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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 |

Answer: Option C

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

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

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

cost of mixture per kg = total cost/total quantity

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

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

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%.

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