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Circular gears P and Q start rotating at the same time at [#permalink]

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20 Mar 2012, 09:04

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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 ?

Circular gears P & 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 ?

A. 6 B. 8 C. 10 D. 12 E. 15

Aany idea how to solve this?

Note that we are given revolutions per minute and asked about revolutions in seconds. So we should transform per minute to per second.

Gear P makes 10 revolutions per minute --> gear P makes 10/60 revolutions per second; Gear Q makes 40 revolutions per minute --> gear Q makes 40/60 revolutions per second.

Let \(t\) be the time in seconds needed for Q to make exactly 6 more revolutions than gear P --> \(\frac{10}{60}t+6=\frac{40}{60}t\) --> \(t=12\).

Re: Circular gears P and Q start rotating at the same time at [#permalink]

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25 Oct 2013, 23:48

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enigma123 wrote:

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 ?

A. 6 B. 8 C. 10 D. 12 E. 15

Work done by Faster gear Q is 6 more than slower gear P in same time

i.e W(q) = W(p) + 6 Rq X t = Rp X t + 6 40t=10t + 6 30t = 6 t =1/5 hrs t = 12 mins

Circular gears P & 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 ?

A. 6 B. 8 C. 10 D. 12 E. 15

Aany idea how to solve this?

Note that we are given revolutions per minute and asked about revolutions in seconds. So we should transform per minute to per second.

Gear P makes 10 revolutions per minute --> gear P makes 10/60 revolutions per second; Gear Q makes 40 revolutions per minute --> gear Q makes 40/60 revolutions per second.

Let \(t\) be the time in seconds needed for Q to make exactly 6 more revolutions than gear P --> \(\frac{10}{60}t+6=\frac{40}{60}t\) --> \(t=12\).

Answer: D.

Hope it's clear.

Is there any theory for this?

I remember from college days, the Key point in solving is that the tangential Velocity should be same at the point of contact?
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Re: Circular gears P and Q start rotating at the same time at [#permalink]

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21 Nov 2013, 02:42

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My approach was the following (similar to what LalaB wrote):

Difference between gears Q and P is 30 revolutions per min (40-10). To make 6 more rotations, the faster gear needs (6 rev / 30 rev/min = 1/5 min). Translate to seconds: 1/5 x 60 = 12.

Re: Circular gears P and Q start rotating at the same time at [#permalink]

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26 Jan 2014, 23:25

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enigma123 wrote:

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 ?

A. 6 B. 8 C. 10 D. 12 E. 15

All the above answers are excellent and I would like to one little tweak that has helped me in quite a few number of Quant problems. This is applicable for people who love to find answers by plugging in the options given

As soon as you arrive at the 2 numbers "6 seconds and 1.5 seconds" (Time taken to complete one rotation) , look at the options given. The best options(9 out of 10 times) are the ones that are a multiple of the above two numbers. Hence you will possibly start with option A and then move to option B and voila.

NOTE: Use this method in scenarios such as this example wherein the number of options that fall under the "multiple umbrella"are quite low. If all the options are multiples it is better to switch on to one of the above strategies.

Re: Circular gears P and Q start rotating at the same time at [#permalink]

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31 Aug 2014, 00:23

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enigma123 wrote:

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 ?

A. 6 B. 8 C. 10 D. 12 E. 15

Another approach: P makes 10 R(evolutions) in 60 secs, so 1 R in 6 Sec. Similarly, as Q makes 40 R in 60 Sec or 4 R in 6 Sec. We can extrapolate the above details to state that P would make 2 R in 12 Sec and Q would make 8 R in 12 sec. Hence Q needs additional 12 Sec to overtake P by 6 Revolutions.

Hence in 12 seconds Gear P completes 2 rev and Gear Q completes 8 rev Difference =8-2= 6 revolutions Therefore in 12 sec Gear Q revolves 6 times ahead than Gear P

Re: Circular gears P and Q start rotating at the same time at [#permalink]

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22 Oct 2015, 14:01

enigma123 wrote:

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 ?

A. 6 B. 8 C. 10 D. 12 E. 15

Gear P: 10 rev. per minute => 10 rev. per 60 seconds Gear Q: 40 rev. per minute => 40 rev. per 60 seconds

After 60 seconds, Gear Q - Gear P = 30 rev. -------x--seconds,-------------------------3 rev => x= (3*60)/30 = 12 rev. Ans. D

Circular gears P and Q start rotating at the same time at [#permalink]

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29 Mar 2016, 03:08

P makes 10 rev per min so in 1 sec it will make 10/60=1/6 rev per sec Similarly Q makes 40 rev per min in 1 sec it will make 40/60 rev=2/3 rev per sec. Now you can multiply by answer choices which are no of seconds to arrive at an answer. Eg if it takes 6 secs 1/6x6=1 rev made by P in 6 sec 2/3x6=4 rev made by Q in 6 sec. Difference is 3 rev If we multiply by 12 the difference is 6.

Re: Circular gears P and Q start rotating at the same time at [#permalink]

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09 Apr 2016, 14:20

enigma123 wrote:

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 ?

A. 6 B. 8 C. 10 D. 12 E. 15

P does 1/6 rotations per second Q does 4/6 rotations per second

difference 3/6 rotations per second we need to have difference 6 rotations. 6 is 36/6 now 36/6 divide by 3/6 36*6/6*3 = 12.

Re: Circular gears P and Q start rotating at the same time at [#permalink]

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23 May 2016, 13:59

Amit0507 wrote:

enigma123 wrote:

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 ?

A. 6 B. 8 C. 10 D. 12 E. 15

Another approach: P makes 10 R(evolutions) in 60 secs, so 1 R in 6 Sec. Similarly, as Q makes 40 R in 60 Sec or 4 R in 6 Sec. We can extrapolate the above details to state that P would make 2 R in 12 Sec and Q would make 8 R in 12 sec. Hence Q needs additional 12 Sec to overtake P by 6 Revolutions.

---- Why cant I use the following approach?

P makes 1R in 6s. Q makes 1R in 1.5s.. Q revolutions = P revolutions + 6 revolutions 1.5s = 6s + 6 - 6 = 4.5s

Re: Circular gears P and Q start rotating at the same time at [#permalink]

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17 Aug 2016, 11:47

Hi everybody,

I'm having some trouble with this question. The answers posted so far are easily understandable. But the more in will gear Q have made exactly 6 more revolutions than gear P ? is throwing me off.

The way I understand is as follows:

Gear P makes the revolutions at 1.5, 3, 4.5, 6, 7.5, 9, 10.5 and 12 seconds Gear Q makes the revolutions at 6 and 12 seconds Both gears do a revolution at the same time during second 6. The sixth revolution that gear P does alone is during second 10.5

The way I understand it, the question is asking for the revolutions that Gear P made alone (kinda like a XOR relationship, since when they make the revolution together, it does not count as one more revolution for Q) so the answer would be 10.5 seconds (which is not even listed). Obviously, the testers had the algebraic solution in mind when they created the question and accordingly, the solution is 12 seconds. Did anybody also think of this problem like this? Does my reasoning make sense? Or what am I getting wrong?