The Revised GRE is a long, grueling test. To get a perfect score on the quantitative section (a 170) you will not only have to answer the difficult questions correctly, but you will also have to answer all of the easy ones. Nonetheless, I’ve focused on the difficult questions for the 170-challenge. I can’t promise a perfect correlation, but something tells me that if you can answer all five of these questions correctly in less than 10 minutes, you are well on your way to a perfect score.

Of course missing a few hardly precludes a perfect score. But make sure you learn from the ones you miss to make sure you don’t make similar mistakes in the future. And if time is a problem, remember the Magoosh GRE product has plently of questions to help you hone your chops—both from a pacing and a conceptual standpoint.

Good luck, and feel free to post your answers. Let’s see who the first perfect score will be!

1. If s and t are both primes, how many positive divisors of v are greater than 1, if v is an integer?

(A) two

(B) three

(C) five

(D) six

(E) eight

2. A quadrilateral has a perimeter of 16. Which of the following alone would provide sufficient information to determine the area of the quadrilateral. Choose ALL that apply.

[A] The quadrilateral contains equal sides

[B] The quadrilateral is formed by combining two isosceles right triangles

[C] Two pairs of congruent angles are in a 2:1 ratio

[D] The width is 4o% of the length and all angles are of equal measure

[E] If the perimeter was decreased by 50%, the area would decrease to 25% of the original

3. Product Question: Triangle in a Parabola

http://gre.magoosh.com/questions/2192

4. If x is an integer, and 169 < < 324, which of the following is the sum of all values of x?

(A) 61

(B) 62

(C) 75

(D) 93

(E) None of the above.

5. x = 350,000

y = 45,000

Column A | Column B |
---|---|

The total number of positive divisors of x | The total number of positive divisors of y |

- The quantity in Column A is greater
- The quantity in Column B is greater
- The two quantities are equal
- The relationship cannot be determined from the information given

[Note from our intern, Dylan: “I can do multivariable calculus, and got 2/5 on these in 10 minutes. These guys are tricky. :P”]

## Explanation Video #1

https://www.youtube.com/watch?v=BxkB8R7esB4

## Explanation Video #2

https://www.youtube.com/watch?v=jZdxqajkZtw

## Explanation Video #3

http://gre.magoosh.com/questions/2192

## Explanation Video #4

https://www.youtube.com/watch?v=pwzlclrvqWU

## Explanation Video #5

https://www.youtube.com/watch?v=–U8aDmLNzE

The area does not reduce BY 25%, it reduces TO 25% of the original as per your explanation of Q2.

Hi Ashu,

Thanks for pointing this out! You are absolutely right, the question should read “reduces to 25% of the original.” I’m going to send this along to our content improvement team so that they can take a look at it 🙂 Thanks again!

Hi Ashu! Just wanted to let you know that we made this change 🙂

B is correct.

Suppose the the two isosceles right triangles have sides of (x, x and xsqrt2) and (y, y and ysqrt2) [This is the only ratio that the sides in an isosceles right triangle can be]. Then the only ways that a quadrilateral can be formed by combining them are-

1) The two hypotenuse are equal, i.e, xsqrt2 = ysqrt2 which implies that x = y and the quadrilateral will be a square and we will be able to find the area.

2) Side x is the same size as the hypotenuse ysqrt2 of the other triangle (or the other way round forming a trapezium). In this case x = y sqrt2 —–eqn 1

And, summing all sides, x + xsqrt2 + y + y = 16 ——eqn 2

Solving these equations will give us x and y and then we can find the areas of the two triangles and thus the quadrilateral.

Your two possibilities both work. The problem, of course, is that there are

twopossibilities. So without knowing which possibility we’re dealing with, B alone doesn’t provide us enough information to determine the area of the quadrilateral. It only narrows things down to 2 possibilities.1) B

2) A,B,D

3) 1/2

4)E

5) A

Very easy questions. I am a 9th grader, all of them took me 7 mins. I got all correct.

