GRE Question of the Day

Where are the 4 quadrants?

(+,+) is 1 (-,+) is 2 (-,-) is 3 (+,-) is 4

equation for standard deviation

(Xmean - Xfor all values)/N Then take the square root of the final value

The leg length to leg length to hypotenuse length ratio in a 30°-60°-90° triangle?

1 to sq root of 3 to 2

Area of a triangle

1/2 x base x height

All of the angles within a triangle must add to equal?

180

The sum of the degrees of the interior angles of a polygon (equation)

180(n - 2) n= number of sides

Surface area of a cylinder

2 π r^ 2 + 2 π r h

Water flows into an empty 128-gallon drum at opening X and out of the drum at opening Y. If the rate of flow through X is 4 gallons per hour, how many gallons per hour must flow out at opening Y so that the drum is full in exactly 192 hours?

If we want to fill up this drum in 192 hours and it holds 128 gallons, that means that we want to fill this at a rate of less than one gallon per hour, which should help you figure out the ratio we need to calculate. If we fill 128/192 gallons per hour, we will reach our goal. So 128/192 => simplify this by dividing by 64 to get a rate of 2/3 gallons per hour. So we need 2/3 gallons to remain per hour after the 4 is pumped in. That means that we want the out flow to be (4 - 2/3) or 31/3 or 10/3. Answer D

Celia has exactly 3 times as many French as non-French photographs in her collection. Which of the following CANNOT be the number of photographs in Celia's collection? 100 84 76 69 32

Since we know that 3 out of every 4 of Celia's photographs is a French photograph, and photographs do not come in fractions, we know that the amount of photographs in the collection must be evenly divisible by 4.

Celia has exactly 3 times as many French as non-French photographs in her collection. Which of the following CANNOT be the number of photographs in Celia's collection? 100 84 76 69 32

Since we know that 3 out of every 4 of Celia's photographs is a French photograph, and photographs do not come in fractions, we know that the amount of photographs in the collection must be evenly divisible by 4. answer is 69 (not divisible by 4).

A certain gallery is hanging a series of 7 paintings. All the paintings will be exhibited in a row along a single wall. Exactly 2 of the paintings are on panel, the remainder are on canvas. In how many ways can the paintings be exhibited if the works on panel must be the second and sixth in the row?

The correct answer is 240. Here, we have a total of 7 paintings, of which 5 are on canvas and 2 are on panel. The 2 panels must occupy the second and sixth positions in the row, so we'll have a pattern like this: C P C C C P C. Remember, if you have to figure out the number of permutations of two groups that have fixed positions relative to one another, figure out the number of permutations of each group separately and then multiply them. So we have a group of 5 canvases and a group of 2 panels. The number of permutations of the canvases will be 5! or 120. The number of permutations of the panels will be 2! or 2. So the number of permutations of the two groups together will be 120 x 2 = 240.

A certain bank vault contains 600 safe deposit boxes. A thief possesses a sack containing 700 keys, only 300 of which correspond to the safe deposit boxes, with no duplicates. If the thief randomly selects a box and then a key, what is the probability that the thief will be able to open the box? 1/1400 1/600 1/7 1/2 6/7

There are 600 boxes, so the possible outcomes are 600. How many desired outcomes are there? Since the thief has keys for only 300 boxes, then there 300 desired outcomes. So that's a 1/2 probability of of picking one of the boxes with keys. Once he's picked a box, the thief has to pick the right key. Since he has 700 keys, there are 700 possible outcomes. Only 1 key will open the box, though, so there is only 1 desired outcome, for a probability of 1/700 of selecting the right key. Now the probability of these two events both happening is (1/2 x 1/700 = 1/1400)

If x2 + y2 = 16 - 2xy, then (x + y)4 =

This is a great GRE example of things you should know. Knowing how to factor a quadratic equation quickly and identifying a frequently recurring pattern will serve you well. Also, knowing the squares of the first 20 or so numbers will save you time here. The first step is to recognize that x2 + 2xy + y2, which factors to (x + y)2, is lurking in the first equation: x2 + y2 = 16 - 2xy => x2 + 2xy + y2 = 16 => (x + y)2 = 16 Rewrite (x + y)4 as ((x + y)2)2 and replace the (x + y)2 with 16: 162 = 256

If a + c = -15 and b - c = 10, then a + b = -25 -5 5 25

This is a system-of-equations problem. As you can see below, when we add the two equations, the c's cancel out and we are left with a + b. a + c = -15 + b - c = 10 a + b + c - c = -5 or a + b = -5 The correct answer is B.

