GRE Math practice questions copied

a^2 − 2ab + b^2= what

(a − b)^2

If you see "at least" and " probability"

1- the probability of something not happening

If Eugene can complete a project in 4 hours and Steve can complete the same project in 6 hours, how many hours will it take Eugene and Steve to complete the project if they work together?

1/ total time of A + 1/ total time of B \= fraction -> flip the problem - or" entire job/ hours spent"

STEVE'S PROPERTY TAX IS $140 LESS THAN PATRICIA'S PROPERTY TAX. IF STEVE'S PROPERTY TAX IS $1,960, THEN STEVE'S PROPERTY TAX IS WHAT PERCENT LESS THAN PATRICIA'S PROPERTY TAX, TO THE NEAREST 0.1 PERCENT?

1940+140=2100 2100/140 = .066....

what is the probability of drawing 2 blue marbles in a row if you replace the first marble after taking it out of the bag if the bag contains 5 marbles, 4 blue, and 1 red

4/5 *4/5 = 16/25

what is the probability of drawing 2 blue marbles in a row if you don't replace the first marble after taking it out of the bag if the bag contains 5 marbles, 4 blue, and 1 red

4/5*3/4

1/[4^(-1/2) ] =

4^1/2 = square root of 4

A clothing store discounts the price of a jacket by 20%. After the discount, the price of the jacket is $18.00. What was the price of the jacket before the discount?

Here is a direct solution. Let's say that the original price of the jacket was x dollars. This price is discounted by 20%. The decimal equivalent of 20% is 02. So the price, in dollars, after the discount, is x-0.2x-08r. The discounted price is $18.00. So we have the equation 0.8r 18. Dividing both sides of this equation by 0.8, we have X= 18/.8 = 18*10/.8*10= 180/8 = 22.5. The price of the jacket before the discount was $22.50.

A toy company makes 8 different types of toys. If 2/5 of the total toys produced are dolls, and 5/12 of the remaining toys are balls, what fraction of the total production of the company are the remaining 6 types of toys?

Since the fractions in the question stem contain the denominators 5 and 12, say that the total number of employees at the company is 5 × 12 = 60 as 60 is the lowest common denominator for both the numbers 5 and 12 - 60 *(2/5) = 24 dolls - 60-24 = 36 toys - 36*(5/12) = 15 balls - 36-15 =21 neither dolls nor balls - 21/60 = 7/20 = answer

a line inside a square connect from the corner point of DAB to a point on the other side of the square on Side CB. This line is AE in the sqaure ABCD. Segment BE (inside segment CB) is one-third the length of side BC, and AE has a length of 10. What is the length of a side of the square?

Use the Pythagorean theorem to determine the length of AB, which is also a side of the square. Set the length of the short side of the triangle to x. Since that is one-third the length of the side of the square, 3x is that length. Now set up the Pythagorean theorem and solve for x.

two groups sets formula

one group + one group - both groups + neither group

two complicated quantities quantitative comparison

simplify the two quantities

A CERTAIN TRAIN IS TRAVELING AT A CONSTANT RATE. IF THE TRAIN TRAVELS S MILES IN T HOURS, IN HOW MANY HOURS WILL THE TRAIN TRAVEL Y MILES?

t*y/s

T IS A LIST OF 100 DIFFERENT NUMBERS THAT ARE GREATER THAN O AND LESS THAN 50. THE NUMBER X IS GREATER THAN 60 PERCENT OF THE NUMBERS IN T, AND THE NUMBER Y IS GREATER THAN 40 PERCENT OF THE NUMBERS IN T. QUANTITY A ( x-y) QUANTITY B (20)

- they said numbers not integers - so can be fractions - exclsuvie = exclude 0 and 50 - there is an infinite of numbers between 49 and 50 - we can pick , because there is an inifite number between each of them, we can pick between 0 and 50 EXOne : 49.5 = x and y = 1 o 49.5-1= 48 EX two: x = 49.5 and y=40 o 49.5-40 = 9.5 - becuase 48 is bigger but 9.5 is smaller, the answer is d

George is 3 years older than Sydney, who is 4 years younger than Carl. QUANTITY A( The average ,arithmetic mean, age of George, Sydney, and Carl) QUANTITY B ( The median age of George, Sydney, and Carl)

The problem is essentially asking us to compare the average of three numbers to its median. As we are not given the actual ages of the three people, let us use Picking Numbers to make this problem easier. If Sydney is 10, then George is 10 + 3 = 13 and Carl is 10 + 4 = 14. The median of 10, 13, and 14 is 13. In order for the average to equal the median, 13 must be exactly halfway between the other two numbers. As 13 is closer to the larger number, it must be larger than the average, meaning Quantity B is larger than Quantity A. Let us try one more set of numbers just to be sure. If Sydney is 100, then George is 100 + 3 = 103 and Carl is 100 + 4 = 104. Because 103 is still closer to 104 than it is to 100, the median is still larger in this case.

