# Inverse Proportions on the GRE: The Flip It Method

Dealing with percent and word problems is always difficult. First off you have to understand what the question is asking for, then you have to set up what you believe is an equation that will help you quickly derive the answer.

Each of the questions below tests the following concept: if one thing increases by a certain percent, another thing will decrease by a certain percent. That is two things are in inverse proportion to each other.

To illustrate, let’s take a relatively straightforward example:

The product of positive integers b and c is 60. If b is doubled, then c must decrease by what percent so that the product of the two will still equal 60?

(A)  10

(B)  25

(C)  50

(D)  200

(E)  Cannot be determined from information given.

One way to solve this is to set up an equation:

If

And , what must x equal?

Using substitutions we take bc = 60 and we plug it into the bottom equation:

or 50% (C).

You can probably see that this is an unpleasant and cumbersome approach. Nonetheless, many students feel they have to set up some equation, however unwieldy it may be. Instead, think of the above problem in the following way:

If one element doubles, then the other element must be halved in order for the product to remain unchanged. Therefore the answer is 50%.

It’s also important to note that the fact that the product is 60 is irrelevant to the problem. The product could have been any number. Don’t fall for answer (E)!

Now let’s take the concept above and apply it to a different problem.

Mike drives 45 mph to work and arrives in 20 minutes. If he wants to complete the trip in 15 minutes, how fast must he drive?

(A)  42

(B)  50

(C)  54

(D) 57

(E)  60

Solution: The logic of inverse relationships is that one number increases while the other decreases. There must also be a constant number, such as 60 in the preceding problem.

In this case, the constant number is the distance Mike drives to work. What varies is the speed and his time.

We can set up an equation (but we saw where that got us last time), or we can think of it this way—Mike wants to arrive at work at 15/20 of the time. Therefore, he must increase his speed by the reciprocal. Or as I like to say Flip It! 15/20 becomes 20/15. 20/15 x 45 = 60.

Indeed, we did the exact same on the preceding problem: 2/1 (the amount we increase b) is flipped to give us ½, the amount we decrease c by.

## Practice Questions

1. If X decreases 40%, by what percent must it increase to return to its original value?

(A)  40

(B)  50

(C)  66 2/3%

(D) 75%

(E)  80%

2. Driving at a constant speed of 60/mph, Bob can reach his work in 15 min. If he wants to reach work three minutes faster, by how many miles per hour must he increase his speed?

(A)  12 mph

(B)  15 mph

(C)  20 mph

(D) 75 mph

(E)  80 mph

3. 15 workers can paint 3 fences in 2 hours. Assuming each worker works at the same rate, how long will 5 workers take to paint 6 fences?

(A)  3 hrs

(B)  6 hrs

(C)  7.5 hrs

(D) 9 hrs

(E)  12 hrs

1. We can choose the number 100 for X. If X drops 40%, it is at 60. That is we multiply the 100 by 6/10. The flip it method shows us that all we have to do is flip the 6/10 to get 10/6. 10/6  = 1 2/3. We discount the ‘1’ because that is already the value of X after it has dropped 40%. Therefore it only needs to increase by 2/3 to reach its original value. Therefore the answer is (C) 66 2/3.

2. We are multiplying the time it takes him to get to work (15 minutes) by a number (x), which will yield 3 minutes faster (12 minutes). 15x = 12; x = 4/5. Using the flip it method, we multiply 60 by 5/4 to get 75. The question asks how much he must increase his speed by. So be careful not to pick (D), but to subtract 60 from 75, to get 15. Answer (B).

3.5 workers means that painting the fence will take 3 times as long. Because those five workers have to paint twice as many fences, we have to multiply the original time by 3 and by 2: 2 x 3 x 2 = 12, Answer (E).

### 24 Responses to Inverse Proportions on the GRE: The Flip It Method

1. Tee February 9, 2016 at 10:00 pm #

Never mind! I was able to solve #1 after watching the Sequential Percent Changes video and applying the multipliers rule for % decrease and % increase!

Thanks,

Tee

2. Tee February 9, 2016 at 5:29 pm #

Hi Magoosh Team,
I’m still stuck on question #1. Can someone please clarify for me? I watched Mike videos on Percent Increases and Decreases and tried to use the method of multipliers but I am still not making a headway!

Thanks,

Tee

3. Michelle March 18, 2013 at 11:35 am #

Question 2 – Shouldn’t the answer be B)15 because you are asking for how much he must increase his speed by? He must go 75mph, but originally traveled at 60mph; therefore, he must increase his speed 75-60=15mph

• Chris Lele March 20, 2013 at 9:16 am #

Hi Michelle,

Thanks for pointing that out! It looks like I got tricked by my own question :). I’ll make the necessary edits.

4. Daniela March 1, 2013 at 8:35 am #

I think I found a little mistake… question 2′ answer says that the answer is (B) and you get 75, but the option with that outcome is (D).