Great job, Rohan! You are a math maven 🙂

I don’t understand why B is not an option for question two. Even if it was a rhombus, you would be able to figure out the hypotenuse as a*square root of 2. Then find a by solving 2*(a+a*root2) equal perimeter of 16. Then you know the height and length of the rhombus. Where is my error?

You don’t have an error per se in your immediate calculations, Sindhu. You’re correct that if we know for a fact that the quadrilateral is a rhombus, and we can see the angles of the rhombus, (B) is sufficient. However, you’ve made a mistake on a broader level– you’ve forgotten that we don’t actually know if the shape is a rhombus!

You can see a good breakdown of this in the first two minutes of the answer explanation video for this GRE Quant question. Basically, A and B are both insufficient for the same reason.

1 – B

2 – A, D

3 – 0.25

4 – B

5 – C

The only question I was stuck on beyond 2 mins was the first as I thought assigning the two different variable s and t meant they were different values (different prime numbers). I thought this was the case in gre where variables of different letters have different values?

Hi Chris,

Won’t the answer of question 1 depend on the choice of numbers,

Take x = 3, y = 27 => x^3*y^3 = 3^12 , hence v= 3^6

So there are 6 factors except 1 in this case. Similarly, y can be 3^(3k) and v can attain higher values.

The no. of factors would be different in each case.

Reread it. The numbers are both primes. So it won’t work. 🙂

Hi Chris,

I heard that you need to answer all quant questions correctly in under 30 minutes instead of the 35 minutes in order to get a 170. Is that true?

Thank you!

Best,

Nina

Nina,

That is a very interesting question. I’m curious as to where you heard that.

I’ve actually never heard any such thing, but it might actually be true. I hear this because apparently people have answered every question in quant correctly but not received a perfect score. The thinking is that it is difficulty of questions within that specific batch of questions that determine the ceiling. Or so the thinking goes.

Well, let me know where you heard that rumor. I really do want to find out if there is any truth behind it 🙂

Hi Chris,

thanks for the reply. I’ve read it on various forums on the internet. I actually just emailed ETS and will comment here if they get back to me. If this was really the case, it seems like it would make more sense to do the paper-based (non-adaptive and without such time constraints) to make a 170 most likely, right? Assuming someone had the choice between the PBT and CBT.

Best,

Nina

Nina,

Thanks for the quick reply! I’ll have to do some snooping around the forums on my own. Looking forward to seeing what ETS has to say on the issue. Thanks for keeping us posted 🙂

Hi Chris,

they say its only the number of questions answered correctly that counts – with possible differences arising due to question difficulty (which seems really unfair to me).

Nina

Well thanks to getting to the bottom of that.

And I agree with the question difficulty thing — totally unfair 🙁

1 d

2 a,d

3 0.25

4 b

5 b

Answer to 5 is c

thanks to GMAT quant I got 4 out of 5 in 9 mins

Yes, working with GMAT quant makes these questions a lot easier!

Congrats on an impressive performance!

Hi Chris,

I don’t find any solution for Question # 3.

Can you please either add a video or provide a solution in a reply to my post?

………

BR

Mandeep

You can find the explanation here: http://gre.magoosh.com/answers/22698583?&prompt_id=2192#text_explanation

Thanks for posting that Sonamata! 🙂

I was wondering something about question #2 (like everyone else). I know trig is not required in the GRE, but regardless, wouldn’t you be able to determine the area in choice B knowing it’s two isosceles triangles?

From that given information, you know that the parallelogram that results has side ratios of sqrt(2)x + sqrt(2)x + 1x + 1x, since the sides of a right isosceles triangle are in sqrt(2) :1:1 ratio. You can then use algebra to calculate out the size of the edges by finding x: sqrt(2)x + sqrt(2)x + x + x = 16. From there you can find the height of the parallelogram, and thus you can find the area, which is base*height.

Technically, you CAN find the area with enough work, correct? Thanks for this little quiz, though, it’s nice brushing up on my quantitative.

