Mixture and Alligation

Mixture and Alligation

Hello Readers,

Below in the post ,We shall discuss about Mixture and Alligation of the Quant section.  Now a days these topics  have became an important part of the Quant test ,So Here we will help you in this.We will  provide short tricks and quant quiz.

TRICKS  FOR SOLVING QUESTIONS RELATED TO MIXTURE AND ALLIGATION :
Mixture and Alligation Shortcut Formula :

 Mixture : Aggregate of two or more than two type of quantities gives us a mixture.

Alligation: It is a method of solving arithmetic problems related to mixtures of ingredients. This rule enables us to find the ratio in which two or more ingredients at the given price must be mixed to produce a mixture of desired price.

Alligation Rule: When two elements are mixed to make a mixture and one of the elements is cheaper and other one is costlier then,

 {Quantity of Cheaper} / {Quantity of Costlier} = {C.P. of Costlier - Mean Price} / {Mean Price - C.P. of Cheaper}

                        C.P. of cheaper                                C.P. of costlier

                        quantity (c)                                         quantity (d)

                                                          Mean Price

                                                                (m)

                         (d - m)                                                       (m - c)

Thus, (Cheaper quantity) : (Dearer quantity) = (d - m) : (m - c).

Let a container contains x units of liquid from which y units are taken out and replaced by water.

The quantity of pure liquid after 'n' operations is equal to

                                     = [ x ( 1 - y/x}^n] 

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. 26 litres

B. 29.16 litres

C. 28 litres

D. 28.2 litres

2. A vessel is filled with liquid, 3 parts of which are water and 5 parts syrup. How much of the mixture must be drawn off and replaced with  water so that the mixture may be half water and half syrup?

A. 1/3

B. 1/4

C. 1/5

D. 1/7

3. 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.182.50

B. Rs.170.5

C. Rs.175.50

D. Rs.180

 4.A can contains a mixture of two liquids A and B is the ratio 7 : 5. When 9 litres of mixture are drawn off and the can is filled with B, the ratio of A and B becomes 7 : 9. How many litres of liquid A was contained by the can initially?

A. 10

B. 20

C. 21

D. 25

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

A. 3: 4

B. 4 : 3

C. 9 : 7

D. 7 : 9

6. In what ratio must a grocer mix two varieties of pulses costing Rs. 15 and Rs. 20 per kg respectively so as to get a mixture worth Rs. 16.50 kg?

A. 3 : 7

B. 5 : 7

C. 7 : 3

D. 7 : 5

7.8 litres are drawn from a cask full of wine and is then filled with water. This operation is performed three more times. The ratio of the quantity of wine now left in cask to that of the water is 16 : 65. How much wine did the cask originally hold?

A. 30 litres

B. 26 litres

C. 24 litres

D. 32 litres

8. A dishonest milkman professes to sell his milk at cost price but he mixes it with water and thereby gains 25%. The percentage of water in the mixture is:

A. 4%

B. 6 1/4%

C. 20%

D. 25%

9. A jar full of whiskey contains 40% alcohol. A part of this whisky is replaced by another containing 19% alcohols and now the percentage of alcohol was found to be 26%. The quantity of whisky replaced is

A. 4/3

B. 3/4

C. 3/2

D. 2/3

10. A jar full of whisky contains 40% alcohol. A part of this whisky is replaced by another containing 19% alcohol and now the percentage of alcohol was found to be 26%. The quantity of whisky replaced is:

A. 1/3

B. 2/3

C. 2/5

D. 3/5

ANSWER AND SOLUTION :
1( B)Explanation :
Assume that a container contains x of liquid from which y units are taken out and replaced by water. After n operations, the quantity of pure liquid
=x(1-y/x)^n
Hence milk now contained by the container = 40(1-4/40)^3
=40(1-1/10)^3
=40×9/10×9/10×9/10 =(4×9×9×9)/100 =29.16

2(C)Explanation:
Suppose the vessel initially contains 8 litres of liquid.
Let x litres of this liquid be replaced with water.
Quantity of water in new mixture = (3 - 3x/8 + x) litre
Quantity of syrup in new mixture = (5 - 5x/8) litres
So  (3 - 3x/8 + x)  = (5 - 5x/8) litres
=> 5x + 24 = 40 - 5x
=>10x = 16
=> x = 8/5 .
So, part of the mixture replaced = (8/5 x 1/8) = 1/5

3(C)Explanation :
Tea worth Rs. 126 per kg and Rs. 135 per kg are mixed in the ratio 1 : 1
So their average price = (126+135)/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 the rule of alligation, we can write as

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
=>x - 153 = 22.50
=> x = 153 + 22.5 = 175.5

