Class 9th math 3.2 solution english PCTB

Question 3: consider the sets \(P=\{x:x\ \text{is a prime number and } 0< x\leq 20\}\) and \( Q=\{x:x\text{ is a divisor of } 210\ \text{and} \ 0< x\leq 20\}\)

\((i)\) \(\;\) Find \(P \cap Q \)

Solution:

\(P=\{x:x\ \text{is a prime number and } 0< x\leq 20\}\)

\(P=\{2,3,5,7,11,13,17,19\}\)

\( Q=\{x:x\text{ is a divisor of } 210\ \text{and} \ 0< x\leq 20\}\)

\( Q=\{1,2,3,5,6,7,10,14,15\}\)

\(P \cap Q \)=\(\{2,3,5,7,11,13,17,19\}\)\(\cap\)\(\{1,2,3,5,6,7,10,14,15\}\)

\(P \cap Q \)=\(\{2,3,5,7\}\)


Explanation:

\(P=\{x:x\ \text{is a prime number and } 0< x\leq 20\}\)

Write all the prime numbers greater than \(0\) and less than or equal to \(20\).

\( Q=\{x:x\text{ is a divisor of } 210\ \text{and} \ 0< x\leq 20\}\)

To find the divisors of \(210\) up to \(20\), start dividing \(210\) by the numbers \(1, 2, 3, …, 20\).

Use a calculator or long division, and select only those numbers that divide exactly — meaning they leave no remainder.

Intersection \(\cap\) means the common elements in both sets.

Tips for Students:

  • Use a highlighter or circle common elements in both lists.
  • Always write sets in increasing order for easier comparison.

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