimensions were 16 inches and 9 inches. This would yield a perimeter of 50 inches. Double the width and height to double the perimeter.

If Nash had 12 grandchildren and three times as many granddaughters as grandson, how many granddaughters did he have? a) 3 b) 4 c) 6 d) 8 e) 9

Rather than assignung separate variables to the granddaughters and grandsons, define them both in terms of the same unknown multiplier, based on the ratio given: Number of granddaughters = 3m Number of grandsons = m Note that you are solving for 3m, not of m! 3m + m = 12 4m = 12 m = 3 3m = 9 Alternatively, suppose that Nash had exactly one grandson and three granddaughters. That would sum ti four grandchildren altogether. Triple the number of grandsons and granddaughters to triple the number of grandchildren.

*****Question 15 (Test 2/Quant 2) (Test 2/Quantitative 2/Question 11)

Read the chart and question carefully!!! The chart shows the number of two pointers made, not the score from 2007-2008 season (2,986 * 2 = 5,972) (5,972/8,245) * 100 = 72.4%

*****Which of the following is equal to ((10^-8)(25^7)(2^16))/((20^6)(8^-1))?

Reorder terms so that they all have positive powers. (25^7)(2^16)(8) / (20^6)(10^8) Break all the numbers into their prime factors and cross of common terms. Numerator: (25^7): (5 * 5)^7 => (5^2)^7 => 5^14 (2^16): 2^16 (8): 2^3 Denominator: (20^6): (5 * 2 * 2)^6 => (5 * (2^2))^6 => (5^6) * (2^12) (10^8): (5 * 2)^8 => (5^8) * (2^8) Note what you can cross off (be careful): (2^3)/(2^4) = 8/16 = 1/2

S = {2, 4, 8, x, 13, 16, 20} x is the median of the numbers in set S. a) The average (arithmetic mean) of the numbers in set S b) x

Seeing that x is the median we know that it must be greater than or equal to 8 or less than or equal to 13. The reasoning for this is because, for x to be the median it could not be any number outside of this range otherwise it would be in a different position. For example, if x were 18, then 13 would be the median and not x, while if x were 3, then 8 would be the median. But we are told that x must be the median. To figure out whether x is greater or if the arithmetic mean is greater, take the two extreme end points from 8<=x<=13, so 8 and 13 and put them into the arithmetic mean equation. (2 + 4 +8 + 8 + 13 + 16 + 20)/10 appoximatly 10.14 > 8 (2 + 4 + 8 + 13 + 13 + 16 + 20)/10 approximately 10.86 < 13 as a result the answer is D) the solution can not be determined

a > 1 a) 3/(a/a+1)+(1/−1+a) b) 3

Simplify a so that it becomes 3 * ((a^2) - 1)/((a^2) + 1). Since the numerator will always be smaller than the denominator, we know that we will always be taking a percentage of three, making the answer for a always smaller than 3, hence why (B) is correct. Better explanation Test 2/Quant 3/Question 4. TAKEAWAY: When determining the value of complex fractions, simplify as much as possible before making your comparison.

abc is a three-digit number in which a is the hundreds digit, b is the tens digit, and c is the units digit. Let &(abc)& = (2^a)(3^b)(5^c). For example, &(203)& = (2^2)(3^0)(5^3) = 500. For how many three-digit numbers abc does the function &(abc)& yield a prime number? - Zero - One - Two - Three - Nine

Since a prime number has only two factors, 1 and itself, (2^a)(3^b)(5^c) cannot be prime unless the digits a, b, and c are such that two of the digits are 0 and the third is 1. For instance, (2^0)(3^1)(5^0) = (1)(3)(1) = 3 is prime. Thus, the only three values of abc that would result in a prime number &(abc)& are 100, 010, and 001. However, only one of those three numbers (100) is a three-digit number.

If a, b, and c are multiples of 3 such that a > b > c > 0, which of the following values must be divisible by 3? Indicate all such values. * a + b + c * a - b + c * abc/9

Since all the values are multiples of three we can say 3a > 3b > 3c > 0. Substitute the values into possible answer choices. * 3a + 3b + 3c = 3(a + b + c), the sum of a, b, and c is just some number. If multiplied by three, it is guaranteed to be a multiple of three. * 3a - 3b + 3c = 3(a - b + c), once again a, b, and c are just some number. When multiplied by three we get a number divisible by 3. * (3a)(3b)(3c) = 27abc, 27 is a factor of 3 and as a result, if you were to multiply 27 times any number, abc in this instance, you would get a number divisible by 3. All three are correct.

a>b>0 a) 1/((b/a)+1) b)(ab-((a)^2))/((a^2)-(b^2))

Since both quantities contain similar variables, let's try using the strategy of Making the Quantities Look Alike by simplifying. Quantity B: (-a(a - b))/(a + b)(a - b) -a/(a + b) So, Quantity B is the negative of Quantity A. Since a and b are positive, Quantity A is a positive value and Quantity B is a negative value. Therefore, choice (A) is correct.

