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SubjectsMathematics
अपरिमेय संख्याएँ
In this Class 9 Mathematics topic from the Number Systems chapter, students learn that irrational numbers cannot be written in the form p/q, where p and q are integers and q is not zero. They explore familiar examples such as √2 and π, understand their non-terminating, non-repeating decimal expansions, and distinguish them from rational numbers. The topic also develops skills for representing irrational numbers on the number line and understanding their place within the real number system.
Correct answer: A. \(r+x\) is an irrational number.
Explanation: If \(r+x\) were rational, subtracting the rational number \(r\) would make \(x\) rational, a contradiction. Hence A is correct. Option B is also false because the difference remains irrational. Use contradiction in such questions.
Correct answer: A. (\sqrt{16})
Explanation: Since (14<16<18), (\sqrt{16}) lies between them. For positive square roots compare the numbers inside.
Correct answer: A. \(a+bx\)
Explanation: If \(a+bx\) were rational, then \(bx\) would be rational; since \(b\ne0\), \(x=\frac{bx}{b}\) would be rational, a contradiction. But \(x^2\) can be rational, for example when \(x=\sqrt2\). Exam tip: adding a rational number to an irrational number keeps it irrational.
Correct answer: A. (4)
Explanation: (\sqrt{200}=10\sqrt{2}) and (\sqrt{72}=6\sqrt{2}), so the numerator is (4\sqrt{2}). Dividing gives (4).
Correct answer: A. \(11+\sqrt{30}\)
Explanation: The square of a square root gives the number inside. So \(\left(\sqrt{11+\sqrt{30}}\right)^2=11+\sqrt{30}\).
Correct answer: A. \(\sqrt{2}\times\sqrt{8}=4\)
Explanation: Both \(\sqrt{2}\) and \(\sqrt{8}\) are irrational, but \(\sqrt{2}\times\sqrt{8}=\sqrt{16}=4\), which is rational. Hence, the word “always” makes the statement false. Exam tip: one counterexample is enough to disprove an “always” statement.
Correct answer: A. \(\frac{\sqrt{18}}{\sqrt{2}}\)
Explanation: \(\frac{\sqrt{18}}{\sqrt{2}}=\sqrt{9}=3\), which is rational. Hence, a quotient of two irrational numbers need not be irrational. Option B simplifies to \(\sqrt{3}\), which is irrational. Exam tip: combine the radicals first, then simplify the number inside the square root.
Correct answer: A. (\frac{13+2\sqrt{22}}{9})
Explanation: Multiplying by the conjugate gives numerator (13+2\sqrt{22}) and denominator (11-2=9). So the simplified form is (\frac{13+2\sqrt{22}}{9}).
Correct answer: A. The sum \(a+r\) is irrational
Explanation: If \(a+r\) were rational, then \(a=(a+r)-r\) would also be rational, contradicting that \(a\) is irrational. Hence the sum is irrational. Exam tip: adding or subtracting a rational number preserves irrationality.
Correct answer: B. (11\sqrt{5})
Explanation: (\sqrt{605}=11\sqrt{5}), (\sqrt{245}=7\sqrt{5}), and (\sqrt{125}=5\sqrt{5}). Therefore the result is (9\sqrt{5}).
Correct answer: A. (192)
Explanation: The direct answer is A, 192. Simplify the radicals: √27 = √(9×3) = 3√3 and √75 = √(25×3) = 5√3. Their sum is 8√3. Squaring gives (8√3)^2 = 64×3 = 192, so option A is correct. Option B, 128, uses the wrong final multiplication and does not equal the expression. Option C, 16√3, is not the square of 8√3; squaring removes the remaining radical and gives a rational number. Option D, 300, results from an incorrect expansion. As a second check, (√27+√75)^2 = 27+75+2√(27×75) = 102+2√2025 = 102+90 = 192. Memory cue: when both radicals contain the same remaining factor, add their coefficients first, then square.
Correct answer: A. (\frac{\sqrt{14}-\sqrt{5}}{3})
Explanation: Multiplying by the conjugate makes the denominator (14-5=9). So (\frac{3(\sqrt{14}-\sqrt{5})}{9}=\frac{\sqrt{14}-\sqrt{5}}{3}).
Correct answer: B. (9)
Explanation: (\sqrt{18}=3\sqrt{2}) and (\sqrt{72}=6\sqrt{2}) so the numerator is (9\sqrt{2}). Dividing by (\sqrt{2}) gives (9).
Correct answer: C. (3\sqrt{3})
Explanation: (\sqrt{12}=2\sqrt{3}), (\sqrt{75}=5\sqrt{3}), and (\sqrt{48}=4\sqrt{3}). Therefore (a-b=3\sqrt{3}).
Correct answer: B. \(\sqrt{12}=2\sqrt{3}\), so \(\sqrt{12}-\sqrt{3}=\sqrt{3}\), which is irrational.
Explanation: Since \(12=4\times3\), \(\sqrt{12}=2\sqrt{3}\). Thus the difference is \(2\sqrt{3}-\sqrt{3}=\sqrt{3}\), which is irrational. Exam tip: never use \(\sqrt a-\sqrt b=\sqrt{a-b}\).
Correct answer: B. (\frac{\sqrt{13}+\sqrt{5}}{2})
Explanation: Multiplying by the conjugate makes the denominator (13-5=8). So (\frac{4(\sqrt{13}+\sqrt{5})}{8}=\frac{\sqrt{13}+\sqrt{5}}{2}).
Correct answer: B. (5)
Explanation: Area is ((\sqrt{11}+\sqrt{6})(\sqrt{11}-\sqrt{6})=11-6=5). Conjugate dimensions can give a rational area.
Correct answer: D. (\sqrt{12}\times\sqrt{3})
Explanation: (\sqrt{12}\times\sqrt{3}=\sqrt{36}=6), which is rational. Check whether multiplication creates a perfect square.
