An epicyclic gear train is shown in the figure below. The number of teeth on the gears A, B and D are 20, 30 and 20, respectively. Gear C has 80 teeth on the inner surface and 100 teeth on the outer surface. If the carrier arm AB is fixed and the sun gear A rotates at 300 rpm in the clockwise direction, then the rpm of D in the clockwise direction is
The speed of the gears in epicyclic gear train can be analyzed from the following table,
If the external toothed gear is meshing with the internal toothed gears then the direction of the velocity remains same, and the velocity ratio between two mating gears is given by
\(\frac{N_2}{N _1}=\frac{T_1}{T_2}\)
Where N is the angular velocity in rpm and T is the number of teeth
Conditions of motion 
Revolutions of elements 

Arm 
Gear A 
Gear B 
Gear C 
Gear D 

Arm is fixed, wheel A rotates +1 revolutions 
0 
+1 
\( \frac{{{T_A}}}{{{T_B}}}\) 
\( \frac{{{T_A}}}{{{T_B}}} \times \frac{{{T_B}}}{{{T_{Ci}}}} =  \frac{{{T_A}}}{{{T_{Ci}}}}\) 
\(\frac{{{T_A}}}{{{T_{Ci}}}} \times \frac{{{T_{Co}}}}{{{T_D}}} = \frac{{{T_A}}}{{{T_D}}}\frac{{{T_{Co}}}}{{{T_{ci}}}}\) 
Arm is fixed, wheel A is rotated trough +x revolution 
0 
+x 
\( x\frac{{{T_A}}}{{{T_B}}}\) 
\( x\frac{{{T_A}}}{{{T_{Ci}}}}\) 
\(x\frac{{{T_A}}}{{{T_D}}}\frac{{{T_{Co}}}}{{{T_{ci}}}}\) 
Add +y revolution to all 
y 
Y + x 
\(y  x\frac{{{T_A}}}{{{T_B}}}\) 
\(y  x\frac{{{T_A}}}{{{T_C}}}\) 
\(y + x\frac{{{T_A}}}{{{T_D}}}\frac{{{T_{Co}}}}{{{T_{ci}}}}\) 
Calculation:
Given: T_{A} = 20, T_{B} = 30, T_{D} = 20, T_{Ci} = 80, T_{Co} = 100, N_{Arm} = 0, N_{A} = 300 rpm, N_{D} =?
N_{Arm} = 0 = y
N_{A} = 300 rpm = y + x ⇒ x = 300
\({N_D} = y + x\frac{{{T_A}}}{{{T_D}}}\frac{{{T_{Co}}}}{{{T_{ci}}}} = x\frac{{{T_A}}}{{{T_D}}}\frac{{{T_{Co}}}}{{{T_{ci}}}} = 300 \times \frac{{20}}{{20}} \times \frac{{100}}{{80}} = 375\;rpm\)
Mistake: Take care of the inner and outer teeth of gear C.