For how many positive integer values of x is 65/x an integer?

1,5,13, and 65 (do not forget the identity (1 * 65).

a and b are both greater than 4 but less than 10. c and d are both greater than 2 but less than 5. a) a + b b) c + d

Since we are dealing with ranges for the variables, the sums of these variables will also be ranges. Since 4 < a < 10 and 4 < b < 10, 4 + 4 < a + b < 10 + 10, and then 8 < a + b < 20. Since 2 < c < 5 and 2 < d < 5, 2 + 2 < c + d < 5 + 5, and then 4 < c + d < 10. Thus, 8 < a + b < 20 and 4 < c + d < 10. Given that these ranges overlap (don't forget that we could pick 4.9 for an and b), either quantity could be greater, or the quantities could be equal. Therefore, the relationship between the quantities cannot be determined. Choice (D) is correct. Use smart numbers if needed, specifically their endpoints (do not forget the values could be decimals). If both an and b are 3 then the left endpoint, excluding decimals, will be 3 and the right endpoint, if using 9 and 9, will be 18. For c and d, test endpoints, excluding decimals, 1 and 1 and 4 and 4 to get a range of 2 and 8. Notice that the range for an and b, 6 to 18, and the range for c and d, 2 to 8, overlap. This means the quantities can be equal, greater than, or less than each other.

If c cherries cost d dollars, then how many dollars would 50 cherries cost? a) 50cd b) (50d)/c c) d/(50c) d) (50c)/d e) c/(50d)

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. Or figure out the formula by using dummy variables. Say that 10 cherries cost $5 ($5 = 10 cherries). This means that 10 cherries/$5 = $2.00 per cherry. Which ultimately means 50 cherries would cost 50 * 2 = $100.

a) (3^(19)) + (3^(19)) + (3^(19)) b) (3^(20))

The values are a bit too difficult to compute, so try making the quantities look alike by factoring out a 319 in Quantity A: 319 + 319 + 319 = 319(1 + 1 + 1) = 319(3) = 319(31) = 319 + 1 = 320. This is the same as Quantity B, so the correct answer is choice (C). We used the law of exponents that says that babc = ba + c, when we said above that 319(31) = 319 + 1.

A car traveled 24 miles, at an average speed of 40 miles per hour. What is the average speed, in miles per hour, at which the car must travel another 48 miles if the total travel time for both distances is to be 116 minutes? a) 26 b) 28 c) 30 d) 36 e) 48

We will be using the distance formula, which is Distance = Rate × Time. The car traveled the first 24 miles at an average speed of 40 miles per hour. So using the formula in the rearranged form Time=DistanceRate, the time the car took to travel the first 24 miles was 24 miles/40 miles-hour = 24/40 hours = 35 hours = 0.6hours. There are 60 minutes in an hour, so in 0.6 hours there are 0.6(60) = 36 minutes. The car took 36 minutes to travel the first 24 miles. Since the total time of travel was 116 minutes, the amount of time in which the car is to travel the remaining 48 miles is 116 - 36 = 80 minutes. Thus, the car is to travel the remaining 48 miles in 80 minutes. Since we want to find the rate in miles per hour, let's convert 80 minutes to hours. There are 60 minutes in an hour, so in 80 minutes, there are 80/60 = 8/6 = 4/3 hours. Let's keep the fraction 43 in this form, since we will be dividing by 4/3 hours. Now let's use the distance formula in the rearranged form Rate = Distance/Time to find the car's average speed over the remaining 48 miles. The car's average speed is to be 48 miles(4/3hours) = 48 (4/3) miles per hour. Now let's find the value of 48/(4/3). We have 48/(4/3)=48×3/4 = 12 × 3 = 36. So, the car's average speed will have to be 36 miles per hour for the remainder of the trip. Choice (D) is correct. Note, you could have also done this problem working in minutes and then converting the final answer to hours (specifically: going from miles per minute to miles per hour).