a = a_{1 }+ a_{2}

For pulley A, Apply F = ma

T – 20 = (20/10) a_{1} , take g = 10m/sec^{2}

T – 20 = 2a_{1} ...(2)

For pulley C, 40 – 2T = (40/10)a_{40} – 2T = 4a ...(3)

For pulley B, T – 30 = (30/10) a_{2} T – 30 = 3a_{2} ...(4)

From equation (2) and (4)

2a_{1} – 3a_{2} = 10 ...(5)

Equation (3) can be rewritten as

40 – 2T = 4(a_{1} + a_{2}) ...(6)

Now (6) + 2 (4)

40 – 2T + 2T – 2 X 30 = 4(a_{1} + a_{2}) + 6a_{2}

–20 = 4a_{1} + 10a_{2} ...(7)

Solving equation (5) and (7), we get

**a**_{1} = 5/4 m/sec^{2}

**a**_{2} = -5/2m/sec^{2 }

Acceleration of ‘C’ = a = a_{1} + a_{2}

**= 5/4 – 5/2 = –1.25m/sec**^{2}