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Formulae

1. Alligation:
It is the rule that 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.

2. Mean Price:
The cost of a unit quantity of the mixture is called the mean price.

3. Rule of Alligation:
If two ingredients are mixed, then
\( {{Quantity of cheaper} \over {Quantity of dearer}} \) = \( {{C.P. of dearer - Mean Price} \over {Mean price - C.P. of cheaper}} \)

We present as under:

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

4. Suppose 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 = \( [x {({1 - {y \over x}})^n}] \) units.

Exercise : Alligation or Mixture MCQ Questions and Answers

Question 1

Milk and water are mixed in a vessel A as 4:1 and in vessel B as 3:2. For vessel C, if one takes equal quantities from A and B, find the ratio of milk to water in C.

A.  
B.  
C.  
D.  

Correct Answer : D. 7:3

Description :
Ratio of Milk and water in a vessel A is 4 : 1
Ratio of Milk and water in a vessel B is 3 : 2
Ratio of only milk in vessel A = 4 : 5
Ratio of only milk in vessel B = 3 : 5
Let 'x' be the quantity of milk in vessel C
Now as equal quantities are taken out from both vessels A & B
=> \( 4 \over 5 \) : \( 3 \over 5 \)
        x
\( 3 \over 5 \) - x x - \( 4 \over 5 \)

=> x = \( 7 \over 10 \)
Therefore, quantity of milk in vessel C = 7
=> Water quantity = 10 - 7 = 3
Hence the ratio of milk & water in vessel 3 is 7 : 3

Question 2

A milkman claims to sell milk at its cost price, still, he is making a profit of 30% since he has mixed some amount of water in the milk. What is the % of milk in the mixture?

A.  
B.  
C.  
D.  

Correct Answer : D. 76.92 %

Description :
Let the milk he bought is 1000 ml
Let C.P of 1000 ml is Rs. 100
Here let he is mixing K ml of water
He is getting 30% profit
=> Now, the selling price is also Rs. 100 for 1000 ml
=> 100 : K%
= 100 : 30
10 : 3 is ratio of milk to water
=> Percentage of milk = 10 x 100/13 = 1000/13 = 76.92 %

Question 3

The milk and water in two vessels A and B are in the ratio 4:3 and 2:3 respectively. In what ratio the liquids in both the vessels be mixed to obtain a new mixture in vessel c consisting half milk and half water?

A.  
B.  
C.  
D.  

Correct Answer : D. 7:5

Description :
Milk in 1-litre mixture of A = 4/7 litre.
Milk in 1-litre mixture of B = 2/5 litre.
Milk in 1-litre mixture of C = 1/2 litre.

By rule of alligation we have required ratio X:Y

   X                  :                 Y  

  4/7                                2/5  

           \                      /

              (Mean ratio)
                   (1/2) 

           /                      \

 (1/2 – 2/5)     :       (4/7 – 1/2)  

    1/10                      1/1 4 

So Required ratio = X : Y = 1/10 : 1/14 = 7:5

Question 4

A vessel full of orange juice contains 40% orange pulp. A part of juice is replaced by another juice containing 19% orange pulp and now the percentage of orange pulp is found to be 26%. What quantity of juice is replaced?

A.  
B.  
C.  
D.  

Correct Answer : B. 2:3

Description :
Concentration of orange pulp in 1st vessel = 40% Concentration of orange pulp in 2nd vessel = 19% After the mixing, Concentration of orange pulp in the mixture = 26% By rule of alligation,

Concentration of orange pulp in 1st          Concentration of orange pulp in 2nd
          Vessel                                         Vessel
           40%                                           19%

                                       26%
	
         26-19=7                                      40-26=14

Hence ratio of 1st and 2nd quantities = 7 : 14 = 1 : 2 i.e., 2/(1+2)=2/3 part of the juice is replaced.

Question 5

An alloy contains zinc, copper and tin in the ratio 2 : 3 : 1 and another contains copper, tin and lead in the ratio 5 : 4 : 3. If equal weights of both alloys are melted together to form a third alloy, then the weight of lead per kg in new alloy will be:

A.  
B.  
C.  
D.  

Correct Answer : A. \( {1 \over 8} \) kg

Description :
Ratio of Zinc, Copper and Tin is given as, Z : C : T = 2 : 3 : 1.
Now, let the first alloy be 12 kg (taken as 4 kg Zinc, 6 kg Copper and 2 Kg Lead). Weight of second alloy = 12 kg as, C : T : L = 5 : 4 : 3. (taken as 5 kg Copper, 4 kg Tin and 3 Kg Lead.)
Alloys are mixed together to form third alloy. Then the ratio of content in it,
Z : C : T : L = 4 : (6 + 5) : (2 + 4) : 3
Weight of third alloy = 12 + 12 = 24 Kg.
So, weight of the Lead = \( {3 \over 24} \) = \( {1 \over 8} \) kg.

Question 6

A mixture is made by mixing alcohol and water in the ratio 9 ∶ 7. If ‘x’ litres of alcohol and ‘3x’ litres of water is mixed in 80 litres of mixture, the new ratio becomes 13 ∶ 14. Find the quantity of new mixture.

