*Note: the scores discussed in this post refer to the old (pre-August 2011) GRE

**Meet the Word Problem**

The GRE Math contains a medley of word problems. You can see probability, interest, rates, combinations and even geometry problems. In general, the problems are worded in such a way that they can be vague or even downright confusing. To add to all that, some of the more difficult word problems have a little twist, in which if you’re not reading carefully you’ll get trapped.

One strategy students employ when dealing with word problems is to read the question over and over again, until their heads begin spinning. Inducing vertigo, however, is never a good strategy, especially in the test center. That’s why you want to prepare yourself for even the most convoluted and diabolical of word problems by following the steps below.

**Breaking it Down**

After reading the problem, you want to break the problem apart into manageable pieces. For instance, if a word problem is comprised of three sentences, make sure to understand a sentence at a time, instead of trying to “get it” all in one reading. Following this process is much like eating—you wouldn’t try to down a whole meal in one swallow. Likewise, you don’t want to try to understand the entire problem, with all its different parts, in one read (refer to vertigo in preceding paragraph).

**Put in to Your Own Words**

When breaking down a problem into its various parts, another effective strategy is putting the problem into your own words. Doing so will help you break a word problem down to its essence and basic meaning. Note: The longer the problem is, the more applicable this strategy tends to be.

You may notice that breaking down a complex sentence is a strategy very similar to that employed on Text Completion or Sentence Equivalence Questions. Again, throughout the GRE, a great strategy is thinking through a problem by breaking it up in your own words (versus staring at the screen and trying to read through the same turgid verbiage over and over again).

**Read Carefully!**

Even if you follow all the above steps, you can still get a word problem wrong if you do not do the following: always remember what is being asked. It is very easy to lose sight of the forest and get lost in the trees, so to speak, and forget what the question was originally asking for. A good idea is to write down on scratch paper what the question is. That way, after you’re done interpreting the word problem and calculating an answer, you’ll remember what the original goal was.

Finally, watch out for any twists in the way the question is worded. Many times a question will throw in one twist—such as the word approximate—so that if you are not careful you will either choose the wrong answer, or wonder why your answer choice is not there (the latter is often an indication that you missed the twist).

To see an example of convoluted word problems with a twist, as well as to practice the strategies above, try the following five word problems. These word problems are mostly in the 600 – 700 range and, just like the real GRE, cover a range of topics.

So get ready, set your timers and let’s see if you can finish these word problems in 10 minutes or less!

1. Positive integer n is equal to the difference of the squares of x and y, where x and y are integers. If n is equal to 21, then which of the following could be the sum of x and y?

(A) 0

(B) 11

(C) -1

(D) -11

(E) 121

2. Bikesville is 200 miles from Restsville. Jasmine can complete the entire trip in 8 hours, and Monte takes two hours longer than Jasmine to complete the entire trip. If one begins in Bikesville and the other in Restville, then approximately how long will it take them to meet?

(A) 2 hrs

(B) 2.5 hrs

(C) 4 hrs

(D) 5 hrs

(E) 8 hrs

3. A lottery game consists of the host removing one ball at a time from an opaque jar. Each ball has one of the digits (0 – 9) written on it, and no two have the same number on them. If the host removes three balls without replacement, what is the probability that the sum of the numbers written on the balls equals 24?

(A) 1/3

(B) 3/10

(C) 1/720

(D) 1/120

(E) 1/40

4. The area of isosceles right triangle XYZ equals 9. Which of the following could be the perimeter?

I. 12

II. 13 + √145

III. 6√2 + 6

(A) I only

(B) I and III

(C) III only

(D) I and II

(E) All of the above

5. A brown flask contains a concoction that is 20% alcohol. A red flask contains a concoction with an unknown amount of alcohol. An equal amount of both flasks is poured into an empty jug. If the resulting mixture has an alcohol concentration that is 30% greater than the alcohol concentration of the brown flask, what percent alcohol did the red flask originally contain?

(A) 23%

(B) 25%

(C) 32%

(D) 33%

(E) 40%

**Answers: **

1. C **Hint:** Don’t forget negative numbers. Also note use of phrase “could be”.

2. C **Hint:** It’s asking for an approximation

3. D **Hint:** Don’t forget: there are 10 balls.

4. C **Hint:** III is 45:45:90. You can solve for x.

5. C **Hint:** The empty jug does not contain 30%

**How Did You Do?**

5: You are well on your way to the 800-level. Definitely spend your time prepping with the more challenging material. In fact, you may want to take a look at some of the previous challenge problems on Clemmonsdogpark.

4: Not bad at all! You may have missed one through a careless oversight. Regardless, with a little more practice you can ace word-problems and be on your way to an impressive GRE quant score.

