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      Question

      A 330 gm mixture contains two components 'P' and 'Q' such

      that the quantity of 'P' is 20% more than that of 'Q'. After adding 'K' gm of 'P', the ratio of 'P' to 'Q' becomes 4:3. Find K% of the final mixture weight.
      A 65 gm Correct Answer Incorrect Answer
      B 88 gm Correct Answer Incorrect Answer
      C 25 gm Correct Answer Incorrect Answer
      D 70 gm Correct Answer Incorrect Answer
      E None of these Correct Answer Incorrect Answer

      Solution

      ATQ,

      Let quantity of 'Q' = 15y

      Then quantity of 'P' = (6/5) × 15y = 18y

      So total = 15y + 18y = 33y

      33y = 330 ⇒ y = 10

      Initial 'Q' = 15 × 10 = 150 gm

      Initial 'P' = 18 × 10 = 180 gm

      ATQ:

      (180 + K) : 150 = 4 : 3

      Or, 3 × (180 + K) = 4 × 150

      Or, 540 + 3K = 600

      Or, 3K = 60

      Or, K = 20

      Final mixture = 330 + 20 = 350 gm

      Required value = 0.2 × 350 = 70 gm

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