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10 Jan 2021, 07:32
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Question Stats:
57% (02:46) correct
43%
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wrong
based on 132
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A boat travels from City A to City B upstream and returns back to City A downstream. If the average speed of the total journey is 48 mph and the speed of the stream is 10 mph, what is the upstream speed of the boat?
A. 38
B. 40
C. 42
D. 44
E. 46
A. 38
B. 40
C. 42
D. 44
E. 46
rocky620
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Joined: 10 Nov 2018
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GMAT 1: 590 Q49 V22
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10 Jan 2021, 07:54
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Let,
Distance (one way) = d; so, distance (to & fro) = 2d
Speed of boat = s
Given,
Avg speed (to & fro) = 48
Stream Speed = 10
So, time taken for complete journey = \(\frac{Total distance}{ Avg. speed}\) = \(\frac{2d}{48}\) = \(\frac{d}{24}\)......Eq-1
When we go upstream our actual speed = Speed - Speed of the stream (and vice versa for downstresm)
So, Total time = Time to travel upstream + Time to travel downstream = \(\frac{d}{s-10}\)+\(\frac{d}{s+10}\).......Eq-2
Equating, EQ-1 = Eq-2
\(\frac{d}{s-10}\)+\(\frac{d}{s+10}\) = \(\frac{d}{24}\)
\(\frac{1}{s-10}\)+\(\frac{1}{s+10}\) = \(\frac{1}{24}\)
48s = \(s^2\)-100
\(s^2\)-48s-100 = 0
s = -2, 50 (neglect -ve value as speed cannot be -ve)
s=50
So, upstream speed = 50-10=40
Option-B
Distance (one way) = d; so, distance (to & fro) = 2d
Speed of boat = s
Given,
Avg speed (to & fro) = 48
Stream Speed = 10
So, time taken for complete journey = \(\frac{Total distance}{ Avg. speed}\) = \(\frac{2d}{48}\) = \(\frac{d}{24}\)......Eq-1
When we go upstream our actual speed = Speed - Speed of the stream (and vice versa for downstresm)
So, Total time = Time to travel upstream + Time to travel downstream = \(\frac{d}{s-10}\)+\(\frac{d}{s+10}\).......Eq-2
Equating, EQ-1 = Eq-2
\(\frac{d}{s-10}\)+\(\frac{d}{s+10}\) = \(\frac{d}{24}\)
\(\frac{1}{s-10}\)+\(\frac{1}{s+10}\) = \(\frac{1}{24}\)
48s = \(s^2\)-100
\(s^2\)-48s-100 = 0
s = -2, 50 (neglect -ve value as speed cannot be -ve)
s=50
So, upstream speed = 50-10=40
Option-B
16 Feb 2021, 01:03
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Let the Speed of the Boat in still water = R
Speed Upstream = R - 10
Speed Downstream = R + 10
Since the distance covered in each part is constant/the same (from A to B = from B to A):
The average speed = Harmonic Mean of the 2 speeds traveled at over the 2 equal parts
B = (2)(A)(C) / (A + C)
Where B = average speed = 48
And A and C are the Speed Upstream and the Speed Downstream.
48 = (2)(R + 10)(R - 10) / (R + 10 + R - 10)
48 = (2)(R + 10)(R - 10) / 2R
——the 2 cancels and the NUM is the Difference of Squares template——-
48 = (R^2 - 100) / R
(R)^2 - 48R - 100 = 0
(R - 50) (R + 2) = 0
R = 50 kmph = speed of boat in still water
Speed Upstream = 50 - 10 = 40
B
Posted from my mobile device
Speed Upstream = R - 10
Speed Downstream = R + 10
Since the distance covered in each part is constant/the same (from A to B = from B to A):
The average speed = Harmonic Mean of the 2 speeds traveled at over the 2 equal parts
B = (2)(A)(C) / (A + C)
Where B = average speed = 48
And A and C are the Speed Upstream and the Speed Downstream.
48 = (2)(R + 10)(R - 10) / (R + 10 + R - 10)
48 = (2)(R + 10)(R - 10) / 2R
——the 2 cancels and the NUM is the Difference of Squares template——-
48 = (R^2 - 100) / R
(R)^2 - 48R - 100 = 0
(R - 50) (R + 2) = 0
R = 50 kmph = speed of boat in still water
Speed Upstream = 50 - 10 = 40
B
Posted from my mobile device