3: Pretty good. These questions can be confusing, and as long as you review your basic concepts and continue prepping assiduously you should do well.

2: Don’t fret. Learning to decipher these types of problems takes time. The more you practice, the more confident you’ll become!

0-1: Word problems are the bane of many students. The good news is I’ve seen students go from fear and trembling to uber-confident when it comes to longer word problems. The key is to learn your fundamentals and then do problems to help reinforce concepts. When you begin to get questions right, step up the level of difficulty just enough so that you are being challenged, but not intimidated.

**Explanations**

1. We can rewrite the information as x^2 – y^2 = 21 = n. A good little trick to know regarding the difference of squares is that the difference of squares of consecutive integers will always be x + x + 1, where x is the lesser of the two integers. To illustrate:

5^2 – 4^2 = 9, x = 4 + 4 + 5 = 9

So what two integers, when summed, equal 21?

x + x + 1 = 21, 2x = 20, x = 10 (10, 11).

We are looking for the sum, which would be 21. However this is not an answer choice. Do not despair: we can also use -10 and -11, because each number squared will yield the same number as its positive counterpart.Playing around with these numbers we get -11 and 10. The sum of -11 and 10 is -1, answer (C).

5^2 – 2^2 would also yield a sum of 21. But remember the question is asking for which of the following COULD be the sum. Out of the five answer choices only (C) works.

2. Jamie can finish the trip in 8 hrs: 200/8 = 25 mph. Monte takes 2 hrs longer, or 10 hrs: 200/10 = 20. We want their combined rates, since they are pedaling towards each other: 200/45 = 4 4/9 , which is a little less than 4.5. The question is asking for an approximation, so that gives us: (C) 4hrs.

3. For this question, we have determine how many ways we can use three distinct numbers (0 – 9) to sum to 24. 7, 8, 9 is the only way we can do so, but remember we can rearrange those three balls in 3! = 6 ways. (8,9,7, etc.).

The next step is to determine how many total ways can we choose from 10 balls (don’t get tricked by thinking there are nine balls!). 10 x 9 x 8 = 720.

Using probability we want to divide the total number of ways by desired outcomes: 6/720, 1/120, which is (D).

4. A 45:45:90 triangle gives us x^2 = 18, x = 3√2. Solving for the perimeter we get III. Notice that answer (C).

5. The brown flask has 20% alcohol. The combined (brown + red) has 30% more than 20% (26%). We mixed equal parts so we can reason that the red flask had to have 6% more than the mixed (remember the brown flask has 6% less). This gives us 32%, or (C).

In question 1

x^2-y^2=21

which is (x+y)(x-y)=21

which means either the sum of the numbers is 21 or the difference is 21

the question is what could be the sum of x and y?

why can’t I say the sum can be 21 in this case?

You actually

cansay the sum would be 21. And if 21 were in the answer choices, you could select it. However, because 21 is not in the answer choices, you need to select a different possibility.For question 2 , to calculate time to meet two people is more that 4 hours so , why its approximated to 4. It should be 5 hours.

Because after 4 hours , A covered 125 mile and B covered 80 miles.

So to meet them, it will take more that 4 hours , so I marked 5 instead of 4.

Is’t it correct way to think ?

This isn’t quite the correct approach, Rohit. But you’re half right. If a number is more than four but less than five, it can be estimated as 4 or estimated as 5. It all comes down to the rules for rounding. If a number is 4 1/2 or larger (aka 4.5 or larger), it can be rounded up to five. If a number is less than 4 1/2 (or less than the decimal equivalent of 4.5), it should be rounded down to 4. 4 4/9 = 4.44. This is less than 4.5. Since it’s greater than 4 but less than the half point between 4 and 4, you get 4 if your round 4 and 4/9 to the nearest whole number.

for 5. why can’t the answer be 6% i.e resulting -brown flask(in case it was one of the options)

Happy to clarify 🙂

In this question, we’re mixing equal volumes of two solutions: the 20% alcohol solution in the brown flask and the x% alcohol solution in the red flask. Then, we’re told that the final solution is a higher concentration that the solution in the brown flask. This means that the solution in the red flask must be higher than the concentration in the brown flask. If not, then when we combined the two solutions, the concentration of the new solution would not be higher than the solution in the brown flask.

Let’s look at what would happen if the red flask has a 6% alcohol solution. We’re told we add equal volumes of the two solutions, so let’s say we add 10L of both solutions together.

