ps

There are two ways you can solve this problem.

First method

P = 10 res./min => 1/6 res./sec
Q = 40 res./min =>2/3 res./sec

Increasing or decreasing rate between P and Q = (2/3) - (1/6) = 1/2 resolutions/sec

Now we want this rate to be 6, therefore,

(1/2)*x=6
x=12

Second method

P = 10 res./min => 1/6 res./sec
Q = 40 res./min =>2/3 res./sec

Insert the option to find which is correct.

I started with 10 because it is the middle value

(1/6)*10 = 10/6=5/3
(2/3)*10 =20/3 = 6.66
eliminate this

(1/6)*12 = 2
(2/3)*12 = 8

difference between the two is 6, Hence D is the answer.


Let me know if you still have any doubts.

Thanks a lot...!

I'm not the greatest in math but here is how I did it:

gear P rotate at a rate of 1 rev/ 6 sec
gear Q rotate at a rate of 2 rev / 3 sec

If you look @ the answers, you see that a & d are both divisible by 3 & 6
Backsolve : ans d)12 gear P will rotate 2 times
gear Q will rotate 8 times

Hope it helps....I know the true math guru's will shed a better light.

My way:
P: 10/60 seconds is also 1/6 seconds
Q: 40/60 seconds is also 4/6 seconds

After 6 seconds 4-1= 3 more revolutions for Q
So after 12 seconds there are 6 more revolutions for Q