Gear L -- 10 rotations per 60 seconds -- 1 rotation per 6 seconds.
Gear R -- 40 rotations per 60 seconds -- 4 rotations per 6 seconds.

First 6 seconds -- Gear L makes 1 rotation. -- Gear R makes 4 rotations -- Net difference -- 3 rotations

Hence every 6 seconds the difference between the number of rotations of R and L gear is 3 units.

Required net difference should be 6 rotations => 2 (6 seconds later) ==> 12 seconds.

Answer: 12 seconds (Option D).

this is excellent quetion for backsolving....

L = 10 rev/min = (10/60) rev/sec = (1/6) rev/sec
R = 40 rev/min = (40/60) rev/sec = (2/3) rev/sec

L in 1 sec makes (1/6) revolutions
L in 2 sec makes (1/6)*2=(1/3) revolutions
L in 3 sec makes (1/6)*3=(1/2) revolutions

R in 1 sec makes (2/3) revolutions
R in 2 sec makes (2/3)*2=(4/3) revolutions
R in 3 sec makes (2/3)*3=(2) revolutions

x = seconds after the gears start to rotate will gear R have made exactly 6 more revolutions than gear L
(2/3)*x=(1/6)*x + 6
(4/6-1/6)*x=6
(1/2)*x=6
x=12 seconds
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As we can see, Gear R makes 30 more revolutions per minute than Gear L does. So, we can set up the following proportion to solve for the number of seconds for Gear R to make 6 more revolutions than Gear L does:

30 revolutions/1 minute = 6 revolutions/x seconds

We need for both denominators to have the same units. Since 1 minute = 60 seconds, we can re-express the proportion as:

30/60 = 6/x

Cross multiplying, we have:

30x = 360

x = 12

Alternate Solution:

Gear L makes one revolution in 60/10 = 6 seconds and gear R makes one revolution in 60/40 = 1.5 seconds. If we let n the time (in seconds) that passes when gear R makes 6 more revolutions than gear L, we can create the following equation:

n/1.5 - n/6 = 6

Let’s multiply each side by 6:

4n - n = 36

3n = 36

n = 12

Answer: D
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L = 10 Rev/min
R = 40 Rev/min
R/L = 4 ; R = 4 L
How many second, R has made exactly 6 more revolutions than gear L
Needed R = L+6 ; 4L=L+6 ; 3L=6; L=2.
Since L does 10 Rev in 1 min. It will do 2 Rev in 60/5 = 12 seconds.