During a trip on an expressway, Don drove a total of x miles. His aver

Question

During a trip on an expressway, Don drove a total of x miles. His average speed on a certain 5-mile section of the expressway was 30 miles per hour, and his average speed for the remainder of the trip was 60 miles per hour. His travel time for the x-mile trip was what percent greater than it would have been if he had traveled at a constant rate of 60 miles per hour for the entire trip?

A. 8.5%
B. 50%
C. x/12%
D. 60/x%
E. 500/x%

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A. 8.5% 
B. 50% 
C. x/12% 
D. 60/x% 
E. 500/x%

Bunuel · Accepted Answer

During a trip on an expressway, Don drove a total of x miles. His average speed on a certain 5-mile section of the expressway was 30 miles per hour, and his average speed for the remainder of the trip was 60 miles per hour. His travel time for the x-mile trip was what percent greater than it would have been if he had traveled at a constant rate of 60 miles per hour for the entire trip?

A. 8.5%
B. 50%
C. x/12%
D. 60/x%
E. 500/x%

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A. 8.5% 
B. 50% 
C. x/12% 
D. 60/x% 
E. 500/x%

For such questions I personally prefer plug in method. For this particular problem this method should give you the answer in less than a minute.

Say x = 5 miles (so no remainder of the trip).

Time to cover x = 5 miles at 30 miles per hour = (time) = (distance)/(rate) = 5/30 = 1/6 hours = 10 minutes.
Time to cover x = 5 miles at 60 miles per hour = (time) = (distance)/(rate) = 5/60 = 1/12 hours = 5 minutes.

(Or simply, half rate will result in doubling the time.)

So, we can see that the time to cover x = 5 miles at 30 miles per hour (10 minutes) is 100% greater than the time to cover x = 5 miles at 60 miles per hour (5 minutes).

Now, plug x = 5 miles into the answer choices to see which one yields 100%. Only answer E works.

Answer: E.

Hope it's clear.

ywilfred · Answer

total dist = x miles

time take to clear 5 mile section at 30miles/hr = 5/30 = 1/6 hr

time taken to clear x-5 miles section at 60 miles/hr = x-5/60 hr

Total time = 1/6 + x-5/60 = x+5/60 hr

Total time to clear x miles if he traveled at constant rate of 60 miles/hr = x/60 hr

Extra time = 5/60 = 1/12 hr

Percentage greater = (1/12)/(x/60) * 100% = 500/x %

v1rok · Answer

You were on the right track:

(x+5)/x = 1 + 5/x

So, increase as a ratio = 5/x
increase as percent = 500/x (%)

(E)

skovinsky · Answer

Long word problem with a variable in the choices - perfect situation for picking numbers.

Let's let x=15 to keep things simple.

So, he drove 5 miles at 30mph and 10 miles at 60mph.

time = dist/rate, so he spent 5/30 + 10/60 = 1/6 + 1/6 = 1/3 of an hour = 20 minutes total.

If he had travelled at 60mph for the entire 15 miles, his time would have been 15/60 = 1/4 hour = 15 minutes.

The question is "His travel time for the x mile trip was what percent greater", so we use the percent change formula:

% change = (amount of change / original amount) * 100%

= (20-15)/15 * 100% = 5/15 * 100% = 1/3 * 100% = 33 1/3 %

Plugging x=15 into the choices:

A) 8.5 %... nope
B) 50%... nope
C) x/12%.... 5/4 of 1%... nope
D) 60/x%... 4%... nope.
E) 500/x%... 500/15 = 100/3 = 33 1/3.. yay! Choose (E).

maddy2u · Answer

Time when droven x miles in 60 mph  t1 =  x/60 
time when droven 5 miles in 30mph and x-5 miles in 60mph  t2 =  5/30  +  (x-5)/60  =  (x+5)/60

Percentage increase = 100 * (t2 - t1)/t1  =  100  *  (x+5-x)/x =  500/x  %

fluke · Answer

I think the question should be:

During a trip on an expressway, Don drove a total of x miles. His average speed on a certain 5-mile section of the expressway was 30 miles per hour, and his average speed for the remainder of the trip was 60 miles per hour. His travel time for the x-mile trip was what percent greater than it would have been if he had traveled at a constant rate of 60 miles per hour for the entire trip?

