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# Speed

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Senior Manager
Joined: 08 Jan 2009
Posts: 326

Kudos [?]: 176 [0], given: 5

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10 Sep 2009, 19:33
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A boat crossed a lake from North to South at the speed of 4 km/h, entered a river and covered twice as much distance going upstream at 3 km/h. It then turned around and stopped at the south shore of the lake. If it averaged 3.8 km/h that day, what was its approximate downstream speed?

[Reveal] Spoiler:
5

I have some word translation issues. Hope some one can there also.

Husband is seven times older than his son ? does this translate to W=7S or W - S = 7S.

Kudos [?]: 176 [0], given: 5

Manhattan Prep Instructor
Joined: 28 Aug 2009
Posts: 152

Kudos [?]: 491 [1], given: 6

Location: St. Louis, MO
Schools: Cornell (Bach. of Sci.), UCLA Anderson (MBA)

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11 Sep 2009, 11:47
1
KUDOS
tkarthi4u wrote:
A boat crossed a lake from North to South at the speed of 4 km/h, entered a river and covered twice as much distance going upstream at 3 km/h. It then turned around and stopped at the south shore of the lake. If it averaged 3.8 km/h that day, what was its approximate downstream speed?

This one is really difficult to do algebraically in 2 minutes! I just did that solution and wouldn't recommend it to anyone. Not only are you solving for some variable R (the unknown downstream rate), but you have some unknown distances: X = distance across the lake and 2X = distance upstream = distance downstream. In addition, the RT = D or T = D/R formula means we will encounter fractions in our algebra--not ideal!

Since X is not specified, and it must drop out of the algebra in order to solve for R, why not pick a number for X? If we pick strategically, we can avoid dealing with too many fractions.

Avg. Speed = Total Distance/Total Time.
3.8km/h = (X + 2X + 2X)/Total Time.

Total Time = Time to cross lake + Time to go upstream + Time to go downstream.
In general, Time = Dist/Rate, so let's pick distance X such that it is divisible by both given rates: 3km/h and 4km/h. Thus, X = 12km.
If the lake is 12km across, Time to cross lake = 12km/4km/h = 3h.
If the travel upstream is 24km, Time to go upstream = 24km/3km/h = 8h.
If the travel downstream is 24km, Time to go downstream = 24km/R. (Remember, we are solving for this R)

So,
3.8 = 5X/(3 + 8 + 24/R) = 60/(11 + 24/R)
3.8(11 + 24/R) = 60
38(11 + 24/R) = 600 (multiplied both sides by 10, just to get rid of decimals)
418 + (38)(24)/R = 600
(38)(24)/R = 182
R = (38)(24)/182 = (19)(24)/91 = 456/91 = approx 5.

tkarthi4u wrote:
Husband is seven times older than his son ? does this translate to W=7S or W - S = 7S.

I think it could be interpreted either way. To avoid this confusion, I think the actual GMAT would present the constraint with different wording. Something like "A father is seven times as old as his son..." (F = 7S)
_________________

Emily Sledge | Manhattan GMAT Instructor | St. Louis

Manhattan GMAT Discount | Manhattan GMAT Course Reviews | Manhattan GMAT Reviews

Kudos [?]: 491 [1], given: 6

Intern
Joined: 03 Sep 2009
Posts: 29

Kudos [?]: 15 [0], given: 1

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11 Sep 2009, 15:59
X = distance across the lake and 2X = distance upstream = distance downstream
y: downstream speed
=> time to across the lake = X/4
time of upstream = 2X/3
time of downstream = 2X/y

Total time (for across + upstream + downstream)= X/4 + 2X/3 + 2X/y
Average speed = (X + 2X + 2X) / (total time) = 5X / (X/4 + 2X/3 +2X/y) = 3.8
=> 5/3.8 - 1/4 - 2/3 = 2/y => y ~ 2/0.4 = 5

=> we don't need to know X or give X a value to solve the problem

esledge wrote:
tkarthi4u wrote:
A boat crossed a lake from North to South at the speed of 4 km/h, entered a river and covered twice as much distance going upstream at 3 km/h. It then turned around and stopped at the south shore of the lake. If it averaged 3.8 km/h that day, what was its approximate downstream speed?

This one is really difficult to do algebraically in 2 minutes! I just did that solution and wouldn't recommend it to anyone. Not only are you solving for some variable R (the unknown downstream rate), but you have some unknown distances: X = distance across the lake and 2X = distance upstream = distance downstream. In addition, the RT = D or T = D/R formula means we will encounter fractions in our algebra--not ideal!

