A bucket contains a mixture of two liquids a and b in the proportion 7:5.if 9 litres of the mixture is replaced by 9 litres of liquid b then the ratio of the two liquids become 7:9.how much of the liquid a was there in the bucket ??
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Solution:-
Let us assume that the bucket initially contains 7x and 5x liters of the mixture of two liquid A and B respectively.
When 9 liters of mixture is drawn off, the quantity of liquid A in the mixture left = {7x - (7/12)*9}
= {7x - (21/4)} liters
Similarly, the quantity of liquid B in the mixture left = {5x - (5/12)9}
= {5x - (15/4)} liters
Therefore, the two ratios are
∴ {7x - (21/4)}/{5x - (15/4) + 9} = 7/9
⇒ (28x - 21)/(20x + 21) = 7/9
⇒ 252x - 189 = 140x + 147
⇒ 252x - 140x = 147 + 189
⇒ 112x = 336
x = 336/112
x = 3
Therefore, the quantity of liquid A in the bucket is 7 × 3 = 21 liters.
Answer.
Let us assume that the bucket initially contains 7x and 5x liters of the mixture of two liquid A and B respectively.
When 9 liters of mixture is drawn off, the quantity of liquid A in the mixture left = {7x - (7/12)*9}
= {7x - (21/4)} liters
Similarly, the quantity of liquid B in the mixture left = {5x - (5/12)9}
= {5x - (15/4)} liters
Therefore, the two ratios are
∴ {7x - (21/4)}/{5x - (15/4) + 9} = 7/9
⇒ (28x - 21)/(20x + 21) = 7/9
⇒ 252x - 189 = 140x + 147
⇒ 252x - 140x = 147 + 189
⇒ 112x = 336
x = 336/112
x = 3
Therefore, the quantity of liquid A in the bucket is 7 × 3 = 21 liters.
Answer.
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