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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 litresB. 28 litres
C. 28.2 litresD. 26 litres
Discuss
answer with explanation

Answer: Option A

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.175.50B. Rs.180
C. Rs.182.50D. Rs.170.5
Discuss
answer with explanation

Answer: Option A

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 teaCost 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 litresB. 4litres, 8 litres
C. 6litres, 6 litresD. 7litres, 4 litres
Discuss
answer with explanation

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 canCP 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 : 9B. 3: 4
C. 9 : 7D. 4 : 3
Discuss
answer with explanation

Answer: Option A

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 litre
mixture from
vessel A
CP of 1 litre
mixture from
vessel 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. 17B. Rs. 18
C. Rs. 19D. Rs. 16
Discuss
answer with explanation

Answer: Option B

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 materialCP of Type 2 material
1520
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

Comments(67)

profileveeksha
2013-04-12 14:00:20 
i personally like this whole idea of organising different topics to diff. strata... gr8 work career bless!!
like 1 dislike 0 reply
profilePriya
2013-04-11 03:34:54 
please explain Q. 8, how is the required quantity 2/3 . 
like 0 dislike 0 reply
profileSooraj
2013-04-11 20:20:57 
Initially we had liquid-1

Then this liquid-1 is mixed with liquid-2.
We got that liquid-1 is mixed with liquid-2 in the ratio 1 : 2

This means, in the mixture of liquid-1 and liquid-2, quantity of liquid-2 will be in the ratio 2/(2+1) = 2/3

Hence the quantity of whisky replaced will be in the ratio 2/3
like 0 dislike 0
profileshweta
2013-04-06 18:25:23 
der r sum ques in which ans is different & explanation is different which is confusing...............
like 0 dislike 0 reply
profileSupport Team, careerbless.com
2013-04-07 13:07:32 
Hi Shweta,

We have tried most attention to each and every problem mentioned here.  But if you feel some problems and explanations are are confusing, please mention specifically and we are happy to correct if any
like 0 dislike 0
profileJayan
2013-03-03 00:39:00 
Dear Vivek,

In question 1, it is said that 4 litres of milk was taken out and replaced by water => 1 time

This process was repeated further two times => 2 times

Total => 3 times and so n = 3

Hope this will clear
like 0 dislike 0 reply
profilevivek
2013-03-02 18:42:08 
can anybody plz clear a doubt regarding question 1, that in question value of n is 2 but in solution they have taken n=3. plz elaborate ??
like 0 dislike 0 reply
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