Since X is not specified, and it must drop out of the algebra in order to solve for R, why not pick a number for X? If we pick strategically, we can avoid dealing with too many fractions.

Avg. Speed = Total Distance/Total Time.
3.8km/h = (X + 2X + 2X)/Total Time.

Total Time = Time to cross lake + Time to go upstream + Time to go downstream.
In general, Time = Dist/Rate, so let's pick distance X such that it is divisible by both given rates: 3km/h and 4km/h. Thus, X = 12km.
If the lake is 12km across, Time to cross lake = 12km/4km/h = 3h.
If the travel upstream is 24km, Time to go upstream = 24km/3km/h = 8h.
If the travel downstream is 24km, Time to go downstream = 24km/R. (Remember, we are solving for this R)

So,
3.8 = 5X/(3 + 8 + 24/R) = 60/(11 + 24/R)
3.8(11 + 24/R) = 60
38(11 + 24/R) = 600 (multiplied both sides by 10, just to get rid of decimals)
418 + (38)(24)/R = 600
(38)(24)/R = 182
R = (38)(24)/182 = (19)(24)/91 = 456/91 = approx 5.

tkarthi4u wrote:
Husband is seven times older than his son ? does this translate to W=7S or W - S = 7S.

I think it could be interpreted either way. To avoid this confusion, I think the actual GMAT would present the constraint with different wording. Something like "A father is seven times as old as his son..." (F = 7S)

Kudos [?]: 15 [0], given: 1

SVP
Joined: 29 Aug 2007
Posts: 2472

Kudos [?]: 841 [0], given: 19

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14 Sep 2009, 22:06
tkarthi4u wrote:
A boat crossed a lake from North to South at the speed of 4 km/h, entered a river and covered twice as much distance going upstream at 3 km/h. It then turned around and stopped at the south shore of the lake. If it averaged 3.8 km/h that day, what was its approximate downstream speed?

[Reveal] Spoiler:
5

The question is not crystal clear.

4n + 8n + 8n = 3.8 (n + 8n/3 + 8n/k) ........... where n = number of hours and k is speed in km/h
20k = 3.8 (3k + 8k + 24)
60k = 3.8 (11k) + 3.8 (24)
60k - 41.8k = 3.8 (24)
k = 3.8 (24)/41.8
k = 5.011 km/h

tkarthi4u wrote:
I have some word translation issues. Hope some one can there also.

Husband is seven times older than his son ? does this translate to W=7S or W - S = 7S.

This sentence probably wanted to say, imo, h = 7s but still another very unclear.
_________________

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GT

Kudos [?]: 841 [0], given: 19

Manager
Joined: 10 Jul 2009
Posts: 125

Kudos [?]: 182 [0], given: 60

Location: Ukraine, Kyiv

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20 Sep 2009, 07:45
my solution is long, but still I managed to solve this. good question, besides! Thanks.

x, 2x, 2x distances. total distance 5x.
x/4+2x/3+2x/y = total time (11xy+24x)/12y

main equation:

5x/[(11xy+24x)/12y]=3.8
solving it we get 5 as the answer.
(5x*12y)/(11xy+24x)=60xy/(11xy+24x)=3.8
60xy=41.8xy+91.2x
18.2xy=91.2x
y=5
_________________

Never, never, never give up

Kudos [?]: 182 [0], given: 60

Manager
Joined: 04 Sep 2009
Posts: 53

Kudos [?]: 29 [0], given: 7

WE 1: Real estate investment consulting

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20 Sep 2009, 12:22
GMAT TIGER wrote:

4n + 8n + 8n = 3.8 (n + 8n/3 + 8n/k) ........... where n = number of hours and k is speed in km/h
20k = 3.8 (3k + 8k + 24)
60k = 3.8 (11k) + 3.8 (24)
60k - 41.8k = 3.8 (24)
k = 3.8 (24)/41.8
k = 5.011 km/h

This is good!

Got there via different road - almost identical to the previous post, but did not use any x.

$$\frac{5}{1/4+2/3+2/x}=3.8$$

The rest of the principle is the same

Kudos [?]: 29 [0], given: 7

Manager
Joined: 27 Oct 2008
Posts: 185

Kudos [?]: 164 [0], given: 3

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24 Sep 2009, 11:32
Question not put forth clearly. After some trouble, i arrived at the same equation as you guys and got the answer.

5x/(x/4 + 2x/3 + 2x/y) = 3.8
x cancels out and thus we can solve for y which is the downstream speed.

Kudos [?]: 164 [0], given: 3

Re: Speed   [#permalink] 24 Sep 2009, 11:32
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