Over the last few weeks, we have started writing content again for the blog. A lot of our readers asked us to write something on Venn Diagrams based Maxima-Minima questions. Well guess what, we have come up with something which would provide you with some insights.
Disclaimer
This is a post on the advanced techniques and question types that one comes across in Venn Diagrams. For those of you who are not very well versed with the concept of Venn Diagrams, I would advise that you wait for the prequel that I would be penning down soon. The basic assumption while writing this post is that readers are fairly confident of plugging in data in two factor and three factor Venn Diagrams.
Now, the funny thing here is that, these questions are not very difficult to understand or to solve using dear old hit-until-you-break-through method. But the accuracy in these is ridiculously low compared to other straightforward DI sets. There is no formula as such (well there is something with variables but I won’t exactly call it a ‘formula’ or a ‘theorem’) which makes it a tad difficult to explain to students and for students to remember. I have tried to keep this a bit theoretical, with a few examples and not make this a 20 questions-and-answers compendium. Hope that it is of some help to all of you.
Venn Diagrams based Maxima-Minima questions for CAT 2016
The difference between participants and incidences
This is the cornerstone of many a maxima and minima based set. Say there is a bunch of 10 kids few of whom have Buzz Lightyear action figures and few who have Sheriff Woody toys. Let’s say there are 4 kids with Buzz Lightyear action figures and 7 with Sheriff Woody toys. Let the kids with both these toys be 3 in total. The representation would be as follows:
The participants here are the 10 kids that I have mentioned in the scenario. The number of incidences is nothing but unique participant -> object pair that we can make. In this scenario, it simply translates to the number of toys that can be distributed. This is an important distinction and one needs to understand this fully. If you haven’t understood this, let the 10 kids be A, B, C, D … J. The one with only Buzz Lightyear be A, the ones with only Sheriff Woodys be B, C, D and E and the ones with both the toys be F, G and H. As is obvious, I and J do not have any toys. So, if you are someone who is distributing these toys, you would have to purchase 1 + 4 + 2*3 = 11 toys in total.
In other words, if the participants are defined as:
a – Number of kids with exactly 1 toy
b – Number of kids with exactly 2 toys
x – Number of kids with no toys then,
a + b + x = 10 and
a + 2b = 11
For a three factor figure, it would be a + b + c + x and a + 2b + 3c. Generally, you will not have anything in x (in short, everybody gets at least one item) and so, it will be easy to manipulate the remaining variables.
Two factor Venn Diagrams based Maxima-Minima questions
Let’s start with a simple example.
In a survey it was found that 80% like Friends whereas 70% like HIMYM. If all the people surveyed like at least one of the two shows, how many people like both the shows?
According to the above mentioned concept,
Let a be the number of people who like exactly 1 show
Let b be the number of people who like both the shows
So, we get
a + b = 100
a + 2b = 150
Remember that we are talking in terms of percentages here. We can also take numbers (say 100 or 200) and then use them. Just remember to keep it constant throughout and do not take percentages at one place and absolute values at another.
Solving the two equations, we get b = 50 and a = 50. Hence, we can easily say that the number of people who like both the shows is b = 50%.
Simple till now? Let’s try the next one.
In a survey it was found that 80% like Friends whereas 70% like HIMYM. What is the maximum and minimum number of those who like both?
Now this is similar to the first question minus a small phrase. This will bring the variable x into the picture.
Let a be the number of people who like exactly 1 show
Let b be the number of people who like both the shows
Let x be the number of people who do not like either of the shows
Hence,
a + b + x = 100
a + 2b = 150
b – x = 50 and a + 2x = 50
Now, we have to find the minimum possible value of b. To minimize b, we have to maximize a and consequently minimize x (as b = x + 50). So, if we look at a + 2x = 50, maximum value that a can take is 50 and x is 0. So, b minimum will be 50 as well.
To find the maximum value of b, we will simply have to get the lowest of the percentages. So, there could be a 70% overlap with the 10% of the people who like Friends going into the part marked by an ‘f’.
Let’s take it another step forward
80% of the CAT aspirants spend time solving Quant whereas 60% spend time solving Verbal. If only those aspirants who spend time solving both the sections crack CAT, what is the maximum and the minimum percentage of the aspirants who will not crack CAT?
Now, we can skip the diagram altogether.
Let a be the percentage of aspirants who spend time on exactly one section
Let b be the percentage of aspirants who spend time on both sections
Let x be the percentage of aspirants who do not spend time on either of the sections
a + b + x = 100
a + 2b = 140
b – x = 40 and a + 2x = 60
To minimize b, we need to minimize x and so, maximize a. So, a = 60, x = 0 and b = 40%
To maximize b, we need to find the smaller of the two values i.e. 60%
Now, we go for getting the complement of the above findings. So, the minimum percentage of aspirants who will crack CAT is 40 and so, the maximum amount of people who won’t crack CAT will be 60%. Similarly, the minimum amount of people who won’t crack CAT will be 40%.
