Real Numbers

Correct answer: B. The decimal expansion will be non-terminating recurring because the denominator 120, in lowest form, also has 3 as a factor.

Explanation: The fraction \(\frac{13}{120}\) is already in lowest terms because 13 and 120 have no common factor. A rational number has a terminating decimal expansion only when, in lowest form, its denominator has no prime factors other than 2 and/or 5. Here, \(120=2^3\times3\times5\); the factor 3 makes the decimal expansion non-terminating recurring. Option A is incorrect because the mere presence of 2 and 5 is not enough; no other prime factor may occur. Exam tip: first reduce the fraction, then prime-factorise its denominator.

Correct answer: C. 10

Explanation: \(a^2=(\sqrt{3}+\sqrt{2})^2=5+2\sqrt{6}\). Also, \((\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})=1\), so \(\frac{1}{a}=\sqrt{3}-\sqrt{2}\) and \(\frac{1}{a^2}=5-2\sqrt{6}\). Therefore, \(a^2+\frac{1}{a^2}=(5+2\sqrt{6})+(5-2\sqrt{6})=10\). The \(2\sqrt{6}\) terms cancel out. Exam tip: use the conjugate surd first to find the reciprocal in such questions.

Correct answer: A. 21

Explanation: Substituting \(x=-7\), we get \(x^2=(-7)^2=49\). Hence, \(\sqrt{x^2}=\sqrt{49}=7\), since the principal square root is non-negative. Also, \(-2x=-2(-7)=14\). Therefore, \(\sqrt{x^2}-2x=7+14=21\). Taking \(-7\) for the square root would be incorrect because \(\sqrt{x^2}=|x|\), not always \(x\). Exam tip: Rewrite \(\sqrt{x^2}\) as \(|x|\) to avoid sign errors.

Correct answer: A. ( \sqrt{20}-3 )

Explanation: The direct answer is option A: \(\sqrt{20}-3\). Since \(\sqrt{20}=2\sqrt{5}\) and \(20>9\), we have \(\sqrt{20}>3\). Therefore \(3-\sqrt{20}\) is negative. The absolute-value rule says \(|y|=-y\) when \(y<0\), so \(|3-\sqrt{20}|=-(3-\sqrt{20})=\sqrt{20}-3\). Option A has exactly this value. Option B forgets to change the sign. Option C changes a minus into a plus, which is not what absolute value does. Option D has the same positive terms but in the wrong order, so it represents \(\sqrt{20}+3\), a different number. Always check whether the expression inside absolute value is positive or negative first.

Correct answer: A. The sum of two irrational numbers is irrational.

Explanation: Statement A is not always true. For example, \(\sqrt{2}\) and \(-\sqrt{2}\) are both irrational, but their sum is \(0\), which is rational. Statement B is always true: if the sum of an irrational and a rational number were rational, subtracting the rational number would make the irrational number rational, which is impossible. Similarly, multiplying or dividing an irrational number by a non-zero rational number keeps it irrational. Exam tip: Never assume that the sum or difference of two irrational numbers must be irrational.

Correct answer: A. Both are equal

Explanation: ( \sqrt{27}+\sqrt{12}=3\sqrt{3}+2\sqrt{3}=5\sqrt{3} ). Also ( \sqrt{75}=5\sqrt{3} ), so both are equal.

Correct answer: A. यह दशमलव प्रसार असांत और अनावर्ती है, इसलिए यह अपरिमेय संख्या है।

Explanation: In this decimal, the number of zeros between successive 1s keeps increasing, so no fixed block of digits repeats. Therefore, its decimal expansion is non-terminating and non-repeating, which identifies an irrational number. Using only 0 and 1 does not make a number rational. Exam tip: A rational number has a decimal expansion that is either terminating or non-terminating recurring.

Correct answer: A. 2√2

Explanation: The governing concept is combining fractions with conjugate denominators and applying the difference-of-squares identity. Use the common denominator (√3−√2)(√3+√2). The product of these conjugates is (√3)² − (√2)² = 3 − 2 = 1. The numerator becomes (√3+√2) − (√3−√2) = √3 + √2 − √3 + √2 = 2√2. Hence the expression equals (2√2)/1 = 2√2, so option A is correct. Option B comes from failing to cancel the two √3 terms. Option C confuses multiplication of radicals with the subtraction required here, and option D omits the factor √2. The original denominators are valid because √3 and √2 are unequal, so neither difference nor sum creates a zero denominator.

