Irrational numbers

Correct answer: B. (-10)

Explanation: The direct answer is option B, −10. Simplify the expression from inside the brackets. First, √40 = √(4 × 10) = 2√10. Hence 3√10 − 2√40 = 3√10 − 2(2√10) = 3√10 − 4√10 = −√10. Now multiply by the outside factor: √10(−√10) = −(√10)² = −10. Option A, 10, misses the negative sign and is therefore wrong. Option B, −10, is correct because the bracket becomes negative and the product of √10 with √10 is 10. Option C, 30 − 4√10, is wrong because it comes from incomplete or incorrect expansion and is not the simplified value. Option D, −2√40, is wrong because it leaves the expression unsimplified and also does not include the contribution of the first term correctly. The key rule is to simplify radicals before multiplying: take out perfect-square factors such as 4 from under √40.

Correct answer: C. (\frac{6\sqrt{19}}{19})

Explanation: Multiplying numerator and denominator by (\sqrt{19}) gives (\frac{6\sqrt{19}}{19}). Rationalisation removes the radical from the denominator.

Correct answer: C. Both are equal

Explanation: (\sqrt{12}=2\sqrt{3}) and (\sqrt{108}=6\sqrt{3}), so the sum is (8\sqrt{3}). Find simplified forms before comparing.

Correct answer: C. Irrational

Explanation: It has no fixed repeating block, so it is irrational. Identifying non-terminating non-repeating decimals is important.

Correct answer: A. (\frac{5-\sqrt{6}}{19})

Explanation: The direct answer is option A, (5 − √6)/19. To rationalise a denominator means to remove the square root from the denominator. The conjugate of 5 + √6 is 5 − √6, so multiply numerator and denominator by it: 1/(5 + √6) × (5 − √6)/(5 − √6). The denominator becomes (5 + √6)(5 − √6) = 25 − 6 = 19. Thus the result is (5 − √6)/19. Option A is correct because its denominator is rational and the fraction is equivalent to the original. Option B, (5 + √6)/19, is wrong because the conjugate sign was not used correctly; direct multiplication does not produce that numerator. Option C, 5 − √6, is wrong because it omits the denominator 19. Option D, 1/(5 − √6), is only the reciprocal-looking conjugate expression; it is not equal to the original fraction and does not rationalise its denominator. The exam cue is: for a denominator a + √b, multiply by a − √b, and use the difference of squares.

Correct answer: A. Its decimal expansion is non-terminating and non-repeating, so it is irrational.

Explanation: The number of zeros between successive 1s increases as 1, 2, 3, 4... Hence, no fixed block repeats and the number is irrational. Exam tip: a non-terminating, non-repeating decimal is irrational.

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

Explanation: (\sqrt{48}=4\sqrt{3}) and (\sqrt{27}=3\sqrt{3}), so the sum is (7\sqrt{3}). Simplify before adding like radicals.

Correct answer: B. It is irrational because the increasing number of zeros prevents any fixed repeating block.

Explanation: The numbers of zeros between successive 1s are 1, 2, 3, 4, …, so no fixed digit block can repeat forever. A non-terminating, non-repeating decimal is irrational. In exams, check repetition, not merely whether the decimal is infinite.

Correct answer: C. (√6 − 1)/5

Explanation: The governing concept is rationalisation of a denominator containing a surd. To remove √6 + 1 from the denominator, multiply the numerator and denominator by its conjugate, √6 − 1. Thus 1/(√6 + 1) × (√6 − 1)/(√6 − 1) = (√6 − 1)/[(√6)^2 − 1^2]. Applying the difference-of-squares identity gives the denominator 6 − 1 = 5, so the rationalised form is (√6 − 1)/5. Therefore option C is correct. Option A uses the original expression instead of the conjugate, so it does not produce the required rational denominator. Option B leaves the denominator irrational, and option D is not the equivalent result of multiplying by the appropriate conjugate. The value is unchanged; only its representation has been altered.

Correct answer: A. \(\sqrt{36}=6\)

Explanation: Since \(\sqrt{36}=6=\frac{6}{1}\), it is rational, so Riya’s statement is false. \(2\), \(5\), and \(11\) are not perfect squares, so their square roots are irrational. Exam tip: check for perfect squares first.