Hi Daniel,

So (2) is really tricky!

See, you can arrange two isosceles right triangles into a quadrilateral by arranging them into a rhombus. If you do so, you’ll find a rhombus with a perimeter of 16 only has an area of 8, not 16, the way a square with a perimeter of 16 does.

In general, almost all of the questions on the GRE will be easier than this one. There will only be one or two (in my humble opinion :)) that will be comparable in difficulty.

Hope that helps!

If you arrange two isosceles right triangles side by side, you would get a parallelogram – not a rhombus. The hypotenuse of the two right triangles would be two sides of the quadrilateral and hence longer than the other two sides. So all four sides will not be of equal length.

As Daniel suggests, if the isosceles triangle have the equal sides to be of length x units, then the perimeter of the quadrilateral formed would be 2x(1 + sqrt(2)).

From the above information, the value of x can be determined as 8 ( sqrt(2) – 1) and the area computed to be x^2 = 10.98.

But the essence of the matter is that we still can’t one single answer for the area, therefore option B must be wrong!

Wow, you are right! I can’t believe no one has caught that after all these years (myself included!). Yes, it’s definitely a parallelogram. And I’ll definitely have to change this question!

Thanks again, Sriram, for your valuable input!

I guess it can form a square as well if you just superimpose the hypotenuse. In which case all sides wud be equal and each angle 90.

I’m confused.. So this means that with what the current question is, B is wrong because even though you can calculate the area, we get two possible values for it. So since the question asks whether you can determine the area or not, the answer is ‘no’ because there are two possible values for it.

Sounds like you’re not so confused after all— you are exactly right that (B) can’t be the right answer, because there is more than one possible value for the area of the choice. You want to only select answer choices that lead to one specific result, with no other results being possible.

For question #2, Since you know the angle and the length of any one side, can’t you use trig to find the area?

Hi Kai,

First off, the test wouldn’t require trig., beyond knowing your 30:60:90 ratios. With question #2, you don’t necessarily know the length of any one side, given the perimeter and the measurement of the angles. A rhombus could have the same angles and perimeter as very long parallelogram–but a very different area.

Hope that helps!

How can I get access to the explanation videos? the links on this page are not active.

Thanks,

TT

Hmmm…that is a good question! I didn’t even notice that they had disappeared. I’m going to find out what happened, and I will get back to you :)!

Any updates on the videos?

Hi, Eric

They should be back in now! 🙂 Let me know if you have any trouble watching the videos on YouTube!

Best,

Margarette

Thanks! 🙂

just post the answers atleast

Hey Rahman,

You can find the answers in the video explanations at the bottom of the post. 🙂

Which is the best GRE Prep material out there that provides practice questions with the level of difficulty of these 5 questions?

Hi Lena,

I’d say there are only a few out there, besides Magoosh.

Manhattan GRE

Nova’s

GMATprep (Official Guide and GMATprep test)

Hope that helps!

rhombus is also one type of quadrilateral , then why should u neglect option A in second question

Let’s say you have a square with sides 4. Thus you have an area of 16. Now let’s say you have a rhombus with sides 4 but with two angles equal to 150 degrees each and the other two angles equal to 30 degrees each. Notice how this “squashed figure” would have a much smaller area than 14. Thus, (A) is not sufficient to answer the question.

Hope that helps!

did you mean to write by 75%*

Hi Chris,

I’m a little confused in your explanation about 3 answer choice e.

It says decrease the area by 25%, but you appear to be decreasing the area by 75%. Did you mean to write to 75% in the question?

Thanks!

(P.S. I’m watching this without sound so forgive me if i missed something vital)

Great catch! Thanks for noticing that :). It should be decreased by 75%.

I did not get any. I do not understand them. Do not know how to solve them. An explanation for all would be very helpful.

Thanks.

Julia

No problem, I will provide a video explanation. Coming very soon :).

Plz put up the explanations for all the questions. I got Q1 & Q2 wrong!

Okay, I will put up explanations for all of the questions :).

