A tin of oil was frac{4}{5} full when 6 bottl | vedclass

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(D) Let the total capacity of the tin be $x$ bottles.Initially,the tin was $\frac{4}{5}$ full,so the amount of oil was $\frac{4}{5}x$.When $6$ bottles

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$40$

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(D) Let the total capacity of the tin be $x$ bottles.Initially,the tin was $\frac{4}{5}$ full,so the amount of oil was $\frac{4}{5}x$.When $6$ bottles were removed,the amount became $\frac{4}{5}x - 6$.Then,$4$ bottles were added,making the amount $\frac{4}{5}x - 6 + 4 = \frac{4}{5}x - 2$.According to the problem,this final amount is $\frac{3}{4}$ of the total capacity,so $\frac{4}{5}x - 2 = \frac{3}{4}x$.Rearranging the equation: $\frac{4}{5}x - \frac{3}{4}x = 2$.Finding a common denominator $(20)$: $\frac{16x - 15x}{20} = 2$.$\frac{x}{20} = 2$.$x = 40$.Therefore,the tin can contain $40$ bottles of oil.

$A$ tin of oil was $\frac{4}{5}$ full when $6$ bottles of oil were taken out. Again,$4$ bottles of oil were poured into it,and it became $\frac{3}{4}$ full. How many bottles of oil can the tin contain?

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