1. A vessel contains 100 litres mixture of milk and water in the respective ratio of 22 : 3. 40 litres of the mixture is taken out from the vessel and 4.8 litres of pure milk and pure water each is added to the mixture. By what percent is the quantity of water in the final mixture less than the quantity of milk?
Write Comment
Comments
By: anil on 05 May 2019 01.37 am
Quantity of milk in vessel = $$frac{22}{25} imes 100 = 88$$ litres => Quantity of water = $$100 - 88 = 12$$ litres 40 litres of the mixture is taken out, i.e., $$frac{40}{100} = (frac{2}{5})^{th}$$
=> Milk left = $$88 - frac{2}{5} imes 88 = 52.8$$ litres Water left = $$12 - frac{2}{5} imes 12 = 7.2$$ litres Now, 4.8 lires of milk and water are added. => Quantity of milk in the vessel = 52.8 + 4.8 = 57.6 litres Quantity of water in the vessel = 7.2 + 4.8 = 12 litres $$ herefore$$ Required % = $$frac{57.6 - 12}{57.6} imes 100$$ = $$frac{475}{6} = 79 frac{1}{6} \%$$
Terms And Service:We do not guarantee the accuracy of available data ..We Provide Information On Public Data.. Please consult an expert before using this data for commercial or personal use
=> Milk left = $$88 - frac{2}{5} imes 88 = 52.8$$ litres Water left = $$12 - frac{2}{5} imes 12 = 7.2$$ litres Now, 4.8 lires of milk and water are added. => Quantity of milk in the vessel = 52.8 + 4.8 = 57.6 litres Quantity of water in the vessel = 7.2 + 4.8 = 12 litres $$ herefore$$ Required % = $$frac{57.6 - 12}{57.6} imes 100$$ = $$frac{475}{6} = 79 frac{1}{6} \%$$