GRE review- Quantitative reasoning

If y is an integer, which of the following must be an odd integer? y2 + 4y + 4 y2 + 3y + 8 y2 - 7y + 3 y2 - 11y - 10 y2 + 8y - 3

Because there are variables in the answer choices, we can use Picking Numbers to determine the correct answer. Let's keep in mind that when we use 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 choose. First, let's substitute 2 for y in all the expressions in the answer choices. If the resulting value is even, then we can eliminate that answer choice: (A) gives us 16. Eliminate. (B) gives us 18. Eliminate. (C) gives us -7. Leave. (D) gives us -28. Eliminate. (E) gives us 17. Leave. Next, we should use an odd number to eliminate (C) or (E). Because 1 is the smallest positive odd number, it will be the easiest to work with. (E) gives us 6. Eliminate. Choice (C) is the only remaining answer. Therefore, choice (C) must be correct.

x < 0 < y + z z ≠ 0 y+zx yz Quantity A is greater. Quantity B is greater. The two quantities are equal. The relationship cannot be determined from the information given.

Let's first consider Quantity A. The numerator is specified to be positive , and the denominator is specified to be negative. The quotient of a positive number divided by a negative number is always negative, but the magnitude of Quantity A cannot be determined. You may have been tempted to immediately declare Quantity B to be larger, because the y + z is positive, but be careful. While y + z > 0, there's nothing stopping either y or z alone from being negative. This makes it impossible for us to know whether Quantity B is positive, negative, or even 0. Therefore, the relationship cannot be determined from the information given. Choice (D) is correct.

The number of square units in the area of circle O is equal to 16 times the number of units in its circumference. Which of the following are diameters of circles that could fit completely inside circle O ? Indicate all such diameters. 25364248546670

Let's say that the radius of circle O is r. The area of circle O is πr2, the diameter of circle O is 2r, and the circumference of circle O is 2πr. Start by finding the diameter of circle O: πr2πr2r2r2r=====16(2πr)32πr32r3264 So the diameter of circle O is 64 and any circle with a diameter less than 64 would fit completely inside. The correct answers are choices (A), (B), (C), (D), and (E).

a > 0 and b < 0. Quantity A: 1/a^2 Quantity B: 1/b^2 when d Quantity A is greater. Quantity B is greater. The two quantities are equal. The relationship cannot be determined from the information given.

Picking numbers quickly shows that the relationship cannot be determined. If a = 1 and b = -2, then Quantity A is 112=11=1 and Quantity B is 1(−2)2=14. In this case Quantity A is greater. If a = 2 and b = -1, then Quantity A is 122=14 and Quantity B is 1(−1)2=11=1. 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.

probability formula is Probability

Probability=Number of desirable outcomes/Number of possible outcomes

A jar contains 3 red, 5 green, 2 blue, and 6 yellow marbles. A marble is chosen at random from the jar. After replacing the first marble, a second marble is chosen. A: The probability of choosing a green marble and then a yellow marble B:11/16

Since 5 is the number of green marbles in the jar and 16 is the total number of marbles in the jar, the probability of selecting a green marble is 516. Applying the same formula to the yellow marble for the second selection, the probability of selecting a yellow marble is 616. To find the probability of selecting 1 green and then 1 yellow marble, we take the product of these two probabilities: 516×616=30256 Without reducing the fraction, it is clear that 30256<12 and 1116>12. The correct answer is choice (B).

The integers x and y are both positive and xy > 84. The remainder when xy is divided by 84 is 36. Quantity A:The remainder when y is divided by 12 Quantity B: 4 Quantity A is greater. Quantity B is greater. The two quantities are equal. The relationship cannot be determined from the information given.