10L of 20% solution + 10L of 6% solution

The concentration can also be described as [volume solvent]/[volume of total solution]. So, we have

Brown flask: 20 L alcohol / 100 L solution

Red flask: 6 L alcohol / 100 L solution

And to find out how much alcohol we add together when we have 10 L of solution, we can multiply both of these fractions by 10 L solution:

Brown flask: (20 L alcohol / 100 L solution)*(10 L solution) = 2 L alcohol

Red flask: (6 L alcohol / 100 L solution)*(10 L solution) = 0.6 L alcohol

The sum of these two values is 2 + 0.6 = 2.6 L alcohol. And now to find the concentration of the new solution, we can divide this value by the total volume of the solution, 20 L:

2.6L alcohol / 20L solution = 0.13 = 13%

As you can see, this value is less than the original concentration of the solution in the brown flask. However, the problem tells us that the final concentration is higher than the original concentration of the solution in the brown flask! So, we can see that the concentration of the solution in the red flask cannot be 6%.

Hope this clears up your doubts 🙂

Hi there,

Great post! I have a quick question about question 3: the answer and explanation were clear to me when I thought about it as a counting question (in how many ways can we get 24), but the question asked about probability. Isn’t it the case that the probability is the same regardless of order? Whether I get 8 then 9 then 7, my odds are 1/10, 1/9, 1/8, so don’t we just multiply those?

Thanks!

I think that the difference between the square of two consecutive numbers is not equal to ( x+x+1). Actually it equals to (x+x-1) based on

x^2-(X-1)^2= x^2- (x^2-2x+1) = x^2-x^2+2x-1

= 2x-1=x+x-1

thanks

Sam,

Thanks for brining that to my attention :). I think the confusion results from not being clear what ‘x’ is. If ‘x’ is the greater of the two consecutive integers, then your equation is correct. If ‘x’ is the lesser of the two integers, then my version is correct. Even with the equation, it all depends on which of the two numbers ‘x’ is.

In question no. 1, where is it specifically mentioned the x and y are consecutive integers?

Deepa,

x and y are not necessarily consecutive integers, but they could be. When testing cases for this problem, that’s one potential case to test. And as it turns out, the possibility that x and y could be consecutive integers is what makes (C) a possible sum of x and y, a sum that “could be,” to quote the wording of the problem.

Hi Chris, I think the answer for the question # 4 is C (III only), but you have mentioned the answer choice D (I and II).

Yeah…it seems in the answer choices that (D) slipped in there somehow. In the explanation, the answer is written as (C). Let me change that.

Thanks for spotting it :)!

Hey Chris,

Is it me or does question 1 not specify x and y are consecutive numbers? Because 5^2-2^2 is also 21.

Hi Sudha,

Yes, you are correct. As you’ve mentioned, “5^2 – 2^2” would also yield a sum of 21. But remember the question is asking for which of the following COULD be the sum. Out of the five answer choices only (C) works.

Is there an algebraic way to solve question 5? I simply cannot imagine how this has come about despite reviewing the answers on your blog. Are there any other sources to find questions and explanations similar to the mixture question akin to 5?

Hi Colin.

This will be a longer method, but it will provide an insight on how to solve mixture problems. What Chris mentioned was a shortcut which you will understand when you play around with mixture problems. If you are a Clemmonsdogpark Premium subscriber, they have got a very good lesson on how to solve basic mixture problems which is excellent.

Now for the solution:

Lets take easy values for % like 100.

Brown flask contains 100 ml solution. Therefore, alcohol content = 20ml

Red flask contains 100ml solution[equal amount]. Alcohol content = x[assumed]

Now we mix them to get resulting solution of 200[100+!00]ml. This solution contains[Very important: we are dealing in percents and concentration not absolute values of volume] 30% MORE alcohol conc than 20% in brown flask which is = 130% of 20 [% of alcohol]

= 26% of alcohol conc.

I deduced the value 130% because by saying 30% more, we are including the concept of percentage increase[Again refer to Clemmonsdogpark lessons or online resources for understanding the concept]

Since, the resultant solution is 200 ml and contains 26% alcohol, therefore vol. of alcohol = 52 ml

Vol of alcohol in red flask= 52[Total vol of alcohol in resulting sol] – 20[Brown flask]

= 32 ml

Therefore concentration in red flask is 32% since it contains 32ml alcohol/100 ml.

Dear Aditya,

Thank you so much for clearing this up. Perfect explanation in all regards! I truly appreciate your time regarding my inept math abilities. Best of luck test day!

Cheers,

Colin

For question #1, I am quite happy to learn that the difference of squares of consecutive integers can be written as x+x+1, however, the question does not state that the integers in question are consecutive. Why would I assume that the integers are consecutive for this problem? How would I solve the problem if the integers are not consecutive?

sorry, I see this question was already asked. I will check it out….

