You might have prepped for months, using some of the toughest material out there. Nonetheless, come test day, there will be a question (or two!), that leaves you flummoxed.

How you solve complicated math problems plays a very important part in how you solve them. Reading the question over and over again, sweat cascading down your brow with each futile attempt, is probably not the best way to go about it.

Below are some important strategies to help you approach tough math questions. Then, below these strategies, are a few tough math questions for you to apply what you’ve learned.

1. Read it one piece at a time

Many tough problems are word problems that need to be digested one chunk of information at the time. Rushing towards the end, hoping the meaning of the question will suddenly cohere in your mind, is not a wise strategy.

2. Come back if nothing clicks after three reads

This is very important point. If you are still uncertain what the question is asking after you’ve read it three times, move on. First off, there are no extra points for tough questions. You will want to spend time making sure you are able to answer easy questions correctly. Secondly, the GRE allows you to return to questions.

3. Remember the question

After sifting through oodles of information and embarking upon an epic question to solve the problem, you may end up forgetting what the question was asking for in the first place. A good strategy is to write down—in short hand—the question on the scratch paper.

4. Check your work

Once you’ve made sure you’ve read the question correctly, look back over to see if your work is correct. Also make sure that you’ve looked over the important parts of the question. Of course this step only makes sense if you have some time left and aren’t at the very beginning of the test (see point #2).

5. Think logically and eliminate

Oftentimes questions are just so difficult that we are able to make little headway. That doesn’t mean we have no recourse but to throw our hands up in despair. Think about what the question is asking and what the possible range of answer choices can be. You might also want to use the answer choices to “back solve” the problem.

1. A cylindrical cup is used to fill a bowl, which is in the shape of a hemisphere (half of a sphere), with milk. If the radius of the cup is half the radius of the hemisphere, and the height of the cup is double its radius, what is the maximum number of cups that can fill the bowl without overflowing it?

- 2 cups
- 3 cups
- 5 cups
- 6 cups
- 9 cups

2. IF n = 100! x 100!, what is the largest value of prime number t, such that n/t is an integer?

- 7
- 19
- 97
- 99
- 101

3. An airport conveyor belt is 120 meters long, and moves at a constant rate of 2m/s. A man carrying two suitcases walks at a constant rate of 4m/s. How long will it take him to cross from one end of the conveyor belt to the other, if the man only walks for half the time he is on the conveyor belt?

- 25 seconds
- 30 seconds
- 42 seconds
- 45 seconds
- 54 seconds

## Answers and explanations

1. There are two ways to work this question: one is by using algebra, the other is by plugging in a convenient number.

Using ‘r’ for the radius, we can arrive at the volume of the bowl by using the formula for volume of a sphere: . Since the bowl is half of the sphere, we end up getting . (Go back and look at point #3 from the strategies above).

To find the volume of the cup, we will use the formula for the cylinder. Note that ‘r’ is half the ‘r’ of the bowl. Thus (1/2r) is the radius of the cup. The height is equal to twice the radius, giving us ‘r’ for the height. The formula for a cylinder is, which gives us .

Now we set the equations equal to each other, but multiply by ‘x’, where ‘x’ stands for the number of cups. Solving for ‘x’ will get us very close to the answer. . Solving for ‘x’ yields: x = 2.6667. At this point, it is very tempting to round up and choose (B). But remember, we want the maximum number of cups without overflowing the bowl. Three cups would overflow the bowl. Therefore, the answer is (A).

2. This is an example of a question that sounds daunting—and one we are prone of reading and re-reading till her head starts spinning. Really, all the question is asking is what is the largest prime number that divides 100! x 100! without leaving a remainder.

The conceptual jump is to note that the answer won’t be affected by the extra 100!. Because we are looking for a prime number, we don’t want to draw from both 100!s (which would lead to numbers that are not prime). So what’s the largest prime number in 1x2x3…x99x100? The answer is (C) 97.

3. This one is tricky!

Note that the man has to spend an equal amount time walking and not walking. His rate walking on the conveyor belt is 6m/s (we have to account for the conveyor belt). Not walking on the conveyor belt, he goes at a rate of 2m/s.

Let’s think conceptually for the next step. If he spends an equal time walking and not walking, out of the 120 meters he walks 90 meters and does not walk for 30 meters. Notice how that ratio between those distances is 3:1, just as the ratio between 6m/s and 2m/s is 3:1.

Now using the rate formula, we can see he spends 15 seconds at 90 meters (90/15) and an equal time at 30 seconds. This gives us answer (B).

Going back to point #4, a good way to look at the problem is to know that the answer must fall between 20 seconds (if we walks the entire way) and one minute (if he doesn’t walk at all). Just like that, we can eliminate (A) and (E). Even (D) seems a little bit slow, since it is closer to (E) than (A). The man’s speed walking is three times that not walking, so you know the answer has to be closer to (A).