I hope it’s a bit clear now. Let’s move on to 3 sector Venn Diagrams.
Three factor Venn Diagrams based Maxima-Minima questions
Let’s consider an example here:
In a survey it was found that 40% like tea, 50% like coffee and 60% like milk. Every person likes at least one of the three items tea/coffee/milk. What are the maximum and minimum possible values of those who like all three?
Similar to what we did above, in case of three factor Venn Diagrams, we will have three types of participants:
a who drink only one item
b who drink exactly two items
c who drink all three items
a + b + c = 100
a + 2b + 3c = 40 + 50 + 60 = 150
So, b + 2c = 50 and a – c = 50
Now, to find the minimum possible value of c, we have to maximize b which will be 50% and so, minimum value of c will be 0.
To find the maximum possible value of c, we have to minimize b and so, the maximum value of c will be 25. Remember that each and every person likes at least one of the three items and so, there is no variable x as was there before.
Let’s look at another question.
There are 3 electives offered to the students class of 92 (the students have a choice of not choosing any electives). 60 students opted for marketing, 80 for finance and 50 for systems, 40 students opted for both marketing and systems, 50 for both marketing and finance and 45 for both finance and systems. What is the maximum and the minimum possible number of students who opted for all 3 electives?
This is easy as we already know the overlaps between two sectors. Diagrammatical representation of the above data would be:
Now, we will assume variables as above:
a be the number of students who took only 1 elective
b be the number of students who took exactly 2 electives
c be the number of students who took all 3 electives
x be the number of students who didn’t take any electives
a + b + c + x = 92
a + 2b + 3c = 190
b = 135 – 3c or b + 3c = 135
So, a + b = 55 and c + x = 37
So, maximum value of c will be 37.
Minimum value of c will be given if we fill the remaining parts of the diagram. So, only marketing will c – 30, only finance will be c – 15 and only systems will be c – 35. So, c has to be 35 at minimum.
Cases with at least and at most in Venn Diagrams based Maxima-Minima questions
According to a survey, at least 70% of people like apples, at least 75% like bananas and at least 80% like cherries. What is the minimum percentage of people who like all three?
Now, in this case the maximum extent of overlap will of course be 100% (as everything is in terms of at least and so, the maximum limit for each case is 100%). So, whenever at least is given to you, minima will be asked. And to get to minima, you have to assume the values to be at their least.
Again, we assign variables as always
a be the number of people who like one fruit
b be the number of people who like exactly two fruits
c be the number of people who like all three fruits
a + b + c = 100
a + 2b + 3c = 225
b + 2c = 125
Also, b + c <= 100
We can observe that if b + c goes below 100, c starts increasing. So, ideally b + c should be 100. Subtracting it from the first equation, we get c = 25.
Probably an easier way to do it is by distributing the values and making sure that the overlap is as less as possible. So, let the people be ranked from 1 – 100. Ranks 1 – 70 like apples. Now, to make sure that the overlap is as less as possible, we say that Ranks 71 – 100 like bananas and distribute the remaining people who like bananas from Ranks 1 – 45. Now, we have 45% people who like both apples and bananas. Let the remaining 55% like cherries. So, we are left with distribution of 25% people who like cherries which will be in the zone that has people who like both apples and bananas. So, 25% is the minima. Diagrammatically speaking, we get:
Venn Diagrams based Maxima-Minima questions – Multiple sectors
A survey was conducted among a group of 50,000 people about preferred newspapers. The following were the results: 90% liked Times of India, 85% liked Economic Times, 80% liked Hindustan Times, 82% liked Mumbai Mirror, 75% liked Daily News and Analysis, and 4000 did not like any of these newspapers. Find the minimum number of people who liked all the five newspapers.
Again, the graphical method would work fine here. 85 split as 2 and 83, 80 as 9 and 71, 82 as 21 and 61, 75 as 31 and 44. So, it should be 44% or 22,000.
Also, we can use the algebraic method from above:
a + b + c + d + e = 92
a + 2b + 3c + 4d + 5e = 412
b + 2c + 3d + 4e = 320
Now, if d were 92, the total would have been 4*92 = 368. But, the total is 412 which is 44 greater than what we have got. Now, if we push one percent from d to e, the equation will become 4*91 + 5*1 = 369, if we push two percent, we get 4*90 + 5*2 = 370 and so on.
So, for every item pushed from d to e, we get an increment of 1. We have to bring it up by 44 and so, we need to transfer 44% to e. So, minimum e will be 44% or 22,000.
I hope it clears a few doubts regarding Venn Diagrams based Maxima-Minima questions. You may use either of the above methods (graphical or algebraic) to solve Venn Diagrams based Maxima-Minima questions. Trial and error is something that I am not a big fan of, personally as it leads to a lot of unnecessary confusion and there is no coming back from a blind spot in case you hit it.
I will be covering a few basic concepts on two, three and four factor Venn diagrams for those who find it difficult to plot values and form equations in direct questions.
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