Question
A vessel contains 120 liters of a mixture in which oil
and water are in the ratio 2:3. If 40 liters of this mixture are taken out and 'x' liters of oil are added, then the oil becomes 10 liters less than the water in the vessel. Find 'x'.Solution
ATQ, Quantity of oil initially = 120×2/5=48 liters Quantity of water initially = 120×3/5=72 liters Quantity of oil removed = 40×2/5=16 liters Quantity of water removed = 40−16=24 liters After removal, oil left = 48−16=32 liters; water left = 72−24=48 liters Let 'x' be the liters of oil added. The equation for the final mixture with oil being 10 liters less than water: 48-(32+x)=10 Solving, 32+x =38 Thus, x =6 liters
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