Since the remainder when xy is divided by 84 is 36, and xy > 84, an equation can be written as xy = 84N + 36, where N is a positive integer. If N = 1, then xy = 84N + 36 = 84(1) + 36 = 84 + 36 = 120. Thus, xy could be equal to 120. If x = 3, then y=1203=40. When y, which is 40, is divided by 12, the quotient is 3 and the remainder is 4. In this case, both Quantity A and Quantity B are 4, so the quantities are equal. However, if xy = 120, x could also be equal to 4, in which case y=1204=30. When y, which is 30, is divided by 12, the quotient is 2 and the remainder is 6. In this case, Quantity A is 6 and Quantity B is 4, and Quantity A is greater. Since different relationships between the quantities are possible, (D) is correct.

Values in List P are distinct from values in List Q. List R consists of all the values in list P and all the values in list Q. Quantity A: The average (arithmetic mean) of list R Quantity B: The median of list R Quantity A is greater. Quantity B is greater. The two quantities are equal. The relationship cannot be determined from the information given

The question has tables listing the frequency of the values of list P and list Q. A third list, list R, is defined as the combination of lists P and Q. The comparison to be made is between the mean and median of list R. There are 12 + 13 + 9 + 10 + 16 = 60 values in list P and 6 + 4 + 5 + 9 + 16 = 40 values in list Q, so the number of values in list R is 60 + 40 = 100. The total of all the values in list P is 2(12)+ 3(13) + 4(9) + 5(10) + 6(16), which is 24 + 39 + 36 + 50 + 96 = 245. The total of the values in list Q is 5(6) + 6(4) + 7(5) + 8(9) + 9(16). This simplifies to 30 + 24 + 35 + 72 + 144 = 305. So, list R is made up of 100 values that sum to 245 + 305 = 550. The mean of list R is thus 550100=5.5. Since there are 100 values in that list, the median will be the average of the 50th and 51st values. Because some values appear in both lists P and Q, both lists will have to be considered to identify the median value. There are 12 + 13 + 9 = 34 values less than 5 in list P. The value 5 appears in both lists P and Q a total of 10 + 6 = 16 times. So there are exactly 50 values of 5 or less and 50 values of 6 or more. Thus, the median is the average of 5 and 6, which is 5.5. The correct choice is (C).

The table above shows the distribution of randomly distributed variable X. Quantity A:The probability that X≤ 4 Quantity B:50% Quantity A is greater. Quantity B is greater. The two quantities are equal. The relationship cannot be determined from the information given.

The table accompanying the question lists the frequency distribution of a randomly distributed variable, X, and asks you to compare the probability that X ≤ 4 with the value 50%. There are 24 total values, 12 of which are less than or equal to 4. Thus, there is a 50% probability that a random value of X will be less than or equal to 4. The correct choice is (C). To confirm your answer, check your logic and that you used the correct values from the table.

The only people in an auditorium are sophomores, juniors, and seniors. The ratio of the number of sophomores to the number of juniors in the auditorium is 4 to 3. The ratio of the number of juniors to the number of seniors in the auditorium is 5 to 7. The total number of people in the auditorium is 448. How many juniors are in the auditorium?

There are sophomores, juniors, and seniors and no one else. The question provides the ratios of sophomores to juniors and of juniors to seniors. It's worth noting that juniors appear in both ratios. In addition, the question gives the total number of students. The answer choices are potential numbers of juniors in the room. Identify the task Relate the number of total students to the ratios of groups of students to arrive at the number of juniors. Approach strategically First relate the two ratios by way of their common element, the juniors. The juniors are represented by 3 in one ratio and 5 in the other, so find the least common multiple of 3 and 5, which is 15. Then state both ratios with juniors represented by 15. sophomoresjuniors=43=2015 juniorsseniors=57=1521 Now the ratio of sophomores to juniors to seniors is 20:15:21. Since each unit in the ratio represents a "basket" with a certain number of people in it, this can be written as 20x:15x:21x, where x represents the number of people in each "basket." Since the total number of people is 448, 20x + 15x + 21x = 448; 56x = 448; x = 8. Since the number of juniors is 15x, the number of juniors is 15 × 8 = 120. Choice (B) is correct. Confirm your answer Double-check that you accurately copied the numbers from the question and set up the ratios correctly, and make sure your arithmetic is correct. Also, make sure to answer the right question; (C) is the number of sophomores, and (D) is the number of seniors.

