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.

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:

100% (05:14) correct
0% (00:00) wrong based on 1 sessions

Hi everybody.

A boat travels up the river and down the river the same distance. If the average relative speed of the boat 48 mph and the speed of the river is 10 mph, what is the upriver speed of the boat?

My solution: Up stream the boat must be 10 mph slower than average. So I came up with 38 mph.

However, the correct answer is: 40 mph.

Can anybody explain? _________________

Consider giving kudos, if you like my thread!

FYI: Questions are from the workbook of my prep course.

Re: Speed Problem 4 [#permalink]
25 Apr 2010, 12:02

2

This post received KUDOS

gmatt14 wrote:

Hi everybody.

A boat travels up the river and down the river the same distance. If the average relative speed of the boat 48 mph and the speed of the river is 10 mph, what is the upriver speed of the boat?

My solution: Up stream the boat must be 10 mph slower than average. So I came up with 38 mph.

However, the correct answer is: 40 mph.

Can anybody explain?

40 is correct, average speed = total Distance/total Time

Let speed of boat = x upstream speed = x-10 downstream speed = x+10

total distance covered = 2y time taken upwards = \frac{y}{{x-10}} time taken downwards =\frac{y}{{x+10}}

total time = \frac{y}{{x-10}} + \frac{y}{{x+10}}

average speed = \frac{2y}{{{y/(x-10)}+ y/(x+10)}} = \frac{{x^2-100}}{x}= 48

=> x^2 - 48x - 100 = 0 = (x-50)(x+2) = 0, x cannot be -ve thusx=50

upstream speed =x-10 = 50-10 = 40

PS: 48 is the average relative speed. _________________

Re: Speed Problem 4 [#permalink]
25 Apr 2010, 13:26

1

This post received KUDOS

gmatt14 wrote:

Ok. I can follow you. However, I have difficulties internalizing it. Can you recommend a source where I could review this topic?

You can buy Manhattan guides which are very good though I have not done for maths( Followed Manhattan SC only). with every book you will get good question bank and 6 Cats( CATs are common for all and bank is book specific)

Re: Speed Problem 4 [#permalink]
25 Apr 2010, 18:29

gurpreetsingh +1kudos for you. Thanks, your solution helped me too. I see that you have been actively participating in the forum since few months, when are you planning for the actual exam? are you still a student?

Re: Speed Problem 4 [#permalink]
25 Apr 2010, 21:07

zz0vlb wrote:

gurpreetsingh +1kudos for you. Thanks, your solution helped me too. I see that you have been actively participating in the forum since few months, when are you planning for the actual exam? are you still a student?

Yes I m still preparing for the GMAT, actually I need a lot of improvement in my Verbal. I can not think of giving the exam without improving verbal.

I believe we should give the exam only when we are well prepared. Once prepared any one can beat this exam. _________________

Re: Speed Problem 4 [#permalink]
26 Apr 2010, 03:17

Gurpreet, thanks for your explanation! +1 _________________

But there’s something in me that just keeps going on. I think it has something to do with tomorrow, that there is always one, and that everything can change when it comes. http://aimingformba.blogspot.com

Re: Speed Problem 4 [#permalink]
26 Apr 2010, 07:06

Hi,

Just an alternate solution to get the answer faster. Gurpreet's answer explains the problem step by step, this is just a formula that can get you straight to the equation.

Since the distance traveled both upstream and downstream is same we can use the average speed formula

Average speed = 2xy/x + y Let assume speed of boat = s Speed of stream Speed upstream = x = s-10 Speed downstream = y = s+10 Average speed = 48

Plug in the values we get 2*(s-10)(s+10)/s-10+s+10 = 48