Math, asked by arjunmakilari8039, 1 year ago

There are two vessels a and b containing mixtures of milk and water in the ratio of 5:x and 8:15 respectively. vessel a contains 48l of mixture. 12l of mixture from vessel a is replaced by 23l mixture of vessel b. final ratio of milk to total mixture in vessel a become 23:59. find the value of x?

Answers

Answered by santy2
3

Mixture A :

Ratio of milk to water = 5 : x

When 12 litres are withdrawn we remain with : 48 - 12 = 36 litres

The amount of milk and water is :

Milk = (5 × 36)/(x + 5)

= 180/(x + 5)

Water = (x × 36)/(x + 5)

= 36x/(x +5)

Mixture B:

The ratio of milk to water = 8 : 15

If we withdraw 23 litres and add it to mixture A, the amount of milk and water added is :

Milk = 8/23 × 23 = 8 litres

Water = 15/23 × 23 = 15 litres

We add this to mixture A and get the new amounts of milk and water:

Milk :

180/(x + 5)  +  8

Water:

36x/(x + 5)  + 15

The total amount of mixture A after the addition is :

23 + 36 = 59 litres

Now the ratio of the amount of milk to the total amount is 23 : 59

Expressing this we have :

{180/(x + 5)  + 8} / 59 = 23/59

59{180/(x + 5)  +  8}  = 23 × 59

10620/(x + 5)  + 472 = 1357

10620/(x + 5) = 1357 - 472

10620 / (x + 5) = 885

10620 = 885(x + 5)

10620 = 885x + 4425

10620 - 4425 = 885x

885x = 6195

x = 6195/885

x = 7

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