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Question 1: 70% of the people like Coffee, 80% of the people like Tea; then at least what % of people like both?
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?
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 (n1) times 100 from it to get the result. Let’s take another example to highlight this method.
Question 3: 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.
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