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# A man travels by a motor boat down a river to his office and back

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Manager
Joined: 27 May 2010
Posts: 200
A man travels by a motor boat down a river to his office and back  [#permalink]

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19 Jun 2019, 00:06
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00:00

Difficulty:

55% (hard)

Question Stats:

55% (02:40) correct 45% (02:58) wrong based on 20 sessions

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A man travels by a motor boat down a river to his office and back. With the speed of the river unchanged, if he doubles the speed of his motor boat, then his total travel time gets reduced by 75%. The ratio of the original speed of the motor boat to the speed of the river is:

A) 6^0.5 : 2^0.5
B) 7^0.5 : 2^0.5
C) 2 x 5^0.5 : 3^0.5
D) 3:2
E) 7^0.5 : 2
Manager
Joined: 07 Aug 2017
Posts: 84
Location: India
GPA: 4
WE: Information Technology (Consulting)
Re: A man travels by a motor boat down a river to his office and back  [#permalink]

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19 Jun 2019, 02:22
prashanths wrote:
A man travels by a motor boat down a river to his office and back. With the speed of the river unchanged, if he doubles the speed of his motor boat, then his total travel time gets reduced by 75%. The ratio of the original speed of the motor boat to the speed of the river is:

A) 6^0.5 : 2^0.5
B) 7^0.5 : 2^0.5
C) 2 x 5^0.5 : 3^0.5
D) 3:2
E) 7^0.5 : 2

Let $$Sb$$ be the speed of boat and $$Sr$$ be the speed of river.
When the man is travelling downstream, his speed is = $$Sb + Sr$$
and when he is travelling upstream, his speed is = $$Sb - Sr$$

The average speed of both upstream and downstream is:
$$\frac{(2*(Sb+Sr)(Sb-Sr))}{(Sb+Sr+Sb-Sr)}$$

$$=\frac{(Sb^2-Sr^2)}{Sb}$$

Let D be the distance and t be the time

$$\frac{(Sb^2-Sr^2)}{Sb} = \frac{D}{t}$$

When the speed of boat is doubled -
Downstream speed = $$2Sb+Sr$$
Upstream speed = $$2Sb-Sr$$

Average speed = $$\frac{2*(4Sb^2-Sr^2)}{4Sb}$$

$$t1=0.25t$$
Distance will remain same.

$$\frac{(4Sb^2-Sr^2)}{2Sb} = \frac{D}{0.25t}$$

$$\frac{(4Sb^2-Sr^2)}{8Sb} = \frac{D}{t}$$

Equating the two equations -

$$\frac{(Sb^2-Sr^2)}{Sb} = \frac{(4Sb^2-Sr^2)}{8Sb}$$

$$8*(Sb^2-Sr^2) = 4Sb^2-Sr^2$$
$$4Sb^2 = 7Sr^2$$
$$Sb^2/Sr^2 = 7/4$$
$$Sb/Sr = 7^\frac{1}{2}/2$$

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Re: A man travels by a motor boat down a river to his office and back  [#permalink]

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21 Jun 2019, 11:57
prashanths wrote:
A man travels by a motor boat down a river to his office and back. With the speed of the river unchanged, if he doubles the speed of his motor boat, then his total travel time gets reduced by 75%. The ratio of the original speed of the motor boat to the speed of the river is:

A) 6^0.5 : 2^0.5
B) 7^0.5 : 2^0.5
C) 2 x 5^0.5 : 3^0.5
D) 3:2
E) 7^0.5 : 2

We can let b = the normal speed of the boat and r = the speed of the river. Thus, we need to determine b/r. Furthermore, we can let d = the one-way distance and t = total time when the boat is traveling at its normal speed. We can create the equations:

d/(b + r) + d/(b - r) = t

and

d/(2b + r) + d/(2b - r) = t/4

Multiplying the second equation by 4, we have:

4d/(2b + r) + 4d/(2b - r) = t

Now equating this with the first equation, we have:

d/(b + r) + d/(b - r) = 4d/(2b + r) + 4d/(2b - r)

Dividing the equation by d, we have:

1/(b + r) + 1/(b - r) = 4/(2b + r) + 4/(2b - r)

(b - r + b + r)/[(b + r)(b - r)] = (8b - 4r + 8b + 4r)/[(2b + r)(2b - r)]

2b/(b^2 - r^2) = 16b/(4b^2 - r^2)

Dividing both sides by 2b, we have:

1/(b^2 - r^2) = 8/(4b^2 - r^2)

4b^2 - r^2 = 8b^2 - 8r^2

7r^2 = 4b^2

7/4 = b^2/r^2

√(7/4) = b/r

b/r = √(7)/2

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Re: A man travels by a motor boat down a river to his office and back   [#permalink] 21 Jun 2019, 11:57
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