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A milk vendor has 2 cans of milk. The first contains 25% water and the rest milk. The second contains 50% water. How much milk should he mix from each of the containers so as to get 12 litres of milk such that the ratio of water to milk is 3 : 5?

A. 4 litres, 8 litres

B. 6 litres, 6 litres

C. 5 litres, 7 litres

D. 7 litres, 5 litres

Answer: Option B

Solution(By Examveda Team)

Let the cost of 1 litre milk be Rs. 1
Milk in 1 litre mixture in 1st can = $$\frac{{3}}{{4}}$$ litre, C.P. of 1 litre mixture in 1st can Rs. $$\frac{{3}}{{4}}$$
Milk in 1 litre mixture in 2nd can = $$\frac{{1}}{{2}}$$ litre, C.P. of 1 litre mixture in 2nd can Rs. $$\frac{{1}}{{2}}$$
Milk in 1 litre of final mixture = $$\frac{{5}}{{8}}$$ litre, Mean price = Rs. $$\frac{{5}}{{8}}$$
By the rule of alligation, we have:
Alligation mcq solution image
∴ Ratio of two mixtures = $$\frac{{1}}{{8}}$$ : $$\frac{{1}}{{8}}$$ = 1 : 1
So, quantity of mixture taken from each can = $$\left( {\frac{1}{2} \times 12} \right)$$  = 6 litres

This Question Belongs to Arithmetic Ability >> Alligation

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Comments ( 3 )

  1. Shahriar Nayan
    Shahriar Nayan :
    3 years ago

    Let x and (12−x) litres of milk be mixed from the first and second container respectively
    Amount of milk in x litres of the the first container =.75x
    Amount of water in x litres of the the first container =.25x
    Amount of milk in (12−x) litres of the the second container =.5(12−x)
    Amount of water in (12−x) litres of the the second container =.5(12−x)
    Ratio of water to milk =[.25x+.5(12−x)]:[.75x+.5(12−x)]=3:5
    ⇒(.25x+6−.5x)(.75x+6−.5x)=35
    ⇒(6−.25x)(.25x+6)=35
    ⇒30−1.25x=.75x+18
    ⇒2x=12
    ∴x=6

  2. Rajib Mahmud
    Rajib Mahmud :
    4 years ago

    @Selina Akter, Here we got that 1:1 (equal) so every can will take 12/2 = 6 litres

  3. Selina Akter
    Selina Akter :
    4 years ago

    What is the explanationon last line calculation. 1/2*12=6

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