A. 8.5%
B. 50%
C. x/12%
D. 60/x%
E. 500/x%

Attachments

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\frac{\frac{x}{60}-\frac{x-5}{60}}{\frac{x}{60}}*100 = \frac{500}{x}%

Ans: "E"

gmat1220 · Answer

Let x = 15
5 miles @ 30 mph time = 1/6
10 miles @ 60 mph time = 1/6
Total time = 1/3
15 miles @ 60 mph, time = 1/4
% change 
(1/3 - 1/4) /1/4 = 1/3 or 100/3% = 33.33%

500/x = 500/15 = 33.33%. Hence E

nave · Answer

I assumed x=10

time taken for 5 miles at 30mph = 10min
time taken for the remaining 5 miles at 60mph = 5min
total time taken to travel 10miles = 15min = T1

If travelled at 60mph constantly for the entire 10 miles then total time taken = 10min = T2

Percentage change in time =  \frac{(T1-T2)}{T2}*100  = 50%

Now this is just one of the answers, it is supposed to change if the total distance changes, as 5 miles becomes different fraction of the total distance (so  answer must contain the total distance x , thereby eliminating choice A & B).

By plugging in with answer choices to see which results in 50% will get the answer i.e E

fozzzy · Answer

An alternative solution

Don’s time for 5 miles is  \frac{5}{30} >>  \frac{10}{60}

Don's time for X-5 miles is  \frac{X-5}{60}

Total time is  \frac{10}{60}  +  \frac{X-5}{60}  >> \frac{5+x}{60}

to find percentage change  \frac{{5+x-x}}{60}  divided by  \frac{X}{60}  * 100

=  \frac{500}{X}  %

Answer E

SVaidyaraman · Answer

We can plug in any value for x because we can see that x is not tied to a specific value. Let us plug in a convenient value.

1. Assume x=10 miles. Total time taken when traveling at different speeds is 5/30 hr + x-5 /60 hr. = 5/30 hr + 5/60 hr = 10 min + 5min = 15 min
2. Total time taken at 60 miles /hr = (5 + x-5)/60 = 10/60 hr = 10 min
3. (1) greater than (2) by 50 %.
4. Substitute the assumed value of x in the choices . Only E gives 50 %

B is not correct because if you change the value of x, the percentage increase will change.

OptimusPrepJanielle · Answer

During a trip on an expressway, Don drove a total of x miles. His average speed on a certain 5-mile section of the expressway was 30 miles per hour, and his average speed for the remainder of the trip was 60 miles per hour. His travel time for the x-mile trip was what percent greater than it would have been if he had traveled at a constant rate of 60 miles per hour for the entire trip?

A. 8.5%
B. 50%
C. x/12%
D. 60/x%
E. 500/x%

Attachments

Untitled.jpg [ 41.45 KiB | Viewed 43144 times ]

A. 8.5% 
B. 50% 
C. x/12% 
D. 60/x% 
 E. 500/x%

JeffTargetTestPrep · Answer

We are given that Don drives a total of x miles, his average speed on a 5-mile section of the expressway was 30 mph, and his average speed for the remainder of the trip, or x - 5, miles was 60 mph.

Since time = distance/rate, the time for the first 5-mile section is 5/30 = 1/6 of an hour and the time for the remainder of the trip is (x-5)/60 hours.

Thus, the total time is 1/6 + (x-5)/60 = 10/60 + (x-5)/60 = (x + 5)/60 hours.

Had he traveled at a constant rate of 60 miles per hour for the entire trip, then his time would have been x/60 hours.

We need to determine the percent by which (x + 5)/60 is greater than x/60.

/(x/60) * 100%

(5/60)/(x/60) * 100%

5/x * 100%

500/x%

Answer: E

BrentGMATPrepNow · Answer

Don drove a total of x miles. His average speed on a certain 5-mile section of the expressway was 30 miles per hour, and his average speed for the remainder of the trip was 60 miles per hour.   
Total time = (time spent driving 30 mph) + (time spent driving 60 mph)
 time = distance/speed 
So, total driving time = 5/30 + (x - 5)/60
= 10/60 + (x - 5)/60
= (10 + x - 5)/60
=  (x + 5)/60

Hypothetically speaking, Don could have driven the entire x miles at a speed of 60 mph  
 time = distance/speed 
Total driving time =  x/60

His travel time for the x-mile trip was what percent greater than it would have been if he had traveled at a constant rate of 60 miles per hour for the entire trip?   
In other word:   (x + 5)/60  is what percent greater than  x/60 ?

Percentage = 100 (x + 5)/60  -  x/60 ]/( x/60 )
= 100 /(x/60)
= 100(5/60)(60/x)
= 500/x

Answer: E

Hoozan · Answer

IanStewart  I have a question regarding percent greater than.

After solving we get  \frac{5+x}{60}  is what percent greater than \frac{ x}{60}
 
As mentioned in one of your books we can say  \frac{x}{60}  is increased by what percent to get  \frac{5+x}{60}

So we can say % increase or % greater =  \frac{ (larger value - smaller value) }{ smaller value }  x 100%

by doing the above we land up with  \frac{500}{x}  %

BUT please could you help me understand when do we include the % sign and when do we not? Since percent means per 100 can I say that if we divide by 100 we drop the % sign and if we multiply by 100 we include the % sign?