A.  
B.  
C.  
D.  

Correct Answer : D. 108 litres

Description :
Sum of first ratio = 9 + 7 = 16 Quantity of alcohol in 80 litres of mixture = 9/16 × 80 = 45 litres
Quantity of water in 80 litres of mixture = 7/16 × 80 = 35 litres
When ‘x’ litres of alcohol and ‘3x’ litres of water is added,
Quantity of alcohol in new mixture = 45 + x
Quantity of water in new mixture = 35 + 3x
But, ratio of new mixture = 13 ∶ 14
⇒ (45 + x)/(35 + 3x) = 13/14
⇒ 14(45 + x) = 13(35 + 3x)
⇒ 630 + 14x = 455 + 39x
⇒ 39x – 14x = 630 – 455
⇒ x = 175/25 = 7

∴ Total quantity of new mixture = 45 + x + 35 + 3x = 80 + 4(7) = 80 + 28 = 108 litres

Question 7

Two solutions S1 and S2 contain whisky and soda in the ratio 2 : 5 and 6 : 7 respectively. In what ratio these solutions be mixed to get a new solution S3, containing whisky and soda in the ratio 5 : 8 ?

A.  
B.  
C.  
D.  

Correct Answer : A. 7:9

Description :
Let the amount taken from S1 be 7x
And amount taken from S2 be 13y

(2x + 6y)/(5x + 7y) = 5/8
16x + 48y = 25x + 35y
9x = 13y
x/y = 13/9

Actual ratios of amounts
= 7x/13y
= (7/13) * (13/9)
= 7/9

Question 8

A mixture of 150 liters of wine and water contains 20% water. How much more water should be added so that water becomes 25% of the new mixture?

A.  
B.  
C.  
D.  

Correct Answer : A. 10

Description :
Number of liters of water in150 liters of the mixture = 20% of 150 = 20/100 * 150 = 30 liters.
P liters of water added to the mixture to make water 25% of the new mixture.
Total amount of water becomes (30 + P) and total volume of mixture is (150 + P).
(30 + P) = 25/100 * (150 + P)
120 + 4P = 150 + P => P = 10 liters.

Question 9

The ratio of milk and water in three samples is 1 ∶ 3, 3 ∶ 5 and 11 ∶ 5. A mixture containing equal quantities of all three samples is made. What will be the ratio of milk and water in the new mixture?

A.  
B.  
C.  
D.  

Correct Answer : C. 7 : 9

Description :
Method 1 -
Let ‘x’ quantity of each of three mixtures to be mixed
Quantity of milk in the new mixture = (1/4)x + (3/8)x + (11/16)x = 21x/16
Quantity of water in the new mixture = (3/4)x + (5/8)x + (5/16)x = 27x/16
∴ Required ratio = (21x/16) ∶ (27x/16) = 7 ∶ 9

Method 2 -
1 + 3 = 4, 3 + 5 = 8, 11 + 5 = 16
Let the quantity of each sample LCM(4, 8, 16) = 16
Milk in the first sample
⇒ 16 × 1/(1 + 3)
⇒ 4
Water in the first sample = 16 - 4 = 12
Milk in the second sample
⇒ 16 × 3/8
⇒ 6
Water in the second sample = 16 - 6 = 10
Milk in the third sample
⇒ 16 × 11/16
⇒ 11
Water in the second sample = 16 - 11 = 5
Ratio of milk and water in the new mixture
⇒ ( 4 + 6 + 11)/(12 + 10 + 5)
⇒ 21/27 = 7/9
∴ Required ratio is 7 : 9.

Question 10

The ratio of Iron and Zinc in an alloy is 4 : 5. In another alloy, the ratio of Iron, Copper and Zinc is 3 : 2 : 7. Equal amounts of the two alloys are molten and mixed together. What will be the ratio of Iron, Copper and Zinc in the resultant alloy?

A.  
B.  
C.  
D.  

Correct Answer : A. 25 : 6 : 41

Description :
In first alloy, ratio of iron and zinc is 4 : 5
In second alloy, ratio of iron, copper and zinc is 3 : 2 : 7
Suppose, amount T of both alloys is taken.
⇒ Amount of iron in first alloy = (4/ (4 + 5)) × T = (4/9) T
Amount of zinc in first alloy = (5/ (4 + 5)) × T = (5/9) T
Amount of iron in Second alloy = (3/ (3 + 2 + 7)) × T = (1/4) T
Amount of Copper in second alloy = (2/ (3 + 2 + 7)) × T = (1/6) T
Amount of zinc in second alloy = (7/ (3 + 2 + 7)) × T = (7/12) T
After mixing
Total amount of Iron = (4/9) T + (1/4) T = (25/36) T
Total amount of copper = (1/6) T = (6/36) T
Total amount of zinc = (5/9) T + (7/12) T = (41/36) T
⇒ Ratio of Iron, Copper and Zinc = (25/36) T : (6/36) T : (41/36) T
= 25 : 6 : 41

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