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
Ratio Problem
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5:1
speed of the man --> x
of the wind --> y
speed when in the direction of the wind --> x + y
and the time taken is say t1
speed when against the direction of the wind --> x - y
and the time taken is say t2
t1/t2 = 2/3
Since time is inversely proportional to the speed,
(x - y) / (x + y) = 2/3
(x2 - xy) / (x2 + xy) = 2/3
3(x2 - 20) = 2 (x2 + 20)
x2 = 100
x = 10
so y = 2
x:y = 5:1
Please suggest if there is a simpler solution.
Thanks,
Ravi
speed of the man --> x
of the wind --> y
speed when in the direction of the wind --> x + y
and the time taken is say t1
speed when against the direction of the wind --> x - y
and the time taken is say t2
t1/t2 = 2/3
Since time is inversely proportional to the speed,
(x - y) / (x + y) = 2/3
(x2 - xy) / (x2 + xy) = 2/3
3(x2 - 20) = 2 (x2 + 20)
x2 = 100
x = 10
so y = 2
x:y = 5:1
Please suggest if there is a simpler solution.
Thanks,
Ravi
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Since the problem asks for a RATIO, we can plug in values.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
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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Hi Mitch,
Just need one clarification! In case of two given routes (here, cycling against the wind and cycling with the wind),we do consider rate and time as inversely proportional to each other when the distance is constant in both cases.Am i correct? Here, the distance is not mentioned.
Just need one clarification! In case of two given routes (here, cycling against the wind and cycling with the wind),we do consider rate and time as inversely proportional to each other when the distance is constant in both cases.Am i correct? Here, the distance is not mentioned.
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Correct!kalpita123 wrote:Hi Mitch,
Just need one clarification! In case of two given routes (here, cycling against the wind and cycling with the wind),we do consider rate and time as inversely proportional to each other when the distance is constant in both cases.Am i correct? Here, the distance is not mentioned.
As long as the distance in each direction is THE SAME, the time ratio and the rate ratio will be RECIPROCALS.
Whether the distance is 1 mile or 1000 miles has no effect on this reciprocal relationship.
Other problems:
https://www.beatthegmat.com/abstract-spe ... 09462.html
https://www.beatthegmat.com/tricky-question-t115243.html
https://www.beatthegmat.com/rate-work-an ... 14760.html
https://www.beatthegmat.com/rate-work-pr ... 16533.html
https://www.beatthegmat.com/a-work-problem-t123114.html
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Hi Kalpita,kalpita123 wrote:Hi Mitch,
Just need one clarification! In case of two given routes (here, cycling against the wind and cycling with the wind),we do consider rate and time as inversely proportional to each other when the distance is constant in both cases.Am i correct? Here, the distance is not mentioned.
If the distance is not the same, then the info in this question is not enough to solve the problem.
Thanks,
Ravi
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GMATGuruNY wrote:Correct!kalpita123 wrote:Hi Mitch,
Just need one clarification! In case of two given routes (here, cycling against the wind and cycling with the wind),we do consider rate and time as inversely proportional to each other when the distance is constant in both cases.Am i correct? Here, the distance is not mentioned.
As long as the distance in each direction is THE SAME, the time ratio and the rate ratio will be RECIPROCALS.
Whether the distance is 1 mile or 1000 miles has no effect on this reciprocal relationship.
Other problems:
https://www.beatthegmat.com/abstract-spe ... 09462.html
https://www.beatthegmat.com/tricky-question-t115243.html
https://www.beatthegmat.com/rate-work-an ... 14760.html
https://www.beatthegmat.com/rate-work-pr ... 16533.html
https://www.beatthegmat.com/a-work-problem-t123114.html
Does that mean, although not explicitly mentioned here in the above question, we have considered the distance to be the same?
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Yes.kalpita123 wrote:Does that mean, although not explicitly mentioned here in the above question, we have considered the distance to be the same?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
An actual GMAT problem would state explicitly that the route is the same in each direction; otherwise, the man's speed and the wind's speed could be ANY COMBINATION with a product of 20.
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This is not a realistic practice question. It needs to make clear the distance traveled is the same in each direction, it needs to make clear that the speed of the cyclist (ignoring the wind) is the same in each direction, and the phrase "along the wind" does not mean what the question writer intends it to mean. It also tests something the GMAT would never test - physics. We need to assume that in one direction we add the windspeed to the cyclist's speed, and in the other direction we subtract. Finally the information about the product of the speeds is superfluous if we only need a ratio, and GMAT questions almost never include superfluous information.
In any case, if the speed of the cyclist in no wind is s, and the speed of the wind is w, then if the ratio of the times in each direction is 2/3, the ratio of the speeds is 3/2. So the ratio of s+w to s-w is 3 to 2. You could then just solve the equation (s+w)/(s-w) = 3/2 to find that s/w = 5/1. Or you can use the answer choices. The answers are in the form s:w, so we just need to find an answer where s+w : s-w is 3:2, and so you can see very quickly that C is correct.
In any case, if the speed of the cyclist in no wind is s, and the speed of the wind is w, then if the ratio of the times in each direction is 2/3, the ratio of the speeds is 3/2. So the ratio of s+w to s-w is 3 to 2. You could then just solve the equation (s+w)/(s-w) = 3/2 to find that s/w = 5/1. Or you can use the answer choices. The answers are in the form s:w, so we just need to find an answer where s+w : s-w is 3:2, and so you can see very quickly that C is correct.
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