- 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









Mixture contain 99% milk and 1% water. By what % milk should be taken out from the mixture so resultant mixture contains only 98% of milk only ?
Mixture contain 99% milk and 1% water. By what % milk should be taken out from the mixture so resultant mixture contains only 98% of milk only ? Help me solve it by shortCut
hi
Plz tell hw to solve this problem with the help of able technique.
A vessel is filled with liquid, 3 parts of which are water and 5 parts syrup. How much of the mixture must be drawn off and replaced with water so that the mixture may be half water and half syrup?
1. 1/3 2. 1/4 3. 1/5 4. 1/6
thnx a lot………..grt job
Thank you
how to solve :
A dishonest milkman professes to sell his milk at cost price but he mixes it with water and thereby gains 25%. The percentage of water in the mixture is
by mixing two varities of tea costing rs 65 and rs 115 per kg and selling the mixture at the rate of rs 100 kg , a seller made a profit of 15% in what ratio did he mix the two varities options a)101:29 b)100:113c)129:101
by above mentioned method
how this method can be used when ratios are given..Example..In a mixture of milk and water of volume 30 litre the ratio of water and milk is 3:7.How much quantity of water will be added to the mixture to make the ratio of milk and water 1:2…
Excellent Concept bro
A 4OL mixture of milk and water contains 10% water. HOW much water must be added to this mixture so that it contains 20% water?….sir please explain with simple way…
dint understand the 3rd case i.e. add water to decrease concentration
any on plez explain
which part u didn’t u/s ?
80%-75%=v
75%-0=75lit
as the value of can be calculated with frst eqn and the sec eq i.e 75-0=75 has no “v” so “v” will be calculated by the eq 80-75=v and v comes out to be 5
Reply to archana about 3rd case: add water to decrease concentration
1.read question properly that there is already mixture of water and wine that is 80% and is 75ltr.
2.Now we have to make a different mixture which says decrease concentration to 75% it means we have to add water instead of wine
3.concentration/item/volume=[(0/water/v) (75/new mixture/75+v) (80/old mixture/75)]
4. Now write eqn. = [(80-75)/(75-0) =(v/75)]
5. implies: v(volume of water )=5
good
1.
A vessel is filled with liquid, 3 parts of which are water and 5 parts syrup. How much of the mixture must be drawn off and replaced with water so that the mixture may be half water and half syrup?
A. 1/3
B. 1/4
C. 1/5
D. 1/7
ans is 1/5
water 3 part & syrup 5part
total 8 part
let us take out X amount of mix
so now in (8-X) we have 3(8-x)/8 amount of water
now add x amount of water (as the amount of mixture removed is equal to the amoun tof water added.
so the new ration for the water is {3{8-X)/8 + X}/8 =1/2
from here X=1.6
SO the part of the mixture removed is 1.6/8 = 1/5 (ans)
in our visual method
I don’t understand this answer.
please, explain this answer with visulize method.
c
The table in Step IV (Divide and Rule) is missing. Please fix.
Actually helpful.
An ore contains 5% of aluminiun. To get aluminium, the quantity of the ore required is
A.520 kg
B.1040kg
C.780kg
D.1560kg
E.260kg
??
please, complete the question.
Sorry
To get 50 kg aluminium the quantity of ore required is
Sorry
To get 52 kg aluminium the quantity of ore required is
1040 Kg
Assume total 100kg ore,
out of which 5% of al, take 5% as 5kg
for 5kg of Al we need 100kg ore.
For 52kg of Al we need xkg ore
(52*100)/5=1040kg of ore
A bottle is full of dettol. One third of it is taken out and then an equal amount is poured into the bottle to fill it. This operation is done four times. Find the final ratio of dettol and water in the bottle
A.49:65
B.65:16
C.9:16
D.16:65
E.9:25
16:65
May I know solution for this??
prabhat ..you got the approach to such question ?? Please help ..
For such question formula to be used is
reduced concentration = Initial concentration (1 – (amount replaced each time / total amount) )^n
where n= no. of process
So, applying this formula in the question,
let initial concentration of dettol is ‘X’
reduced concentration of dettol after 4 process = X(1-(1/3))^4
= 16X/81
concentration of water after 4 process = X-16X/81
= 65X/81
Ratio of dettol to water = (16X/81):(65X/81)
= 16:65
D
sir how can i solve problem like this :-A DISHONEST MILKMAN PROFESS TO SELL HIS MILK AT COST PRICE BUT HE MIXES IT WITH WATER AND THEREBY GAIN 25%. THE PRECENTAGE OF WATER IN THE MIXTURE IS
Suppose,
Actual price of milk=100Rs
For 25% gain, actual price of mixture should be = 80Rs [ which is sold for Rs 100]
Let there is total 100Lit mixture , out of which x Lit is water, then equation becomes
100-80/80-0= x/(100-x)… x=20
so % of water in mixture is 20%.
