Math, asked by nkjain033, 1 month ago

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

Answered by amitnrw
1

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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Answered by RvChaudharY50
2

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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