Math, asked by ImRitz, 1 year ago

We have 100 litres of a mixture of milk and water which is 10% water. How much more pure milk should be added so that the new mixture has only 5% water (Answer - 4 litres).

#Mathematics

#RatioAndProportion


Golda: Your answer (4 litres) is not correct. Please check your answer.

Answers

Answered by Golda
7
Solution :-

Total mixture of water and pure milk = 100 liter

Quantity of water in the mixture = 10 %

Quantity of water in liters = 10 liters

Quantity of pure milk in liters = 100 - 10 = 90 liters

Let 'x' liters of pure milk is added to the mixture to make water 5 % and Pure milk 95 %

Total amount of pure milk becomes (90 + x) liter and total amount of mixture becomes (100 + x) liters.

⇒ (90 + x) = (95/100)*(100 + x)

⇒ (90 + x) = (19/20)*(100 + x)

⇒ (90 + x)*20 = 19*(100 + x)

⇒ 1800 + 20x = 1900 + 19x

⇒ 20x - 19x = 1900 - 1800

⇒ x = 100 liter

So, 100 liters of pure milk should be added so that the new mixture (200 liters) has only 5 % water.

Let us check our answer :-

Old mixture = 100 liters

Quantity of water in old mixture = 10 liters

Quantity of pure milk in old mixture = 90 liters

Quantity of pure milk added to the old mixture = 100 liters

Total quantity of new mixture after adding 100 liters of pure milk = 100 + 100 

= 200 liters

Now, 

Quantity of water after adding 100 liters of pure milk to the old mixture should be 5 % and quantity of water in liters will not not be changed. It will be 10 liters only.

Quantity of water in the new mixture = 10 liters

Therefore, rest is the quantity of pure milk in the new mixture i.e. 200 - 10 = 190 liters

Percentage of water = (10*100)/200

= 5 % 

So, the quantity of water in the new mixture will not be changing and it will be 5 % of the new mixture.

Hence proved.

rishilaugh: thank you sir
Answered by arnav134
8

Quantity of water in the mixture = 10 %

Quantity of water in liters = 10 liters

Quantity of pure milk in liters = 100 - 10 = 90 liters

Let 'x' liters of pure milk is added to the mixture to make water 5 % and Pure milk 95 %

Total amount of pure milk becomes (90 + x) liter and total amount of mixture becomes (100 + x) liters.

⇒ (90 + x) = (95/100)*(100 + x)

⇒ (90 + x) = (19/20)*(100 + x)

⇒ (90 + x)*20 = 19*(100 + x)

⇒ 1800 + 20x = 1900 + 19x

⇒ 20x - 19x = 1900 - 1800

⇒ x = 100 liter

So, 100 liters of pure milk should be added so that the new mixture (200 liters) has only 5 % water.

Let us check our answer :-

Old mixture = 100 liters

Quantity of water in old mixture = 10 liters

Quantity of pure milk in old mixture = 90 liters

Quantity of pure milk added to the old mixture = 100 liters

Total quantity of new mixture after adding 100 liters of pure milk = 100 + 100 

= 200 liters

Now, 

Quantity of water after adding 100 liters of pure milk to the old mixture should be 5 % and quantity of water in liters will not not be changed. It will be 10 liters only.

Quantity of water in the new mixture = 10 liters

Therefore, rest is the quantity of pure milk in the new mixture i.e. 200 - 10 = 190 liters

Percentage of water = (10*100)/200

= 5 % 

So, the quantity of water in the new mixture will not be changing and it will be 5 % of the new mixture.

Hence proved.

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