Irrational numbers

Correct answer: A. (11)

Explanation: Area is ((4+\sqrt{5})(4-\sqrt{5})=16-5=11). Multiplying conjugate dimensions can give a rational area.

Correct answer: A. (7\sqrt{3})

Explanation: (\sqrt{300}=10\sqrt{3}), (\sqrt{192}=8\sqrt{3}), and (\sqrt{75}=5\sqrt{3}). Therefore the result is (7\sqrt{3}).

Correct answer: A. (\sqrt{5})

Explanation: The first term becomes (\frac{\sqrt{5}-\sqrt{3}}{2}) and the second becomes (\frac{\sqrt{5}+\sqrt{3}}{2}). Their sum is (\sqrt{5}).

Correct answer: A. \((\sqrt{2},-\sqrt{2})\)

Explanation: Both \(\sqrt{2}\) and \(-\sqrt{2}\) are irrational, but \(\sqrt{2}+(-\sqrt{2})=0\), which is rational. Hence Reena’s “always” claim is false. Exam tip: one valid counterexample is enough to disprove an “always” statement.

Correct answer: A. \(\sqrt{2},\ 3-\sqrt{2}\)

Explanation: \(\sqrt{2}\) is irrational, and \(3-\sqrt{2}\) is also irrational; otherwise subtracting it from 3 would make \(\sqrt{2}\) rational. Their sum is \(3\), which is rational. Exam tip: check the sum first in such pairs.

Correct answer: A. Dividing an irrational number by a non-zero rational number gives an irrational result

Explanation: \(3+\sqrt{2}\) is irrational. Dividing it by the non-zero rational number 5 keeps the result irrational. Exam tip: if \(x\) were rational, then \(5x-3=\sqrt{2}\) would be rational, which is a contradiction.

Correct answer: A. (\sqrt{11})

Explanation: Since (10<11<12), (\sqrt{11}) lies between them. For positive square roots compare the numbers inside.

Correct answer: A. \(3\)

Explanation: \(\frac{\sqrt{18}}{\sqrt{2}}=\sqrt{\frac{18}{2}}=\sqrt{9}=3\), which is rational. The quotient of two irrational numbers need not be irrational. Exam tip: combine the radicals before deciding the number type.

Correct answer: A. (3)

Explanation: (\sqrt{125}=5\sqrt{5}) and (\sqrt{20}=2\sqrt{5}), so the numerator is (3\sqrt{5}). Dividing gives (3).

Correct answer: A. \(7+\sqrt{10}\)

Explanation: The square of a square root gives the number inside. So \(\left(\sqrt{7+\sqrt{10}}\right)^2=7+\sqrt{10}\).

Correct answer: A. \(p+q\) is always irrational

Explanation: If \(p+q\) were rational, then \(p=(p+q)-q\) would be the difference of two rational numbers and hence rational, a contradiction. Thus \(p+q\) is irrational. Exam tip: adding or subtracting a rational number from an irrational number remains irrational.

Correct answer: A. \(\sqrt{19}-\sqrt{19}\)

Explanation: In option A, \(\sqrt{19}\) and \(\sqrt{19}\) are both irrational, and their difference is \(\sqrt{19}-\sqrt{19}=0\). Zero is a rational number. In option B, \(\sqrt{2}-\sqrt{3}\) is irrational; option C simplifies to \(\sqrt{5}-\sqrt{20}=-\sqrt{5}\), and option D to \(\sqrt{7}-\sqrt{28}=-\sqrt{7}\), both irrational. Exam tip: the difference of identical irrational numbers is zero, which is rational.

Correct answer: C. योग परिमेय या अपरिमेय, दोनों हो सकता है।

Explanation: The sum of two irrational numbers need not have a fixed type. For example, \(\sqrt{2}+(-\sqrt{2})=0\) is rational, whereas \(\sqrt{2}+\sqrt{3}\) is irrational. Hence, the sum can be rational or irrational. In exams, test claims using “always” with a counterexample.

