Wednesday, April 4th, 2018
Questions based on Set Theory can get immensely complicated when you add conditions based such as ‘At least 2’ ‘At most 3’ etc. in them. To be honest, you should be absolutely comfortable with questions dealing with 3 sets and know enough to draw the Venn-diagram in case of 4 sets. We dealt with the idea of Minima and Maxima in Set Theory in this post.
Beyond that not much is required, at least for the CAT exam.
However, there is one particular type of question that can be asked in MBA entrance exams and that is what this short Blogpost is all about. So, let’s start
Question 1: 70% of the people like Coffee, 80% of the people like Tea; then at least what % of people like both?
Answer: 70 + 80 – 100 = 50 %
The reason behind this is pretty straight forward. If you consider a total of 100 individuals, 70 of them like coffee. The remaining 30 don’t like coffee. Let us assume that all 30 of these like tea but we still need 50 more people to like tea. Where are those 50 going to come from? They are going to come from the coffee drinkers because there is no one else left. The surplus of 50 needs to be adjusted inside the coffee drinkers. That is the reason that 50% of people like both Tea and Coffee.
Question 2: 70% of the people like Coffee, 80% of the people like Tea, 85% of the people like Milk; then at least what % of people like all three?
Answer: Least % of people who like both Tea and Coffee = 70 + 80 – 100 = 50 %
Least % of people who like Tea, Coffee and Milk = 50 + 85 – 100 = 35%
The explanation for the first part is similar to what we had discussed in question 1. Now, on one side (let’s call it the left) we have 50 people who like both Tea and Coffee and on another side (let’s call it right) we have 50 people who do not. We need a total of 85 who like milk. We can pick them from anywhere but our end goal is to have the number of people who like all three (Tea, Coffee, and Milk) to be minimum. So, what we are going to do is that we are first going to select the 50 people on the right (who do not like both Tea and Coffee). We still need 35 more people. These 35 people are going to come from the left (who like both Tea and Coffee). So, whatever arrangement we make – we are going to end up with at least 35 people who like all three.
Another method to solve this:
Least % of people who like both Tea and Milk = 80 + 85 – 100 = 65 %
Least % of people who like Tea, Coffee and Milk = 65 + 70 – 100 = 35%
I have shown this method just to highlight the fact that the order we take is of no significance.
Another method to solve this:
Least % of people who like tea, coffee and milk = 70 + 80 + 85 – 2*100 = 35%
This is a shortcut that we can use to get the answer to such questions quickly. All you need to do is to add up the values individually and subtract (n-1) times 100 from it to get the result. Let’s take another example to highlight this method.
Question 3: 70% of the people like Coffee, 80% of the people like Tea, 85% of the people like Milk, 90% of the people like Vodka, 95% of the people like Whisky, 98% of the people like Beer; then at least what % of people like all six?
Shortcut method discussed above: Least % of people who like all six = 70+80+85+90+95+98 – (6-1)*100 = 18%
Let’s solve it properly.
Least % of people who like Coffee and Tea = 70 + 80 – 100 = 50
Least % of people who like Milk and Vodka = 85 + 90 – 100 = 75
Least % of people who like Coffee, Tea, Milk, and Vodka = 50 + 75 – 100 = 25
Least % of people who like Whisky and Beer = 95 + 98 – 100 = 93
Least % of people who like all six = 25 + 93 – 100 = 18%
Once again, the order does not matter. You could have taken these in any order that you liked.
Question 4: Applicants for the doctoral programmes of Ambi Institute of Engineering (AIE) and Bambi Institute of Engineering (BIE) have to appear for a Common Entrance Test (CET). The test has three sections: Physics (P), Chemistry (C), and Maths (M). Among those appearing for CET, those at or above the 80th percentile in at least two sections, and at or above the 90th percentile overall, are selected for Advanced Entrance Test (AET) conducted by AIE. AET is used by AIE for final selection.
For the 200 candidates who are at or above the 90th percentile overall based on CET, the following are known about their performance in CET:
1. No one is below the 80th percentile in all 3 sections.
2. 150 are at or above the 80th percentile in exactly two sections.
3. The number of candidates at or above the 80th percentile only in P is the same as the number of candidates at or above the 80th percentile only in C. The same is the number of candidates at or above the 80th percentile only in M.
4. Number of candidates below 80th percentile in P: Number of candidates below 80th percentile in C: Number of candidates below 80th percentile in M = 4:2:1.
BIE uses a different process for selection. If any candidate is appearing in the AET by AIE, BIE considers their AET score for final selection provided the candidate is at or above the 80th percentile in P. Any other candidate at or above the 80th percentile in P in CET, but who is not eligible for the AET, is required to appear in a separate test to be conducted by BIE for being considered for final selection. Altogether, there are 400 candidates this year who are at or above the 80th percentile in P.
What best can be concluded about the number of candidates sitting for the separate test for BIE who were at or above the 90th percentile overall in CET?
I hope that after going through this blogpost, you will be able to solve such questions in Set Theory pretty quickly.
Best of luck for the upcoming exams!
Set Theory- Maximum and Minimum Values
Permutation and Combination – Fundamental Principle of Counting
Permutation and Combination – Distribution of Objects
How to find Rank of a Word in Dictionary (With or Without Repetition)
Basic Probability Concepts for CAT Preparation
Sequence and Series Problems and Concepts for CAT 2017 Exam Preparation
All questions from CAT 2017 Quantitative Aptitude – Modern Maths
Quantitative Aptitude – Modern Maths – Progressions – Q1: If a1 = 1/(2*5), a2 = 1/(5*8), a3 = 1/(8*11),……, then a1 + a2 +……..+ a100 is
Quantitative Aptitude – Modern Maths – Progressions – Q2: An infinite geometric progression a1, a2, a3,… has the property that an = 3(a(n+ l) + a(n+2) +….) for every n ≥ 1. If the sum a1 + a2 + a3 +……. = 32, then a5 is
Quantitative Aptitude – Modern Maths – Progressions – Q3: Let a1, a2, a3, a4, a5 be a sequence of five consecutive odd numbers. Consider a new sequence of five consecutive even numbers ending with 2a3.
Quantitative Aptitude – Modern Maths – Progressions – Q4: Let a1, a2,……..a3n be an arithmetic progression with a1 = 3 and a2 = 7. If a1 + a2 + ….+a3n = 1830, then what is the smallest positive integer m such that m (a1 + a2 + …. + an ) > 1830?
Quantitative Aptitude – Modern Maths – Progressions – Q5: If the square of the 7th term of an arithmetic progression with positive common difference equals the product of the 3rd and 17th terms, then the ratio of the first term to the common difference is
Quantitative Aptitude – Modern Maths – P&C – Q1: How many four digit numbers, which are divisible by 6, can be formed using the digits 0, 2, 3, 4, 6, such that no digit is used more than once and 0 does not occur in the left-most position?
Quantitative Aptitude – Modern Maths – P&C – Q2: In how many ways can 8 identical pens be distributed among Amal, Bimal, and Kamal so that Amal gets at least 1 pen, Bimal gets at least 2 pens, and Kamal gets at least 3 pens?
Quantitative Aptitude – Modern Maths – P&C – Q3: In how many ways can 7 identical erasers be distributed among 4 kids in such a way that each kid gets at least one eraser but nobody gets more than 3 erasers?
Quantitative Aptitude – Modern Maths – P&C – Q4: Let AB, CD, EF, GH, and JK be five diameters of a circle with center at O. In how many ways can three points be chosen out of A, B, C, D, E, F, G, H, J, K, and O so as to form a triangle?
Quantitative Aptitude – Modern Maths – P&C – Q5
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