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Exercise 1.1 · Q3

Q.Write the following sets in roster form:

(i) A = {x : x is an integer and –3 ≤ x < 7}
(ii) B = {x : x is a natural number less than 6}
(iii) C = {x : x is a two-digit natural number such that the sum of its digits is 8}
(iv) D = {x : x is a prime number which is divisor of 60}
(v) E = The set of all letters in the word TRIGONOMETRY
(vi) F = The set of all letters in the word BETTER
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Roster form lists every element explicitly inside curly braces. For each set, we identify the members by interpreting the given condition — integers, natural numbers, digit sums, prime divisors, or distinct letters — and write them out separated by commas.


1. Set A: integers from –3 up to (but not including) 7

The condition says xx is an integer and −3≤x<7-3 \le x < 7.

That means we start at −3-3 and include every integer until we reach 66 (since 77 is not included).

List them: −3,−2,−1,0,1,2,3,4,5,6-3, -2, -1, 0, 1, 2, 3, 4, 5, 6.

Watch out

A common mistake is to forget that 00 is an integer — it belongs here. Also, note that 77 is not included because the inequality is x<7x < 7, not x≤7x \le 7.

So

A={−3,−2,−1,0,1,2,3,4,5,6}A = \{-3, -2, -1, 0, 1, 2, 3, 4, 5, 6\}


2. Set B: natural numbers less than 6

Natural numbers are the counting numbers: 1,2,3,4,5,…1, 2, 3, 4, 5, \dots

“Less than 6” means x<6x < 6, so we take 1,2,3,4,51, 2, 3, 4, 5.

Tip

In most Indian exam contexts, natural numbers start from 1. If the problem had said “whole numbers”, we would include 0. Here it’s clear: natural numbers less than 6.

Thus

B={1,2,3,4,5}B = \{1, 2, 3, 4, 5\}


3. Set C: two-digit natural numbers whose digits sum to 8

A two-digit number has a tens digit tt (from 1 to 9) and a units digit uu (from 0 to 9).

We need t+u=8t + u = 8.

Let’s go through possible tens digits:

  • t=1⇒u=7t = 1 \Rightarrow u = 7 → number 17
  • t=2⇒u=6t = 2 \Rightarrow u = 6 → 26
  • t=3⇒u=5t = 3 \Rightarrow u = 5 → 35
  • t=4⇒u=4t = 4 \Rightarrow u = 4 → 44
  • t=5⇒u=3t = 5 \Rightarrow u = 3 → 53
  • t=6⇒u=2t = 6 \Rightarrow u = 2 → 62
  • t=7⇒u=1t = 7 \Rightarrow u = 1 → 71
  • t=8⇒u=0t = 8 \Rightarrow u = 0 → 80
  • t=9⇒u=−1t = 9 \Rightarrow u = -1 → not possible

So the numbers are: 17, 26, 35, 44, 53, 62, 71, 80.

Note

80 is a two-digit number (tens digit 8, units digit 0) and 8+0=88+0=8, so it qualifies.

Therefore

C={17,26,35,44,53,62,71,80}C = \{17, 26, 35, 44, 53, 62, 71, 80\}


4. Set D: prime divisors of 60

First, find all divisors of 60. Then pick only the prime ones.

Divisors of 60: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60.

From these, the primes are: 2, 3, 5.

Watch out

1 is not prime. Also, 60’s prime factors are 2, 3, 5 — but the question asks for prime divisors, not prime factors repeated. So we list each prime only once.

Thus

D={2,3,5}D = \{2, 3, 5\}

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