two capacitors A and B are connected in series with a 100V batteryand it is observedthat p.d. across them are 40V and 60V respectively.A capacitor of capacitance 6mirofarad is now connected in parallel with B and as a result the p.d. across B falls to
Charge Q will be same on both the capacitors that are in series.
For capacitor A
Q = C1(40V )
For capacitor B
Q = C2(60V)
So C1/C2 = 3/2
C1 = 3x and C2 = 2x
On connecting 6 microfarad with B the potential drop is equal to that across A, so the capacitance of this combination is equal to the capacitance of A.
Now you can write an equation in variable x and solve it to find C1 and C2.