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How to use the unit problem bank for Sets, Relations and Functions like a real JEE paper: mixed problems in no order of topic, a timer of about 2 minutes for easy, 3 to 4 for medium and 6 to 8 for hard problems, the write-your-answer-first rule, and two solved problems on De Morgan's law and divisibility counting.
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0:00Introduction
Hello everyone! Okay, tell me one thing. In the JEE paper, does any question come with a label saying, this one is from sets, use De Morgan's law here? Never, right? The paper mixes everything, and the clock keeps running. So today, I will show you how to use our unit problem bank exactly like the real exam: mixed problems, a timer, and one golden rule. And then we will solve two problems together, with the clock on. So, let's begin.
0:31How this chapter works
So, first of all, what is this chapter? In this unit, you learnt sets, then relations, then functions, one chapter at a time, okay? But here, the problems mix all the ideas of the unit together. And inside each part, there is no particular order of topic. Now why would we do that? Why not keep all the De Morgan questions together? Yes, exactly! Because the examination does the same thing. In the real paper, nobody tells you which idea to use. Spotting the idea is your job. The chapter has four parts. Part A is sets, Part B is relations, and Part C is functions, okay? But even inside one part, the next problem can need a completely different idea. So stay alert, got it? And Part D is assertion reason and concept checks. There, you decide whether a statement is true or false, and give a one line reason or a counterexample. Small questions, but they test whether your concepts are really clear.
1:36Treat each part as a timed test
Now the most important instruction of this chapter. Treat each part as a timed test. Every problem carries a badge: easy, medium or hard. So tell me, how much time should you give an easy problem? About two minutes, that's it. Read it, spot the idea, write the answer. Really quick. Medium problems get three to four minutes. More thinking, so a little more time, okay? And hard problems get six to eight minutes. Look, hard does not mean impossible. It only means you need more time, so give it that time, and don't give up after one minute. Alright, now the trap. You read a problem, you feel a little stuck, and you think, let me just peep at the solution, right? You read it, you nod, and you feel, yes, I knew this! No, no, careful! That feeling is the trap. Reading a solution is not the same as solving the problem. So the rule is simple. Write your answer before reading the solution. Your own answer, on paper, right or wrong. Only then open the solution and compare. That is how this chapter will really help you. Got it?
2:51Topper's trick: time the whole part
Now let me share a small trick that toppers use. Before you start a part, count the badges. Take Part A, the sets part. It has three easy problems, six medium ones and two hard ones, okay? So, if Part A is one timed test, how long should that test be? Let's add it up together. At the lower end: easy, three times two, six minutes. Medium, six times three, eighteen minutes. Hard, two times six, twelve minutes. Six plus eighteen plus twelve: thirty six minutes. At the upper end: six minutes for the easy ones, six times four, that is twenty four, for the medium ones, and two times eight, that is sixteen, for the hard ones. Forty six minutes. So set a timer for about thirty six to forty six minutes, and sit as if it is the real paper. Fully serious.
3:47Example 1: neither A nor B with De Morgan's law
Alright, now let's practise exactly the way this chapter says. This one is easy, so your timer is two minutes. The universal set U has seven hundred elements. n of A is two hundred, n of B is three hundred, and n of A intersection B is one hundred. Find n of A prime intersection B prime, the number of elements in neither set. Pause the video, write your answer, and then come back. Done? Okay, let's check. Many students do this: seven hundred, minus two hundred, minus three hundred, that is two hundred. Right? No, no, careful! The one hundred elements in the middle belong to both sets, so they got subtracted twice. So, use De Morgan's law. And notice, the question never said De Morgan. Spotting it was your job, just like I told you, right? See, A prime intersection B prime is the complement of A union B: everything outside both circles, the shaded part, okay? First, n of A union B: two hundred plus three hundred, minus the overlap of one hundred. Four hundred. In the diagram, one hundred are only in A, one hundred are in both, and two hundred are only in B, okay? So neither is seven hundred minus four hundred: three hundred. Two short lines, and the answer, well inside two minutes. Clear?
5:13Example 2: divisible by 2 or 3 but not by 5
Now a medium one, JEE style, so your timer is three to four minutes. Find the number of integers from one to three hundred that are divisible by two or three, but not by five. Pause the video, write your answer first, and then come back. Done? Okay, let's check. First, divisible by two or by three. Multiples of two up to three hundred: one hundred and fifty. Multiples of three: one hundred. But some numbers are in both lists. Which ones? Yes, the multiples of six, and there are fifty of them. Subtract them once. One hundred and fifty, plus one hundred, minus fifty: two hundred. Okay, now let's throw out the ones divisible by five. There are sixty multiples of five up to three hundred. So two hundred minus sixty, one hundred and forty, right? No, no, careful! This is the trap. We must remove only those of our two hundred that are also divisible by five. Five, twenty five, thirty five: these are multiples of five, but they were never in our list! So think: a number divisible by two and by five is divisible by ten, and one divisible by three and by five is divisible by fifteen. That is what the little arrow on screen means: divisible by five as well, so divisible by ten or by fifteen, okay? Multiples of ten: thirty. Multiples of fifteen: twenty. And the numbers in both lists are the multiples of thirty, ten of them. Thirty plus twenty minus ten: forty. So the answer is two hundred minus forty: one hundred and sixty. See? The same inclusion and exclusion idea, used twice in one problem. That is what a mixed problem feels like.
7:06Quick revision: mixed problems and the timer for each level
So, quick revision. The problems are mixed, with no order of topic, just like the exam. Time yourself: about two minutes for easy, three to four for medium, and six to eight for hard. And always write your answer before reading the solution. Now open Part A, start your timer, and begin. You can do this, I promise. All the best!
More videos in Unit 1 · Sets, Relations & Functions
- What is a Set?
- Representation of Sets
- Types of Sets
- Subsets, Intervals and Power Set
- Universal Set and Venn Diagrams
- Operations on Sets
- Laws of the Algebra of Sets
- Counting: Number of Elements in Unions
- Ordered Pairs and Cartesian Product
- Chapter Summary: Mixed Solved Problems on Sets
- What is a Relation?
- Inverse of a Relation
- Types of Relations
- Equivalence Relations
- Equivalence Classes and Partitions
- Composition of Relations and Partial Orders
- Chapter Summary: Mixed Solved Problems on Relations
- What is a Function?
- Domain of Real Functions
- Finding the Range
- Standard Real Functions and Their Graphs
- Algebra of Real Functions
- One-one and Many-one Functions
- Onto and Into Functions; Bijections
- Counting Special Functions
- Composition of Functions
- Inverse of a Function
- Even and Odd Functions
- Simple Functional Equations
- Part A: Sets
- Part B: Relations
- Part C: Functions
- Part D: Assertion–Reason and Concept Checks