4(C)Explanation:
Suppose the can initially contains 7x and 5x of mixtures A and B respectively.
Quantity of A in mixture left = (7x - 7/12 x 9)litres
 = (7x - 21/4) litres.
Quantity of B in mixture left = (5x - 5/12  x 9) litres
= (5x - 15/4)  litres.
So (7x - 21/4)/((5x - 15/4) +9) = 7/9
=> (28x - 21)/(20x + 21) = 7/9
=> 252x - 189 = 140x + 147
=> 112x = 336
=> x = 3.
So, the can contained 21 litres of A.
.
5(D)Explanation :
Let Cost Price(CP) of 1 litre spirit be Rs.1
Quantity of spirit in 1 litre mixture from vessel A = 5/7
Cost Price(CP) of 1 litre mixture from vessel A = Rs. 5/7
Quantity of spirit in 1 litre mixture from vessel B = 7/13
Cost Price(CP) of 1 litre mixture from vessel B = Rs. 7/13
Quantity of spirit to be obtained in 1 litre mixture from vessel C = 8/13
Cost Price(CP) of 1 litre mixture from vessel C = Rs. 8/13 = Mean Price
By the rule of alligation, we can write as

CP of 1 litre mixture     CP of 1 litre mixture                                                                
 from vessel A (5/7)      from vessel B (7/13)
                                                 

                            Mean Price
                               (8/13)


8/13 - 7/13 = 1/13                 5/7 - 8/13 = 9/91

=> Mixture from Vessel A : Mixture from Vessel B = 1/13 : 9/91 = 7 : 9 = Required Ratio

6(C)Explanation:
By the rule of alligation:
Cost of 1 kg pulses of       Cost of 1 kg pulses of
1st kind Rs.  (15)               2nd kindRs. (20)

                                Mean Price
                                Rs. (16.50)

   (3.50)                                    (1.50)

 Required rate = 3.50 : 1.50 = 7 : 3.

7(C)Explanation :
Let initial quantity of wine = x litre
After a total of 4 operations, quantity of wine = x(1-y/x)^n=x(1-8/x)^4
Given that after a total of 4 operations, the ratio of the quantity of wine left in cask to that of water = 16 : 65
Hence we can write as (x(1-8/x)^4)/x =16/81
(1-8/x)^4 = (2/3)^4
(1-8/x) = 2/3
(x-8/x) = 2/3
3x-24=2x
x=24

8(C)Explanation:
Let C.P. of 1 litre milk be Re. 1
Then, S.P. of 1 litre of mixture = Re. 1, Gain = 25%.
C.P. of 1 litre mixture = Re.(100/125 x 1) = 4/5
By the rule of alligation, we have:

C.P. of 1 litre of milk       C.P. of 1 litre of water
    Re. (1)                                     (0)

                          Mean Price
                             Re. 4/5

    4/5                                       1/5

 Ratio of milk to water = 4/5 : 1/5 = 4 : 1.
Hence, percentage of water in the mixture = (1/5 x 100)% = 20%


9(D)Explanation :
Concentration of alcohol in 1st Jar = 40%
Concentration of alcohol in 2nd Jar = 19%
After the mixing, Concentration of alcohol in the mixture = 26%
By the rule of alligation,
Concentration of alcohol in  Concentration of alcohol in                
   1st Jar(40%)                                 2nd Jar (19%)

                                           Mean
                                           (26%)

      (7)                                             (14)

Hence ratio of 1st and 2nd quantities = 7 : 14 = 1 : 2

10(B)Explanation:
By the rule of alligation, we have:

Strength of first jar                Strength of 2nd jar
  (40%)                                     (19%)

                              Mean
                            Strength
                              (26%)

    (7)                                             (14)

So, ratio of 1st and 2nd quantities = 7 : 14 = 1 : 2
 Required quantity replaced = 2/3

Mixture and Alligation

Important Concepts and Formulas - Mixture and Alligation1. AlligationAlligation is the rule which enables us to find the ratio in which two or more ingredients at the given price must be mixed to produce a mixture of a specified price.
2. Mean PriceMean price is the cost price of a unit quantity of the mixture

3. 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
$=x{(1-\frac{y}{x})}^n$ units.

4. Rule of AlligationIf two ingredients are mixed, then
$\left(\frac{\text{Quantity of cheaper}}{\text{Quantity of dearer}}\right)$ $=\left(\frac{\text{CP of dearer - Mean Price}}{\text{Mean price - CP of cheaper}}\right)$

The above formula can be represented with the help of the following diagram which is easier to understand.


=> (Cheaper quantity) : (Dearer quantity) = (d - m) : (m - c)

Proof of Rule of Alligation formula

Let cost price of unit quantity of cheaper ingredient = c
cost price of unit quantity of dearer ingredient = d

Suppose 'a' units of cheaper ingredient and 'b' units of dearer ingredient are mixed.

Total quantity of the mixture = a + b
Total cost price of the mixture = (ac + bd)

cost price of unit quantity of the mixture (mean price) = $\frac{\text{ac + bd}}{\text{a+b}}$

Suppose mean price = m
=> $\frac{\text{ac + bd}}{\text{a+b}}$ = m
=> am + bm = ac + bd
=> am - ac = bd - bm
=> a(m - c) = b(d -m)
=> a/b = (d -m)/(m - c)
=> a : b = (d -m):(m - c)