Correct answer: B. \(\sqrt{3}+(-\sqrt{3})=0\)

Explanation: Both \(\sqrt{3}\) and \(-\sqrt{3}\) are irrational, but their sum is \(0\), a rational number. Hence the word “always” makes the claim false. Exam tip: one counterexample is enough to disprove a universal statement.

Correct answer: C. (2\sqrt{2})

Explanation: (\sqrt{98}=7\sqrt{2}) and (\sqrt{50}=5\sqrt{2}), so the difference is (2\sqrt{2}). First take out perfect-square factors.

Correct answer: C. (6\sqrt{7})

Explanation: (\sqrt{28}=2\sqrt{7}) and (\sqrt{63}=3\sqrt{7}), so the total is (6\sqrt{7}). Convert all terms to like radicals.

Correct answer: D. Irrational

Explanation: This decimal has no fixed repeating block, so it is irrational. Identify non-terminating non-repeating decimals.

Correct answer: A. \(6\)

Explanation: Using the quotient property of square roots, \(\frac{\sqrt{108}}{\sqrt{3}}=\sqrt{\frac{108}{3}}=\sqrt{36}=6\). Therefore, option A is correct. Option D represents only \(\sqrt{108}=6\sqrt{3}\) and does not complete the division by \(\sqrt{3}\). Exam tip: For positive radicands, use \(\frac{\sqrt{a}}{\sqrt{b}}=\sqrt{\frac{a}{b}}\) to simplify such expressions.

Correct answer: B. (15\sqrt{2})

Explanation: (\sqrt{162}=9\sqrt{2}) and (\sqrt{72}=6\sqrt{2}), so the sum is (15\sqrt{2}). Add coefficients of like radicals.

Correct answer: C. (\sqrt{25})

Explanation: Since (21<25<30), (\sqrt{25}) lies between them. Compare square roots using the numbers inside.

Correct answer: A. It is irrational because the number of zeros between successive 1s keeps increasing, so no fixed repeating block is formed.

Explanation: The zero blocks between successive 1s have lengths 1, 2, 3, 4, …, so no fixed block repeats. It is a non-terminating, non-recurring decimal and hence irrational. Exam tip: check repetition, not just the digits used.

Correct answer: A. \(0.\overline{27}\)

Explanation: \(0.\overline{27}=27/99=3/11\), so it is rational despite being non-terminating. Only non-terminating, non-repeating decimals are irrational. In exams, convert a repeating decimal into a fraction.

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

Explanation: Multiplying numerator and denominator by (\sqrt{3}) gives (\frac{7\sqrt{3}}{3}). Rationalisation removes the radical from the denominator.

Correct answer: B. 27

Explanation: The governing concept is simplification of surds before applying an exponent. First, √12 = √(4 × 3) = 2√3. Therefore √12 + √3 = 2√3 + √3 = 3√3. Squaring gives (3√3)^2 = 3^2(√3)^2 = 9 × 3 = 27, so option B is correct. The result can also be checked by using (a + b)^2 = a^2 + 2ab + b^2, but simplifying the radicals first is more efficient and avoids unnecessary expansion. Option A may result from adding the radicands, option D may result from forgetting that the radical is also squared, and option C is only an unsimplified or incorrect partial expression, not the value of the complete square. Thus 27 is the unique correct value.

Correct answer: C. It is irrational because its decimal expansion is non-terminating and non-repeating.

Explanation: The zeros between 1s increase as 1, 2, 3, 4, …, so no fixed block repeats. A non-terminating, non-repeating decimal is irrational. Exam tip: check repetition, not merely which digits occur.

Correct answer: A. (9\sqrt{3})

Explanation: (\sqrt{147}=7\sqrt{3}), (\sqrt{75}=5\sqrt{3}) and (\sqrt{27}=3\sqrt{3}), so the answer is (9\sqrt{3}). Add and subtract coefficients of like radicals.

Correct answer: A. \(\frac{1}{3}=0.333\ldots\)

Explanation: \(\frac{1}{3}=0.333\ldots\) has an infinite decimal expansion, but the digit 3 repeats, so it is rational. \(\sqrt{2}\) is non-terminating and non-repeating. Exam tip: recurring decimals are rational.

Correct answer: C. Irrational

Explanation: The square root of a positive integer is irrational when it is not a perfect square. Check for a perfect square first.