Welcome to AssamSchool. This article contains 30 Extra Questions from Number Systems from Class 9 Mathematics Chapter 1. These Number System Class 9 Extra Questions are helpful to test your understanding and knowledge of the chapter. These additional questions will not only help you to excel in the chapter but also will give you a strong foundation in Mathematics.
We have also provided step-by-step answers to the questions for reference. So, let’s start…
30 Number Systems Class 9 Extra Questions
1. Classify the number 0.3030030003… as rational or irrational. Justify.
Solution:
The number 0.3030030003… is irrational because its decimal expansion is non-terminating and non-repeating. It does not recur in a fixed pattern, so it cannot be expressed as \(\frac{p}{q}\) where \(p\) and \(q\) are integers.
2. Find three rational numbers between \(\sqrt{2}\) and \(\sqrt{3}\).
Solution:
Approximate values:
\(\sqrt{2} \approx 1.414\), \(\sqrt{3} \approx 1.732\).
Three rational numbers between them are:
\(1.5 = \frac{3}{2}\), \(1.6 = \frac{8}{5}\), \(1.7 = \frac{17}{10}\).
3. Simplify \((\sqrt{5} + 2\sqrt{3})(\sqrt{5} – \sqrt{3})\).
Solution:
\((\sqrt{5} + 2\sqrt{3})(\sqrt{5} – \sqrt{3})\)
= \((\sqrt{5})^2 – \sqrt{5}\sqrt{3} + 2\sqrt{3}\sqrt{5} – 2(\sqrt{3})^2 \)
= \((5 – \sqrt{15} + 2\sqrt{15} – 6\))
= \((5 – 6) + \sqrt{15}\)
=\(-1 + \sqrt{15}\)
4. Rationalize the denominator of \(\frac{1}{\sqrt{7} + \sqrt{5}}\).
Solution:
Multiply numerator and denominator by \(\sqrt{7} – \sqrt{5}\):
\(\frac{1}{\sqrt{7} + \sqrt{5}} \times \frac{\sqrt{7} – \sqrt{5}}{\sqrt{7} – \sqrt{5}}\)
= \(\frac{\sqrt{7} – \sqrt{5}}{(\sqrt{7})^2 – (\sqrt{5})^2}\)
= \(\frac{\sqrt{7} – \sqrt{5}}{7 – 5}\)
= \(\frac{\sqrt{7} – \sqrt{5}}{2}\)
5. Express \(5^{\frac{3}{2}}\) in radical form and evaluate.
Solution:
\(5^{\frac{3}{2}} = \sqrt{5^3} = \sqrt{125} = 5\sqrt{5}\).
6. Prove that the sum of a rational and an irrational number is irrational.
Solution:
Assume \(r\) is rational and \(s\) is irrational.
Suppose \(r + s = t\) is rational.
Then \(s = t – r\), which is rational (difference of two rationals).
This contradicts \(s\) being irrational.
Hence, \(t\) must be irrational.
7. Convert \(0.\overline{7}\) into a fraction.
Solution:
Let \(x = 0.\overline{7}\).
Then \(10x = 7.\overline{7}\).
Subtract: \(10x – x = 7.\overline{7} – 0.\overline{7} \implies 9x = 7 \implies x = \frac{7}{9}\)
8. If \(a\) and \(b\) are rational and \(\sqrt{a} + \sqrt{b}\) is irrational, prove \(\sqrt{a} – \sqrt{b}\) is irrational.
Solution:
Suppose \(\sqrt{a} – \sqrt{b}\) is rational.
Then (\(\sqrt{a} + \sqrt{b}) + (\sqrt{a} – \sqrt{b}) = 2\sqrt{a}\) is rational \(\implies \sqrt{a}\) is rational.
Similarly, \(\sqrt{b}\) is rational, contradicting \(\sqrt{a} + \sqrt{b}\) being irrational.
Hence, \(\sqrt{a} – \sqrt{b}\) is irrational.
9. Locate \(\sqrt{5}\) on the number line using geometric construction.
Solution:
- Draw a number line. Mark \(O(0)\), \(A(2)\), and \(B(2.236)\).
- Construct a right triangle \(OAB\) with \(OA = 2\), \(AB = 1\).
- By Pythagoras, \(OB = \sqrt{2^2 + 1^2} = \sqrt{5}\).
- Use a compass to mark \(\sqrt{5}\) on the number line.
