When two gears are mashed the revolutions per minute rpm are

When two gears are mashed, the revolutions per minute (rpm) are inversely proportional to the number of, as shown below. In one machine the rpm ratio of the large gear to the small gear is 2.3. The small gear has 114 teeth, in a second machine the large gear involves at 500 rpm and has 315 teeth. The small gear revolves at 900 rpm. Find the number of teeth on the large gear of the first machine. Find the number of teeth on the small gear of the second machine.

Solution

In one machine
rpm ratio of large gear to small gear = 2:3 = 2/3
the small gear has = 114 teeth

In the second machine
the large gear revolves at = 500 rpm and has 315 teeth
The small gear revolves at 900 rpm

rpm = k/number of teeth , where k is a proportionality constant

=> rpm of big gear/rpm of small gear = 2/3   ---------->(1)
rmp of small gear = k/number of teeth = k/114 , --------->(2)

from (1) and (2)
rpm of big gear = 2/3*k/114 = k/171

now rpm of big gear = k/number of teeth of big gear

=> k/171 = k/number of teeth
=> number of teeth of the big gear in machine 1 is = 171

b>

number of teeth of the small gear on machine two :

rpm of large gear = p/number of teeth of the large gear
=> p = 500*315

and rpm of small gear = 900 = p/number of teeth of small gear

900 = 500*315/number of teeth of small gear

=> number of teeth of small gear = 175