We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.More informationAgree
menu ×
Google
Custom Search
cancel
search
FacebookTwitterLinkedIn×
share
page location
×
ad

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)

profileRaj
2015-02-09 19:56:14 
Let cost of 1 litre of honey = x
Cost of 5 litre of water = x.  => Cost of 1 litre of water = x/5

Cost of 1 litre of honey  : Cost of 1 litre of water = x : x/5 = 5 : 1
ie, amount should be in the ration 5 : 1 to purchase equal quantity of honey and water
Amount spent on honey = 1530  * 5/6 =Rs.1275
like 0 dislike 0
profileRatan
2015-01-24 02:30:43 
How many kilograms of sugar must be added to 20 kilograms of 20% solution of sugar and water to increase the concentration of sugar to 50% of solution?
like 0 dislike 0 reply
profileArun Kumar
2016-07-21 10:50:55 
The concentration of sugar in solution =20*20% = 4 kg
rest solution is 16 kg

Now solution is 16: 4 (kgs)

Need to increase concentration of sugar up to 50%, means equal quantity of sugar in solution.
4 kg already, +12 kg add

Answer: 12 kg
like 0 dislike 0
profileJay
2015-01-24 11:24:55 
Percentage concentration of
sugar in pure sugar(100)
Percentage concentration of
sugar in the solution
(20)

50


50-20=30

100-50=50

Quantity of sugar : Quantity of solution should be 30 : 50 = 3:5

Quantity of sugar : 20 = 3:5

Quantity of sugar = 20*3/5=12 kg
like 0 dislike 0
profileChandra
2015-01-05 11:29:33 
The weight of an empty bottle is one sixth of the bottle full of water. Certain percentage of water is removed from the bottle and weighed. The weight of the bottle is now one third of the bottle full of water. What is the percentage of water removed?
like 0 dislike 0 reply
profileJay
2015-01-11 19:45:25 
Let weight of the full bottle is 60 Kg
Then, weight of the empty bottle is 1/6 of 60 which 10 Kg
Weight of water = 60-10=  50 Kg

After removing some amount of water, weight of the bottle = 1/3 of 60 kg which is 20 Kg
Weight of water present in the bottle = 20 -10 = 10 Kg

Weight of water removed = 50-10 = 40 Kg
Required percentage = 40*100/50  = 80%
like 0 dislike 0
profilemahender
2014-12-24 20:40:31 
A sells milk which contains 5% of water. What quantity of pure milk should be added to 2 litre of milk(contains 5% water) so that proportion of water becomes 4%?
like 0 dislike 0 reply
profileAnkita Gupta
2014-12-28 14:37:55 
Water component in 2lts of milk(contains 5% water)  is 2*5/100 = 0.1 litre
and milk component is 2-0.1 = 1.9 litre

Let required quantity of pure milk added is x litre

Total quantity of pure milk will be (1.9+x)  litre
Total quantity of the solution will be (2+x) litre
Total quantity of water will be same as 0.1 litre
(2+x) * 4/100 = 0.1
2+x = 2.5
x = 0.5 litre
like 0 dislike 0
profilearun kumar
2016-07-21 11:50:54 
Let's do it in alligation rule

Apply alligation rule on those components are added or subtracted.

Here is Milk
(old)       (every thing itself 100%)
95%        100%

       96% (New proportion)

4              1
Divided by 2 by making old quantity 2 litre.
2             .5      

Added milk is 0.5 litre
like 0 dislike 0
profileSaurabh
2014-12-24 07:33:49 
Pure milk costs 16 per litre. After adding water the milkman sells the mixture
15 per litre and thereby makes a profit of 25%. In what respective ratio does
he mix milk with water?  
like 0 dislike 0 reply
Previous1234567Next
21-30 of 67 comments

Add Your Comment

(use Q&A for new questions)
?
Name
cancel
preview