Answer

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**Hint:**The problem deals in volume of Oil. Some volume of oil was taken from the tin and some poured into the tin.

**Complete step by step solution:**Let’s assume the total volume of the tin be ${V_t}$

Initially the volume of oil in the tank ${V_i}$

According to the question,

${V_i} = \dfrac{4}{5}{V_t} \cdots \left( 1 \right)$

From the initial volume, 6 bottles of oil were taken out. Let’s assume the volume of the oil bottles is $v$.

As 6 bottles were taken out volume of oil in the tin will reduce by a factor of $6v$. Now the volume of oil in the tin becomes,

$

{V_{im}} = {V_i} - 6v \\

{V_{im}} = \dfrac{5}{4}{V_i} - 6v \cdots \left( 2 \right) \\

$

Again the question is saying that after taking out 6 bottles of oil, 4bottles of oil of the same volume of the bottle were poured in it. Final volume of the bottles becomes as

${V_f} = \dfrac{5}{4}{V_t} - 6v + 4v \cdots \left( 3 \right)$

The value of the final volume is given as ${V_f} = \dfrac{3}{4}{V_t}$ . Substitute it in equation (3),

$\dfrac{3}{4}{V_t} = \dfrac{5}{4}{V_t} - 6v + 4v \cdots \left( 4 \right)$

Solving equation (4) expression in terms of and , will give the volume of tin in terms of volume of the bottle.

$

\dfrac{3}{4}{V_t} = \dfrac{4}{5}{V_t} - 2v \\

2v = \dfrac{4}{5}{V_t} - \dfrac{3}{4}{V_t} \\

2v = \dfrac{{16{V_t} - 15{V_t}}}{{20}} \\

2v = \dfrac{{{V_t}}}{{20}} \\

{V_t} = 40v \cdots \left( 5 \right) \\

$

The initial volume in tin can be calculated by substituting the value of total volume of tin in equation (1) from equation (5),

$

{V_i} = \dfrac{4}{5}\left( {40v} \right) \\

{V_i} = 32 \\

$

The initial volume in the tin is 32 times the volume of the bottle of oil.

Thus, the correct option is (C).

**Note:**The important thing is to understand that the initial volume in the tin is to be expressed as the volume of the bottle. The main step is the substitution of the final volume of tin in terms of volume of glass bottle in the initial volume of oil in the tin.

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