See if you can finish this rate problem in less than two minutes.

*A transcontinental jet travels at a rate of x – 100 mph with a headwind and x + 100 mph with a tailwind between Wavetown and Urbanio, two cities 3,200 miles apart. If it takes the jet 2 hr 40 minutes longer to complete the trip with a headwind, then what is the jet’s rate flying with a tailwind?*

*(A) 500*

*(B) 540*

*(C) 600*

*(D) 720*

*(E) Cannot be determined by the information given.*

**Did you Plug in?**

You may have been tempted to put together an algebraic equation. If you are adept at doing so, and can usually get the answer quickly, then I encourage you to go ahead and make an equation. Most students, however, find this strategy cumbersome and problematic. Even if they set up the right equation, which they can’t be quite sure of unless they get the answer, they may very well make a mistake in solving the equation. If this scenario describes what just happened when you attempted the problem above, know that there is a better strategy: Plugging-In.

**When and Where to Plug-In**

Once we’ve decided to plug in, where do we start? Do we plug in our own numbers, or the answer choices? First, we want to look at the answer choices. Are they numbers? If so, plug them in. If they are variables, you will need to come up with your own numbers, as long as those numbers conform to the information provided in the question.

Here we have answers, so let’s try plugging them in. The first place to start is the middle. The logic is if the number is too low (or too slow, in this case), you need to pick a larger number. Note: if the middle answer is a weird number like 625, then I would recommend plugging in (B) or (D).

Luckily, we can work with answer choice (C) 600. If the speed with a tailwind is 600 mph, then the speed with the headwind is 400. The distance between the two cities is 3200 miles. Using d = rt, where d stands for distance, r stands for rate, and t stands for time, we find that the time it takes to fly with a tailwind is 5 hr 20 min, and the time with a headwind is 8 hours. The difference in time is 2 hr 40 min. And there is our answer. Just like that.

If that seems too easy, that’s not a bad thing. Plugging-In can look like magic, in that it can make a seemingly intractable problem fall into place, just like that.

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hey!

So the value of head wind is assumed by us. And the tailwind value is taken from the multiple choices. Is there an easier way to directly plug in the right value rather than sit and try other values? or have i gotten this all wrong? i understand the rest of the problem except the choosing of the HW value.

please explain!

thanks!

Hi Siya,

In this case, an algebraic expression would be super complicated, so instead we try to plug in! The headwind and the tailwind are connected by the the equations x-100 and x+100–so as soon as we have a value for the velocity with tailwind (given by the answer choices) we subtract 200 from that to get the value of the velocity with headwind. So when we plug in 600, we first subtract 100 to find the value of x, and then subtract another 100 to find the velocity with headwind. If we were to start with another value, say (A) 500 as the velocity with tailwind, we would again subtract 200 from that to get the velocity with headwind. Does that make sense? So the value of the velocity with headwind would change depending on which value you choose to plug in! Since this is the only information we have regarding the headwind and tailwind, we can’t directly plug in a correct number. 🙂

The algebraic equation is actually not that difficult after which it is easy to plug-in.

The time of a journey with a headwind is T1=3200/x-100, whereas with a tailwind T2=3200/x+100. We also know that it takes 2.67 hours longer to complete the trip with a headwind, thus T1=2.67+T2. By doing that we have prepared ourselves a cute-looking equation: 3200/x-100 = 3200/x+100 + 2.67. Now we can start plugging numbers for X, and to our luck, A is 500 which works perfectly. Therefore, the answer is C), since X=500, and 500+100 is the velocity of the jet travelling with a tailwind.

How you got 3200/x-100 = 3200/x+100 + 2.67 ? T1 = T + 2.67 wheres T2 = T

T1 not equal to T2. Would you please explain ?

Hey Nick Pl, quote the line: “If it takes the jet 2 hr 40 minutes longer to complete the trip with a headwind”, so it makes sense. So T1=T2+(8/3).

How did you make calculation for ‘hw’ as 400 from ‘tw’ 600?

If the speed with a tailwind is 600 mph, then the speed with the headwind is 400?