Also, could we also say let "p" be the percent greater than value so:

\frac{5+x}{60}  =  \frac{1 + p}{100}  x  \frac{x}{60}

Thus p =  \frac{500}{x} %

Further, what if we were asked  \frac{5+x}{60 }  is  \frac{500}{x}  % greater than what value? How would we represent this question using the above two methods?

(1)  \frac{5+x}{60 } = 1 +  \frac{500}{100x }  x (?) OR

(2)  \frac{5+x}{60}  - (?) / (?) x 100% =  \frac{500}{x } %

While writing version (2) I often get confused when to add the % and when to not

IanStewart · Answer

There are quite a few ways to think about this, and if others read my post, they might be thinking about percents differently (when I get to the algebra), and if they're getting right answers, there's no need to think about it the way I'm explaining things here. But the symbol "%" means "per 100", or algebraically, "/100", so "25%"  means  "25/100" -- they're the same thing. And if you have a number like 0.4, then we can always multiply this by "100%" without changing its value, because 100% = 100/100 = 1. So

0.4 = 0.4 * 100% = (0.4*100) % = 40%

So that's why what you've said in the quote above is correct.

I think there's an issue with the mathematical typesetting in what I quote above -- I'm sure you meant to write:

\frac{5+x}{60} = \left( 1 + \frac{p}{100} ight) \left( \frac{x}{60} ight)

which is perfectly correct. Because in this equation you put in "p/100" which is "p %", when you solve for p, you're done -- that's the percentage you're looking for, and you don't want to multiply or divide by 100. I often do these problems slightly differently, just to avoid fractions: instead I'll solve it like a ratio problem (so I'm finding the ratio of the increase to the smaller value), and then as the final step, I convert that ratio into a percentage so it becomes the answer to a percent increase problem. So you can also solve this way:

\frac{5+x}{60} = \left( 1 + r ight) \left( \frac{x}{60} ight)

then solve for r, knowing r is  not  a percentage, then finally multiplying by "100%" at the end (imitating what I did with "0.4" above) to make it into a percentage.

There are a few ways, all perfectly good, to think about a question like that. If (5 + x)/60 is (500/x)% greater than some value V, then (5 + x)/60 is equal to that value V, plus (500/x)% of V. That's where your equation (1) comes from, though again there's an issue with the mathematical typesetting -- it should read:

\frac{5 + x}{60} = V + \left( \frac{500}{100x} ight) V = \left( 1 + \frac{500}{100x} ight) V

Whenever we're working with percents in an equation where you'll genuinely need to do some algebra (adding or subtracting on both sides, say), you'll want to avoid including "%" symbols anywhere, so you'd want to do what you did here -- replace "500/x %" with "500/100x" first, at which point we're completely done with the "%" sign, and then work with that new fraction. Otherwise you'd need to learn a whole set of unnecessary rules about what you're allowed and not allowed to do when you have a "%" sign in an equation.

It can be a bit faster to learn that when we increase something by z%, we are always multiplying it by 1 + (z/100), so for example, if we increase something by 60%, we are multiplying it by 1 + 60/100 = 1.6. You could use that principle here to produce the same equation above in one less step.

As for your equation (2), I think that's a potentially confusing way to think about situations like this. The danger in these abstract percent questions is that you can easily produce an answer that has too many zeros, or not enough, so you want a method where you can be completely certain if you're looking at a ratio or looking at a percentage. I think your equation (2) is incomplete, but in principle, you could use it: when we say one value is p% greater than another, what we mean is that the ratio of the difference in the two values to the smaller value is p%, or p to 100. So here, if (5+x)/60 is (500/x)% greater than V, that means the ratio of the difference in the two values, which is (5+x)/60 - V, to the smaller value, which is V, is equal to (500/x)%, which is 500/100x, or 5/x. So you get this equation:

\begin{align}
\frac{\frac{5 + x}{60} - V}{V} &= \frac{5}{x} \
x \left( \frac{5 + x}{60} - V ight) &= 5V \
(x) \left( \frac{5 + x}{60} ight) &= 5V + xV \
(x) \left( \frac{5 + x}{60} ight) &= V(5 + x) \
V &= \frac{x}{60}
\end{align}

I find that a confusing way to think through the problem though, and it also leads to messier algebra than the other method, so at least in an algebraic question like this one, I wouldn't use it (though in simpler situations I might). Notice though that I converted the percentage into a ratio immediately, so I had an ordinary number to work with, and I never needed to include any percent symbols in the equation anywhere -- I think it would be very easy to end up with the wrong number of zeros trying to write one or both sides of the equation as a percentage rather than as a simple ratio.