• Chris Lele March 1, 2013 at 11:36 am #

Hi Daniela,

Thanks for point that out :). I’m fixing it now.

5. Sai July 5, 2012 at 6:03 pm #

Hi Chris,

I need a quick clarification on problem 1. I seem to be getting 2 different answers depending on the way I think.

1) If I ignore percents and think in terms of value in some unit.

It says, “total value of the brand does not change”. So if we choose 100(some unit) as the past value of Mustland , the current value of Mustland would be at 60(some unit). This means the total value decreased by 40 (not percent but some unit). To ensure that the total value remains constant, the value of Barnigate must increase to 140 (some unit). This shows an increase of 40%.

2) If I look at percents:

The value of Mustland decreased by 40% (4/10). This means that it is (6/10) of what it used to be worth. To compensate, Barnigate must increase by the inverse or 10/6, which is 66.67%.

Can you please tell me what I did wrong in case 1?

• Sammy July 6, 2012 at 8:50 am #

You are facing the same problem I had, the stocks should be multiplied to get a total final value. I believe your mistake lies in the fact that you are adding the two stocks.

• Chris July 6, 2012 at 5:10 pm #

So I changed the problem to a different one altogether. I think it was a little too ambiguous whether we should add or multiply the stocks. Thus I have laid the Mustland problem to rest :).

• Chris July 6, 2012 at 1:58 pm #

Hi Sai,

Thank you for clearly pointing this out. I’d gone back and taken a look at it and realized the problem was not valid due to the ambiguity you pointed out. I will make a change to the problem so that it is clearer.

Thanks 🙂

6. anisha July 5, 2012 at 10:23 am #

Please can you explain question 1 in detail.Thanks

• Chris July 6, 2012 at 2:00 pm #

Hi Anisha,

There was a problem with number 1. I’m making the necessary changes. Sorry for any confusion :).

7. Brad July 2, 2012 at 6:39 pm #

NVM. Figured it out. ‘Flip it’ rule applies to 3, just with one extra step at the end.

• Chris July 3, 2012 at 5:35 pm #

Cool, no problem. Glad you caught it :).

8. Brad July 2, 2012 at 3:28 pm #

So I noticed that you didn’t use the the ‘flip it’ method for question 3. Is it possible to use that method for question 3, or does question 3 represent a separate inverse proportions question type?

9. Pemdas@BTG July 2, 2012 at 1:58 pm #

i found question#3 ambiguous: the rate of 15 men cannot be split into 5 men’s rate. The issue involved is we are not told about the average rate of all men, neither know about work/speed capacity of men.

According to the common logic, yes it’s 15 men->3 fences->2 hours + Another 15 men->3 fences-the same 2 hours = 30 men->6 fences->2 hours OR 30/6 men->6 fences->2*6 hours. But this is true if 5 men’s speed aggregated as 15 men speed translates into the proportional capacity of work.

• Chris July 3, 2012 at 5:21 pm #

Hi Pemdas,

Yes, I think this can be remedied with a quick change, ‘assuming all men work at the same rate.’

Thanks for your keen eyes 🙂

10. Sammy July 2, 2012 at 10:10 am #

Chris,
For practice problem 1, are you multiplying the values of the stocks to get the total value or are you adding them to get a total? (This does not refer to your explanation of the solution; it is a general question). I am trying to solve it by the cumbersome & unwieldy way 🙂

• Chris July 3, 2012 at 5:02 pm #

Hi Sammy,

Yes, you need to multiply the two stocks. I encourage trying it my method :).

• Sana July 4, 2012 at 3:49 pm #

Hi Chris,
I am not able to solve practice question 1 and also the explanation is a bit tricky.
I will be thankful if you make it more clear.

Thanks.

• Chris July 9, 2012 at 2:17 pm #

There was a slight problem with Q1, so I changed it. Sorry for any confusion :).

• Prashant July 6, 2012 at 1:01 pm #

That seems counterintuitive. Isn’t the overall value of the portfolio the sum of the components of the portfolio? By that logic, the increase in the other stock should be a 40% as well.

Stock A : 100 (original) , 60 (depreciated)
Stock B : 100 (original) , x (appreciated)

60 + x = 100 + 100
–> x = 140
–> percentage change = 40%

Please point out the logical inconsistency that I seem to be overlooking here. This thing is really stuck in my head.

• Pemdas@BTG July 6, 2012 at 5:06 pm #

@prashant, IMOH soft skills are pernicious for hard sciences such as math

if devise to go for econ MA or PhD then don’t even mention in your SoP business admin or CFA or financial portfolios 🙂

this is simple percentage conversion concept

the words applied here appre-ciate, depre-ciate are increase in %, decrease in %

e.g. 15% increase is 1.15 15 decrease is 0.85

again 🙂 be hard rather soft with GRE math too

• Chris July 9, 2012 at 2:19 pm #

Hi Prashant,

I made a mistake with this problem. You are exactly right: a portfolio is the sum of its individual stocks. Sorry for any confusion :).

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