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.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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