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Solved Examples(Set 1) - Mixture and Alligation

 1. A container contains 40 litres of milk. From this container 4 litres of milk was taken out and replaced by water. This process was repeated further two times. How much milk is now contained by the container? A. 29.16 litres B. 28 litres C. 28.2 litres D. 26 litres

Explanation:

Suppose a container contains $x$ units of a liquid from which $y$ units are taken out and replaced by water. After $n$ operations, quantity of pure liquid
$=x\left(1-\dfrac{y}{x}\right)^n$ units.

milk contained by the container now
$=40\left(1-\dfrac{4}{40}\right)^3\\= 40\left(1-\dfrac{1}{10}\right)^3 \\=40×\dfrac{9}{10}×\dfrac{9}{10}×\dfrac{9}{10}\\= \dfrac{4×9×9×9}{100}\\= 29.16$

 2. Tea worth Rs. 126 per kg and Rs. 135 per kg are mixed with a third variety of tea in the ratio 1 : 1 : 2. If the mixture is worth Rs. 153 per kg, what is the price of the third variety per kg ? A. Rs.175.50 B. Rs.180 C. Rs.182.50 D. Rs.170.5

Explanation:

Tea worth Rs. 126 per kg and Rs. 135 per kg are mixed in the ratio 1 : 1
So their average price $=\dfrac{\left(126 + 135 \right)}{2} = 130.5$

Hence let's consider that the mixture is formed by mixing two varieties of tea, one at Rs. 130.50 per kg and the other at Rs. $x$ per kg in the ratio 2 : 2, i.e., 1 : 1. Now let's find out $x$.

By rule of alligation,

 Cost of 1 kg of 1st kind of tea Cost of 1 kg of 2nd kind of tea 130.50 $x$ Mean Price 153 $(x-153)$ 22.50

$(x-153):22.5=1:1\\\Rightarrow x-153=22.50\\\Rightarrow x=153+22.5=175.5$

 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.

 4. Two vessels A and B contain spirit and water in the ratio 5 : 2 and 7 : 6 respectively. Find the ratio in which these mixtures be mixed to obtain a new mixture in vessel C containing spirit and water in the ratio 8 : 5 ? A. 7 : 9 B. 3: 4 C. 9 : 7 D. 4 : 3

Explanation:

Let Cost Price(CP) of 1 litre spirit be Rs.1

Quantity of spirit in 1 litre mixture from vessel A $=\dfrac{5}{7}$
Cost Price(CP) of 1 litre mixture from vessel A = Rs. $=\dfrac{5}{7}$

Quantity of spirit in 1 litre mixture from vessel B $=\dfrac{7}{13}$
Cost Price(CP) of 1 litre mixture from vessel B = Rs. $=\dfrac{7}{13}$

Quantity of spirit to be obtained in 1 litre mixture from vessel C $=\dfrac{8}{13}$
Cost Price(CP) of 1 litre mixture from vessel C(Mean Price) = Rs. $=\dfrac{8}{13}$

By rule of alligation,

 CP of 1 litremixture fromvessel A CP of 1 litremixture fromvessel B $\dfrac{5}{7}$ $\dfrac{7}{13}$ Mean Price $\dfrac{8}{13}$ $\dfrac{8}{13}-\dfrac{7}{13}=\dfrac{1}{13}$ $\dfrac{5}{7}-\dfrac{8}{13}=\dfrac{9}{91}$

=> Mixture from Vessel A : Mixture from Vessel B
$=\dfrac{1}{13}:\dfrac{9}{91}=7:9$

 5. The cost of Type 1 material is Rs. 15 per kg and Type 2 material is Rs.20 per kg. If both Type 1 and Type 2 are mixed in the ratio of 2 : 3, then what is the price per kg of the mixed variety of material? A. Rs. 17 B. Rs. 18 C. Rs. 19 D. Rs. 16

Explanation:

Solution 1

Cost Price(CP) of Type 1 material is Rs. 15 per kg
Cost Price(CP) of Type 2 material is Rs. 20 per kg

Type 1 and Type 2 are mixed in the ratio of 2 : 3.

Hence Cost Price(CP) of the resultant mixture
$=\dfrac{(15 × 2)+(20 × 3)}{(2+3)}$
$=\dfrac{\left(30 +60\right)}{5} = \dfrac{90}{5} = 18$

=> Price per kg of the mixed variety of material = Rs.18

Solution 2

Cost Price(CP) of Type 1 material is Rs. 15 per kg
Cost Price(CP) of Type 2 material is Rs. 20 per kg
Let Cost Price(CP) of resultant mixture be Rs.$x$ per kg

By rule of alligation,

 CP of Type 1 material CP of Type 2 material 15 20 Mean Price $x$ $(20-x)$ $(x-15)$

=> Type 1 material : Type 2 material $=(20-x):(x-15)$

Given that Type 1 material : Type 2 material = 2 : 3
$\Rightarrow (20-x) : (x-15) = 2 : 3\\\Rightarrow \dfrac{(20-x)}{(x-15)} = \dfrac{2}{3}\\\Rightarrow 3(20-x) = 2(x-15)\\\Rightarrow 60 - 3x = 2x - 30\\\Rightarrow 90 = 5x\\\Rightarrow x = \dfrac{90}{5} = 18$

=> price per kg of the mixed variety of material = Rs.18

Ajay
2013-11-04 18:26:17
Hence ratio of 1st and 2nd quantities = 7 : 14 = 1 : 2

if x:y is the ratio,to get the quantity of x,the formula is x/(x+y),and to get the quantity of y,the formula is y/(x+y).

Here, second quantity replaces the first . Hence, the 2/3 is the answer
0 0
Ashish
2013-08-18 12:07:21
Can anyone explain ques  12 by the rule of allegation ?
shashank
2013-08-31 21:06:49
ashish i will explain what is the concept behind this.
suppose there is no such restriction of the word drawn off means it will become simple alligation  question, like in a mixture of milk and water which in the ratio of 7:5, 9 litre of water is added and it becomes 7:9, find the initial volume of mixture, so this is straight forward question answer is 27 litre. now apply restriction of the word remember 9 litre were drawn off initially so it will be 27+9=36 litre of mixture. and you have to find amount of p which is (36*7)/(7+5)=21 litre.
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palak
2013-06-27 17:56:42
a milkman mixes 10 litres of water to 50 litres of milk of rs.16 per litre then the cost price of mixture per litre is what?
i want answer of above question
Nayana
2013-06-30 14:34:49
Cost price of water = 0
Cost price of 10 L water = 10 * 0 = 0

Cost price of milk = 16 per litre
Cost price of 50 L milk = 16 * 50

Cost price of the mixture = 0 + (16 * 50) = (16 * 50)
cost price of mixture per litre = (16 * 50)/60 = Rs. 13.33

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Manju
2013-05-06 04:48:41
The ratio of milk and water in a mixture is 4:3. If 14 litres of water is added to the mixture, the ratio of milk and water is 3:4. The quantity of milk in the mixture is:?
shashank
2013-08-31 20:35:44
0 0
2013-04-14 19:34:58
in what proportion must water be mixed with spirit to gain 16 2/3% by selling at its cost price?
shashank
2013-08-31 20:40:05
6:1 ratio
suppose initial cost price is x rs. and we are mixing the water has cost price is equal to 0 rs.
because there is a gain of 16.67% means 1/6 overall becomes 7/6 for selling price, and this will become 6/7 foe cp and now then you have all value to apply method of alligation

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Support Team, careerbless.com
2013-04-15 21:49:02
Dear Prasad,  Its  almost the same  as question 19. Please revert for clarification
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