a previous comment show a similar question in a simple manner..the ratio of water to milk can be obtained by converting the profit % into fraction, here, 1:4. Thus, if the total mixture is 100 l, then water = 20 l or 20% for that matter..:)
Divide 25 by 100 and you will get the ratio of water and milk
Water/Milk =1/4
Now Water = 1/5 *100 =20%
let CP=1
gvn SP=CP=1
allegation
M————–W
1—————0
——4/5——
4/5———1/5
4:1
1/5*100=20%
A vessel is filled with liquid which has 3 parts water and 7 parts milk . How much part of liquid must be drawn off and replaced with water so that mixture will have 50 % milk
ans is 2/7
water 3 part & milk 7part
total 10 part
let us take out X amount of mix
so now in (10-X) we have 3(10-x)/10 amount of water
now add x amount of water (as the amount of mixture removed is equal to the amount of water added.
so the new ration for the water is {3{10-X)/10 + X}/10 =1/2
from here X=20/7
SO the part of the mixture removed is 20/7/10 = 2/7 (ans)
alternative:
(1-x/y)*7/10=1/2(i.e. 50%)
=> 1-x/y = 5/7
=> x/y = 2/7
allegation
3/10————1
———–1/2—-
1/2————-2/10
5:2
or 2/7
Mrunal sir, this article is extremely helpful. Before reading this article i was afraid of attempting aligations/mixtures questions…now i will be looking out for them :D
a grain dealer cheats to the extent of 10% while buying as well as selling by using false weights . what is his total percentage gain?
overall profit: 22.22%
CP=100/11
SP= 100/9
(100/9-100/11)*100/(100/11)=22.22%
CP=90
SP=110
P%=110-90/90*100
22.22%
Be your own boss with pleasure builder dot com
Thnx sir
If with five shirts three shirts are free ? how much total discount i would get ?
actual no of shirts(CP)=8
price paid for (SP)=5
Discount %= (8-5)/8=37.5%
Sir, the link for 2nd part of this article on Alligations is not working. Plz post it again
https://mrunalmanage.wpcomstaging.com/2012/07/aptq-wine-water.html
A vessel is filled with liquid, 3 parts of which are water and 5 parts syrup. How much of the mixture must be drawn off and replaced with water so that the mixture may be half water and half syrup?
help pls
Answer = 1/5 is it correct?
Mixture drawn is 1.6 and part of mixture drawn is 1/5
1/5
Consider the liquid is of 8L (3+5). And 1/1 be the final ratio. Then,
out of 3/5 some liquid (x) is taken off and water (‘x’ qnty.) is added (liq. taken off n water will be of same quantity (x) ).
so, 3/(5+x) = 1/1
now just cross multiply…..
3 = 5+x
and x = 2 (approx)
.
qnty of liq. taken off (drawn) = a value around 2 (units).
a barrel contains a mix of wine and water in ratio 3:1.how much fraction of mixture must be drawn off and substituted by water so that the ratio of wine andwater in the resultant mixture in the barrel becomes 1:1. my ans is 4/3 .but in book it is 1/3 .please explain
let it be 100% so
Syrup Water
75 25
X
50 50
________________________
75 – 50 = 25
Formula = Abs. Change / Initial Value = 25 / 75 = 1/3
allegatn
1/4————1
——–1/2—–
1/2———1/4
4:2
2:1 or 1/3
A person purchase 90 clocks & sells 40 clocks at a gain of 10% and 50 clocks at a gain of 20%.If he sold all of them at a uniform profit of 15%,then he would have got Rs. 40 less.The CP of each clock is:
a)50
b)60
c)80
d)90
There are two methods to Solve :
Let the Cost price be X.
so 40 at 10% = 40+40*10% = 44x
and 50@ 20 % = 50 + 50*20% = 60X
______________________________________
Total cp of 90 = 104 x
______________________________________
if sold 90 at 15% then = 90 + 90*15% = 103.5x
diffrence = .5 x =40 rs
x = 80 Rs
Method 2:
Theory is given as bakchodi 40@ 10% and 50@20% so see it as : 40@ 10% + 40@20% + 10 @ 20% so he sold 80 clocks @15% and 10 at 20%. So difference is only for 10 clocks.
Now calculate for 10. diffence is 20% – 15% = 5%
5% = 40
100% = 800 for 10 clocks
80Rs.