- 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 www.mrunal.org/aptitude

## 262 Comments on “[Aptitude] Alligations,Mixtures,Alloys: Water,Milk,Wine mixing: Weighted Average Made Easy for CSAT,CAT,CMAT,IBPS Aptitude”

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.

2/5 * (2/3 + 3/4) = 17/30

2*(1:2) +3*(1:3)=5:13

silver=13/18 =72.2%

hello sir,

sir in this article i am not able to see the images that u provided regarding divide and rule,so plz sir correct it as soon as possible i will b thankful 2 u nd god bless u for providing such great guidance without any cost,love u sir :)

What is your Message? Search before asking questions & confine discussions to exams related matter only.

pls provide shortcuts for geometry question

i am not able to see the images. please resolve it asap.

Images not loading. Please make it right ASAP

The images in the solutions explained are not visible at all. I tried explorer, chromer and mozilla. If it is explianed somewhere, give me that link pls.

17:34

hmm and percent would be 71.5 i think

A butler stole wine from a butt of sherry which contained 80% of sprirt and he replaced it by wine containing only 32% sprirt.Then , the butt was of 48% strength only.How much of the butt did he steal?

is it 2/3 ?

the strength of good wine = 80% (given)

the strength of bad wine = 32% (given)

mean strength = 48% (given)

So, post mixing, the ratio, Quantity of good wine : Quantity of bad wine = (48 – 32) : (80 – 48) . This comes to 16 : 32 Or simply 1 : 2. But what do we understand by 1 : 2 ? It simply means that finally in the butt 1 part is good wine while 2 parts is bad wine. Therefore, earlier 2 out of 3 parts of good wine in the butt must have been replaced by bad wine, leaving behind only 1 part of good wine, which still remains. So, the answer is 2 : 3

# with allegation

80——32

—-48——-

16:32

1:2

or 2/3

Mrunal Sir,

Can we solve problems when item gets replaced using the same method? If yes, how do we do so?

Question below:

A jar contains a mixture of two liquids A and B in the ratio 4:1. When 10L of the mixture is replaced with liquid B, the ratio becomes 2:3. How many litres of liquid A was present in the jar earlier?

Ans: 16L

Suppose initially contains 4x and 1x as A & B.

Quantity of A in mixture= 4x-4/5*10= (4x-8)L

Quantity of B in mixture=1x-1/5*10=(x-2)L

4x-8/(x-2)+10=2/3

12x-24=2x+16

12x-2x=16+24

10x=40

x=40/10

x=4

hence in A there is 4x….put value of x in it…4*4=16

Sir, im not able to see the images, hence facing problem in understanding the concept

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?

allegation

3/8———1

——-1/2—

1/2:1/8

4:1

or 1/5

if you consider the removed content to be 8 ltrs (considering its simpler for calculations, given ratio being 3:5), you come to a conclusion that initial mixture was 16 litres and the removed liquid is indeed 8 liters! so if options are given, this could be a quick one.

Two vessels contains mixture of spirit and water,,,in fst vessel the ratio of spirit to water is 8:3 & and in second vessel 5:1. if 35 litre is filled with this,,,it contains 4:1 mixture..how many litres are taken from first vessel????

8/11———5/6

——-4/5———

1/30——-4/55

11:24

35=35l

or first vessel 11 L

Two vessels contains mixture of spirit and water,,,in fst vessel the ratio of spirit to water is 8:3 & and in second vessel 5:1. if 35 litre is filled with this,,,it contains 4:1 mixture..how many litres are taken from first vessel????

problem with images

Sir plz answer it

A milkman mixes water with milk that costs 60 rs . he sells it at 75 making a profit of 37.5% . in what ratio had he mixed milk and water

M—————–W

60——————0

—–600/11———-

600/11——60/11

600:60

or 10:1

Sir please rectify the problems with images… 🙄🙄🙄😔😔

hello sir, i am unable to see the images. i tried by opening in different browsers(chrome, explorer) but there is no use. kindly resolve the error please.

Mrunal sir, the link you have given for the images is not working, what should i do now to open it?

is this answer ok for this que.

(5-(5x/8))/(3+(5x/8)) equals to 1/1

then by solving….

x equal to 8/5….comment pz

Images not showing???

Sir, the links given in the article for images are not working and causing a great difficulty in understanding the problem. Please rectify it.

TWO LIQUIDS X AND Y ARE MIXED IN THE RATIO OF 3:2 AND TE MIXTURE IS SOLD AT RS 11 PER LITRE AT A PROFIT OF 10% TFA E LIQUID X COSTS 2 RS MORE PER LITRE THAN Y THE COST OF X PER LITRE IS?

8.80rs

sir it is not displaying images.?

Images can’t be seen. Please provide an alternative.

Images are not pinged ….

image not dispalyed sir , plzz provide alternative any source sir.

If there is 60 litrs of milk in a container.when one take out 6 litr milk from it and add 6 litr water to it repeat this process one more time and after that show the ratio of water and milk in the container. Plz show me the solution

81:19

A solution contains alcohol and water in the ratio 3 : 1. 16 litres of the solution is drawn and 12 litres of water is added. Now 11 litres of the mixture is replaced by 11 litres of the water. If the final ratio of alcohol to water in the solution is 9 : 13, find the initial quantity of the solution.

Hi SIr,

Is there any plan to fix the problem with image? Eagerly waiting for it to get fixed.

Please rectify the image issue. Not able to see any image and hence not able to understand the concept :( :( :(

the ratio of milk:water in a 50 litre mixture is 7:3.howmuch more water should be added to flip the ratio.

the images don’t work