Circular gears P and Q start rotating at the same time at co

[Guest]

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 gear P?

GMAT PREP EXAM 1

The correct answer is 12.

Could someone solve this problem so I can compare to what I got. I just want to make sure I understand this problem. Your input is greatly appreciated...

[guest]

P-> 10 rev/ min, i.e. 10 rev/60 sec = 1 rev in 6 secs or it takes 6 secs to take 1 rev.
Q-> 40 rev/ min, i.e. 60 rev/60 sec = 4 rev in 6 secs or it takes 6 secs to take 4 rev.

Please see that in 6 secs Q takes a lead of 3 revs when compared to A. Therefore in 12 secs, the lead will be by 6revs. Hence answer is 12secs :wink:

[Guest]

(40/60 -10/60)*T =6
1/2 T =6
T =12

[RonPurewal]

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 gear P?

GMAT PREP EXAM 1

The correct answer is 12.

Could someone solve this problem so I can compare to what I got. I just want to make sure I understand this problem. Your input is greatly appreciated...

the solutions posted above are good - one intuitive, one algebraic - and i don't have much to add to them. if you need further explanation, especially of the genesis of the algebraic solution, post back and i can flesh it out a little bit.

with numbers like this, you could also just 'grind' the problem until you find numbers that work.
specifically:
gear p makes ten revolutions per minute, so that's one revolution every 6 seconds.
gear q rotates four times as fast, so that's four revolutions every 6 seconds.

just make a table until you get a difference of 6 revs:

seconds ... revs P ... revs Q
6 ............... 1 ........... 4
12 .............. 2 ........... 8
we have a winner.

this method has obvious limitations, and will crash and burn if the problem contains numbers less friendly than the ones here, but it works admirably in this particular case.

[srivatsan.rangan]

Hi Ron,

Can you flesh the algebra part as well?

is T the time difference in seconds?

Cheers
Vatsan

[Ben Ku]

We want everything in seconds because the answer asks for seconds.
The rate for gear P is 10 revolution/1 minute, or 10 revolutions / 60 seconds. The rate for gear Q is 40 revolutions/1 minute or 40 revolutions / 60 seconds. Obviously gear Q is faster than gear P.

The number of revolutions P makes will be r*t = (10/60)*T, while the number of revolutions Q makes will be r*t = (40/60)*T. If Q makes 6 more revolutions than P, then (40/60)T - (10/60)T = 6.


The rest can be solved algebraically as done above.
(40/60 -10/60)*T =6
1/2 T =6
T =12

Hope that helps.

[gkhan]

We want everything in seconds because the answer asks for seconds.
The rate for gear P is 10 revolution/1 minute, or 10 revolutions / 60 seconds. The rate for gear Q is 40 revolutions/1 minute or 40 revolutions / 60 seconds. Obviously gear Q is faster than gear P.

The number of revolutions P makes will be r*t = (10/60)*T, while the number of revolutions Q makes will be r*t = (40/60)*T. If Q makes 6 more revolutions than P, then (40/60)T - (10/60)T = 6.


The rest can be solved algebraically as done above.
(40/60 -10/60)*T =6
1/2 T =6
T =12

Hope that helps.

Ben,

Thanks for your explanation, I got the same answer but negative can you tell me why my algebraic expression is incorrect?

RATE TIME DIST
P 1/6 T d
Q 2/3 T 6+d

1/6T=2/3T+6
T=-36/3
T=-12 seconds

[zchampz]

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 gear P?

I guess the important information here is gears are rotating at constant speeds.
P -> 10/60 seconds
Q -> 40/60 seconds
Overall, Q gains 30 revolutions over 60 seconds -> 1 revolution for every 2 seconds
So, to gain 6 revolutions Q needs 12 seconds.

Algebraically,
30/60 = 6/x
=> x = 6 * 60/30
=> x = 12
Therefore, Q needs 12 seconds to gain 6 revolutions.

Hope this helps.

-Champ

[RonPurewal]

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 gear P?

I guess the important information here is gears are rotating at constant speeds.
P -> 10/60 seconds
Q -> 40/60 seconds
Overall, Q gains 30 revolutions over 60 seconds -> 1 revolution for every 2 seconds
So, to gain 6 revolutions Q needs 12 seconds.

Algebraically,
30/60 = 6/x
=> x = 6 * 60/30
=> x = 12
Therefore, Q needs 12 seconds to gain 6 revolutions.

Hope this helps.

-Champ

you can do this, too. this falls under the category of "relative rates" (as discussed in the strategy guide).

note that the strategy guide discusses "relative rates" for linear motion (i.e., things moving in a straight line), but it works just as well for examples of circular motion (such as this example).

nicely done.

[focusgmat]

Can anyone tell me if the following line of reasoning correct :

Say x= no.of revolutions made by P in time 't'
Then the question asks that what is 't' (in sec) in which Q does 'x+6' revolutions(i.e. 6 more revolutions than P).

W R T
X 10/60 = 1/6 t
X+6 40/60=2/3 t


x(1/6)= (x+6)2/3
x=2

Hence, t = W/R = x/R= 12sec

[tim]

Yes, this is just fine..

[focusgmat]

Thank You , Tim !

[RonPurewal]

lots of good solutions in this thread.

as a closing remark, notice that this should be the way you approach problems: try not to be satisfied with just one solution; try to find as many solutions as you possibly can.

this principle has added importance because -- don't forget -- the problems on the official exam are NOT going to be like the problems on your practice tests. therefore, it's more important than it otherwise would be to find alternate solutions: the alternate solutions may seem like a nuisance here, but they may be the key to your solving another problem down the road (on which your first attempt doesn't work).

this includes the idea that, even if you find a problem terribly easy, you should still go back and see whether you can find other ways of solving it.

[i.ankurjaini]

Another Approach with a little Higher Mathetimatics : Concept of Relative Speed.
Concept : If two objects are moving with speeds "u" and "v", then the relative speed is :
1) If travelling in opposite direction (approaching each other) : "u + v"
2) If travelling in same direction, then relative speed is : "u - v " or "v - u" (depending which is greater).

Thus, using the concept of Relative speed, the relative speed of Gears P & Q is ( 40 - 10) = 30 rev/min = 30rev/60 sec
i.e,
30 rev --> 60 sec
6 rev --> 60 * 6 / 30 = 12 sec.

Hope this helps.
The concept of relative speed is prominent in case of river problems (upstream and downstreams)

[tim]

thanks!