thank you for your help on this one...
Circular gears P and Q start rotating at the same time at constant speeds. Gear P makes 10 revolutions per minute and gear Q makes 40 revolutions per minute. How many seconds after the gears start rotating will gear Q have made exactly 6 more revolutions than geat P ?
A. 6
B. 8
C. 10
D.12
E.15
See attached picture: Can I say that one does "P does "S" spins and Q does "S + 6" spins", then figure out the time and set equal as shown?
Is this wrong and I just got lucky, or is this actually the right approach?
Secondly, can I say:
Combined rate is 1/2, therefore to do six extra spins, it would be 6/1/2 = 12?
P and Q Circular gears rate problem... can I do it this way?
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I think your approach looks good, but I'm not 100% certain what some of your numbers and expressions represent. Can you offer a little more rationale.benjiboo wrote:thank you for your help on this one...
Circular gears P and Q start rotating at the same time at constant speeds. Gear P makes 10 revolutions per minute and gear Q makes 40 revolutions per minute. How many seconds after the gears start rotating will gear Q have made exactly 6 more revolutions than geat P ?
A. 6
B. 8
C. 10
D.12
E.15
See attached picture: Can I say that one does "P does "S" spins and Q does "S + 6" spins", then figure out the time and set equal as shown?
Is this wrong and I just got lucky, or is this actually the right approach?
Secondly, can I say:
Combined rate is 1/2, therefore to do six extra spins, it would be 6/1/2 = 12?
In the meantime, here's one possible approach:
First rewrite speeds as revolutions per second (since the question uses these units)
Gear P makes 10 revolution per minute, in other words 10 revolutions per 60 seconds.
To determine the number of revolutions per 1 second, divide 10 by 60, to get 10/60 revolutions per second (a.k.a. 1/6 revolutions per second)
Gear Q makes 40 revolution per minute (or 40 revolutions per 60 seconds).
To determine the number of revolutions per 1 second, divide 40 by 60, to get 40/60 revolutions per second (a.k.a. 2/3 revolutions per second)
Now let t = the time in seconds
The number of revolutions gear P makes in t seconds = (1/6)t
The number of revolutions gear Q makes in t seconds = (2/3)t
We need to determine the number of seconds it takes such that gear Q makes exactly 6 more revolutions than gear P.
So, we want to know the value of t such that:
(Q's revolutions) - (P's revolutions) = 6
Or . . . (2/3)t - (1/6)t = 6
To solve, first multiply both sides by 6 to get: 4t - t = 36
3t = 36
t = 12
It will take 12 seconds, so the answer is D
Cheers,
Brent
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Brent@GMATPrepNow wrote:I think your approach looks good, but I'm not 100% certain what some of your numbers and expressions represent. Can you offer a little more rationale.benjiboo wrote:thank you for your help on this one...
Circular gears P and Q start rotating at the same time at constant speeds. Gear P makes 10 revolutions per minute and gear Q makes 40 revolutions per minute. How many seconds after the gears start rotating will gear Q have made exactly 6 more revolutions than geat P ?
A. 6
B. 8
C. 10
D.12
E.15
See attached picture: Can I say that one does "P does "S" spins and Q does "S + 6" spins", then figure out the time and set equal as shown?
Is this wrong and I just got lucky, or is this actually the right approach?
Secondly, can I say:
Combined rate is 1/2, therefore to do six extra spins, it would be 6/1/2 = 12?
In the meantime, here's one possible approach:
First rewrite speeds as revolutions per second (since the question uses these units)
Gear P makes 10 revolution per minute, in other words 10 revolutions per 60 seconds.
To determine the number of revolutions per 1 second, divide 10 by 60, to get 10/60 revolutions per second (a.k.a. 1/6 revolutions per second)
Gear Q makes 40 revolution per minute (or 40 revolutions per 60 seconds).
To determine the number of revolutions per 1 second, divide 40 by 60, to get 40/60 revolutions per second (a.k.a. 2/3 revolutions per second)
Now let t = the time in seconds
The number of revolutions gear P makes in t seconds = (1/6)t
The number of revolutions gear Q makes in t seconds = (2/3)t
We need to determine the number of seconds it takes such that gear Q makes exactly 6 more revolutions than gear P.