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

Explanation: Taking a common denominator gives \((\sqrt{11}+3)(\sqrt{11}-3)=11-9=2\). The numerator is \((\sqrt{11}-3)+(\sqrt{11}+3)=2\sqrt{11}\). Hence, the value is \(\frac{2\sqrt{11}}{2}=\sqrt{11}\). The option \(2\sqrt{11}\) results from forgetting to divide by the denominator 2. Exam tip: use \((a+b)(a-b)=a^2-b^2\) to simplify such expressions quickly.

Correct answer: A. (26+15\sqrt{3})

Explanation: The direct answer is option A: \(26+15\sqrt{3}\). Use \((a+b)^2=a^2+2ab+b^2\): \((2+\sqrt{3})^2=4+4\sqrt{3}+3=7+4\sqrt{3}\). Multiply this by \(2+\sqrt{3}\): \((7+4\sqrt{3})(2+\sqrt{3})=14+7\sqrt{3}+8\sqrt{3}+12=26+15\sqrt{3}\). Option A matches. Option B is only the first-term cube expansion and misses cross terms. Option C has incorrect coefficients. Option D has the correct constant term but the radical coefficient is too small. The safe method is to square first and then multiply, collecting rational and radical terms separately.

Correct answer: A. (45-29\sqrt{2})

Explanation: The direct answer is A: \(45-29\sqrt2\). Use \((a-b)^3=a^3-3a^2b+3ab^2-b^3\), with \(a=3,b=\sqrt2\). Then \(3^3=27\), \(-3(3^2)(\sqrt2)=-27\sqrt2\), \(3(3)(\sqrt2)^2=9\cdot2=18\), and \(-(sqrt2)^3=-2\sqrt2\). Combining rational terms gives \(27+18=45\); combining surd terms gives \(-27\sqrt2-2\sqrt2=-29\sqrt2\). Thus A is correct. Option B, \(27-6\sqrt2\), is only a partial or incorrectly expanded result. Option C, \(33-11\sqrt2\), does not follow from the cube expansion. Option D, \(45-27\sqrt2\), includes the rational part correctly but misses the extra \(-2\sqrt2\) from the final term. Another check is to first square: \((3-sqrt2)^2=11-6\sqrt2\), then multiply by \(3-sqrt2\). Memory cue: in a cube, do not forget the fourth term \(-b^3\).

Correct answer: C. Non-terminating and non-repeating decimal expansion

Explanation: An irrational number has a non-terminating, non-repeating decimal expansion: its digits continue indefinitely without a fixed repeating pattern. Therefore, option C is correct. In option B, the decimal expansion is non-terminating but recurring, which represents a rational number. Exam tip: Decimal expansions of rational numbers are either terminating or non-terminating recurring.

Correct answer: A. \(225<m<256\)

Explanation: Since \(\sqrt{m}\) is positive, squaring every part of the inequality preserves the order. Squaring gives \(15^2<m<16^2\), i.e. \(225<m<256\). Option C (\(196<m<225\)) would correspond to squaring \(14<\sqrt{m}<15\) and so is incorrect; option B simply restates the original range for \(\sqrt{m}\) rather than for \(m\); option D corresponds to squaring \(16<\sqrt{m}<17\). Exam tip: when all terms are nonnegative you may square an inequality termwise — then compute the numerical squares to get the correct interval for the variable inside the square.

Correct answer: A. When (a) and (b) are both positive

Explanation: Generally ( \sqrt{a}+\sqrt{b} \neq \sqrt{a+b} ). For example, ( \sqrt{4}+\sqrt{9}=5 ), not ( \sqrt{13} ).

Correct answer: A. (a)

Explanation: Multiplying the square root of the same non-negative number by itself gives the number. Therefore the value is (a).

Correct answer: B. \(\sqrt{5}+(-\sqrt{5})\)

Explanation: Both \(\sqrt{5}\) and \(-\sqrt{5}\) are irrational, but \(\sqrt{5}+(-\sqrt{5})=0\), and 0 is rational. Hence, the sum of two irrational numbers need not always be irrational. The sums in A, C and D are irrational; in particular, \(\sqrt{2}+\sqrt{8}=3\sqrt{2}\), which is still irrational. Exam tip: To test an ‘always’ statement about irrational numbers, try a number and its additive inverse.

