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Speed Problem 4

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gmatt14
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Speed Problem 4 [#permalink] New post   25 Apr 2010, 12:53
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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?



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Re: Speed Problem 4 [#permalink] New post   25 Apr 2010, 13:02
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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 thus\(x=50\)

upstream speed =\(x-10 = 50-10 = 40\)

PS: 48 is the average relative speed.
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Re: Speed Problem 4 [#permalink] New post   25 Apr 2010, 13:44
Took me some time, but I figured out your approach. I got stuck at your average speed equation.

Although I understand it now, I don't know wheather I could have done it on the GMAT in 2 min.
Is there a simpler way??????? :shock:
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Re: Speed Problem 4 [#permalink] New post   25 Apr 2010, 13:48
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It can be done in 2 min if you understood the concept. Otherwise we could have used POE on the answer choices.

You can remember this result

if a and b are two speeds for same distance( say x) traveled then the average speed for 2x is 2ab/(a+b)

this way you can put a = x-10 , b = x+10 and 2ab/(a+b) = 48
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Re: Speed Problem 4 [#permalink] New post   25 Apr 2010, 14:10
Ok. I can follow you. However, I have difficulties internalizing it.
Can you recommend a source where I could review this topic?
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Re: Speed Problem 4 [#permalink] New post   25 Apr 2010, 14:26
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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)

Also be regular in math forums and do read GMAT CLUB math book.
gmat-math-book-87417.html

PS: Manhattan Word translation might cover the above topic. Its number property book is also very nice.
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Re: Speed Problem 4 [#permalink] New post   25 Apr 2010, 14:48
Thx for your advices. I appreciate it.
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Re: Speed Problem 4 [#permalink] New post   25 Apr 2010, 19:29
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?
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Re: Speed Problem 4 [#permalink] New post   25 Apr 2010, 22:07
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.
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Re: Speed Problem 4 [#permalink] New post   26 Apr 2010, 04:17
Gurpreet, thanks for your explanation! +1
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Re: Speed Problem 4 [#permalink] New post   26 Apr 2010, 08:06
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

s^2-100/s = 48
s^2-100 - 48s = 0

Solve the quadratic we get
s = 50 or -2

So speed upstream = s-10 = 50-10 = 40 mi/hr



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Re: Speed Problem 4 [#permalink]
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