- 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%
Final Destination for Nagaland PSC Notes and Tests, Exclusive coverage of NPSC Prelims and Mains Syllabus, Dedicated Staff and guidence for NPSC NPSC Notes brings Prelims and Mains programs for NPSC Prelims and NPSC Mains Exam preparation. Various Programs initiated by NPSC Notes are as follows:-- NPSC Mains Tests and Notes Program
- NPSC Prelims Exam 2017- Test Series and Notes Program
- NPSC Prelims and Mains Tests Series and Notes Program
- NPSC Detailed Complete Prelims Notes