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# Mixture and Alligation #3

 3. 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. 5litres, 7 litres B. 4litres, 8 litres C. 6litres, 6 litres D. 7litres, 4 litres

Explanation:

Solution 1

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
$=[.25x+.5(12-x)]:[.75x+.5(12-x)]$ $=3:5$

$\Rightarrow\dfrac{\left(.25x+6-.5x\right)}{\left(.75x+6-.5x\right)}=\dfrac{3}{5}\\\Rightarrow\dfrac{\left(6-.25x\right)}{\left(.25x+6\right)}=\dfrac{3}{5}\\\Rightarrow 30-1.25x=.75x+18\\\Rightarrow 2x=12\\\Rightarrow x=6$

Since $x=6, 12-x=12-6=6$

Hence 6 and 6 litres of milk should mixed from the first and second container respectively.

Solution 2

Let cost of 1 litre milk be Rs.1
Milk in 1 litre mix in 1st can $=\dfrac{3}{4}$ litre
Cost Price(CP) of 1 litre mix in 1st can = Rs. $\dfrac{3}{4}$

Milk in 1 litre mix in 2nd can $=\dfrac{1}{2}$ litre.
Cost Price(CP) of 1 litre mix in 2nd can = Rs. $\dfrac{1}{2}$

Milk in 1 litre of the final mix $=\dfrac{5}{8}$
Cost Price(CP) of 1 litre final mix =Rs. $\dfrac{5}{8}$
=> Mean price $=\dfrac{5}{8}$

By rule of alligation,

 CP of 1 litre mix in 2nd can CP of 1 litre mix in 1st can $\dfrac{1}{2}$ $\dfrac{3}{4}$ Mean Price $\dfrac{5}{8}$ $\dfrac{3}{4}-\dfrac{5}{8}=\dfrac{1}{8}$ $\dfrac{5}{8}-\dfrac{1}{2}=\dfrac{1}{8}$

=> mix in 2nd can :mix in 1st can $=\dfrac{1}{8}:\dfrac{1}{8}=1:1$

ie, from each can, $\dfrac{1}{2} × 12 = 6$ litre should be taken. gunjan
2015-10-27 17:01:52
w             m                w              m                 w      m
25/100    75/100         50/100       50/100          3       5
1/4          3/4               1/2            1/2

only milk so:
3/4    1/2
\      /
5/8
/      \
1/8   1/8

so ratio 1:1 0 0 reply angel brownies
2015-01-11 07:24:24
Cant we slove this without the rule of aptitude by using the formula

cost of mixture per kg = total cost/total quantity 0 0 reply mango_man
2014-02-22 18:18:32
Note average of can 1 and 2 is sum/total = (75+50)/2= 62.5

I can milk               II can milk
75                        50
62.5
(50-62.5              (75-62.5
=12.5)                 =12.5)

mix in 2nd can : mix in 1st can = 12.5 : 12.5 = 1:1

ie, from each can, 1/2 × 12 = 6 litre should be taken 0 0 reply arjun
2015-12-23 17:38:59
We cannot take 62.5 as the average of 50 and 75 because then we would always get the ratio from alligation as 1:1 irrespective of the final concentration (because alligating the avg. of two numbers will always give the ratio 1:1, for eg. try alligating the avg. of 50 and 60 ). So, the correct method is to calculate the new average from the given final ratio, i.e 5/(3+5) = 0.625 or 62.5%. 0 0