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by sunilrawat » Fri Jun 10, 2011 2:44 am
Circular gears P and Q start rotating at constant speeds. Gear P makes 10 rev/min and Q makes 40 rev/min. How many seconds after the gears start rotating will Q have made exactly 6 revolutions more than P.

a. 6
b. 8
c. 10
d. 12
e. 15

D
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by Frankenstein » Fri Jun 10, 2011 2:55 am
Hi,
P - 10rev/min, Q - 40 rev/min
So, difference is 30rev/min.
So, time taken for diff. to be 6 rev is 1/5 min = 12 sec

Hence, D
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by GMATGuruNY » Fri Jun 10, 2011 5:36 am
sunilrawat wrote:Circular gears P and Q start rotating at constant speeds. Gear P makes 10 rev/min and Q makes 40 rev/min. How many seconds after the gears start rotating will Q have made exactly 6 revolutions more than P.

a. 6
b. 8
c. 10
d. 12
e. 15

D

When elements compete, subtract their rates.

Rate for Q - Rate for P = 40-10 = 30.
Thus, every 60 seconds, Q makes 30 more revolutions than P.

Since Q needs to make 6 more revolutions than P, set up the following proportion:
30 revolutions/60 seconds = 6 revolutions/x seconds
x = 12.

The correct answer is D.
Last edited by GMATGuruNY on Tue Jul 12, 2011 6:42 am, edited 1 time in total.
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by Anurag@Gurome » Fri Jun 10, 2011 5:41 am
sunilrawat wrote:Circular gears P and Q start rotating at constant speeds. Gear P makes 10 rev/min and Q makes 40 rev/min. How many seconds after the gears start rotating will Q have made exactly 6 revolutions more than P.
Q makes (40 - 10) = 30 revolutions more than P in 1 minute = 60 seconds.
Therefore, Q makes 6 revolutions more than P in (60/30)*6 = 12 seconds.

The correct answer is D.
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by rarns » Fri Jun 10, 2011 7:27 am
This is one of those problems where I need to slow myself down.

P = 10 revs per min = 10/60 rev per second = 1/6 rev per second
Q = 40 reves per min = 40/60 rev per second = 2/3 rev per second

Need to find amount of seconds it takes for Q to have exactly 6 more revs than P.

Set equations equal to each other.

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

x/6 + 6 = 2x/3

x/6 - 2x/3 = -6

-3x = -36

x = 12
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by cans » Fri Jun 10, 2011 8:00 am
P=1/6
Q=2/3
let x seconds
2x/3 = x/6 + 6
x=12
IMO D
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by manpsingh87 » Fri Jun 10, 2011 10:33 am
sunilrawat wrote:Circular gears P and Q start rotating at constant speeds. Gear P makes 10 rev/min and Q makes 40 rev/min. How many seconds after the gears start rotating will Q have made exactly 6 revolutions more than P.

a. 6
b. 8
c. 10
d. 12
e. 15

D
p=10 rev/min= 10/60 rev/sec.
p=1/6 rev/sec.
q=40 rev/min=40/60 rev/sec.
q=2/3;
q-p=6;
now for (q-p) to become integer, no. of seconds should be multiple of 6 (as p=1/6); therefore out of available options only option A and D are feasible lest work with option A let therefore seconds=6,
p=(1/6)*6;
p=1;
q=(2/3)*6=4; hence q-p=4-1=3;
but we want q-p=6; hence answer should be D,, we can also verify it. but that is left as a home work for the readers...!!! :twisted: :twisted: :twisted: :twisted: :twisted:
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by hirenvpatel.mba » Thu Jun 16, 2011 5:47 am
Let after 't'mins Rq = Rp+6 , Rq = rev by Q, Rp = rev by P

Sp = 10 rpm & Sq = 40 rpm

Rq = Rp + 6
-> 40t = 10t + 6 => t = 6/30 mins => t = 6*60/30 = 12 secs.

IMO: D

Cheers,
Hiren
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by divya23 » Fri Jun 17, 2011 7:40 am
in one sec p makes 1/6 rev and q makes 2/3 rev...
try to put the values now and u get ans = 12 instead of doing algebra :)
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