Hard GRE Math Questions: Practice and Tips

Solution:

Although we may be tempted to actually do the math using the online calculator, we must instead remember the following rule:

(a + b + c)² = a² + b² + c² + 2ab + 2ac + 2bc

Using the above identity, we see that the expansion of (500 + 600 + 900)² will contain the terms 500², 600², and 900², plus several other terms, all of which are positive.

Therefore, without calculating each term, we can safely conclude that (500 + 600 + 900)² is much larger than 500² + 600² + 900².

Thus, Quantity B is greater than Quantity A.

Answer: B

Solution:

We are given that the original price equals N, and we can let the sale price = W. Since the closeout price W is 6% of N, we can re-express 6% as 6/100 and create the following equation:

W = 6N/100

W = 3N/50

50W = 3N

50W/3 = N

We know that N is an integer, so 50W/3 must also produce an integer. Since 50 is not a multiple of 3, W must be. Of our answer choices, the only one that is a multiple of 3 is 33, so the correct answer is A.

Answer: A

Solution:

First, we can rewrite the equation as:

7^x – 7^x-2 = 48(7⁶)

7^x – (7^x)(7^-2) = 48(7⁶)

Next, we can factor out 7^x, giving us:

7^x(1 –7^-2) = 48(7⁶)

7^x(1 –1/49) = 48(7⁶)

7^x(48/49) = 48(7⁶)

7^x = 48(7⁶) * 49/48)

7^x = 7⁶ * 49

7^x = 7⁶ * 7²

7^x = 7⁸

x = 8

Answer: C

Solution:

Note that Quantity A is the expression n + m. This is an important hint: we don’t actually need to try to find individual values for our variables. Rather, we should look for a way to use the two inequalities in the stem to produce the expression n + m.

We are given that 9n – 2m > 0 and 3m – 2 > 8n.

We can rewrite 3m – 2 > 8n as 3m – 8n > 2.

Let’s add the inequalities 3m – 8n > 2 and 9n – 2m > 0. Doing so gives us:

m + n > 2

Thus, Quantity A is greater than Quantity B.

Answer: A

Solution:

We can let Julie’s speed = r. Since Hanna’s speed is 1/4 greater than Julie’s, Hanna’s speed is (1 + 1/4) of Julie’s, so Hanna’s speed = 5r/4.

Since they both went 30 miles, we can determine both the times in terms of 4.

time = distance/rate

So, Julie’s time is:

30/r

Hanna’s time is:

30/(5r/4) = 120/5r = 24/r

Finally, since Hanna arrived 10 minutes (or 1/6 of an hour) before Julie, we can create the following equation:

30/r = 1/6 + 24/r

6/r = 1/6

r = 36

Answer: E

Solution:

Let’s let x be the number of liters of solution A. Thus, we can express the amount of iodine in solution A as 0.30x. Similarly, the amount of iodine in the 30 liters of 10% iodine in solution B is 0.10(30). Finally, the resulting mixture has (x + 30) liters of 18% iodine, and the amount of iodine is 0.18(x + 30). We can express this information in the equation as:

0.30x + 0.10(30) = 0.18(x + 30)

0.30x + 3 = 0.18x + 5.4

0.12x = 2.4

x = 20

Answer: B

Solution:

We can create the following equation in which the number of students in classroom A = A and the number of students in classroom B = B.

120 = (100A + 160B)/(A + B)

120A + 120B = 100A + 160B

20A = 40B

A/B = 40/20 = 2/1

We see that 4 to 2, 8 to 4, and 20 to 10 are all in a 2-to-1 ratio.

Answer: A, D, and F

Solution:

Make sure you have carefully read the question stem, noting that the phrase “at least” indicates 3 possible scenarios, as shown below.

Scenario 1: 2 main chefs and 2 sous-chefs

The number of main chefs can be selected in 5C2 ways:

5C2 = (5 x 4)/2! = 5 x 2 = 10 ways

The number of sous-chefs can be selected in 6C2 ways:

6C2 = (6 x 5)/2! = 3 x 5 = 15 ways

Thus, the number of ways to select 2 main chefs and 2 sous-chefs is 15 x 10 = 150.

Scenario 2: 3 main chefs and 1 sous-chef

The number of main chefs can be selected in 5C3 ways:

5C3 = (5 x 4 x 3)/3! = 5 x 2 = 10 ways

The number of sous-chefs can be selected in 6C1 ways:

6C1 = 6 ways

Thus, the number of ways to select 3 main chefs and 1 sous-chef is 6 x 10 = 60.

Scenario 3: 4 main chefs and 0 sous-chefs

We just need to select 4 main chefs. Those chefs can be selected in 5C4 ways:

5C4 = 5 ways

Thus, the number of ways to select at least 2 main chefs for the crew of 4 is:

150 + 60 + 5 = 215

Answer: C

Solution:

We use the rule that the smallest side of a triangle is always opposite the smallest angle, and the longest side of a triangle is always opposite the largest angle.

Thus, we see that side BC, which is opposite the smallest angle x, is the shortest side of the triangle.

Accordingly, side AC is the longest side of the triangle because it is opposite the largest angle z.

Thus, the length of side AB is between the lengths of BC and AC.

Since side AC = 10 and side BC = 6, the length of side AB must be greater than 6 but less than 10. That is, 6 < AB < 10.

Since we are given the lengths of sides AC and BC and the lengths of three possible perimeters, we can determine the possible lengths of side AB. Let’s go through each answer choice.

A) Perimeter = 22

Perimeter = side AB + side AC + side BC

22 = AB + 10 + 6

AB = 6

We know that 6 is not a possible length of AB, since AB must be between 6 and 10.

B) Perimeter = 24

Perimeter = side AB + side AC + side BC

24 = AB + 10 + 6

AB = 8

We know that 8 is a possible length of AB, since AB must be between 6 and 10.

C) Perimeter = 26

Perimeter = side AB + side AC + side BC

26 = AB + 10 + 6

AB = 10

We know that 10 is not a possible length of AB, since AB must be between 6 and 10.

Thus, the only correct answer is choice B.

Answer: B

Solution:

Let’s let x = the amount (in millions) of income tax that Calvin Signorelli paid in 2014. Therefore, our values for taxes paid are: 6, 8, 10, 12, 12, 14, and x.

It’s important to notice that if there is exactly one mode, that mode must be 12, since there are already two values of 12 in the set and no other value occurs more than once. Since we know the mode is 12, and since we know that the average is equal to the mode, we can set the average of these integers equal to 12.

Average = sum / quantity

12 = (6 + 8 + 10 + 12 + 12 + 14 + x) / 7

84 = 62 + x

22 = x

Thus, Calvin Signorelli paid 22 million in taxes in 2014.

Answer: D