Please help

Hi Chris, Finally got after six hours of ponder 😀

Assume, 600mph

by d=rt (for Tail wind), we got time for tail wind equals to 5hrs 20 mins

Since, question says Headwind takes 2hrs 40mins ‘longer’ so add this time with Tailwind time, the value will be 8hrs which is the time for headwind

Again, d=rt (this time for HW), rate will be 400mph.

Now How to check answer? Whether, the calculation we did right or wrong.

Simple!

Question says,

HW = x – 100

TW = x + 100

Just put 400mph in place of HW and 600mph in place of TW.

If value appears analogous for X then you’re right.

Likewise I did calculation for 500. But x doesn’t come equal.

thanks Chris

I know i’m doing something wrong, but want to know what.

Heres my approach- Head wind rate = distance/time

x -100= 3200/160 (2.40 hrs converted to mins)

x = 120

so when i substitute this for x, i dont get anywhere near the answer.

what is wrong?

thanks.

The time it takes for the plane to fly cross country is not 2hr40min. That is the difference between the amount of time it takes the jet to fly with a head wind and without a head wind, respectively. You would have to set up an equation that accounts for this difference. Doing so is more difficult than plugging in the answers, so it’s best to stick to plugging in – esp. when it works.

ohkay…i didnt read the “longer” part in the question! my bad…

thanks chris.

when we plug in C and speed is 600 with tailwind….how do we determine speed with headwind is 400?

We know that he trip one way is 3200 miles. So we plug in the headwind speed of 400 and ask ourselves: how many hours does it take for a plane going 400 mph to travel 3200 miles. The answer is 3200/400 = 8hrs.

Do the same for the tailwind (600 mph) and you get 3200/600 or 5hr 20 min.

The question is asking for the difference between the two: 8hr – 5hr 20 min = 2hr 40 min.

I am sorry but 3200/600 is 5.33….how do we convert it to 5hr 20 mins?

The answer 5 1/3 is expressed in hours. To convert this time into the form x hours y minutes, we need to determine how many minutes are in 1/3 of an hour:

1/3 hours * 60 minutes/hour = 20 minutes.

And our final answer is therefore 5 hours 20 minutes 🙂

We can verify the units by writing the units of the original dividend and divisor:

3200 miles / 600 miles/hr

From here, we move hr to the numerator, since the denominator is being divided by that unit, to get

3200 miles * hr / 600 miles = 5 1/3 hours

Hey Chris, how do I interpret the word “trip” here? One-way or roundtrip?

Hi Najib,

Trip means one way: either east to west or west to east.

Hope that helps!

I think the biggest problem that I had with this question was the interpretation of “2 hr 40 min longer with a headwind”.

I read this and would not have thought to subtract the headwind time from the tailwind time to find 2 hr 40 min difference.

When I tried to solve the problem on my own, I tried to compare the amount of time it would take the plane to fly with a headwind (5 hr 20 minutes) to the time corresponding to flight without a headwind (6 hr 20 minutes) where x = 500 in the rate. I supposed this would correspond to a crosswind.

How can we tell if we are misinterpreting the problem or overthinking the information given?

Hi Amina,

That’s a good question–and a tough one to answer based on just one problem. With this question, it seems that when you got stuck at the headwind vs. crosswind–and none of the answers seemed to work–you should have looked at the original question again, and looked for a different interpretation. Essentially, you have an instance of “big picture” vs. analyzing details. Knowing when to switch approaches is probably something you can best learn by doing a lot tough word problems. That should give you a feel for knowing when it is time to scrap one approach and try another 🙂

actually that’s a great question. I also didn’t know whether to compare “2 hrs and 40 min longer” to the tailwind time or to the “no wind” time of just ‘x’. I think the question needs to state that. There is no time on the GRE to misinterpret and go back and redo the problem. At the end of the day, either we’ll misinterpret and get it wrong or end up guessing/skipping the question. There’s nowhere on the exam or nobody to turn to on exam day to get this clarification on a poorly worded question.

if x+100 is 600 then x-100 will be 400.

Also its not 2.4 hrs, rather 2+ 40/60 i.e 2.67 hr

Hi Harminder,

Thanks for catching this! Since this discussion thread it a bit long, I just want people to know that Harminders message is in response to ak7012 above who took ‘2.4 hours’ and converted it to miles per hour without properly converting it to a decimal 🙂