FOR EACH POSITIVE INTEGER N, THE NTH TERM OF THE SEQUENCE S IS 1+(-1)". QUANTITY A (THE SUM OF THE FIRST 39 TERMS OF S) QUANTITY B (39)

- negative 1 being in paranthesis is important o 5- 1^5 § if the one is not in parantehissi does not impact the power - anything to the zero power is one - nth term can be used in the equation o if it was like nth term of the seuqnedce is 1+n o ten position is 11 o 1+(-1)^ 5 = - we add one each time o it is going to be 0,2 - some of the first 39 o it has to be an even number because the sume is going to be 0+2 o so first off we know that they can not be equal o it is going to be close or slightly above o 38 or 40 o because of the 0 , there is going to be a lot of 0s and so there is going to be half - once you see a pattern you are close to the answer and can predict it

In a certain orchestra, each musician plays exactly one instrument. If 1/5 of the musicians plays brass instruments, and the number of musicians playing wind instruments is 2/3 greater than the number of musicians playing brass instruments, what fraction of the musicians in the orchestra plays neither brass nor wind instruments?

- pick a number for the total that would be a lowest common demonator for the bottom part of each of the fractions mentioned which would be 15 for 3 and 5 - so 15 *(1/5) = 3 - 3 + (2/3) *(3) = 5 because the lower 3 and the 3 cancel each other out - 15-3- 5 = 7 - the answer is 7/15

A QUALITY CONTROL ANALYST COLLECTED 200 MEASUREMENTS OF LENGTH. THE RANGE OF THE 200 MEASUREMENTS IS 17 CENTIMETERS, AND ONE OF THE MEASUREMENTS IS 49.5 CENTIMETERS. WHICH OF THE FOLLOWING COULD BE ANOTHER ONE OF THE 200 MEASUREMENTS, IN CENTIMETERS? INDICATE ALL SUCH MEASUREMENTS. 31.0 32.0 33.0 34.0

- we know the range is 17 cm - they randomly pick 49.5 which we can also use - subject 17 from the 49.5 - that is one of the possibilities it could be - 49.5-17 = 32.5 - can't be 32 because it has to be bigger - 49.5 + 17 = 66.5 - 32.5 < answer <66.5 - just make sure you know if it is exclusivbe or inclusive o 32.5 would work if it is inclusive but not work if it is exslucive

which ratio is bigger 1:5 or 1:25

1:25 is the bigger ratio

A group of 4 seniors and 3 juniors is to stand in a straight line so that no two seniors have a junior between them and no two juniors have a senior between them. How many possible arrangements are there?

Since the seniors must be kept together on the line and the juniors must be kept together on the line, you first need to find the number of ways the 4 seniors can be arranged in a row and the number of ways the 3 juniors can be arranged in a row. Then multiply those values together. Finally, you'll need to double that number, because the line might start with the 4 seniors and end with the 3 juniors, or it might start with the 3 juniors and end with the 4 seniors. The number of ways to arrange the 4 seniors is 4! = 4 × 3 × 2 × 1 = 24. The number of ways to arrange the 3 juniors is 3! = 3 × 2 × 1 = 6. Therefore, the number of ways to arrange 4 seniors followed by 3 juniors is 24 × 6 = 144, and the number of ways to arrange 3 juniors followed by 4 seniors is 6 × 24 = 144. Thus, the total number of ways to arrange the juniors and seniors is 144 + 144 = 288

y > 0 A train traveled 3y miles without stopping. The train traveled the first y miles at an average speed of 30 miles per hour, it traveled the next y miles at an average speed of 60 miles per hour, and it traveled the final y miles at an average speed of 80 miles per hour. Quantity A The train's average speed, in miles per hour, when it traveled the 3y miles Quantity B 51

Since y is unknown, pick a number for y and find the train's average speed using the average speed formula, total distance/total time. Since you'll need to find the time the train traveled at each speed, you'll be dividing the distance of y miles by 30 miles per hour, by 60 miles per hour, and by 80 miles per hour. Thus, you want y to be a multiple of 30, 60, and 80. The least common multiple of 30, 60, and 80 is 240, so say y = 240.

In how many different ways can the 8 letters A, B, C, D, E, F, G, and H be arranged if the letters A and B must be next to one another, the letters C and D must be next to one another, the letters E and F must be next to one another, and the letters G and H must be next to one another?

