Math, asked by learner551, 3 months ago

Q. A vessel contains milk and water in the ratio 3:5. When 6 litres of this solution is
removed and replaced with water, the ratio of milk and water becomes 5:9. How many
litres of the solution was present in the vessel originally?​

Answers

Answered by amitnrw
3

Given : A vessel contains milk and water in the ratio 3:5. When 6 litres of this solution is removed and replaced with water, the ratio of milk and water becomes 5:9.

To Find : How many litres of the solution was present in the vessel originally?​

Solution:

Let say Solution present originally = 8k  litres

Milk = 3k

water  = 5k

When 6 litres of this solution is removed and replaced with water

milk remains  =  3k  -  (3k/8k)6    = 3k  -  9/4   litre   = (12k - 9)/4  litre

and water   = 5k  - (5k/8k)6  + 6   = 5k  + 9/4   litre  = (20k + 9)/4 litre

ratio of milk and water becomes 5:9.

=> (12k - 9)/4 :  (20k + 9)/4 =  5 : 9

=> 12k - 9 : 20k + 9 = 5 : 9

=> (12 k - 9)/(20 k + 9)  = 5/9

=> 9 (12k - 9) = 5(20k + 9)

=> 108k - 81 = 100k + 45

=> 8k = 126

Solution present originally = 8k  litres

= 126 litres

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Answered by RvChaudharY50
3

Solution :-

we know that,

  • FC = IC(1 - x/V) .
  • FC = final concentration .
  • IC = initial concentration .
  • x = Quantity removed / added.
  • V = Total volume. .

given that,

→ Milk : water = 3 : 5 .

→ Milk = (3/8) = IC .

in final ,

→ Milk = 5 : 9 = (5/14) = FC .

then,

→ (5/14) = (3/8)[1 - (6/V)]

→ (5/14) * (8/3) = 1 - (6/V)

→ (20/21) = 1 - (6/V)

→ 6/V = 1 - (20/21)

→ 6 / V = (1/21)

→ V = 126 Litres. (Ans.)

Hence, 126 Litres of the solution was present in the vessel originally .

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