A fair coin is to be tossed 5 times. What is the probability that exactly 3 of the 5 tosses result in heads?

We could count the number of ways of arranging heads and tails one by one (e.g. HHHTT, HHTHT, HTHHT, etc.), but this method is time-consuming and prone to error. A safer way would be to calculate how many different ways these 3 heads and 2 tails can be arranged in sequence. The formula for calculating the number of ways to arrange n items in a sequence is n!, that is, n(n - 1)(n - 2)...(3)(2)(1). This would make 5! the number of ways we could sequence 5 items, i.e. 5 x 4 x 3 x 2 x 1 = 120. However, in a case like this one, some of the sequences will be repeated in this calculation, as some of the items are identical to each other. For example, if we label the first heads-toss as H1, and the second as H2, then H1-H2-H3-T1-T2 and H2-H1-H3-T1-T2 are two different sequences that are actually identical for the purposes of this question (since both count as HHHTT). In situations like this, we must divide our initial result by m!, where m represents the number of identical items. Since we have 3 identical heads and 2 identical tails, we must calculate 120/(3!)(2!) = 120/(3 x 2 x 1)(2 x 1) = 120/12 = 10. Since there are 10 possible ways to have 3 heads and 2 tails, and 32 total possible outcomes, our answer is 10/32 = 5/16. (32 possible outcomes is calculated by (1/2)^5 since there are 5 coin tosses)

A certain machine produces toy cars in an infinitely repeating cycle of blue, red, green, yellow and black. If 6 consecutively produced cars are selected at random, what is the probability that 2 of the cars selected are red

We need to find possible and desired outcomes. What are the possible outcomes here? Well, we're going to be randomly selecting 6 consecutively produced cars and we want to know what the probability is that 2 of those cars will be red. How many different groups of 6 consecutive cars are there? Well, we've got 5 colors, so any one of them could be the first in the string of 6 that we choose. That gives us 5 different strings of 6. So 5 is the number of possible outcomes here. What about desired outcomes? We need to know how many of those 5 strings have 2 red cars. What does getting two red cars depend on? It depends on the color of the first car of the string of 6. For example, if blue is the first car of the string, then you'd get blue, red, green, yellow, black and then blue again. So, the only way to get 2 red cars is if the first car of the string is red, because then we'd have red, green, yellow, black, blue and then red again as the sixth car. So 1 out of the 5 strings will contain 2 red cars, so that's a probability of 1/5

A certain machine produces toy cars in an infinitely repeating cycle of blue, red, green, yellow and black. If 6 consecutively produced cars are selected at random, what is the probability that 2 of the cars selected are red?

We need to know how many of those 5 strings have 2 red cars. What does getting two red cars depend on? It depends on the color of the first car of the string of 6. For example, if blue is the first car of the string, then you'd get blue, red, green, yellow, black and then blue again. So, the only way to get 2 red cars is if the first car of the string is red, because then we'd have red, green, yellow, black, blue and then red again as the sixth car. So 1 out of the 5 strings will contain 2 red cars, so that's a probability 1/5

How to calculate the diagonal length in a square?

a^2 + b^2 = c^2 (where c is the diagonal)

How do you breakdown an unperfect square root?

for example: square root of 45 a = √45 or √(5*9) This can be simplified further, into 3√5 So, basically take its factors, find the highest number that is a perfect square, take the root, and multiply it by the remaining square-rooted factor

What formula would you use to solve each one of these total combinations? Any buyer of a new sports car has to pick between 2 of 5 options for seat colors and 3 of 4 options for dashboard accessories. How many different combinations of colors and dashboard options are available to this buyer

n!/(r!(n-r)!), r= objs chosen n= total number of objs

Permutation formula: equation. Used for what kind of problem?

nPk = n! / (n-k)! n= total number of groups k= number of groups per set "If each tour must include exactly 4 cities, and if there are exactly 10 cities available, how many different tours are possible?"

What is the equation for finding the volume of a cylinder?

volume = (r^2)(pie)(h)

What is the side rule for the 45-45-90 triangle?

x-x-sqrt2(x)

Area of a circle

π r ^ 2