10. Simplify \(27^{\frac{2}{3}} + 16^{\frac{3}{4}} – 81^{\frac{1}{4}}\).
Solution:
\(27^{\frac{2}{3}} = (\sqrt[3]{27})^2 = 3^2 = 9, \)
\(16^{\frac{3}{4}} = (\sqrt[4]{16})^3 = 2^3 = 8, \)
\(81^{\frac{1}{4}} = \sqrt[4]{81} = 3\)
\(\text{Result: } 9 + 8 – 3 = 14\)
11. Determine if \((2 + \sqrt{3})(2 – \sqrt{3})\) is rational.
Solution:
\((2 + \sqrt{3})(2 – \sqrt{3}) = 2^2 – (\sqrt{3})^2 = 4 – 3 = 1 \quad (\text{Rational})\).
12. Find the decimal expansion of \(\frac{1}{13}\) and state its type.
Solution:
\(\frac{1}{13} = 0.\overline{076923}\), a non-terminating recurring decimal.
13. Show that \(\sqrt{18}\) can be written as \(3\sqrt{2}\).
Solution:
\(\sqrt{18} = \sqrt{9 \times 2} = \sqrt{9} \times \sqrt{2} = 3\sqrt{2}\).
14. Which are irrational: \(\sqrt{16}, \sqrt{12}, 0.454545…, \frac{\pi}{2}\)?
Solution:
- \(\sqrt{16} = 4\) (Rational)
- \(\sqrt{12} = 2\sqrt{3}\) (Irrational)
- \(0.454545… = 0.\overline{45}\) (Rational)
- \(\frac{\pi}{2}\) (Irrational).
Irrational: \(\sqrt{12}, \frac{\pi}{2}\).
15. If \(x = 1 + \sqrt{2}\), find \(x + \frac{1}{x}\).
Solution:
\(x + \frac{1}{x} = (1 + \sqrt{2}) + \frac{1}{1 + \sqrt{2}} \times \frac{1 – \sqrt{2}}{1 – \sqrt{2}}\)
\(= (1 + \sqrt{2}) + \frac{1 – \sqrt{2}}{-1}\)
\( = (1 + \sqrt{2}) – (1 – \sqrt{2}) = 2\sqrt{2}\)
16. Simplify \(4\sqrt{3} – 3\sqrt{12} + 2\sqrt{75}\).
Solution:
\(4\sqrt{3} – 3\sqrt{4 \times 3} + 2\sqrt{25 \times 3}\)
\( = 4\sqrt{3} – 6\sqrt{3} + 10\sqrt{3} = 8\sqrt{3}\)
17. Rationalize \(\frac{1}{\sqrt{3} – \sqrt{2} + 1}\).
Solution:
Let \(a = \sqrt{3} – \sqrt{2}\).
Multiply numerator and denominator by (\(a – 1)\):
\(\frac{1}{a + 1} \times \frac{a – 1}{a – 1}\)
\(= \frac{a – 1}{(a)^2 – 1^2}\)
\( = \frac{\sqrt{3} – \sqrt{2} – 1}{(\sqrt{3} – \sqrt{2})^2 – 1}\)
Further, simplify the denominator and numerator.
18. Express (\0.\overline{123}\) as a fraction.
Solution:
Let \(x = 0.\overline{123}\).
\(1000x = 123.\overline{123}\).
Subtract: \(1000x – x = 123 \implies 999x = 123 \implies x = \frac{123}{999} = \frac{41}{333}\).
19. Prove that \(\sqrt{3}\) is irrational.
Solution:
Assume \(\sqrt{3} = \frac{a}{b}\) (reduced form).
Then \(3b^2 = a^2\).
Thus, \(3\) divides \(a^2 \implies 3\) divides \(a\).
Let \(a = 3k\).
Then \(3b^2 = 9k^2 \implies b^2 = 3k^2 \implies 3\) divides \(b\), contradicting \(a\) and \(b\) being coprime.
Hence, \(\sqrt{3}\) is irrational.
20. Calculate (\(\sqrt{8} + \sqrt{18})^2\).
Solution:
\((\sqrt{8} + \sqrt{18})^2\)
\( = (\sqrt{8})^2 + 2\sqrt{8}\sqrt{18} + (\sqrt{18})^2 \)
\(= 8 + 2 \times \sqrt{144} + 18\)
\(= 8 + 24 + 18\)
\( = 50\)
21. Find two irrational numbers whose product is rational.
Solution:
Let \(a = \sqrt{2}\) and \(b = \sqrt{8}\).
