Question

A water tank holds a total of 95/2 liters of water. If 88/3 liters of water has gone in one time and 53/6 liters in the second time, So how much water is required to fill the tank now?


《Frèe poïnts》​

Answer 1

Originally water Content = $\frac{95}{2}$ litres

Water removed in first and second time = $\frac{88}{3} + \frac{53}{6}$

$\implies$ $\frac{88 \times 2 + 53 \times 1}{6}$$\: \: \: \: \: \: \: \: \: lcm = 6$

$\implies$$\frac{176 + 53}{6}$

$\implies$$\frac{229}{6}$ litres

Now,

water is required to fill the tank now = total water content- water removed

$\implies$$\frac{95}{2} - \frac{229}{6}$

$\implies$$\frac{95 \times 3-229 \times 1}{6}$$\: \: \: \: \: \: \: \: \: lcm = 6$

$\implies$$\frac{285 - 229}{6}$

$\implies$$\boxed{ \frac{56}{6} \: litres}$

Answer 2

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