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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

1. How much coffee of variety A, costing Rs. 5 a kg should be added to 20 kg of Type B coffee at Rs. 12 a kg so that the cost of the two coffee variety mixture be worth Rs. 7 a kg?

 A.   58 kg B.   53 kg C.   54 kg D.   50 kg

Correct Answer : D. 50 kg

Description :
As per the rule of alligation,
Quantity of Dearer: Quantity of Cheaper = (12-7) : (7-5) = 5:2
Quantity of Variety A coffee that needs to be mixed ⇒ 5:2 = x:20
⇒ x =50 kg

2. 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.   7 : 9 B.   4 : 9 C.   7 : 6 D.   9 : 7

Correct Answer : A. 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.

3. 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.   2:3 B.   5:7 C.   3:7 D.   3:4

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.

4. Cost of two types of pulses is Rs.15 and Rs, 20 per kg, respectively. If both the pulses are mixed together in the ratio 2:3, then what should be the price of mixed variety of pulses per kg?

 A.   Rs. 17 per kg B.   Rs. 16 per kg C.   Rs. 18 per kg D.   Rs. 15 per kg

Correct Answer : C. Rs. 18 per kg

Description :
Let the cost of mixed variety of pulse be Rs. x
As per the alligation rule,
2:3 = (20-x) : (x-15)
⇒ 2x+3x = 60+30
⇒ 5x = 90
⇒ x = 18

5. 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.   12 B.   11 C.   15 D.   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.

6. 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.   7:9 B.   7:6 C.   7:3 D.   7:4

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

7. 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.   108 litres B.   180 litres C.   104 litres D.   120 litres

Correct Answer : A. 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

8. 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.   75.92 % B.   74.92 % C.   76.82 % D.   76.92 %

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 %

9. From a homogeneous mixture of salt and sugar containing 4 parts of salt and 5 parts of sugar, 1/4th part of the mixture is drawn off and replaced with salt. The ratio of salt and sugar in the new mixture is?

 A.   6 ∶ 4 B.   7 ∶ 5 C.   5 ∶ 7 D.   8 ∶ 5

Correct Answer : B. 7 ∶ 5

Description :
Let amount of the mixture of salt and sugar is 1 kg
Mixture contains 4 parts of salt and 5 parts of sugar
⇒ Amount of salt in the mixture = 4/9 kg
⇒ Amount of sugar in the mixture = 5/9 kg

After replacing 1/4th part with salt,
⇒ Amount of salt in the new mixture = (4/9 – 4/9 × ¼ + 1/4) kg = (16 – 4 + 9) /36 kg = 7/12 kg
⇒ Amount of sugar in the new mixture = (5/9 – 5/9 × 1/4) kg = 5/12 kg
∴ Required ratio = 7/12 ∶ 5/12 = 7 ∶ 5

10. According to an instruction a mixture of colour and Turpentine containing half part of each is perfect for painting a wall. A painter is provided a mixture, 3 parts of which are colour and 5 parts are Turpentine. How much of the mixture drawn off and replaced with colour so that the mixture becomes perfect?

 A.   1/6th B.   1/2th C.   1/4th D.   1/5th

Description :
Let total amount of the mixture be 1 litre
3 parts of the mixture are colour and 5 parts are Turpentine
⇒ Amount of colour in the mixture = 3/8 litre
⇒ Amount of turpentine in the mixture = 5/8 litre
Let amount of mixture removed by the painter be ‘x’ litre
⇒ Amount of colour in the new mixture = (3/8 – 3x/8 + x) litre
⇒ Amount of turpentine in the new mixture = (5/8 – 5x/8) litre
The mixture will be perfect if, Amount of colour in the new mixture = Amount of turpentine in the new mixture
⇒ 3/8 – 3x/8 + x = 5/8 – 5x/8
⇒ 3 – 3x + 8x = 5 – 5x
⇒ 10x = 2
⇒ x = 1/5
∴ 1/5th part of the mixture should be replaced