Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

To arrive at its destination on time the bus should have [#permalink]

Show Tags

29 Oct 2007, 17:02

2

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

95% (hard)

Question Stats:

42% (02:57) correct
58% (02:04) wrong based on 92 sessions

HideShow timer Statistics

To arrive at its destination on time, a bus should have maintained a speed of \(V\) kmh throughout its journey. Instead, after going the first third of the distance at \(V\) kmh, the bus increased its speed and went the rest of the distance at \((1.2)*V\) kmh. If, as a result, the bus arrived at its destination \(X\) minutes earlier than planned, what was the actual duration of the trip?

Re: To arrive at its destination on time the bus should have [#permalink]

Show Tags

29 Oct 2007, 20:52

beckee529 wrote:

To arrive at its destination on time the bus should have maintained a speed of V kmh throughout the journey. Instead, after going the first third of the distance at V kmh, the bus increased its speed and went the rest of the distance at (1.2)*V kmh. As a result, the bus arrived at its destination X minutes earlier than planned. What was the actual duration of the trip?

1. V = 60 2. X = 20

We have 3 rates: Original rate, 1/3 original rate, and 2/3 new rate.

Let: t be the time for 1/3 rate
T be the time for 2/3 rate
Z be the time for the original rate.
d be the distance

We really want to know what T+t is. But the problem isn't that easy so well need to do some rewording. So here are a few equations well need

:Z- (T+t)=X

riginal time: Z=d/V

:1/3 time: t=(d/3)/V --> (6d/18)/V

:2/3 time: T= (2d/3)/1.2V --> (10d/18)/V

Now we can reword our first equation to: d/V - ((10d/18)/V+(6d/18)/V)

(18d/18)/V - (16d/18)/V --> (2d/18)/V = X --> (d/9)/V=X

Now we can finally look at the statements:

S1: V=60. This doesn't help us. We cannot find out what T+t is.

All we can get is that (16d/18)/60= T+t We still have d in there and there is no way to elim the d without adding another variable. Insuff.

S2: x=20 well now we can refer back to our original Z-(T+t)=X

Convieniently reworked to: (d/9)/V=20 180V=d. Plug this into

T+t--> T+t= (16(180V)/18)/V ---> 160V^2=T+t...

ughhh NOOOOOOO on paper I was able to cancel out the V's now i cant. Arghhh!!!

Re: To arrive at its destination on time the bus should have [#permalink]

Show Tags

29 Oct 2007, 21:49

1

This post received KUDOS

2

This post was BOOKMARKED

beckee529 wrote:

To arrive at its destination on time the bus should have maintained a speed of V kmh throughout the journey. Instead, after going the first third of the distance at V kmh, the bus increased its speed and went the rest of the distance at (1.2)*V kmh. As a result, the bus arrived at its destination X minutes earlier than planned. What was the actual duration of the trip?

1. V = 60 2. X = 20

I got B.

d = total distance
d/v is the original total time
d/(3v) + 2d/(3*(1.2v)) is the new total time
Therefore,
d/v - d/(3v) - 2d/(3.6v) = x
d/v*(1 - 1/3 - 1/1.8) = x

Looking for d/v = ?

(1) Given v, you can find d in term of x. This doesn't help us. Imagine the distance is infinite, there is no way we can find the actual time. INSUFFICIENT

(2) Given x, we can plug in
d/v*(1 - 1/3 - 1/1.8) = x
to obtain d/v
SUFFICIENT

Re: To arrive at its destination on time the bus should have [#permalink]

Show Tags

30 Oct 2007, 03:49

bkk145 wrote:

beckee529 wrote:

To arrive at its destination on time the bus should have maintained a speed of V kmh throughout the journey. Instead, after going the first third of the distance at V kmh, the bus increased its speed and went the rest of the distance at (1.2)*V kmh. As a result, the bus arrived at its destination X minutes earlier than planned. What was the actual duration of the trip?