Correct answer: B. (\frac{5+\sqrt{21}}{2})

Explanation: Multiplying by the conjugate gives numerator (10+2\sqrt{21}) and denominator (4). So the simplified form is (\frac{5+\sqrt{21}}{2}).

Correct answer: A. If \(5-\sqrt{7}\) were rational, then \(\sqrt{7}=5-(5-\sqrt{7})\) would also be rational, which is impossible.

Explanation: \(\sqrt{7}\) is irrational. If \(5-\sqrt{7}\) were rational, subtracting it from 5 would make \(\sqrt{7}\) rational, a contradiction. Exam tip: rational ± irrational is always irrational.

Correct answer: C. (7\sqrt{5})

Explanation: (\sqrt{245}=7\sqrt{5}), (\sqrt{180}=6\sqrt{5}), and (\sqrt{80}=4\sqrt{5}). So the result is (7\sqrt{5}-6\sqrt{5}+4\sqrt{5}=5\sqrt{5}).

Correct answer: A. (128)

Explanation: The direct answer is A, 128. First simplify each radical by taking out the largest perfect-square factor: √18 = √(9×2) = 3√2 and √50 = √(25×2) = 5√2. Therefore their sum is 8√2. Squaring gives (8√2)^2 = 64×2 = 128. Option A is correct because it equals this result. Option B, 64, forgets the factor 2 produced by squaring √2. Option C, 16√2, is not the square of the sum; it is an unsimplified or incorrect form. Option D, 200, comes from an incorrect expansion or multiplication. A useful check is to use (a+b)^2: 18+50+2√(18×50) = 68+2√900 = 68+60 = 128. Memory cue: simplify radicals first, then square the common radical carefully.

Correct answer: B. (\frac{\sqrt{11}-\sqrt{7}}{2})

Explanation: Multiplying by the conjugate makes the denominator (11-7=4). So (\frac{2(\sqrt{11}-\sqrt{7})}{4}=\frac{\sqrt{11}-\sqrt{7}}{2}).

Correct answer: C. (17)

Explanation: (\sqrt{98}=7\sqrt{2}) and (\sqrt{200}=10\sqrt{2}), so (s=17\sqrt{2}). Dividing by (\sqrt{2}) gives (17).

Correct answer: C. (\frac{\sqrt{45}}{\sqrt{9}})

Explanation: The first, second, and fourth options give (5), (5), and (6) respectively. The third gives (\sqrt{5}), which is irrational.

Correct answer: A. \(5\sqrt{2}\) cm; it is irrational

Explanation: Side = \(\sqrt{50}=\sqrt{25\times2}=5\sqrt2\) cm. Since \(\sqrt2\) is irrational, multiplying it by non-zero rational 5 keeps it irrational. Exam tip: take the square root of the area.

Correct answer: A. \(\sqrt{2},\,-\sqrt{2}\)

Explanation: \(\sqrt{2}+(-\sqrt{2})=0\), and 0 is rational. Hence, the sum of two irrational numbers need not be irrational. In option B, the sum is \(3\sqrt{2}\), which is irrational. Exam tip: one counterexample is enough to disprove an “always” statement.

Correct answer: A. (\frac{23+8\sqrt{7}}{9})

Explanation: Multiplying by the conjugate gives denominator (16-7=9) and numerator ((4+\sqrt{7})^2=23+8\sqrt{7}). So the correct form is (\frac{23+8\sqrt{7}}{9}).

Correct answer: A. \(0.\overline{3}\)

Explanation: \(0.\overline{3}=1/3\), so it is non-terminating but repeating and hence rational. In contrast, \(\sqrt{2}\) and \(\pi\) are irrational. Exam tip: a non-terminating repeating decimal is always rational.

Correct answer: A. (5\sqrt{3})

Explanation: (\sqrt{147}=7\sqrt{3}), (\sqrt{75}=5\sqrt{3}), and (\sqrt{27}=3\sqrt{3}). Therefore the result is (5\sqrt{3}).