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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

Neeraj Jain
2014-07-19 21:51:12
 cost ofmilk per litreRs. 10.50 cost ofwater per litre0 Cost of mixture per litreRs. 9 9-0=9 10.5-9=1.5

Quantity of milk : Quantity of water = 9 : 1.5 = 6 : 1
Total quantity = 56 litre
Quantity of water = 56 * 1/7 = 8 litre
0 0
Atul
2014-05-04 23:56:14
Please Provide Solution For This :-
cost price of three types of cashew nuts are Rs. 482 , Rs. 578 , Rs. 698 Respect.
In what proportion these should be mixed , so that the mixture costs Rs. 566 Per Kg ?
Ajay
2014-05-05 15:56:15
Since there are more than two quantities, multiple ratios are possible where the resultant mixture will cost Rs.566

For example, take the required ratio as x:y:x
then, (482x + 578y + 698z)/(x+y+z) = 566
on simplification, this becomes y + 11z = 7x

if we take y=z,
12y = 7x
y=7 and x=12 satisfies this. z=y=7
Required ratio = 12 : 7 : 7

See that (482x + 578y + 698z)/(x+y+z) = 566 for these values of x,y,z
similarly you can get different ratios which also satisfies this.
0 0
andy
2014-02-08 10:04:15
400 gm spirit solution has 40% spirit in it. How may grams of spirit should be added to make it 60 % in the solution?
Jay
2014-02-11 19:01:28
 % concentrationof spirit inthe solution40% % concentrationof spirit inpure spirit100% Mean Concentration60% 100-60=40 60-40=20
i.e., Quantity of the solution : Quantity of spirit = 40 : 20 = 2 : 1
Since the quantity of the solution is 400 gm, amount of spirit needs to be added = 400/2 = 200 gm
0 0
imteyazul
2013-11-15 18:56:39
in the question -30 why 35 is multiplied not 20 ?
Mohan
2013-11-16 11:27:43
20 kg of wheat at the rate of Rs.8.50 per kg and 35 kg at the rate of Rs.8.75 per kg
Hence 20 * 8.50 + 35 * 8.75 is done
0 0
Mononit
2013-11-15 16:42:00
The proportion of 1st and 2nd liquids in a mixture is 2:3 while in another mixture their ratio is 5:4. In what ratio should the two mixture be mixed, so that the new mixture contains equal amount of the two liquids?
Ajay
2013-11-15 18:35:04
Let x liter of the first mixture needs to be mixed with y litre if the second mixture so that the resultant mixture will have both the liquids as 1 : 1

Quantity of 1st liquid in the resultant mixture = 2x/5 + 5y/9
Quantity of 2nd liquid in the resultant mixture = 3x/5 + 4y/9

2x/5 + 5y/9 = 3x/5 + 4y/9
=>18x + 25y = 27x + 20y
=> 9x = 5y
=> x/y = 5/9
=> x:y = 5:9

Required ratio is 5:9
0 0
MammU
2013-11-03 08:57:06
There is a mistake in 8 th prblm.  Ans... Check it