- Alligation
It 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.
- Mean Price
Mean price is the cost price of a unit quantity of the mixture
- 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/x)n]
- Rule of Alligation
If two ingredients are mixed, then
(Quantity of cheaper/Quantity of dearer)=(P. of dearer – Mean Price)/(Mean price – C.P. of cheaper)
| Cost Price(CP) of a unit quantity of cheaper (c) | Cost Price(CP) of a unit quantity of dearer (d) | |
| Mean Price | ||
| (m) | ||
| (d – m) | (m – c) | |
- => (Cheaper quantity) : (Dearer quantity) = (d – m) : (m – c)
Solved Examples
Level 1
| 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 |
Answer : Option 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.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. 43 | B. 34 |
| C. 32 | D. 23 |
Answer : Option 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 1st Jar | Concentration of alcohol in 2nd Jar | |
| 40% | 19% | |
| Mean | ||
| 26% | ||
| 7 | 14 | |
Hence ratio of 1st and 2nd quantities = 7 : 14 = 1 : 2
| 3.In what ratio should rice at Rs.9.30 per Kg be mixed with rice at Rs. 10.80 per Kg so that the mixture be worth Rs.10 per Kg ? | |
| A. 7 : 8 | B. 8 : 7 |
| C. 6 : 7 | D. 7 ; 6 |
Answer : Option B
Explanation :
By the rule of alligation, we have
| Cost of 1 kg rice of 1st kind | Cost of 1 kg rice of 2nd kind | |
| 9.3 | 10.80 | |
| Mean Price | ||
| 10 | ||
| 10.8-10 = .8 | 10 – 9.3 = .7 | |
Required ratio = .8 : .7 = 8 : 7.
| 4.In what ratio must water be mixed with milk costing Rs.12 per litre in order to get a mixture worth of Rs.8 per litre? | |
| A. 1 : 3 | B. 2 : 2 |
| C. 1 : 2 | D. 3 : 1 |
Answer : Option C
Explanation :
By the rule of alligation, we have
| Cost Price of 1 litre of water | Cost Price of 1 litre of milk | |
| 0 | 12 | |
| Mean Price | ||
| 8 | ||
| 12-8=4 | 8-0=8 | |
Required Ratio = 4 : 8 = 1 : 2
| 5.In 1 kg mixture of iron and manganese 20% of manganese. How much iron should be added so that the proportion of manganese becomes 10% | |
| A. 1.5 Kg | B. 2 Kg |
| C. .5 Kg | D. 1 Kg |
Answer : Option D
Explanation :
By the rule of alligation, we have
| Percentage concentration of manganese in the mixture : 20 | Percentage concentration of manganese in pure iron : 0 | |
| Percentage concentration of manganese in the final mixture | ||
| 10 | ||
| 10 – 0 = 10 | 20 – 10 = 10 | |
=> Quantity of the mixture : Quantity of iron = 10 : 10 = 1 : 1
Given that Quantity of the mixture = 1 Kg
Hence Quantity of iron to be added = 1 Kg
| 6.A trader has 1600Kg of sugar. He sells a part at 8% profit and the rest at 12% profit. If he gains 11% on the whole , find the quantity sold at 12%. | |
| A. 1200 Kg | B. 1400 Kg |
| C. 1600 Kg | D. 800 Kg |
Answer : Option A
Explanation :
By the rule of alligation, we have
| % Profit by selling part1 | % Profit by selling part2 | |
| 8 | 12 | |
| Net % Profit | ||
| 11 | ||
| 12 – 11 = 1 | 11 – 8 = 3 | |
=>Quantity of part1 : Quantity of part2 = 1 : 3
Given that total quantity = 1600 Kg
Hence quantity of part2 (quantity sold at 12%) = 1600×3/4=1200
| 7.How many litres of water must be added to 16 liters of milk and water contains 10% water to make it 20% water in it | |
| A. 4 litre | B. 2 litre |
| C. 1 litre | D. 3 litre |
Answer : Option B
Explanation :
By the rule of alligation, we have
| % Concentration of water in pure water : 100 | % Concentration of water in the given mixture : 10 | |
| Mean % concentration | ||
| 20 | ||
| 20 – 10 = 10 | 100 – 20 = 80 | |
=> Quantity of water : Quantity of the mixture = 10 : 80 = 1 : 8
Here Quantity of the mixture = 16 litres
=> Quantity of water : 16 = 1 : 8
Quantity of water = 16×1/8=2 litre
| 8.A merchant has 1000 kg of sugar part of which he sells at 8% profit and the rest at 18% profit. He gains 14% on the whole. The Quantity sold at 18% profit is | |
| A. 300 | B. 400 |
| C. 600 | D. 500 |
Answer : Option C
Explanation :
By the rule of alligation,we have
| Profit% by selling 1st part | Profit% by selling 2nd part | |
| 8 | 18 | |
| Net % profit | ||
| 14 | ||
| 18-14=4 | 14-8=6 | |
=> Quantity of part1 : Quantity of part2 = 4 : 6 = 2 : 3
Total quantity is given as 1000Kg
So Quantity of part2 (Quantity sold at 18% profit) = 1000×3/5=600Kg