Thinking this way allows you to guess from only two answers. Something that isn’t bad at all, as you can move on to other questions not thinking you completely missed the question.

I approached 3 differently. We know that velocities going in the same direction are additive. So when he’s walking, he’s at 6m/s.

We also know that he spends equal time walking and not walking. So the question is: Some amount of time spent walking at rate 2, plus the same amount of time spent walking at rate 6, should total a distance of 120. Therefore:

2x + 6x = 120. Solving for x, we get 15, which we know is half the time. Doubling that gives us a total of 30 seconds.

Nice work, Vincent 🙂 Thanks for sharing your approach!

hey chris..thanx for your help, can you get some more problems like this..

Hi Chris , I’m unable to understand this part of the 1st question’s solution –

To find the volume of the cup, we will use the formula for the cylinder. Note that ‘r’ is half the ‘r’ of the bowl. Thus (1/2r) is the radius of the cup. The height is equal to twice the radius, giving us ‘r’ for the height. The formula for a cylinder is({pi}r^2)h, which gives us {1/4} {pi}r^3.

My understanding is – (pi*r^2)h which comes to

(pi * (r/2)^2) * 2r = 1/2 * pi * r^3

Please explain this further .

Thanks !

Hi Chris,

I didn’t understand this part of the question 3- “Let’s think conceptually for the next step. If he spends an equal time walking and not walking, out of the 120 meters he walks 90 meters and does not walk for 30 meters. Notice how that ratio between those distances is 3:1, just as the ratio between 6m/s and 2m/s is 3:1. ”

How did you determine that he walked 90 meters in half the time and the other half 30 meters? Please elaborate.

Hi Chris,

I din’t get the 2nd question please explain why don’t we consider the 2nd 100! .And what if he asks about the largest composite number,

Sorry if my question is stupid.:D

Thank You.

Hi Srinivas,

We don’t have to worry about the composite number since we are only dealing with primes. Suppose I ask you for the largest prime for 10!. The answer would be 7. Now, suppose I ask you for the largest prime of 10! x 10!. The answer is still 7, since all you’ve done is repeat the exact same numbers, 1 through 10. You haven’t introduced any new primes over 7.

Hope that helps clear things up :)!

Hi Chris,

I approached number 4 like so:

t=half the time to walk 120 meters

2t + (4+2)t = 120

8t = 120

t = 15

2t = 30

Obviously, the math works out. How about the method?

Hi Matt,

Nice approach!

That definitely gets you the answer quickly :).

okay,sorry gotcha! ^_^”

Are these level of detailed questions on the GRE? I mean to a score of 160 maybe?

Hi Rebecca,

Most of the questions on the GRE math won’t be this difficult. Even if you get these very difficult questions wrong–which hopefully you won’t!–you’ll still be able to score around 160.

Hello,

My question is on the part that reads:

“Going back to point #4, a good way to look at the problem is to know that the answer must fall between 25 seconds (if we walks the entire way) and one minute (if he doesn’t walk at all).”

If he walked the entire way, wouldn’t the answer be 20 seconds? His rate on the conveyer belt is 6m/s (conveyer belt speed + walking speed), thus it seems the solution would be 120 m * (s/6m) = 20 s. Is this correct?

Thanks!

Hi Zach,

Thanks for catching that! That’s definitely a typo. I’ve corrected it, so future readers won’t be confused :).

thanks a lot for the sample question! really intriguing and challenging. i think the 3rd question can be improved by adding an answer of 40 seconds for another big tempting trap. as my way to solve the question is:

let the total time the man on the belt as t.

t/2 * (4+2) + t/2 (2) = 120

where at the very first place i failed to heed the first part has to be (4+2) instead of 4, where i arrived to the answer 40sec on the first try!

also for the 2nd question, the answer D is a nice try. as i conceived that all the options would be prime number.

anyway thank you so much to help revoking my maths concept and formula.

I’m glad you enjoyed the questions :).

Yes, 40 is a nice trap on the last question! Many will forget to add the conveyor’s speed to the man’s speed. I’ve gone ahead and changed one of the wrong answers to 40 :).

The explanation for 3rd question is confusing.. I’m not able to understanding the ‘walking’ and ‘not walking’ part and the explanation answer as such. Please help!

I agree the walking thing is confusing!

Imagine not walking on a conveyor belt. Your speed is the speed at which the conveyor belt moves. If you are walking on the conveyor belt, you have to add your walking speed as though you were walking on non-moving ground to the speed of the conveyor belt.

I hope that helps shed some insight :)!