- Easy Wine and Water Mixing Problem
- Two Solutions: Selling Price of Wine and Profit
- Dilution by adding Water in the Wine
- Average Profit / Interest Rate: Jethalal’s Mobile shop
- Average weight of Group : Gokuldham society’s case
- Recommended Booklist
- Previous Articles
Introduction
- Mixture, Alligation and Alloy questions routinely appear in Aptitude exams for Government jobs, Bank PO (IBPS) and· MBA entrance exams (CMAT, CAT).
- The concept is very easy, You can master this concept, after barely 2 hours practice, unless you try to complicate it by yourself by mugging up the formulas from R.S.Agarwal’s (Most bogus) book on Quantitative aptitude.
A typical problem runs like this
- there is a cheap liquid (water), and there is an expensive liquid (Milk, Wine)
- sometimes, instead of liquid, solids are given: rice / wheat of different variety, gold, silver, zinc and iron alloys etc.
- both are given in different quantities, and mixed together. You are asked to calculate the concentration of final mixture or its selling price.
- Or you can be asked to find the amount of water or milk to be added in given mixture to bring the concentration ·to 50% etc.
- The same concept can be extended for finding average speed, average height, interest rates etc.
- Mathematically speaking, this is a problem of “weighted average”
- Let us start with a very basic (and easy) problem
Case: Easy Wine and Water mixture Problem
- you have a big bottle of Blue water 10 lit.
- and you’ve a glass of Pure Red wine 2 lit. (assuming that a glass can be that big !)
- You mix them both, and you get a purple colour solution. (10+2=12 lit)
- (Mixture) what is the concentration of wine in the final mixture?
Step 1: Arrange them in a straight line
- First arrange these three bottles in a straight line, in ascending order of their price / quality / concentration.
- Water is cheapest, and wine is costliest. The Mixture is going to be not as cheap as water and not as costly as wine, so we put it in between these two bottle.
(alternative logic:), since we’ve to find the concentration of wine, : Water bottle has 0% wine in it, so in terms of concentration it has 0% wine. So it goes in the left, wine is pure, so it has 100% wine, and the mixture will have wine between the range of 0 to 100%.
Step 2: Write down the Quality /concentration /price
- On the head (top) of each item, write down its concentration, price, speed (Whatever is given in the sum).
- Since we are asked about wine’s concentration in mixture, it looks like this
- Blue Water =0% wine
- Purple Mixture= m% wine (we’ve to find out)
- Red glass= 100% wine
Step 3: Write down the weight / volume.
Step 4: Divide and Rule!
You don’t need to mug up any formulas for ‘cheaper liquid and expensive liquid’. Just remember this ‘visual-move’
100 minus m = 10 lit…..(i)
Same way

(M minus 0) =2 lit…..(ii)
We cannot do (0 minus M) because M is concentration of wine in the final mixture. It is bigger than 0% and smaller than 100%.
Matter is over. Sum is solved. Game Finished.
Just divide equation (i) from equation (ii) or you can do reverse divide (ii) from (i), you’ll get the value of “m%” in either case. See this image for actual calculation

Final answer= the concentration of Wine in given mixture is 16.66%
Which also means, concentration of Water in this given mixture
=100 – 16.6% ;because conc. Of water + conc. Of wine =100%
=83.4% water.
Ofcourse real exam questions won’t be this easy so lets take a few cases.
Case: Average price and profit after mixing two solutions
16 lit. of Soda is mixed with 5 lit. of Wine. Price of this Soda is Rs.12 / lit and price of wine Rs.33/lit. What is the average price of this mixture.
And if the bartender wishes to make 25% profit on his investment, at what price should he sell this mixture?
Our method remains the same. Arrange them in ascending order, put values on the top and bottom of each item. It’ll look like this