Are you going to post the video explanations?

Thanks.

Julia

Hi Julia,

Next week I will post the video explanation for #2, as that is the question everybody seems to be missing.

1) B

2) B, D

3) 0.25

4) E

5) C

🙂

It took me more time than it should have.

Anshul,

You answered most of them correctly :), except #2, which most people are missing. Take a look at answer choice (B) again.

On #4 I was able to get an answer of E,

because 14, 15, 16, 17 when added together =62

and then I used negative integers

-14,-15,-16,-17=-62

then I added the two sums and came up with 0 which wasn’t listed, so I chose E….I know my answer is right, but is my logic correct in how I got to that answer?

Yep your logic is correct, but there is an even faster way:

If you know that each solution for x has a positive and a negative answer (14, -14, 15…etc, then each will cancel out so you don’t have to worry about adding up all the positive integers than subtracting the sum of the negative ones).

Hope that helps :).

got my mistake

No problem 🙂

Shouldnt the answer to the second question be B and D? since b gives us all the angles and a square is formed and a quadrilateral can only be formed if we join them with a common hypotenuse ( equal in length).

1)B (s,t=2 and v=8)

2) B,C,D (I am not sure about option C)

3) 0.25 ( solve for 0.5 * k * 2*sq(k) = 1/8)

4)D (because k includes negative values as well, adding up to zero)

5) 60000 divisors each

for 350,000 divisors = 5 * 6 * 2 ( 350,000 = 2^4 * 5^5 * 7)

for 45,000 divisors = 5 * 3 * 4 (45,000 = 2^3 * 3^2 * 5^4)

I took 13 minutes but 🙁

Hi Praveen,

All of those are correct, except for #2, which seems to be stumping everyone :).

My clue for that question is try to think of two different shapes that can result from each condition. Just because you can think of one condition doesn’t mean that is the only condition. (Hint: a square + rhombus; rhombus + parallelogram).

Hope that helps 🙂

Let me retype Questions ## 4 and 5 as something went wrong in text after I posted

Question #4) answer E, 13^2<x^2<17^2

14,-14,15,-15,16,-16

Sum=0

Question #5) 350,000=5*7*(2*5)^4=5^5 *2^4 *7, (5+1)(4+1)(1+1)=60

45,000=3^2 *5 *(2*5)^3=2^3 *3^2 *5^4, (3+1)(2+1)(4+1)=60

answer C

Hi Pemdas,

All your answer and explanations are correct – except for #2:

(A) Not an answer

If the quadrilateral is a rhombus, then it will have a different area than if it was a square.

(D) Answer

We have the exact dimensions of quadrilateral: width = .6x, length = x, (x)(.6x) = 16. Solving for ‘x’ gives us only one positive value.

right, the area of rhombus is always less than the area of square. In our case if the quadrilateral is a square we get max.area=(16/4)^2=16, and if we have rhombus the area will be always less than 16. How I could miss it 🙁

Maybe I’ll make this a monthly challenge, so you’ll have another chance to get 5/5. Still, 4/5 is very good :).

Question #5) was merged with Question #4) 🙂

Question #5) answer is C

the whole task took me 9 mins

Question #1) when two primes factorized and raised to the power of 3 is perfect square the only viable case which comes to my mind is s=t=2 and (st)^3=(2*2)^3=64 or 8^2. Hence v=8 and 8=2^3 or (3+1) factors with 1 and 3 factors greater than 1.

answer B.

Question #2) I mark answers A and D

Question #3) answer 0.25, this one I knew answer ans solved earlier as have premium Magoosh account

Question #4) answer E, 13^2<x^2 (5+1)(4+1)(1+1)=60

45,000=3^2 *5*(2*5)^3=2^3 *3^2 *5^4, factors number –> (3+1)(2+1)(4+1)=60

answer C

1. A)

2. C), E)

3. 0.707

4. B)

5. A)

Hi Vanan,

As the post mentioned, these are some tough questions :).

Give it another shot, and see if you can get them right.