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ATQ, Let the initial quantity of diesel = ‘d’ L So, the initial quantity of petrol = (60 – d) L From the question: (d + 8) = (60 – d) × (125/300) 12d + 96 = 300 – 5d d = 12 The initial quantity of diesel = 12 L Quantity of petrol = 60 – 12 = 48 L So, value of p = 48/12 = 4
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