Now do the visual move. And you get two equations
33 minus m = 16
M minus 12 = 5 ; very important. Donot make mistake. M is bigger than 12.
Divide them
(33-m)/(m-12)=16/5
- Manually solving this equation
- (33-m)x5=16(m-12)
- Now you can manually solve this equation to get the value of m, but as You can understand, this can be time consuming method to solve equation because we’ve to multiply 33 with 5 and 16 with 12 and then do addition, subtraction – might make mistake in calculation.
- So better apply the
“Componendo” principle of ratio proportion
Sidenotes
- Answer for average must be between the two extremes: 12 and 33.
- So if you get the answer outside this [12-33] range, know that you’ve made mistake somewhere in calculation.
- Whenever possible, do this componendo method to solve the sum quickly without making mistakes in lengthy multiplications. At time you’ll have to use “Dividendo” principle i.e. same but instead of adding (+) bottom to top, you subtract (-) bottom from top.
Back to the question:
- Price of Final mixture is 17 Rs. Per litre. But this is the ‘cost-price’ for the bartender. HE wants to make 25% profit on this.
- What is 25% of 17? = 17 x (25/100) = Rs.4.25
- So his Selling price = Cost price + profit of 25% =17 + 4.25 = 21.25
Quicker method for profit calculation
- 25% =25/100 = 1/4
- You add this one fourth part to the total one part of given cost price.
- So 1 plus ¼ =5/4 parts.
- Multiply (5/4) with 17 and you get 21.25 = our selling price.
- Final Answer : Bartender should sell this mixture at 21.25 rupees per liter, if he wants to make 25% profit.
- Same way, if he had asked to find selling price for 50% profit, multiply 17 with 3/2. (because 50%=1/2)
Case: Add water to decrease the concentration (dilution)
A 75 liter mixture of wine and water contains 80% wine. How much water should be added to decrease the concentration of wine to 75%?
- We are already given a mixture and we’ve to add water and create a new diluted mixture.
- Assume that purple bottle contains this 75 liter mixture of 80% wine.
- We’ll add ‘v’ liters of pure water into this purple bottle to dilute it and get the middle mixture of 75% concentration.
- The process is same, arrange them in ascending order, and then add values at top and bottom.
Now apply the visual move:
80-75=V lit. –eq.(1)
75-0=75 lit –eq.(2)
Very easy, divide eq. 1 with 2 and you get V=5 liters directly.
Case: Average profit or interest Rate
The owner of Gada Electronics, Jethalal sold total 108 mobiles of two companies last month. Samsung at 36% profit and Nokia at 9% profit. If he made total 17% profit on total sales of these mobile phones, how many Samsung phones did he sell?

It is same wine and water problem, but instead we’ve phones.
108 phones = total volume of final mixture containing wine + water.
Our procedure remains the same, first arrange them in a straight line, in ascending order of their profit (Value or whatever).
Assume that we’ve “V” number of Samsung mobiles
Since total mobiles =108 = Samsung + Nokia,
hence Nokia mobiles = 108-V
Now do the visual method:
36-17=108-V
17-9=V
Divide them and apply componendo principle of ratio

Final Answer= Jethalal sold 32 samsung mobiles.
Which also means he sold 108 minus 32 =76 Nokia phones.
Case: average weight of group
400 people live in Gokuldham society. Average weight of men is 80kg and women is 60 kg. If the average weight of all people combined is 65 kg, how many women live in this society ?
First arrange them in a straight line, in ascending order of their average weight.

- Assume that Number of women = V. and since total residents are 400, men are 400-V.
- The the simple Visual step and division of two equations
- (80-65)/(65-60)=v/(400-v)
- Solve it and you get
- Number of ladies (v)=300.
- This answer seems plausible too, because the average weight 65 is closer to 60, that means more number of women in the weighting scale, so the balance shift towards their side. because Men are only 100.
Case : Average speed.
Will be covered under separate article on TSD (Time, Speed, distance)
For more articles, visit http://www.mrunal.org/aptitude









