Real Numbers

Correct answer: C. (7)

Explanation: Multiplying the same square root by itself gives the number inside. Therefore ( \sqrt{7}\times\sqrt{7}=7 ).

Correct answer: A. 0.125 is rational because \(0.125=\frac{1}{8}\)

Explanation: The decimal expansion of 0.125 terminates. Writing it as \(125/1000\) and dividing by 125 gives \(1/8\), so it is rational. Option B is a misconception: terminating decimals are rational. Exam tip: a terminating or repeating decimal can be expressed as a fraction.

Correct answer: C. It is an irrational number

Explanation: The number is irrational because its decimal expansion is non-terminating and does not repeat in a fixed pattern; the number of zeros between successive 1s keeps increasing. Merely being written with digits does not make a number rational. A rational number has a terminating or recurring decimal expansion. Exam tip: classify an infinite non-repeating decimal as irrational.

Correct answer: C. (3) and (4)

Explanation: (3^2=9) and (4^2=16) so ( \sqrt{11} ) lies between (3) and (4). Use squares to locate square roots.

Correct answer: A. \(7\sqrt{2}\)

Explanation: Since \(98=49\times2\) and 49 is a perfect square, \(\sqrt{98}=\sqrt{49\times2}=\sqrt{49}\times\sqrt{2}=7\sqrt{2}\). Option C, \(2\sqrt{49}\), equals 14, not \(7\sqrt{2}\). To simplify a surd, factor out the largest perfect-square factor first.

Correct answer: B. \(6\sqrt{2}\)

Explanation: Since \(72=36\times2\) and \(36\) is a perfect square, \(\sqrt{72}=\sqrt{36\times2}=\sqrt{36}\times\sqrt{2}=6\sqrt{2}\). Therefore, option B is correct. The coefficients in options A and D are incorrect, while option C does not square to \(72\). Exam tip: factor the number using the largest perfect-square factor before simplifying a square root.

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

Explanation: Since \(63=9\times7\) and \(9\) is a perfect square, \(\sqrt{63}=\sqrt{9\times7}=\sqrt{9}\times\sqrt{7}=3\sqrt{7}\). Option A results from incorrectly treating \(7\) as a perfect square. In exams, identify the greatest perfect-square factor of the number under the radical.

Correct answer: D. 10

Explanation: The governing concept is the product rule for square roots: for non-negative numbers, √a × √b = √(ab). Applying it gives √2 × √50 = √(2 × 50) = √100 = 10, because the principal square root of 100 is 10. Therefore option D is correct. The result can also be checked by simplifying √50 first: √50 = √(25 × 2) = 5√2. Hence √2 × √50 = √2 × 5√2 = 5(√2)² = 5 × 2 = 10. Option A incorrectly treats the factors as if their square roots produced 25. Option B is only an intermediate expression before multiplying the remaining surds, so it is not the final value. Option C incorrectly adds the radicands instead of multiplying them.

Correct answer: A. (4)

Explanation: The quotient of square roots can be simplified by using \\(\frac{\sqrt{a}}{\sqrt{b}}=\sqrt{\frac{a}{b}}\\) when the denominator is positive. Here, both 80 and 5 are positive, so the rule applies directly. Dividing the numbers inside the radical gives \\(\sqrt{80}\div\sqrt{5}=\sqrt{\frac{80}{5}}=\sqrt{16}\\).

The square root of 16 is 4 because \\(4\times4=16\\). Therefore, option A is correct. Another valid route is to write \\(\sqrt{80}=\sqrt{16\times5}=4\sqrt{5}\\), and then divide by \\(\sqrt{5}\\), again obtaining 4. The other numerical choices result from incorrect treatment of the square roots.

Correct answer: B. \(9\sqrt{2}\)

Explanation: Both terms contain the same surd, \(\sqrt{2}\), so their coefficients can be added: \(4\sqrt{2}+5\sqrt{2}=(4+5)\sqrt{2}=9\sqrt{2}\). Option A uses an incorrect sum of coefficients. Option C equals \(9\sqrt{4}=18\), while option D simplifies to \(\sqrt{18}=3\sqrt{2}\), so neither is correct. Exam tip: coefficients can be added or subtracted only when the surd parts are identical.

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

Explanation: An infinite decimal can still be rational if it repeats. Here, the number of zeros between successive 1s keeps increasing, so no fixed repeating block exists. Exam tip: a non-terminating, non-repeating decimal is irrational.

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

Explanation: \(\sqrt{48}=\sqrt{16\times3}=4\sqrt{3}\) and \(\sqrt{27}=\sqrt{9\times3}=3\sqrt{3}\). Hence, \(\sqrt{48}+\sqrt{27}=4\sqrt{3}+3\sqrt{3}=7\sqrt{3}\), so option A is correct. Option B incorrectly treats the sum of two square roots as \(\sqrt{48+27}=\sqrt{75}\). Exam tip: simplify each radical first, then combine terms having the same radical part.

Correct answer: A. A terminating decimal is rational; \(0.375=\frac{3}{8}\)

Explanation: \(0.375=\frac{375}{1000}=\frac{3}{8}\), so it can be written as \(p/q\) and is rational. Option B is wrong because terminating decimals are also rational. Exam tip: terminating or recurring decimals are rational.

