kevs wrote:The time taken by a man to ride the cycle along the wind is two third of the time taken by him when he cycles against the direction of the wind. If the product of the speed of the man and the speed of the wind is numerically equal to 20 (considered in kmph), then what is the ratio of speed of the man to that of the wind?
(a). 5:4
(b). 20:1
(c). 5:1
(d). 4:1
Since the problem asks for a RATIO, we can plug in values.
Let the time cycling AGAINST the wind = 6 hours.
Then the time cycling WITH the wind = (2/3)6 = 4 hours.
Time and rate are RECIPROCALS.
Since the TIME ratio = 6:4, the RATE ratio = 4:6.
The DIFFERENCE between the two rates -- 2 miles per hour -- is caused by the wind.
Thus, the wind's rate = 1 mile per hour, INCREASING the rate by 1 mile per hour in one direction and DECREASING it by 1 mile per hour in the other direction, so that the difference between the two rates is 2 miles per hour.
Since the rate cycling WITH the wind = 6 miles per hour, the man's rate WITHOUT the wind = combined rate - wind's rate = 6-1 = 5 miles per hour.
(Man's rate) : (wind's rate) = 5:1.
The correct answer is
C.
Please note the following:
The problem states that the product of the man's rate and the wind's rate is 20.
This information has no effect on the ratio of the two rates and thus is not needed.
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