The 4 pairs (in either order) AB, CD, EF, and GH can be arranged in 4! = (4)(3)(2)(1) = 12(2) = 24 ways. Each of the pairs AB, CD, EF, and GH can be arranged in 2! = 2 ways. (For example, we can have AB and BA.) So the number of possible arrangements is 24(2)(2)(2)(2) = 24(16) = 384. Choice (A) is correct.

A bag contains only 3 blue disks, 3 green disks, and 4 orange disks. If 3 disks are selected at random from the bag, what is the probability that 1 of the disks will be green and 2 of the disks will be orange?

There are too many ways that the desired result can be obtained to actually list them, so use the combination formula since the order in which the disks are selected does not matter.

The integers a and b are both greater than 5. If each person in a group of a people is given b cookies from a box of cookies, there will be no cookies left in the box. If each person in a group of a − 5 people is given b + 10 cookies from this box of cookies, there will be no cookies left in the box. Which of the following is an equation correctly describing b in terms of a ? 1. b=a-5/2 2. b=a+10/2 3. b = 2a − 10 4. b = 2a + 10 5. b = 2a − 5b + 10

To find the number of cookies in the box, the number of people and the number of cookies each person is given must be multiplied together. After creating an equation that works for both scenarios, that equation can be solved for b.

In a certain school, the ratio of boys to girls is 5:13. If there are more 72 more girls than boys, how many boys are there?

for this problem you only have two elements to look at 1. 5:13 2. 72 more girls than boys - so for this problem, the actual has to match the ratio (actual being the 72) - if there are 72 more girls than boys, and boys = 5 , then 5-13 = 8 = 72 - 72/8 = 9 - now 9*5 = 45 = answer - the 8 here represents the "more girls than boys" as it is 8 more girls than boys in the original ratio which must transltate to the new actual

In a certain room containing 100 students, 72 students are taking geography, 48 students are taking both geography and history, and 12 students are not taking either geography or history. A student was selected at random among those students taking history. What is the probability that the student is NOT taking geography?

he overlapping sets formula is Total = Group 1 + Group 2 - Both + Neither. In this case, that's Total = Geography + History - Both + Neither. Plug in the numbers given here: 100 = 72 + History - 48 + 12. Solve for History and find there are 64 students taking history. Of these, 48 are also in geography, so 64 - 48 = 16 who take only history. The probability that a student who takes history is not also taking geography is 16/64, which can be reduced to 1/4. Choice (A) is correct.

The ratio of petunias to daffodils in Shelby's garden is 3:4. If he were to plant y more petunias, the new ratio of petunias to daffodils would be 7:8. If he instead removed 2y daffodils from the garden, there would be an equal number of petunias and daffodils. Quantity A (y) Quantity B(5)

p/d=3/4 p= petunias and d = daisies Since the fraction pd could be any fraction that reduces to 3/4, the actual numbers may be multiples of 3 and 4. Thus, this information can also be stated as p = 3x and d = 4x, where x is an integer. If y petunias are added, then there would be 3x + y petunias and 4x daffodils. The equation to express the new ratio is 3x+y/4x=7/8. Next, if 2y daffodils were removed, there would be an equal number of daffodils and petunias. So, p = d − 2y, or 3x = 4x − 2y. Thus, x = 2y. Now there are two equations and two variables: 3x+y/4x=78 and x = 2y. Cross multiply the first equation to get 24x + 8y = 28x. Subtract 24x from both sides: 8y = 4x, or 2y = x. It turns out that, although there are two equations, they are identical. Thus, there are a host of different possible values for x and y, so (D) is correct.

A car traveled x miles at an average speed of 30 miles per hour, and it traveled the y miles at an average speed of 40 miles per hour. How many minutes did it take the car to travel the x + y miles?

Another option would be to pick numbers for x and y. If the car traveled x = 60 miles at 30 mph, then it traveled 2 hours at that speed. If it traveled y = 120 miles at 40 mph, it traveled 3 hours at that speed. It thus traveled 2 + 3 = 5 hours, which is 5 × 60 = 300 minutes. Now plug in 60 for x and 120 for y in each of the answer choices to see which one(s) yield 300. Only (C) does, so it is the correct answer. When you pick numbers with variables in the answer choices, if your numbers result in more than one answer equaling the target number, then you'll need to pick different numbers and test those choices again.

How many of the positive divisors of 192 are also multiples of 6 ?