If 5% more is gained by selling an article for Rs. 350 than by selling it for Rs. 340,the cost of the article is :
a)50
b)160
c)200
d)225
Hi Santosh,
The method is Absolute difference method:
5% is giving Rs 10 Profit
100% will be 200 Rs. ie: Cost Price
Sir, please solve the question with above mention method Q. How many Kg of tea worth Rs. 25/kg must be blended with 30 kg of tea worth Rs. 30/kg, so that by selling the blended variety @ Rs. 30/per kg, there should be gain of 10% ?
a) 36 kg b) 40kg c) 32 kg d) 42 kg.
25 30/1.1 30
—————————————-
x 30
by didvide and rule, x=36
36 kg would be the answer.
Let the qty be X.
First find the cp for the new mixture.
That would be 300/11 as per the formula.
25 300/11 30
X 30+X 30
(30 – 300/11)/ (300/11- 25) = X/30
solving it, you will get X=36 kg.
STRUCTURAL FORM OF ABOVE 2 METHODS, (FOR THOSE WHO STILL HAVE CONFUSION….)
Find the C.P from S.P = 30 & P% = 10%
THEN:
25 30
300/11
X 30
NOW:
(30 – 300/11) divided by (300/11 – 25) = X/30
solve it….
Ans: 36 kg.
I am not getting this answer
a mixture is composed of 3 parts of whisky and one part of water. By adding 12 litres of water the mixture contains whisky twice as much as water. The amount of whisky in the mixture is
1) 76 litres 2) 72 litres 3) 84 litres 4) 80 litres
25% 100/3% 100%
x 12
by divide and rule, x=96 litres, but the whisky is 3/4 *96= 72 litres
Sir can u please explain it in more details ? I didn’t get it right.
First take whisky as 3x & water as x (or 1x). And the final rayio is been given as 2/1. Now 12L of ‘water’ is been added to the mixture. So the water content in the mixture altogether becomes 1x+12.
=> 3x/(1x+12) = 2/1
Now just cross multiply and get the value for ‘x’
multiply it with 3, for the vol. of Whisky
.
.
.
3x/(1x+12) = 2/1
(2x+24) = 3x
24 = x.
.
whisky is taken as 3x
so, 3*24 = 72L.
Ans:- option: 2.
(Hope, my method is right)
can we apply the same logic to solve the following sum
question: two alloys of tin and copper first mix has 93.33% tin and second has 86.66% tin what weight of the first alloy should be mixed with some weight of the second alloy so as to create 50kg mass containing 90% tin
BTW. awesome explanation on how to approach alligations i watched many videos but was not able to grab the concept thankyou for this awesome approach and explanation…..
9:1??
THERE IS VERY EASY METHOD TO SOLVE THIS QUESTION WHICH IS GIVEN JUST ABOVE “JETHALAL” QUESTION
QUES-:A 75 liter mixture of wine and water contains 80% wine. How much water should be added to decrease the concentration of wine to 75%?
STEP 1
IN 75 LITERS 80% IS WINE SO SOLVE IT
75X80/100 = 60 LTR
THEN YOU WANT TO MAKE IT 75% BY ADDING SOME WATER. SUPPOSE WE ADD “X” LTR IN 75 LTR. WHICH IS WHEN MULTIPLY TO 75% ,WHICH WILL BE IN ANSWER, QUANTITY OF WINE (WHICH IS SAME ALL THE TIME) IS 60 LTR…..
60 LTR. = 75+x* 75/100
JUST SOLVE IT AND ANSWER IS “5”
I think u can follow this step too for getting the answer if the choice given are whole nos. and are wide apart (u’ll get an approx. ans in a few seconds)….
.
80% of 75 = 60
75% of 75 = 56.25
if 60 units is to be diluted to 56.25, obviously u have to add water. so volume of water will be approximately 60 – 56.25 = 3.75
u will be having an ans. choice of 4 or something around 4 max upto 5 (not more than that) in options. And it will be the answer.
(PLEASE DON’T FOLLOW THIS METHOD IN ALL CASES…. UNLESS U HAVE YOUR CHOICES SO WIDE (LIKE: 1, 4, 8, 10 and all…) ).
[Aptitude Q] Mixture and Alligiation: Change Alcohol concentration from 15% to 32%
link missing
Spectular Job!..u made it so easy to solve such complicated and time consuming questions. Thanks a lot for this incredible work!
Can anyone suggest hindi medium book for ecology and biodiversity
Thank you sir…
There was 60 litres of pure milk in a vessel. 18 litres of pure milk was taken out and replaced with equal amount of water. Again 24 litres of mixture was taken out and replaced with equal amount of water. What is the amount of water in the vessel now ? (in litres) how to solve this problem
34.8 ltr water,25.2 ltr milk remaing in the mixture
FOR 1st PROBLEM YOU MAY TAKE QUANTITY OF WINE = 2/12 * 100= ANS
Please send answer for ramesh sir question…i have also doubt in that question…plz help…
A can contains a mixture of two liquids A and B in the ratio 7:5. When 9 litres of mixture are drown off and the can is filled with B, the ratio of A and B becomes 7:9. How many litres of liquid A was contained by the can initially?
a. 10 b. 20 c. 21 d. 25
(please explain) How to solve this problem by your method?