In simpler questions, where all we're doing is simplifying a fraction that will turn into the answer, I often leave the % symbol in the equation though. So if something increases from $300 to $360, and I want to find the percent increase, I'd often just write

\left(\frac{360 - 300}{300} ight) 	imes 100\%

then I'd leave the '%' symbol alone, and cancel the '100' with the denominator:

\left(\frac{360 - 300}{300} ight) 	imes 100\% = \left( \frac{360 - 300}{3} ight) \% = 20\%

and the '%' symbol here just reminds me that I've already taken care of the conversion to a percentage. But this situation is different from the question in this thread -- in this example, we're not manipulating quantities in an equation and moving things around, and we're just tacking the "%" symbol on at the end. In any more complicated situation, I'd convert percents to ordinary fractions first, and then you can use all the familiar rules of arithmetic and algebra, and then you'd just need to pay attention to whether you need to convert your answer to a percentage at the end.

Hope that clears things up!

Hoozan · Answer

Thanks a lot IanStewart . This was very helpful

UserMaple5 · Answer

Dear experts,

Can someone please point out what I'm doing wrong here? I notice there are significantly better ways to deal with this question but just wanted to get clarify where my math was going wrong in my original approach. Thanks!

Assume x = 100 miles
5 miles at 30 mph so time to cover 5 miles = 1/6 hr
95 miles at 60 mph, so time to cover 95 miles = 19/12 hr
Total time for the trip is: (1/6) + (19/12) = 21 / 12 hr.

At 60 mph, the time needed to cover 100 miles is 5/3 hours

So additional time in actual trip = (21/12) - (5/3) = 1/12 hrs.

At this point I'm getting confused. The wording is getting me confused "His travel time for the x-mile trip was what percent greater than it would have been if he had traveled at a constant rate of 60 miles per hour for the entire trip?"

what should I do next ? I notice if I do (1/12) / 5/3 I get the answer , but I don't understand it. Any help would be appreciated.

what should I do next ? I notice if I do (1/12) / 5/3 I get the answer , but I don't understand it. Any help would be appreciated.

thank you!

Bunuel   JeffTargetTestPrep   IanStewart   BrentGMATPrepNow

Kinshook · Answer

Given: During a trip on an expressway, Don drove a total of x miles. His average speed on a certain 5-mile section of the expressway was 30 miles per hour, and his average speed for the remainder of the trip was 60 miles per hour.

Asked: His travel time for the x-mile trip was what percent greater than it would have been if he had traveled at a constant rate of 60 miles per hour for the entire trip?

For 5-mile section: 
Average speed = 30 mph
Time spent = 5/30 = 1/6 hours = 10 minutes

For remaining trips: 
Distance travelled = (x - 5) miles
Average speed = 60 mph
Time spent = (x-5)/60

Total time spent = 1/6 + (x-5)/60 = (x+5)/60 hours

Time spent with constant rate of 60 miles per hour for the entire trip = x/60

Difference in time spent = 5/60 = 1/12 hours

Percentage difference in time spent = (1/12)/(x/60)*100 = (500/x)%

IMO E

IanStewart · Answer

As you say, there are better ways to solve this question, but everything you've done is perfect. There are just two things left to do -- first we need to decode the language of the question:

"His travel time for the x-mile trip ( you found this was 21/12 = 7/4 hours ) was what percent greater than it would have been if he had traveled at a constant rate of 60 miles per hour for the entire trip ( you found this was 5/3 hours )?"

So the question is really "7/4 is what percent greater than 5/3?" If you've learned a "percent greater" or "percent increase" formula, it will work here. Or you could think about how we calculate "percent greater" when we have simpler numbers: if you're asked "260 is what percent greater than 200", all you need to do is divide the difference, 260-200 = 60, by the smaller value, 200, to get 60/200 = 30/100, then multiply by 100% to express that as a percentage, so the answer would be 30%. We can do the same here or in any other percent greater question: we just want to divide the difference, 7/4 - 5/3 = 1/12, by the smaller value, 5/3, and then multiply by 100% to get a percentage. If you do that you get (1/12) / (5/3) = 1/20, and expressing that as a percentage, the answer is 5%.

So what we've now proven is that when x = 100, the answer must be 5%. So the last step is to plug x = 100 into each answer choice, looking for an answer that turns out to equal 5%. Luckily only answer E gives us 5% as our answer (if two answers both gave us 5%, we'd have to do more work to decide between them, which is one potential disadvantage of using an approach like this).

If you do decide to use this kind of approach on similar questions, before getting deep into calculation, it's worthwhile asking what choice of number will lead to the simplest solution -- I think the later steps here might have become confusing because you had fractions you needed to work with, and if you had started from a simpler total distance you might have found everything easier (you might look at Bunuel and skovinsky's solutions -- they chose starting numbers that lead to nicer calculations).

I hope that all makes sense! If not, just let me know at what point I haven't explained in enough detail.

During a trip on an expressway, Don drove a total of x miles. His aver

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During a trip on an expressway, Don drove a total of x miles. His aver

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

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