A tank holds a mixture of Strawberry, Blueberry, and Raspberry juices in the ratio of 6:9:5 respectively. 40 litres of this mixture is taken out, and then 15 litres of strawberry juice and 10 litres of raspberry juice are added to the tank. If the resultant quantity of blueberry juice is 22 litres more than the resultant quantity of strawberry juice, what was the initial quantity of mixture(approximate) in the tank (in litres)?

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1. Initial parts: Strawberry = $	frac{6}{20}T=	frac{3}{10}T$, Blueberry = $	frac{9}{20}T$, Raspberry = $	frac{5}{20}T$.

2. After removing 40 L, each is multiplied by $\frac{T-40}{T}$. So Blueberry = $	frac{9}{20}(T-40)$ and Strawberry = $	frac{3}{10}(T-40)$.

3. Then add 15 L strawberry and 10 L raspberry. Given Blueberry = Strawberry + 22:

$	frac{9}{20}(T-40)=	frac{3}{10}(T-40)+37$. This gives $	frac{3}{20}(T-40)=37$.

4. Solve: $T-40=\dfrac{37	imes20}{3}=\dfrac{740}{3}$, so $T=\dfrac{860}{3}\approx286.67\approx287$.

FINAL_ANSWER: a

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