A can contains a mixture of two liquids A and B is the ratio 7 : 5. When 9 litres of mixture are drawn off and the can is filled with B, the ratio of A and B becomes 7 : 9. How many litres of liquid A was contained by the can initially?
A. 10 B. 20
C. 21 D. 25
Thanx in Advance !
two alloys of chromium have different percentage of chromium in them. first one weighs 6 kg and second one weighs 12 kg. one piece of equal weight was cut off from the alloys and the first piece was alloyed with the second alloy and the second piece alloyed with the first alloy. as a result, the percentage of chromium became the same in the resulting two new alloys. what was the weight of each piece cut off?
whats the answer
Can anyone solve & explain the logic for the below pbm ?
Soya bean oil costs Rs.90 per kg. After mixing inferior quality oil of Rs. 50 per kg with it, the shopkeeper sells the mixture at Rs. 96 per kg, thereby making a profit of 20%. In what ratio does the shopkeeper mixes the two?
answer- 1:3 (20% profit in 96/-= selling price-80/-)
ratio= (90-80)/ (80-50)= 10/30= 1/3
1:3 is dis d solution
plz……….. explain how to solve below questions?
1)There are two vessels of milk of different prices with the volume of 220 ltr and 180 ltr respectively. Equal amounts of milk were poured off simultaneously from the two vessels and the milk poured from the first vessel was poured into the second vessel and the milk poured off from the second vessel was poured into first vessel. Then the price of milk in both the vessels becomes the same. How much milk was poured from one vessel into the other?
2)Two-fifths of the volume of the mixture of the milk and water is of the ratio of 4:3 is converted into a mixture of ratio of 5:3 by the replacement with milk. The 4:3 mixture was prepared from 4:1 mixture by the method of addition of the substances of the mixture. If the replacement volume is 14 litres, what is the volume of substances added?
3)From a 3:5 solution of milk and water, 20% is taken out and replaced by milk. How many times should this process be done to make the ratio of milk to water as 17:8?
This article z indeed useful…now mixtures and alligations are no more a problem…thanx everyone out dere for ur wonderful posts
4L are drawn from a container full of milk and is,then filled with water. this operation is performed three more times. The ratio of the quantity of milk left in the container and that of water is 16:65 how much milk did the container hold initially? this question is in arihant sbi po book solution is also there but I cant get it how they solved it if any short trick or explaination please suggest
thanks…very informative and made easy
2orngs 3bnanas 4appls cost Rs15.
3orngs 2bnanas 1appls costRs 10.
Hari bought 3orngs 3bananas 3appls. how much did he pay
explain this problem
Two vessels A and B contain spirit and water in the ratio 5:2 and 7:6 respectively. find the ratio in which these mixtures be mixed to obtain a new mixture in vessel C containing spirit and water in the ratio of 8:5.
and the options for above mentioned problem are:
a 4:3
b 3:4
c 5:6
d 7:9
Sir, I am confused on which parameter to take in top and in bottom. For example, Gokuldham society case if I take Number of people in top and weight in bottom row I am getting a wrong answer.
Residents 400-V 400 V
gender Women Mixed Men
Weight 60 65 80
(You have taken weight at the top and number of residents at the bottom)
Pls point out where am I going wrong?
if in question of mobile phones, loss is occured on nokia phone then how will solve the same problem??
Please! please solve it
Ther are 2 vessels P and q.
P containing 120 liter of milk and Q containg 120 liters of water.
In first operation 30 liter is removed from P and poured in Q and then 30 liter is poured back from Q to P.
Like this one more operation take place i.e transfering 30 liter from P to Q and then 30 liter is poured back from Q to P.
What is the ratio of milk and water in P finally
ANSWER:- 17:8
Sir thanks a lot for the great article anfld saving my time! You are simply brilliant. You sir should write your own book.
A barrel contains a mixture of wine and water in the ratio of 3:1.How much fraction of the mixture must be drawn off and substituted by water so that the ratio of wine to water in the resultant mixture in the barrel becomes 1:1 ?
Two alloys A and B contain gold and silver in the ratio of 1:2 and 1:3 respectively. A third alloy C is formed by mixing A and B in the ratio of 2:3 . Find the percentage of silver in the alloy C.