Correct answer: A. ( \sqrt{2}\times\sqrt{8} )

Explanation: ( \sqrt{2}\times\sqrt{8}=\sqrt{16}=4 ), which is rational. The other products do not become square roots of perfect squares.

Correct answer: A. 4

Explanation: Combine the two fractions using a common denominator. The denominator is (2 + √3)(2 − √3) = 4 − 3 = 1. The numerator is (2 − √3) + (2 + √3) = 4 because the surd terms cancel. Therefore the entire expression equals 4, making option A correct. Options B and D incorrectly retain a radical, while C ignores the numerator after rationalisation.

Correct answer: C. (15\sqrt{5})

Explanation: The direct answer is C: \(15\sqrt5\). To simplify a square root, separate the largest perfect-square factor. Factor 1125 as \(225\times5\), and since \(225=15^2\), \(\sqrt{1125}=\sqrt{225\times5}=\sqrt{225}\sqrt5=15\sqrt5\). Option C is correct. Option A, \(25\sqrt5\), would square to \(625\times5=3125\), not 1125. Option B, \(15\sqrt3\), would represent \(\sqrt{675}\), not \(\sqrt{1125}\); the remaining factor is 5, not 3. Option D, \(75\sqrt5\), is five times too large and also squares to the wrong number. A numerical check gives \(15\sqrt5\approx33.54\), while \(\sqrt{1125}\) is also about 33.54. The key rule is \(\sqrt{a^2b}=a\sqrt b\) for positive a. Memory cue: find the largest square factor before simplifying.

Correct answer: A. \(r+x\)

Explanation: \(r+x\) is always irrational. If \(r+x\) were rational, then \(x=(r+x)-r\) would be rational as the difference of two rational numbers. This contradicts the fact that \(x\) is irrational. \(rx\) is not always irrational because it becomes 0 when \(r=0\). Also, for \(x=\sqrt{2}\), \(x^2=2\) is rational, and \(x/x=1\). Exam tip: Adding or subtracting a rational number and an irrational number always gives an irrational number.

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

Explanation: (5\sqrt{20}=10\sqrt{5}), (2\sqrt{125}=10\sqrt{5}), and ( \sqrt{45}=3\sqrt{5} ). Therefore the result is (3\sqrt{5}).

Correct answer: B. (25\sqrt{3})

Explanation: (3\sqrt{75}=15\sqrt{3}), (4\sqrt{48}=16\sqrt{3}), and (2\sqrt{27}=6\sqrt{3}). Therefore (25\sqrt{3}) is correct.

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

Explanation: The student's statement is incorrect. In \(0.101001000100001\ldots\), the number of zeros between successive 1s keeps increasing, so no fixed block of digits repeats. Its decimal expansion is non-terminating and non-repeating; therefore, it is irrational. Having only the digits 0 and 1 does not make a number rational. Exam tip: A rational number has either a terminating or a non-terminating recurring decimal expansion.

Correct answer: A. (66+24\sqrt{6})

Explanation: The direct answer is option A: \(66+24\sqrt{6}\). Use the identity \((a+b)^2=a^2+2ab+b^2\). Here \(a=4\sqrt{3}\) and \(b=3\sqrt{2}\). First, \(a^2=(4\sqrt{3})^2=16\times3=48\). Next, \(b^2=(3\sqrt{2})^2=9\times2=18\). The middle term is \(2ab=2(4\sqrt{3})(3\sqrt{2})=24\sqrt{6}\). Adding gives \(48+24\sqrt{6}+18=66+24\sqrt{6}\). Option A is correct because it contains both square terms and the middle term. Option B, \(30+24\sqrt{6}\), has the wrong constant term; 48 and 18 add to 66, not 30. Option C, \(66+12\sqrt{6}\), halves the middle term incorrectly. Option D, \(48+18\sqrt{6}\), does not correctly combine either the constant and radical parts. Remember: square both terms, calculate twice their product, and then add all three parts.

Correct answer: C. \(x\) is irrational and \(x^2\) is irrational

Explanation: \(x^2=(\sqrt{2}+\sqrt{3})^2=5+2\sqrt{6}\). Since \(\sqrt{6}\) is irrational, \(5+2\sqrt{6}\) is also irrational. Hence, \(x^2\) is irrational. If \(x\) were rational, then \(x^2\) would have to be rational, which is a contradiction. Therefore, \(x\) is also irrational. Option B is the closest distractor, but \(x^2\) is not rational. Exam tip: To test whether a number can be rational, use the fact that the square of every rational number is rational.