Circular gears L and R start to rotate at the same time at the same ra

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

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

Relative Speed = 40-10 = 30 rev per minute
Distance = 6 revs
Time = distance/relative speed= 6/30 = 0.2 minutes
Time in seconds = 0.2 x 60= 12 seconds

this is excellent quetion for backsolving....

let m=# min
m*(40-10 rpm)=6 rev
m=6/30=1/5 min=12 sec

My approach:

Take x to be the number of minutes after which Gear R would have made 6 more revolutions than Gear L:

10x+6=40x
6=30x
6/30=x
--------
6/30 is in minutes, to convert into seconds we multiply by 60
Therefore, 6/30 X 60, which equals 12 seconds.
Although this isn't the "rate and work" way to go, I thought this was quicker.

Gear R makes 30 more revolution per minute than gear L, or, 1/2 revolution per second (30/60).
Lets say that X is the seconds needed to gear R to be exactly 6 revolutions ahead.

1/2*X = 6
X = 6 / 1/2
X = 6 * 2
X = 12

Option D.

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.

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