# 3 Hardest GRE Questions from April 2017

Posted by on April 17, 2017 in GRE | 24 comments

A GRE student with a 170 on the quantitative section last week shared that these were 3 of the hardest types of questions you could see on the test. Can you answer these? I have given the answers below and will share solutions once you give these a try.

1. Mary sold 15 widgets last week. The median sale price of all widgets was \$130 and the average sale price of all widgets was \$150.

Quantity A: \$165
Quantity B: The least possible sale price of the most expensive widget.

A. Quantity A is greater.
B. Quantity B is greater.
C. The two quantities are equal.
D. The relationship cannot be determined from the information given.

2. Each student in a school takes either the French class or Spanish class, but not both. Each class has some male and some female students. The ratio of the number of female students to the number of male students is less for Spanish class than for French class.

Quantity A: The ratio of the number of female students to the number of male students in French class.
Quantity B: The ratio of the number of female students to the number of male students in the school.

A. Quantity A is greater.
B. Quantity B is greater.
C. The two quantities are equal.
D. The relationship cannot be determined from the information given.

3. Function f(100)= the product of all the even integers from 2 to 100, inclusive.

Quantity A: 40
Quantity A: The largest prime factor of f(100)

A. Quantity A is greater.
B. Quantity B is greater.
C. The two quantities are equal.
D. The relationship cannot be determined from the information given.

1. B
2. A
3. B

1. Would you mind explaining these? The fact that I cannot solve these after 4 months of full time prep is kinda scary.

2. Q1:- The sum of 15 widgets CP will be = 15 X 150 = 2250
As median is 130 so there will be 7 items whose CP will be less than or equal to this. To get the least greatest price, assume all are 130 then 8*130 = 1040
The remaining will have a sum of = 7 X A(Avg price)
Now 2250 – 1040 = 7 X A
A = 172

Q2- Assume the ratio of Female to male in Spanish is 1/3 and French is 1/2. This satisfies the question condition.
A= 1/2 and B=2/5 so A is greater.

Q3- A = 40 and B=highest prime factor of f(100)
f(100) = 2 x 4 x 6 x …x 98 x 100

For B, 94 has factors 2, 47. 47 is prime Hence answer is B

• Please explain 47. It is not an even iteger. Does it fall acc to condition? (Product of even integer). 2 is the only even prime factor.47 is prime but not even integer. Plz correct me if i am wrong

• f(100) = 2 x 4 x 6 x …x 98 x 100
For B, 94 has factors 2, 47. 47 is prime Hence answer is B

• What about 97? It is a prime number too

• Yes it is!

• I think we should say that B option is correct bcz of 97 not 47

• Gurpreet and Harron — 97 is not a factor of f(100), so Harron you would be incorrect in saying the answer is B BECAUSE of 97.

This is a fairly simple computation. f(100) = 2 * 4 * 6 * ….. * 98 * 100.
We can extract 2 from every number, since they are all even. So we get
f(100) = 2^(50) * (1 * 2 * 3 * …. * 48 * 49 * 50)
Now all that is required is to select the greatest prime number of the numbers between 1 and 50, which is 47.

• Not 97, it is just the product of all the even numbers from 2 to 100. So 97 is not in the list.

• f(100) is one big product; not a list of integers!! of that product, 47 is a prime factor….

• We are not given that number of females is greater than the number of males. So it can be 2/1 and 3/1. Which gives us D

• For question number 2, assume that there are 7 male students and 4 female in Spanish class. Ratio= 4/7=0.57

assume that there are 5 male and 3 female students in French.
Ratio=3/5=0.6 (Option A)

Total= (4+3)/(5+7)= 0.58 (Option B)

THEREFORE, A>B

–If male students are less compared to female–

assume that there are 3 male and 5 female in Spanish class.
ratio=5/3=1.66

assume that there are 4 male and 7 female in French class.
ratio=7/4= 1.75 (option A)

Total=(5+7)/(3+4)= 1.71 (option B)

THEREFORE, A>B.

WHICHEVER WAY YOU DO THE ANSWER IS CORRECT

• What if :

French class, F/M = 10/1. Spanish class F/M = 1/100.

So that, Entire school: F/M = 11/101.

In this case, choice (B) does not work. Quantity (B) is less than quantity (A).

Unless I misunderstand, choice (B) only works if the F/M class ratios are either both top-heavy or both bottom-heavy. But it isn’t necessarily the case that they are. So it seems the answer should be (D).

• can you please elaborate a little on the second statement “least possible price of the most expensive widget”

3. Determining if f(100) has a prime factor larger than 40 requires going through each and every value from 100 down to 94 in order to find that 47 is indeed a prime factor of 94 and thus of f(100). A more computationally friendly way to do it?

• Yes a simple way will be to identify the largest prime below 50…. since that prime number times 2 will give us the actual number in h(100)…

• I did this:

f(100)=2·4·…·98·100=2^2(2)·6·8…·100=2^3(2·3)·8·…·100

Doing that 50 times:

f(100)=2^50(50!). Then, 2^50 is even, and the largest prime factor will be 47.

Best Regards from Chile!

4. How is 94 the product of even numbers in question 3

• 94=47*2 (product of an even number 2 and a prime number 47)

• Can we suppose that 82 has 2 and 41
So 41 larger than 40

5. 1st question:
The median between these 15 objects is 130, it’s the 8th in the sequence, thus there are 7 items before it and 7 items after it.
As it’s the median, 7 items before it should be less or equal to it and those items after it, should be more or equal to it,
We have average which is 150, we want to know the least possible amount for the last number in the sequence, So for satisfying this, we choose all 7 items before median as 130, thus for having the average 150, all items after the median should have a +20 in average, to compensate the -20 for items 1 to 8 [130 130 130 130 130 130 130 130 , 170,170,,170, 170,170, 170, 170], or they can be any other numbers,
I mean this +20 * 7 should be decided between the last 7 items. And if we want to have the minimum amount for the last item we can set all of them to 170, otherwise the last item will be higher than 170, So it’s at least 170 and the answer is B

6. #1 The rules are that the 8th number in order of size must be 130 with at least 7 numbers less than or equal to 130, and that the 15 widgets sum up to 2,250 to make the average. To get the largest possible number, assume 14 widgets were 130, equaling 130 × 14 or 1,820. That leaves 430 for the largest possible widget sold.

#2 The total school ratio will have to lie between these two numbers, so you can view that the larger one must be bigger. Others here show the calculation, and the logic makes sense.

#3 One only needs to focus on primes begger than 40, and it’s probably that the prime will be multiplied times 2 to make the larger number. 41, 43, and 47 to double to be 82, 86, and 94 respectively. The next prime is 53 and that’s outside of 100