So, we want to know the value of t such that:
(Q's revolutions) - (P's revolutions) = 6
Or . . . (2/3)t - (1/6)t = 6
To solve, first multiply both sides by 6 to get: 4t - t = 36
3t = 36
t = 12
It will take 12 seconds, so the answer is D
Cheers,
Brent
Relative Speed of Q wrt P = 30 revolution per minute-> Q makes 1\2 revolution per second. This follows, to make 6 revolution extra, Q has to take 12 seconds.
Brent is this approach correct??
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When elements COMPETE, determine the DIFFERENCE between the rates.benjiboo wrote:thank you for your help on this one...
Circular gears P and Q start rotating at the same time at constant speeds. Gear P makes 10 revolutions per minute and gear Q makes 40 revolutions per minute. How many seconds after the gears start rotating will gear Q have made exactly 6 more revolutions than geat P ?
A. 6
B. 8
C. 10
D.12
E.15
Q's rate - P's rate = 40-10 = 30 revolutions per minute.
Time for Q to make 6 more revolutions = (number of revolutions)/(rate difference) = 6/30 = 1/5 of a minute = 12 seconds.
The correct answer is D.
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Is this always true? So the time it takes "Train A" to go 10 more miles than "Train B" if "the difference between their rates is 5mph" is simply 10/5 = 2 hours?GMATGuruNY wrote:
When elements COMPETE, determine the DIFFERENCE between the rates.
Q's rate - P's rate = 40-10 = 30 revolutions per minute.
Time for Q to make 6 more revolutions = (number of revolutions)/(rate difference) = 6/30 = 1/5 of a minute = 12 seconds.
The correct answer is D.
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Brent,Brent@GMATPrepNow wrote:
I think your approach looks good, but I'm not 100% certain what some of your numbers and expressions represent. Can you offer a little more rationale.
In the meantime, here's one possible approach:
First rewrite speeds as revolutions per second (since the question uses these units)
Gear P makes 10 revolution per minute, in other words 10 revolutions per 60 seconds.
To determine the number of revolutions per 1 second, divide 10 by 60, to get 10/60 revolutions per second (a.k.a. 1/6 revolutions per second)
Gear Q makes 40 revolution per minute (or 40 revolutions per 60 seconds).
To determine the number of revolutions per 1 second, divide 40 by 60, to get 40/60 revolutions per second (a.k.a. 2/3 revolutions per second)
Now let t = the time in seconds
The number of revolutions gear P makes in t seconds = (1/6)t
The number of revolutions gear Q makes in t seconds = (2/3)t
We need to determine the number of seconds it takes such that gear Q makes exactly 6 more revolutions than gear P.
So, we want to know the value of t such that:
(Q's revolutions) - (P's revolutions) = 6
Or . . . (2/3)t - (1/6)t = 6
To solve, first multiply both sides by 6 to get: 4t - t = 36
3t = 36
t = 12
It will take 12 seconds, so the answer is D
Cheers,
Brent
S= # of spins.
In my photo, filled in the middle column last - by using the right column, and the given rates.
Does this make sense now?
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Ah, yes it does!benjiboo wrote: Brent,
S= # of spins.
In my photo, filled in the middle column last - by using the right column, and the given rates.
Does this make sense now?
Cheers,
Brent
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Correct! Some other problems that I solved by determining the CATCH-UP rate:benjiboo wrote:Is this always true? So the time it takes "Train A" to go 10 more miles than "Train B" if "the difference between their rates is 5mph" is simply 10/5 = 2 hours?GMATGuruNY wrote:
When elements COMPETE, determine the DIFFERENCE between the rates.
Q's rate - P's rate = 40-10 = 30 revolutions per minute.
Time for Q to make 6 more revolutions = (number of revolutions)/(rate difference) = 6/30 = 1/5 of a minute = 12 seconds.
The correct answer is D.
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Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
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