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A Banana Boat travels from Sunniville to Gloomiville upstream and retu 10 Jan 2021, 07:35
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A Banana Boat travels from Sunniville to Gloomiville upstream and returns back to Sunniville downstream. If the average speed of the total journey is 37.5 km/hr and the speed of the stream is 10 km/hr, what is the upstream speed of the Banana Boat?

A. 27.5
B. 30
C. 35
D. 40
E. 50
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B
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A Banana Boat travels from Sunniville to Gloomiville upstream and retu Updated on: 15 Jan 2021, 02:15
Originally posted by SherzodAzamov on 11 Jan 2021, 00:19.
Last edited by SherzodAzamov on 15 Jan 2021, 02:15, edited 1 time in total.
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Bunuel
A Banana Boat travels from Sunniville to Gloomiville upstream and returns back to Sunniville downstream. If the average speed of the total journey is 37.5 km/hr and the speed of the stream is 10 km/hr, what is the upstream speed of the Banana Boat?

A. 27.5
B. 30
C. 35
D. 40
E. 50
Show more

Let's say the distance is S, speed of the boat in still water is V.
average speed is found by dividing total distance to total time:
Tu = \(\frac{S}{(V-10)}\)
Td = \(\frac{S}{(V+10)}\)

(S+S)/(Tu+Td) = 37.5
2S/(S/(V-10) + S/(V+10)) = 37.5
2/(1/(V-10) + 1/(V+10)) = 37.5
2/(2V/(V^2-100)) = 37.5
...
V^2-100 = 37.5V
V^2 - 37.5V - 100 = 0
Solving for V, we get:
V = 40
We need upstream speed, so 40 - 10=30

Answer B
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A Banana Boat travels from Sunniville to Gloomiville upstream and retu 15 Jan 2021, 01:44
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SherzodAzamov
Bunuel
A Banana Boat travels from Sunniville to Gloomiville upstream and returns back to Sunniville downstream. If the average speed of the total journey is 37.5 km/hr and the speed of the stream is 10 km/hr, what is the upstream speed of the Banana Boat?

A. 27.5
B. 30
C. 35
D. 40
E. 50

Let's say the distance is S, speed of the boat in still water is V.
average speed is found by dividing total distance to total time:
Tu = \(\frac{S}{(V-10)}\)
Td = \(\frac{S}{(V+10)}\)

(S+S)/(Tu+Td) = 37.5
2S/(S/(V-10) + S/(V+10)) = 37.5
2/(1/(V-10) + 1/(V+10)) = 37.5
2/(2V/(V^2-100)) = 37.5
...
V^2-100 = 37.5V
V^2 - 37.5V - 100 = 0
Solving for V, we get:
V = 40
We need upstream speed, so 40+10=50

Answer E
Show more

All good except the last line, upstream means 40-10=30
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A Banana Boat travels from Sunniville to Gloomiville upstream and retu 15 Jan 2021, 02:17
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Pedrotri45
SherzodAzamov
Bunuel
A Banana Boat travels from Sunniville to Gloomiville upstream and returns back to Sunniville downstream. If the average speed of the total journey is 37.5 km/hr and the speed of the stream is 10 km/hr, what is the upstream speed of the Banana Boat?

A. 27.5
B. 30
C. 35
D. 40
E. 50

Let's say the distance is S, speed of the boat in still water is V.
average speed is found by dividing total distance to total time:
Tu = \(\frac{S}{(V-10)}\)
Td = \(\frac{S}{(V+10)}\)

(S+S)/(Tu+Td) = 37.5
2S/(S/(V-10) + S/(V+10)) = 37.5
2/(1/(V-10) + 1/(V+10)) = 37.5
2/(2V/(V^2-100)) = 37.5
...
V^2-100 = 37.5V
V^2 - 37.5V - 100 = 0
Solving for V, we get:
V = 40
We need upstream speed, so 40+10=50

Answer E

All good except the last line, upstream means 40-10=30
Show more

I guess I was in a hurry a bit. GMAT does not like hurry) Thanks buddy, for the help.
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A Banana Boat travels from Sunniville to Gloomiville upstream and retu 15 Jan 2021, 03:32
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SOLUTION:

Let speed of the boat be B kmph and one way distance be "d" km.

Total distance in the journey is 2d and the respective upstream and downstream time taken are d/(B-10) kmph & d/(B+10) kmph respectively

Given average speed of the journey is 37.5kmph.

=> 2d / [d/(B-10) + d/(B+10)] = 37.5
=> 2 / [1/(B-10) + 1/(B+10)] =37.5
=> (B-10)(B+10)/B = 37.5

Now use the options and plug in values.

If you use option B and set B-10(Upstream speed)=30=>B=40 then,
at B =40 equation (1) is (40-10)(40+10)/40 =37.5

Option (B)

Hope this helps :thumbsup:
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A Banana Boat travels from Sunniville to Gloomiville upstream and retu 17 Jan 2021, 08:49
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Let speed of the boat be u kmph and one way distance be "d" km.

Total distance in the journey is 2d

We can represent the data in a table.

Given average speed of the journey is 37.5kmph.

=> 2d = 37.5 * [d/(u-10) + d/(u+10)]
=> 2 = 37.5 * [1/(u-10) + 1/(u+10)]
=> 2(u-10)(u+10)= 75* u
=> 2u^2 -75u-200=0
solving the equation, the value of u is 40, - 5/2(not acceptable)

Upstream speed is 40-10=30
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A Banana Boat travels from Sunniville to Gloomiville upstream and retu 10 Apr 2025, 08:38
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