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SubjectsMathematics
संख्या पद्धति: सामान्य अभ्यास
This practice topic helps Class 9 Mathematics students consolidate the ideas from Number Systems through a focused set of questions. Students work with rational and irrational numbers, locate numbers on the number line, interpret decimal expansions, and apply the laws of exponents. The exercises strengthen calculation, comparison, simplification, and mathematical reasoning while encouraging learners to explain why a result is valid. It supports concept revision, self-assessment, and written problem-solving.
Correct answer: B. Every irrational number is a real number
Explanation: Real numbers include both rational and irrational numbers. Therefore every irrational number is real.
Correct answer: B. Irrational number
Explanation: \(\sqrt{2}\) is irrational. If \(r+\sqrt{2}\) were rational, subtracting the rational number \(r\) would make \(\sqrt{2}\) rational, which is impossible. Exam tip: rational ± irrational is always irrational.
Correct answer: A. Rational number
Explanation: The direct answer is A, Rational number. A rational number is any number that can be written as \\(p/q\\), where \\(p\\) and \\(q\\) are integers and \\(q\\ne0\\). In the decimal 1.272727..., the block 27 repeats forever. Step by step, let \\(x=1.272727...\\). Multiplying by 100 gives \\(100x=127.272727...\\). Subtracting the first equation gives \\(99x=126\\), so \\(x=126/99=14/11\\), which is a ratio of integers. Therefore it is rational. Option A is correct. Option B, irrational number, is wrong because irrational decimals do not terminate and do not repeat in a fixed block; this decimal repeats. Option C, natural number, is wrong because 1.272727... is not a counting number such as 1, 2 or 3. Option D, perfect square, is wrong because a perfect square normally refers to a number such as 4, 9 or 16 obtained by squaring an integer; this number is not one of those. Exam cue: a terminating or recurring decimal is rational.
Correct answer: C. Their product can be rational
Explanation: The product of two irrational numbers can be rational. For example, \(\sqrt{2}\times\sqrt{2}=2\), and 2 is rational. Hence, “always irrational” is false. Exam tip: test universal claims using a counterexample.
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