That’s a great point. I think the explanation just kind of jumps into the difference of squares in consecutive integers without explaining how it got there. There is one pair of numbers that works for x and y (5 and 2 = 5^2 – 2^2 = 21). Since that doesn’t work, we have to think of another pair of numbers. Testing numbers all day will take forever — so that’s where the difference in squares of consecutive integers comes in. In other words, every set of consecutive integers will yield an odd number, when you take the difference of their squares. (9 for 5^2 – 4^2, for example). Notice that 4 and 5 add up to 9. Just like 11^2 – 10^2 = 21, and 10 + 11 = 21.

I can’t really see too many questions asking you to endless start plugging in numbers looking for some number that is the difference in squares. I think the main point for this question is to remember that negative numbers, when squared, equal positive integers. Thus, (-11)^2 – (10)^2 = 21.

Hope that helps!

Indeed! Thanks for clearing that up.

You are welcome!

if (x2 – y2) = 21 and x and y are integers then

(x-y) (x+y) = 21

from here we can find number pairs which can be multiply to 21

so, the pairs are {21,1}, {-21,-1}, {3,7}, {-3,-7}

therefore, the equation can be written as

(x-y) (x+y) = 21 * 1

(x-y) (x+y) = -21 * -1

(x-y) (x+y) = 3 * 7

(x-y) (x+y) = -3 * -7

by observation we can understand that the, (x + y) must be in set

{21, 1, -21, -1, 3, 7 , -3, -7}

now check for the options, option (C) matches and hence the answer

there is no need to consider consecutive numbers

I followed the same method and found it much less ambiguous than the consecutive numbers approach!

Your explanation makes it much more clear as well. Thanks!

You are welcome!

Please explain me prob:4 .. i have problem in understanding or doing such problems.

Hi Yamini,

Sure :)! First thing: an isosceles triangle is the same as a 45:45:90 triangle.

With 45:45:90 triangles you always have to remember the relationship between the sides: x, x, and x√2 (the last of these is the hypotenuse, or longest side).

The two equal sides (the ‘x’ and the ‘x’) meet at a right angle, and are thus form the base and the height. Using area of triangle formula, we get x^2/2 = 9, x^2 = 18, x = 3√2. The perimeter is x + x + x√2, or 12 + 6√2. Therefore, answer (C).

Hope that helps!

Hi Chris!

I think you left out something in problem 1. When I first did it, I let x = 5 and y = 2 (so 25-4 = 21). In the solutions though, you say, “So what two consecutive integers, when summed, equal 21?”

In the question, you didn’t mention that x and y were consecutive integers. Although the question is still solvable without that information, just wanted to let you know!

David

Also, shouldn’t the perimeter in number 4 actually be 6 + 6*sqrt(2), since one side is 3*sqrt(2) and the hypotenuse is 6?

Oops, yes, you are right! Thanks for catching that :).

If for number 4 , 6 + 6*sqrt(2) is the perimeter, isn’t the answer: “(C) III only”

rather than (D), as given in “4. D Hint: III is 45:45:90. You can solve for x”?

Am I missing something?

Sorry about that – it was a typo :).

I’ve corrected it.

am getting difficulty in problem 5..

For number 5, the resulting mixture is 26% alcohol. (30% more than the original 20% in the brown flask). If we put an equal amount of the brown flask (20%) and the red flask (x), and get 26%, then how much must originally be in the red flask? 26% will be exactly in the middle of 20% and x, because an equal amount of both flasks are mixed. Therefore x has to be 32%, the amount of alcohol in the red flask.

Note: very little computation was needed to solve this problem. The GRE awards a student who think in this fashion, because that person will solve the question quickly.

Hi Chris,

I still do not understand how you came about 26% in question 5. Kindly break it down please.

Also, if the 2 sides of the isoceles triangle are 3 and the last side is 6, the perimeter can be 12 and the area 18. correct me if I’m wrong please (p.s you haven’t corrected the answer from what I can see)

Hi Soty,

For question #5, my reasoning as follows:

If we have a brown flask that contains 20% alcohol and we mix it with a certain amount of mixture from another flask, the resulting mixture is 26% (notice it says the resulting mix has 20% alcohol than the brown flask; 20% of 20 is 26).

Since we poured equal amounts of the mixture from each flask, we know that the red flask has 6% more alcohol than the resulting mixture. The reason is because the brown flask has 6% less (20% is 6% less than 26%). You can think of the mixing of equal amounts of 20% alcohol solution and 32% alcohol solution will yield a mixture that is right between the two numbers, i.e. 26%.

For the triangle problem it seems that I made the correction. The only possible isosceles right triangle can have sides 3rt2, 3rt2, and 6. The dimensions you gave me – 3,3, and 6 – do not make a right triangle, or indeed a triangle at all. The longest side of any triangle can never be greater than or equal to the sum of the other two sides.

Hope that helps!