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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)

profileJay
2015-02-11 21:28:55 
Quantity of water in the 30 litre mixture = 30*30/100 = 9 litre

After adding 10 litre of water, quantity of water becomes 19 litre and total quantity becomes 40 litre
percentage of water = 19*100/40 = 47.5%
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profileNitasha Mehmi
2015-01-30 11:35:32 
One litre of water was mixed to 6 litres of sugar solution containing 4% of sugar.What is the percentage of sugar in the solution?
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profileSundeep
2015-10-13 14:50:30 
Answer is 3.4% 

first 4% of 6 is 6/25
After the mixture 6+1 = 7 

Concentration of sugar is same 

let x be the percent we have to find 
so, x% of 7 is 6/25

After solving this equation you will get 3.4 %
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profileviksa bansal
2015-08-30 20:45:01 
Suppose 6000ml and sugar is 4%.
So sugar is 240 out of new mixture 7000
so $\dfrac{240}{7000}×100=3.42\%$
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profileRaj
2015-02-09 21:36:16 
6 litres of sugar solution contains $\dfrac{6×4}{100}=0.24$ litre of sugar
quantity of water = $\dfrac{6×96}{100}=5.76$
After adding one litre of water, its quantity becomes 6.76

Required percentage of sugar = $\dfrac{0.24×100}{0.24+ 6.76}=\dfrac{24}{7}=3.42\%$
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profileBlanco
2015-01-26 17:41:17 
Two vessels contain mixture of milk and water in the ratio 5: 1 and 9 :1. They are mixed together in the ratio 1 : 5. Find the ratio of milk and water in the final mixture.
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profileRaj
2015-02-09 20:06:28 
Concentration of milk in these vessels will be 5/6 and 9/10.

quantity of milk in 1 litre of first vessel = 1*5/6 = 5/6
quantity of milk in 5 litres of second vessel = 5*9/10 = 9/2

quantity of milk in 6 litres of the mixture = 5/6+9/2 = 32/6
quantity of water = 6-32/6 = 4/6
Required ratio = 32/6 : 4/6 = 8 : 1
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profilesam
2016-02-04 06:23:19 
@shefali, total quantiy is 6 litre in which quantity of milk is 32/6.
Hence 6-32/6 is the quantity of water
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profileshefali
2016-01-12 05:54:57 
How 6-32/6  please explain
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profilebilu
2015-01-26 17:31:32 
Amount used to purchase 1 litre of honey can be used to purchase 3 l of milk or 5l of water. Total amount spent is 1530  what will be the amt spent on honey if the total amt is spent to purchase equal volumes of honey and water only.
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