Level 2
| 1.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 |
Answer : Option 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
| 2.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. 7litres, 4 litres |
| C. 6litres, 6 litres | D. 4litres, 8 litres |
Answer : Option C
Explanation :
Let cost of 1 litre milk be Rs.1
Milk in 1 litre mix in 1st can = 3/4 litre
Cost Price(CP) of 1 litre mix in 1st can = Rs.3/4
Milk in 1 litre mix in 2nd can = 1/2 litre
Cost Price(CP) of 1 litre mix in 2nd can = Rs.1/2
Milk in 1 litre of the final mix=5/8
Cost Price(CP) of 1 litre final mix = Rs.5/8
=> Mean price = 5/8
By the rule of alligation, we can write as
| CP of 1 litre mix in 2nd can | CP of 1 litre mix in 1st can | |
| 1/2 | 3/4 | |
| Mean Price | ||
| 5/8 | ||
| 3/4 – 5/8 = 1/8 | 5/8 – 1/2 = 1/8 | |
=> mix in 2nd can :mix in 1st can = 1/8 : 1/8 = 1:1
ie, from each can, 12×12=6 litre should be taken
| 3.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 |
Answer : Option 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 from vessel A | CP of 1 litre mixture from vessel B | |
| 5/7 | 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
| 4.How many kilograms of sugar costing Rs. 9 per kg must be mixed with 27 kg of sugar costing Rs. 7 per Kg so that there may be a gain of 10 % by selling the mixture at Rs. 9.24 per Kg ? | |
| A. 60 Kg | B. 63 kg |
| C. 58 Kg | D. 56 Kg |
Answer : Option B
Explanation :
Selling Price(SP) of 1 Kg mixture= Rs. 9.24
Profit = 10%
Cost Price(CP) of 1 Kg mixture= 100×SP /(100+Profit%) =100×9.24/(100+10)
=100×9.24/110=92.4/11=Rs.8.4
By the rule of alligation, we have
| CP of 1 kg sugar of 1st kind | CP of 1 kg sugar of 2nd kind | |
| Rs. 9 | Rs. 7 | |
| Mean Price | ||
| Rs.8.4 | ||
| 8.4 – 7 = 1.4 | 9 – 8.4 = .6 | |
ie, to get a cost price of 8.4, the sugars of kind1 and kind2 should be mixed in the
ratio 1.4 : .6 = 14 : 6 = 7 : 3
Let x Kg of kind1 sugar is mixed with 27 kg of kind2 sugar
then x : 27 = 7 : 3
⇒x/27=7/3
⇒x=(27×7)/3=63
| 5.A container contains a mixture of two liquids P and Q in the ratio 7 : 5. When 9 litres of mixture are drawn off and the container is filled with Q, the ratio of P and Q becomes 7 : 9. How many litres of liquid P was contained in the container initially? | |
| A. 23 | B. 21 |
| C. 19 | D. 17 |
Answer : Option B
Explanation :
Let’s initial quantity of P in the container be 7x
and initial quantity of Q in the container be 5x
Now 9 litres of mixture are drawn off from the container
Quantity of P in 9 litres of the mixtures drawn off = 9×7/12=63/12=21/ 4
Quantity of Q in 9 litres of the mixtures drawn off = 9×5/12=45/12=1/54
HenceQuantity of P remains in the mixtures after 9 litres is drawn off =7x−21/4
Quantity of Q remains in the mixtures after 9 litres is drawn off =5x−15/4
Since the container is filled with Q after 9 litres of mixture is drawn off,
Quantity of Q in the mixtures=5x−15/4+9=5x+21/4
Given that the ratio of P and Q becomes 7 : 9
⇒(7x−21/4):(5x+21/4)=7:9
⇒(7x−21/4)(5x+21/4)=7/9
63x−(9×21/4)=35x+(7×2/14)
28x=(16×21/4)
x=(16×21)/(4×28)
litres of P contained in the container initially = 7x=(7×16×21)/(4×28)=(16×21)/(4×4)=21
| 6.A dishonest milkman sells his milk at cost price but he mixes it with water and thereby gains 25%. What is the percentage of water in the mixture? | |
| A. 25% | B. 20% |
| C. 22% | D. 24% |
Answer : Option B
Explanation :
Let CP of 1 litre milk = Rs.1
Given that SP of 1 litre mixture = CP of 1 Litre milk = Rs.1
Given than Gain = 25%
Hence CP of 1 litre mixture = 100/(100+Gain%)×SP
=100(100+25)×1
=100/125=4/5
By the rule of alligation, we have
| CP of 1 litre milk | CP of 1 litre water | |
| 1 | 0 | |
| CP of 1 litre mixture | ||
| 4/5 | ||
| 4/5 – 0 = 4/5 | 1- 4/5 = 1/5 | |
=> Quantity of milk : Quantity of water = 4/5 : 1/5 = 4 : 1
Hence percentage of water in the mixture = 1/5×100=20%
WBPCS Notes brings Prelims and Mains programs for WBPCS Prelims and WBPCS Mains Exam preparation. Various Programs initiated by WBPCS Notes are as follows:-- WBPCS Mains Tests and Notes Program
- WBPCS Prelims Exam 2020- Test Series and Notes Program
- WBPCS Prelims and Mains Tests Series and Notes Program
- WBPCS Detailed Complete Prelims Notes