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
8 liters of wine is replaced by water from a pot full of wine and repeated this two more times. The ratio of the wine :water left in pot is 8 : 19. How much wine was there in the pot originally?
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Correct Answer : A. 24
Description :
Let initial quantity of wine =x litre After a total of 1+2=3 operations,
quantity of wine =x(1−y/x)n =x(1−8/x)3
⇒x(1−8/x)3 x=8/27
⇒(1−8/x)3=(2/3)3
⇒(1−8/x)=2/3
⇒x=24
View Answer
50/50
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Question 2
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?
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Correct Answer : A. 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
Question 3
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:
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Correct Answer : C. \( {1 \over 8} \) 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 4
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.
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Correct Answer : B. 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 5
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?
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Correct Answer : C. 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
Question 6
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.
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Correct Answer : A. 29.16 litres
Description :
Amount of milk left after 3 operations
\( {(40(1 − {4 \over 40})^3)} = (40 \times {9 \over 10} \times {9 \over 10} \times {9 \over 10}) \) = 29.16 litres
Question 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.
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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 8
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?
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Correct Answer : A. 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 9
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?
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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
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?
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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