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A man can row against the current three-fourth of a kilometre in 15 min and returns same distance in 10 min, then ratio of his speed to that of current is
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A) 3 : 5 |
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B) 5 : 3 |
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C) 1 : 5 |
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D) 5 : 1 |
Correct Answer:
| D) 5 : 1 |
Description for Correct answer:
Let the speed of man and current be x and y km/h, respectively.
Speed upstream = (x - y) km/h
Speed down stream = (x +y) km/h
According to the question,
\( \Large \frac{3 \times 60}{4 \times 15} = x-y \)
= x - y = 3 ...(i)
and \( \Large \frac{3}{4} \times \frac{60}{10}=x+y \)
= x + y = \( \Large \frac{9}{2} \) ...(ii)
On adding Eqs. (i) and (ii), we get
\( \Large 2x = 3+\frac{9}{2} \)
\( \Large 2x = \frac{6+9}{2} \)
\( \Large x = \frac{15}{4} \)
On putting the value of x in Eq. (ii), we get
\( \Large \frac{15}{4}+y=\frac{9}{2} \)
\( \Large y=\frac{9}{2}-\frac{15}{4}=\frac{18-15}{4} \)
= \( \Large y = \frac{3}{4} \)
Hence, speed of man \( \Large x = \frac{15}{4} \) and
Speed of current \( \Large y = \frac{3}{4} \)
Hene, required ratio = \( \Large \frac{15}{4} : \frac{3}{4} = 5 : 1 \)
Let the speed of man and current be x and y km/h, respectively.
Speed upstream = (x - y) km/h
Speed down stream = (x +y) km/h
According to the question,
\( \Large \frac{3 \times 60}{4 \times 15} = x-y \)
= x - y = 3 ...(i)
and \( \Large \frac{3}{4} \times \frac{60}{10}=x+y \)
= x + y = \( \Large \frac{9}{2} \) ...(ii)
On adding Eqs. (i) and (ii), we get
\( \Large 2x = 3+\frac{9}{2} \)
\( \Large 2x = \frac{6+9}{2} \)
\( \Large x = \frac{15}{4} \)
On putting the value of x in Eq. (ii), we get
\( \Large \frac{15}{4}+y=\frac{9}{2} \)
\( \Large y=\frac{9}{2}-\frac{15}{4}=\frac{18-15}{4} \)
= \( \Large y = \frac{3}{4} \)
Hence, speed of man \( \Large x = \frac{15}{4} \) and
Speed of current \( \Large y = \frac{3}{4} \)
Hene, required ratio = \( \Large \frac{15}{4} : \frac{3}{4} = 5 : 1 \)
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