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. 1Milk 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:
∴ 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
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Comments ( 3 )
Related Questions on Alligation
A. $$\frac{{1}}{{2}}$$ kg
B. $$\frac{{1}}{{8}}$$ kg
C. $$\frac{{3}}{{14}}$$ kg
D. $$\frac{{7}}{{9}}$$ kg
A. 81 litres
B. 71 litres
C. 56 litres
D. 50 litres
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
@Selina Akter, Here we got that 1:1 (equal) so every can will take 12/2 = 6 litres
What is the explanationon last line calculation. 1/2*12=6