Then \(a \times b = \sqrt{16} = 4\) (Rational).
22. Simplify (\(\sqrt[3]{5})^6 \times 25^{\frac{1}{6}}\).
Solution:
\((\sqrt[3]{5})^6 = 5^{2}, \quad 25^{\frac{1}{6}} = (5^2)^{\frac{1}{6}} = 5^{\frac{1}{3}} \)
\(\text{Result: } 5^{2} \times 5^{\frac{1}{3}} = 5^{\frac{7}{3}}\)
23. If \(\sqrt{ab}\) is irrational, what can be said about \(a\) and \(b\)?
Solution:
At least one of \(a\) or \(b\) must be irrational. If both were rational, \(\sqrt{ab}\) would also be rational.
24. Write the reciprocal of \(1 + \sqrt{2}\) in simplified form.
Solution:
\(\frac{1}{1 + \sqrt{2}} \times \frac{1 – \sqrt{2}}{1 – \sqrt{2}} \)
\(= \frac{1 – \sqrt{2}}{1 – 2} \)
\(= \sqrt{2} – 1\)
25. Compare \(\sqrt{5} + \sqrt{3}\) and \(\sqrt{7} + \sqrt{2}\) without calculating.
Solution:
Square both:
(\(\sqrt{5} + \sqrt{3})^2 = 8 + 2\sqrt{15} \approx 8 + 7.746 = 15.746\).
(\(\sqrt{7} + \sqrt{2})^2 = 9 + 2\sqrt{14} \approx 9 + 7.483 = 16.483\).
Thus, \(\sqrt{7} + \sqrt{2} > \sqrt{5} + \sqrt{3}\).
26. Simplify (\frac{a^{\frac{2}{3}} \cdot b^{\frac{3}{2}}}{a^{\frac{1}{3}} \cdot b^{\frac{1}{2}}}).
Solution:
Separate the terms and apply the law of exponents \(\frac{x^m}{x^n} = x^{m-n}\):
\(\frac{a^{\frac{2}{3}}}{a^{\frac{1}{3}}} \times \frac{b^{\frac{3}{2}}}{b^{\frac{1}{2}}} = a^{\frac{2}{3} – \frac{1}{3}} \cdot b^{\frac{3}{2} – \frac{1}{2}}\)
Simplify the exponents:
\(a^{\frac{1}{3}} \cdot b^{1} = a^{\frac{1}{3}}b\)
Final Answer: \(\boxed{a^{\frac{1}{3}}b}\)
27. Find \(x\) if \(3^x = 81^{\frac{1}{4}}\)
Solution:
\(81^{\frac{1}{4}} = (3^4)^{\frac{1}{4}} = 3^1 = 3\)
Thus, \(3^x = 3 \implies x = 1\).
28. Show that \(0.999… = 1\)
Solution:
Let \(x = 0.\overline{9}\)
Then \(10x = 9.\overline{9}\).
Subtract: \(10x – x = 9 \implies 9x = 9 \implies x = 1\).
29. If \(a = \sqrt{2} + 1\), find \(a^2 – 2a\).
Solution:
\(a^2 = (\sqrt{2} + 1)^2 = 3 + 2\sqrt{2}\)
\(a^2 – 2a = (3 + 2\sqrt{2}) – 2(\sqrt{2} + 1) = 3 + 2\sqrt{2} – 2\sqrt{2} – 2 = 1\)
30. Prove that between any two distinct real numbers, there exists a rational number.
Solution:
Let \(a < b\).
Choose \(n\) such that \(\frac{1}{n} < b – a\).
Let \(m\) be the smallest integer greater than \(na\).
Then \(\frac{m}{n}\) lies between \(a\) and \(b\).
Hence, a rational number exists.
Related Posts
- Polynomials Class 9 Extra Questions | Mathematics Chapter 2
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Abdur Rohman is an Electrical Engineer from Charaideo, Assam, who wears multiple hats as a part-time teacher, blogger, entrepreneur, and digital marketer. Passionate about education, he founded The Assam School blog to provide free, comprehensive textbook solutions, MCQs (Multiple Choice Questions), and other academic content for students from Class V to XII.