Lines u and v form 4 angles that each measure 90 degrees. The points (−4, 7) and (8, 23) are on line u. The points (−68, −16) and (−40, t) are on line v. What is the value of t ?

When x1 ≠ x2, the slope m of a line going through the points (x1, y1) and (x2, y2) is given by m=y2−y1x2−x1. Then the slope of line u, which goes through the points (−4, 7) and (8, 23), is: 23−78−(−4)=1612=16÷412÷4=43 So the slope of line u is 43. When two lines in a plane form an angle whose measure is 90 degrees, the lines are perpendicular. The slopes of two lines that are perpendicular to each other, where each of the two lines is not parallel to a coordinate axis, are negative reciprocals. If m is a non-zero slope, the negative reciprocal of m is −1m. Since the slope of line u is 43, the slope of line v is the negative reciprocal of this, which is −1(43). Thus, the slope of line v is −34. Now the points (−68, −16) and (−40, t) are on line v. So, the slope of line v is t−(−16)−40−(−68)=t+16−40+68=t+1628. Since the slope of line v is −34, set up the equation t+1628=−34. Solve this equation for t: Thus, t = −37. (B) is correct.

If p + q > 0, ps3 < 0, and q(r + 1)2 < 0, then which of the following must be true? Select all the true statements. p > 0 q < 0 r > 0 s < 0

This is a question that requires familiarity with working with positive and negative numbers. There are a few important facts to remember: The product of two positive numbers is positive. The product of a positive number and a negative number is negative. The product of two negative numbers is positive. Since q(r + 1)2 < 0, q(r + 1)2≠ 0. When the product of a group of numbers is not 0, each number must be non-zero. So q is not equal to 0 and r + 1 is not equal to zero. The square of any non-zero number is positive. So since r + 1 ≠ 0, (r + 1)2 > 0. Since q(r + 1)2 < 0 and (r + 1)2 > 0, q < 0. (B) will thus be part of the answer. Since p + q > 0, p > −q. Since q < 0, −q > 0. Since p > −q and -q > 0, p > 0. So (A) will be part of the answer. Since p > 0 and ps3 < 0, s3 < 0. Any number raised to an odd exponent keeps its original sign; since s3 is less than 0, s is also less than 0, and (D) will be part of the answer. This leaves (C). Having determined that r + 1 ≠ 0, this means that r≠−1. However, r could be equal to other negative numbers, so it isn't necessary to have r > 0. (C) does not have to be true. (A), (B), and (D) are correct.

The table shows selected percentiles and corresponding standard deviations below the mean in a normal distribution. Variables A and B are normally distributed. Variable A has a mean of 50 and a standard deviation of 10. Variable B has a mean of 80 and a standard deviation of 20. If the probability that A is at or below the number x is 0.20, and the probability that B is at or above the number y is 0.40, what is the value of y - x ?

You are provided with a table that shows the frequency of various percentiles on a normal distribution curve in terms of their standard deviations below the mean. The question states the mean and standard deviations of variables A and B, and it further defines x as the value for which the probability that A is at or below that value is 0.20, and y as the value for which the probability that B is at or above that value is 0.40. The question asks you to determine the value of y - x. Percentiles can be converted to probabilities; the 20th percentile is the value with a probability of 0.20 that a randomly selected value will be below (to the left of on a graph) that value. So, x will be 0.84 standard deviations less than the mean of A, meaning that x is 50 - 0.84(10) = 41.6. Similarly, at 40th percentile the standard deviation is 0.25, so y will be 0.25 standard deviations above the mean of B. Use addition because the question says "above" the mean: y is 80 + 0.25(20) = 85. Finally, y - x = 85.0 - 41.6 = 43.4; the correct answer is choice (E). Check your math and logic and that you extracted the correct values from the table.

what is the ratio of the number of workers in the Blue Collar category in 2005 to the number of such workers in 2011

find the % for each pie chart and then divide the numbers to get the fraction.