CBSE Class 9 Mathematics Number System Assignment Set 06

NUMBER SYSTEM - MCQ

Q1. Every natural number is
(a) always a whole number
(b) not a rational number
(c) not a fraction
(d) not an integer
Answer: (a) always a whole number
Natural numbers start from \(1, 2, 3, \dots\) and whole numbers start from \(0, 1, 2, 3, \dots\). Therefore, every natural number is included in the set of whole numbers.
In simple words: All counting numbers are also whole numbers.

Exam Tip: Remember the hierarchy: Natural numbers \(\subset\) Whole numbers \(\subset\) Integers \(\subset\) Rational numbers \(\subset\) Real numbers.

Q2. Select the correct statement from the following
(a) \(\frac{-2}{3} > \frac{-4}{5}\)
(b) \(\frac{-5}{7} < \frac{-3}{4}\)
(c) \(\frac{2}{5} < \frac{3}{8}\)
(d) \(\frac{4}{5} < \frac{7}{9}\)
Answer: (a) \(\frac{-2}{3} > \frac{-4}{5}\)
Converting to common denominators or cross-multiplying:
For option (a): \(-2 \times 5 = -10\) and \(-4 \times 3 = -12\). Since \(-10 > -12\), \(\frac{-2}{3} > \frac{-4}{5}\) is true.
For option (b): \(-5 \times 4 = -20\) and \(-3 \times 7 = -21\). Since \(-20 > -21\), \(\frac{-5}{7} > \frac{-3}{4}\), so statement (b) is false.
For option (c): \(2 \times 8 = 16\) and \(3 \times 5 = 15\). Since \(16 > 15\), \(\frac{2}{5} > \frac{3}{8}\), so statement (c) is false.
For option (d): \(4 \times 9 = 36\) and \(7 \times 5 = 35\). Since \(36 > 35\), \(\frac{4}{5} > \frac{7}{9}\), so statement (d) is false.
In simple words: Cross-multiplication helps compare fractions easily, keeping negative sign rules in mind.

Exam Tip: For negative fractions, the number with the smaller absolute value is larger.

Q3. How many rational numbers are there between two distinct rational numbers
(a) Infinite
(b) Finite
(c) Only 2
(d) only 5
Answer: (a) Infinite
Between any two distinct rational numbers, there exist infinitely many rational numbers (dense property of rational numbers).
In simple words: You can always find a new rational number between any two numbers by taking their average, and repeat this infinitely.

Exam Tip: The set of rational numbers is dense, meaning between any two rational numbers there is always another rational number.

Q4. \(\pi\) is a
(a) Irrational Number
(b) Rational Number
(c) Natural Number
(d) Integers
Answer: (a) Irrational Number
The mathematical constant \(\pi\) is a non-terminating and non-recurring decimal, making it an irrational number.
In simple words: \(\pi\) cannot be written as a simple fraction of two integers.

Exam Tip: Note that \(\frac{22}{7}\) is a rational approximation of \(\pi\), but \(\pi\) itself is irrational.

Q5. If \(x = 2 + \sqrt{3}\) then the value of \(x + \frac{1}{x}\) is
(a) 4
(b) 5
(c) 3
(d) 2
Answer: (a) 4
Given \(x = 2 + \sqrt{3}\),
\(\frac{1}{x} = \frac{1}{2 + \sqrt{3}}\)
Rationalizing the denominator:
\(\frac{1}{2 + \sqrt{3}} \times \frac{2 - \sqrt{3}}{2 - \sqrt{3}} = \frac{2 - \sqrt{3}}{4 - 3} = 2 - \sqrt{3}\)
Now, find \(x + \frac{1}{x}\):
\(x + \frac{1}{x} = (2 + \sqrt{3}) + (2 - \sqrt{3}) = 4\)
In simple words: Find the reciprocal of \(x\) by rationalizing, then add it to \(x\).

Exam Tip: When \(x = a + \sqrt{b}\) and \(a^2 - b = 1\), then \(\frac{1}{x} = a - \sqrt{b}\).

Q6. The rational number between \(\frac{1}{2}\) and \(\frac{1}{3}\) is
(a) \(\frac{5}{12}\)
(b) \(\frac{12}{5}\)
(c) \(\frac{-7}{25}\)
(d) \(0$
Answer: (a) \(\frac{5}{12}\)
The mean of two numbers \(a\) and \(b\) lies between them:
\(\frac{1}{2} \left(\frac{1}{2} + \frac{1}{3}\right) = \frac{1}{2} \left(\frac{3 + 2}{6}\right) = \frac{1}{2} \times \frac{5}{6} = \frac{5}{12}\)
In simple words: The average of two numbers always lies strictly between them.

Exam Tip: You can also convert fractions to decimals to check: \(\frac{1}{2} = 0.5\), \(\frac{1}{3} \approx 0.333\), and \(\frac{5}{12} \approx 0.416\).

Q7. The rationalising factor of \(\frac{1}{\sqrt{3} + \sqrt{2}}\) is

(a) \(\sqrt{3}-\sqrt{2}\)

(b) \(\sqrt{3}+\sqrt{2}\)

(c) \(\sqrt{3}+2\)

(d) \(3-\sqrt{2}\)
Answer: (a) \(\sqrt{3} - \sqrt{2}\)
To rationalize an expression of the form \(a + b\), we multiply by its conjugate \(a - b\). Thus, the rationalizing factor for \(\sqrt{3} + \sqrt{2}\) is \(\sqrt{3} - \sqrt{2}\).
In simple words: Change the middle plus sign to a minus sign to get the rationalizing factor.

Exam Tip: The product of a surd binomial and its rationalizing factor uses the identity \((a + b)(a - b) = a^2 - b^2\).

Q8. The product of \((3 + \sqrt{2})(3 - \sqrt{2})\) is
(a) 7
(b) 1
(c) -1
(d) 11
Answer: (a) 7
Using the algebraic identity \((a + b)(a - b) = a^2 - b^2\):
\((3 + \sqrt{2})(3 - \sqrt{2}) = (3)^2 - (\sqrt{2})^2\)
\(= 9 - 2 = 7\)
In simple words: Apply the difference of squares formula directly.

Exam Tip: Memorize standard algebraic identities to solve product problems quickly.