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?
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.