1. V = 60 2. X = 20

I got B.

d = total distance d/v is the original total time d/(3v) + 2d/(3*(1.2v)) is the new total time Therefore, d/v - d/(3v) - 2d/(3.6v) = x d/v*(1 - 1/3 - 1/1.8) = x

Looking for d/v = ?

(1) Given v, you can find d in term of x. This doesn't help us. Imagine the distance is infinite, there is no way we can find the actual time. INSUFFICIENT

(2) Given x, we can plug in d/v*(1 - 1/3 - 1/1.8) = x to obtain d/v SUFFICIENT

Re: To arrive at its destination on time the bus should have [#permalink]

Show Tags

01 Nov 2011, 15:52

Just showing some more of the steps a little different way....

velocity * time = distance vt = d t = d/v

Thus, if we can find d/v, we can find t.

Now, Let: t = original planned time (hours) x = minutes arrived early; Must divide by 60 to convert to hours v= original velocity (km/h) d = total distance

Then, actual time of the trip (what we are solving for) is (t - x/60) which equals (from substitution from above): (d/v) - x/60; So, if we know d/r and x, we can find the actual time of the trip

For the later 2/3rds of the trip, what is given in the problem is:

Re: To arrive at its destination on time the bus should have [#permalink]

Show Tags

01 Nov 2011, 23:34

Expert's post

beckee529 wrote:

To arrive at its destination on time the bus should have maintained a speed of V kmh throughout the journey. Instead, after going the first third of the distance at V kmh, the bus increased its speed and went the rest of the distance at (1.2)*V kmh. As a result, the bus arrived at its destination X minutes earlier than planned. What was the actual duration of the trip?

1. V = 60 2. X = 20

Here is my take. Let the actual duration of the trip be t. Distance is same in both the cases. So, V*t = Increased Average Speed*(t - X/60)

What is the Increased Average Speed? Average Speed = Total Distance/Total Time = \(\frac{1}{[\frac{\frac{1}{3}}{V} + \frac{\frac{2}{3}}{1.2V}]}\) We will get this increased average speed in terms of V.

When we put it in the equation above, V will get canceled leaving us with t in terms of X. So to get t, we only need X. _________________

Re: To arrive at its destination on time the bus should have [#permalink]

Show Tags

12 Sep 2014, 04:33

Expert's post

1

This post was BOOKMARKED

m10 q08

To arrive at its destination on time the bus should have maintained a speed of \(v\) kilometers per hour throughout the journey. Instead, after going the first third of the distance at \(v\) kilometers per hour, the bus increased its speed and went the rest of the distance at \(1.2v\) kilometers per hour. As a result, the bus arrived at its destination \(x\) minutes earlier than planned. What was the actual duration of the trip?

A bus covered 1/3 of the distance at \(v\) kilometers per hour and the remaining 2/3 of the distance at \(1.2v\) kilometers per hour.

Say the actual duration of the trip is \(t\) and the distance is \(d\);

Then \(t=\frac{(\frac{d}{3})}{v}+\frac{(\frac{d2}{3})}{1.2v}\) --> \(t=\frac{d}{v}*(\frac{1}{3}+\frac{2}{3.6})\) --> \(t=\frac{d}{v}*\frac{8}{9}\)

Also we know that if the speed throughout the journey had been \(v\) kilometers per hour the bus would need \(\frac{x}{60}\) hours more time to cover the same distance: \(t+\frac{x}{60}=\frac{d}{v}\);

Substitute \(\frac{d}{v}\) in the first equation: \(t=(t+\frac{x}{60})*\frac{8}{9}\). So, to get \(t\) we need to know the value of \(x\).

(1) v = 60. Not sufficient. (2) x = 20. Sufficient.

Part 2 of the GMAT: How I tackled the GMAT and improved a disappointing score Apologies for the month gap. I went on vacation and had to finish up a...

Cal Newport is a computer science professor at GeorgeTown University, author, blogger and is obsessed with productivity. He writes on this topic in his popular Study Hacks blog. I was...

So the last couple of weeks have seen a flurry of discussion in our MBA class Whatsapp group around Brexit, the referendum and currency exchange. Most of us believed...