Divisors are the same thing as factors. First, break 192 down into its prime factorization, because from the prime factorization you can get every other possible factor. The prime factorization of 192 is: 3 × 2 × 2 × 2 × 2 × 2 × 2. If we multiply 3 by 2, we get 6, so 6 is a factor of 192 which is also a multiple of 6, and for every additional time we multiply 3 × 2 = 6 by 2 we get another divisor of 192 which is also a multiple of 6. So the divisors are of 192 that are also multiples of 6 are 6, 12, 24, 48, 96, and 192. So there are 6 factors (divisors) of 192 which are multiples of 6. Choice (D) is correct.

discount problems (dealing with % on sale problems)

If a price is discounted by n percent, then the price becomes (100-n) percent of the original price - 1 minus discount automatically and then 1- discount ppernt of original price = price - Dealing with % "on sale" problems, remember that you can subtract the discount from 100% and then solve. So if something is 30% off, 100%-30% = 70%, so the sale price is 70% of the regular price. (Regular Price)(70%)=(Sale Price) <-- Memorize this formula! If something is increased 25%, it is 100% + 25% = 125% of the original price (or whatever it is--these are not just for prices). So let's say you are reselling something for 25% more than you bought it for: (Price You Bought it For)(125%)=(Resale Price)

How many integers between 500 and 600 have both 3 and 7 as prime factors?

Now, here's the shortcut: 500/21 = 23.8 600/21 = 28.5 We ignore the decimal and subtract: 28-23=5 The reason we ignore the decimal is basically because neither 500 nor 600 are divisible by 21. If they were, you wouldn't get the decimal. If the gap isn't too big, you can just brute force (AKA use the first method and count it), but if there's a big gap, use the division method.

The sum of four consecutive even integers is H, and the smallest of these integers is a. What is the sum of four consecutive even integers where the greatest of these integers is a ? a. h-24 b. h-12 c. h-6 d. h-4 e. 3h

Picking numbers would be a good approach. If the first set of integers were 2, 4, 6, and 8, then a = 2. The sum of these consecutive even integers is 2 + 4 + 6 + 8 = 20, so H = 20. Then, for the second set of four consecutive even integers, a would be the greatest, so the sum of the integers would be -4 + -2 + 0 + 2 = -4. Now plug 20 in for H in each of the choices to see which one(s) work: (A) 20 - 24 = -4. That's a match, so keep this one. (B), (C), and (D) are clearly different from (A). They start with H but subtract different values. Eliminate these. (E) 3 × 20 = 60. Eliminate. The correct answer is (A).

At Jefferson High School, 3 out of 5 students and 2 out of 5 teachers leave campus each day for lunch. The ratio of teachers to students is 1:10, and a total of 230 students and teachers do not leave campus for lunch. QUANTITY A (The number of teachers at Jefferson High who leave campus for lunch) QUANTITY B ( 23)

To compare the quantities, you need to find the number of teachers at Jefferson High who leave campus for lunch. Let t be the total number of teachers. The number of teachers who leave for lunch is 2/5 *T - The only actual number provided is 230, which is the number of students and teachers who do not leave campus. Since 2/5 of the teachers do leave campus, the remaining 1−2/5=3/5 do not. And since 3/5 of the students do leave campus, the remaining 1−3/5=2/5 do not. So, 230 is equal to 2/5 of the students plus 3/5 of the teachers.

The probability that independent events M and N will both occur is 3/8 QUANTITY A ( The probability that event N will occur) QUANTITY B (3/4)

We know that if two events are independent, then the probability of them both occurring is the product of the probabilities of each event occurring. The given information tells us that the probability that event M will occur multiplied by the probability that event N will occur is 3/8 3/8 = 3*1/4*2 =3/4*1/2 So, if the probability that event M occurs is 3/4, then the probability that event N occurs is 1/2. In this case, Quantity B is greater. If the probability that event M occurs is 1/2 , then the probability that event N occurs is 3/4. In this case, Quantity A and Quantity B are equal. Because more than one relationship between the quantities is possible, the correct answer is choice

A pizza parlor offers four different toppings and a choice of either thin or thick crust for their pizzas. The pizza parlor does not offer double of the same topping. QUANTITY A ( The number of different two-topping pizzas that can be ordered from the parlor) QUANTITY B (12)

We want to find the number of unique two-topping pizzas, so we care about the different combinations of toppings and crust, but the order in which they're selected does not matter. This is therefore a combinations problem. We will use the combinations formula to choose 2 toppings from 4 possible toppings. Then using the formula with n = 4 and k = 2, here are 6 possible combinations of toppings for a two-topping pizza. Each of those 6 possible topping combinations could have two types of crust, for a total of 6 × 2 = 12 unique two-topping pizzas that can be ordered. Quantity A is equal to Quantity B, so the correct answer is choice (C).

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?

he 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=1/4. 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 1/6. The rolls of the dice are independent. So the probability that both dice show a 1 is 1/6×1/6=1/36. 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 34, 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.

QUANTITY A ( the sum of the integers from 19 to 31) QUANTITY B (The sum of the integers from 22 to 32 inclusive)

quantity a is bigger - piece by piece strategy