Solution A contains acid and water in the natio 5:2 and solution B contains acid and water in the ratio 3:4.
Solution C is prepared by mixing I and B in the ratio 4:5. In 270 mL of solution C. how much water
in ml) should be added so that the ratio of acid and water in C becomes 2:3?
Answers
Given : solution A contains acid and water in the ratio 5:2 and
solution B contains acid and water in the ratio 3:4.
solution C is prepared by mixing A and B in the ratio 4:5.
To Find : In 270ml of solution C. How much water should be added so that the ratio of acid and water in C becomes 2:3
Solution:
solution A contains acid and water in the ratio 5:2
Acid = 5K and water = 2K
solution A contains acid and water in the ratio 3 : 4
Acid = 3K and water = 4K
solution C is prepared by mixing A and B in the ratio 4:5.
Hence Acid (5K)4 + 5(3K) = 35K
Water = (2K)4 + 4K(5) = 28K
35K : 28K
= 5 : 4
270ml of solution C Acid = (5/9)270 = 150 ml
270ml of solution C Water = (4/9)270 = 120 ml
water should be added = W ml
Then Ratio 150 : ( 120 + W) = 2 : 3
=> 150 / (120 + W) = 2/3
=> 120 + W = 225
=> W = 105
105 ml of water to be added
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Solution :-
- A = 5 : 2 => 4
- B = 3 : 4 => 5
→ In C acid = (4*5/7) + (5*3/7) = (35/7) = 5
→ In C water = ((4*2/7) + (5*4/7) = (28/7) = 4
so, In C acid : water = 5 : 4 .
then,
- Acid in 270 litre = 270 * (5/9) = 150 ml .
- Water = 120 litre .
now, let x ml of water is added .
A/q,
→ 150 / (120 + x) = 2/3
→ 75 * 3 = 120 + x
→ x = 225 - 120
→ x = 105 ml (Ans.)
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