From 90 liters of 4: 5 milk-water mixture, 15 liters are taken out and replaced by same amount of water. This process is repeated once more. Now, 9 liters of the mixture is taken out and replaced by same amount of milk. The amount of water at the end of third cycle is how much percent of the amount of water at the end of second cycle? Option 1) 20%Option 2) 90%     Option 3) 201.6%Option 4) 49.6%Option 5) 10%

90 L = 40 L milk

= 50 L milk

15 L = 1/6 taken out

$\inline \left ( 40-\frac{1}{6} \times 40\right )\left ( \left (50-\frac{1}{6} \times 50 \right )+15 \right )$

First process = 100/3 milk, 125/3 +15 = 170/3 water

Second process$\frac{250}{90}$ milk; $\frac{425}{9} + 15 = \frac{560}{9}$ L water

Third process = water $\frac{560}{9} \times \frac{9}{10} = 56$ L;  milk = $34\: L$

To = $\frac{56}{560} \times 9 \times 100= 90 \, ^{0}/_{0}$

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