Correct answer: B. \(\frac{3\sqrt{2}}{2}\)

Explanation: To rationalise the denominator, multiply the numerator and denominator by \(\sqrt{2}\): \(\frac{3}{\sqrt{2}}\times\frac{\sqrt{2}}{\sqrt{2}}=\frac{3\sqrt{2}}{2}\). Therefore, option B is correct. In option A, the factor \(\sqrt{2}\) is incorrectly omitted from the numerator. Exam tip: When the denominator contains \(\sqrt{a}\), multiply both numerator and denominator by \(\sqrt{a}\).

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

Explanation: Multiplying numerator and denominator by ( \sqrt{7} ) gives ( \frac{5\sqrt{7}}{7} ). Rationalising the denominator is a common exam question.

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

Explanation: To rationalise a denominator containing a square root, multiply the fraction by the conjugate of the denominator. The conjugate of \\(2+\sqrt{3}\\) is \\(2-\sqrt{3}\\). Multiplying by \\(\frac{2-\sqrt{3}}{2-\sqrt{3}}\\) does not change the value, but it changes the denominator into a difference of squares.

Thus, \\(\frac{1}{2+\sqrt{3}}\times\frac{2-\sqrt{3}}{2-\sqrt{3}}=\frac{2-\sqrt{3}}{2^2-(\sqrt{3})^2}=\frac{2-\sqrt{3}}{4-3}=2-\sqrt{3}\\). The resulting denominator is 1, so it is rational. Therefore, option D is correct. The conjugate is essential; multiplying by the same expression would not remove the radical from the denominator.

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

Explanation: The conjugate of \(3-\sqrt{5}\) is \(3+\sqrt{5}\). Therefore, multiply the numerator and denominator by \(3+\sqrt{5}\): \(\frac{1}{3-\sqrt{5}}\times\frac{3+\sqrt{5}}{3+\sqrt{5}}=\frac{3+\sqrt{5}}{9-5}=\frac{3+\sqrt{5}}{4}\). Hence, the new numerator is \(3+\sqrt{5}\). Remember that the conjugate is formed by changing only the sign between the two terms; \(8\) is not the denominator, since \(9-5=4\).

Correct answer: B. −1

Explanation: The two factors are conjugates, so the difference-of-squares identity (a+b)(a−b) = a²−b² applies directly. Here a = 2 and b = √5. Thus (2+√5)(2−√5) = 2² − (√5)² = 4 − 5 = −1, making option B correct. Direct expansion confirms the result: 4 − 2√5 + 2√5 − 5. The middle terms cancel because they have equal magnitudes and opposite signs, leaving −1. Option A could arise from using the wrong sign between the squares. Option C ignores the fact that 2² is 4, and option D fails to simplify the conjugate product. The cancellation of the radical term is the key structural idea.

Correct answer: C. \(\pi\)

Explanation: \(\pi\) is irrational because it cannot be expressed as a ratio \(\frac{p}{q}\) of two integers with \(q\neq 0\). In contrast, \(0.25=\frac{1}{4}\), \(\sqrt{49}=7\), and \(-7\) are all rational numbers. Exam tip: the square root of a perfect square is rational, whereas \(\pi\) and, generally, the square root of a non-perfect square are irrational.

Correct answer: C. ( \sqrt{12} )

Explanation: Since (12>3), ( \sqrt{12}>\sqrt{3} ). For positive numbers, a larger radicand gives a larger square root.

Correct answer: C. ( -3 )

Explanation: ( \sqrt{10}\approx3.16 ), so ( -\sqrt{10} ) is less than ( -3 ). The negative number closer to zero is greater.

Correct answer: C. 1.2

Explanation: Since \(1.2^2=1.2\times1.2=1.44\), we have \(\sqrt{1.44}=1.2\). The principal square root is always taken as non-negative. The other options are incorrect because \(0.12^2=0.0144\), \(1.02^2=1.0404\), and \(12^2=144\). As an exam tip, square the proposed answer to check the decimal placement.

Correct answer: A. Any integer n can be written as \(\frac{n}{1}\).

Explanation: A rational number can be written as \(\frac{p}{q}\), where p and q are integers and \(q\ne0\). For example, \(-7=\frac{-7}{1}\), so every integer is rational. Non-terminating, non-recurring decimals are irrational. Exam tip: write an integer over 1 to test it.

Correct answer: C. \(\frac{5}{6}\)

Explanation: Using the square-root rule, \(\sqrt{\frac{25}{36}}=\frac{\sqrt{25}}{\sqrt{36}}=\frac{5}{6}\). Therefore, option C is correct. Option B, \(\frac{6}{5}\), is the reciprocal of the required value, not the value itself. Exam tip: the principal square root is taken as the non-negative value.

Correct answer: A. \(\frac{11}{12}\)

Explanation: Using the property of square roots, \(\sqrt{\frac{121}{144}}=\frac{\sqrt{121}}{\sqrt{144}}\). Since \(\sqrt{121}=11\) and \(\sqrt{144}=12\), the value is \(\frac{11}{12}\). Exam tip: For a positive fraction, take the square root of the numerator and denominator separately.