Real Numbers

The governing concept is rationalisation of a denominator containing a surd. First, √9=3, so the expression becomes 1/(√10+3). Multiply numerator and denominator by the conjugate √10−3. The result is [1(√10−3)]/[(√10+3)(√10−3)]. Using the difference-of-squares identity, the denominator becomes (√10)²−3²=10−9=1. Therefore the fraction simplifies to √10−3, so option A is correct. Option B is only the original denominator, not the rationalised value. Option C introduces the incorrect denominator 19. Option D is an equivalent reciprocal-style expression before rationalisation, but its denominator still contains a surd and it is not the required simplified rationalised form. The calculation also confirms the sign and denominator exactly.

To rationalise the denominator, multiply the numerator and denominator by its conjugate, \(\sqrt{11}+\sqrt{2}\). The denominator becomes \((\sqrt{11}-\sqrt{2})(\sqrt{11}+\sqrt{2})=11-2=9\), while the numerator becomes \(\sqrt{11}+\sqrt{2}\). Hence, the rationalised form is \(\frac{\sqrt{11}+\sqrt{2}}{9}\). Exam tip: use the identity \((a-b)(a+b)=a^2-b^2\).

( \sqrt{98}=7\sqrt{2} ), ( \sqrt{162}=9\sqrt{2} ), and ( \sqrt{50}=5\sqrt{2} ). So the result is (11\sqrt{2}).

Since \(44=4\times11\) and \(99=9\times11\), we have \(\sqrt{44}=2\sqrt{11}\) and \(\sqrt{99}=3\sqrt{11}\). Therefore, \(2\sqrt{44}+3\sqrt{99}=2(2\sqrt{11})+3(3\sqrt{11})=4\sqrt{11}+9\sqrt{11}=13\sqrt{11}\). Hence, option B is correct. Options A and C use incorrect coefficients, while option D changes the radicand incorrectly. Exam tip: simplify each radical first, then combine like radicals.

Since \(112=16\times7\), \(\sqrt{112}=4\sqrt{7}\). Also, \(175=25\times7\), so \(2\sqrt{175}=2\times5\sqrt{7}=10\sqrt{7}\). Therefore, \(4\sqrt{7}+10\sqrt{7}=14\sqrt{7}\), making option B correct. Exam tip: after reducing surds to the same radical, add or subtract their coefficients only.

\(\sqrt{48}=\sqrt{16\times3}=4\sqrt{3}\), so \(5\sqrt{48}=20\sqrt{3}\). Similarly, \(\sqrt{75}=\sqrt{25\times3}=5\sqrt{3}\), so \(3\sqrt{75}=15\sqrt{3}\). Therefore, \(20\sqrt{3}-15\sqrt{3}=5\sqrt{3}\), making option A correct. In such questions, first simplify each surd and then combine only like surd terms.

The governing concept is the product rule for square roots: √a × √b = √(ab) for non-negative a and b. Applying it gives √32 × √50 = √(32 × 50) = √1600. Since 1600 = 40², its principal square root is 40. A second verification is possible by simplifying each radical: √32 = √(16×2) = 4√2 and √50 = √(25×2) = 5√2. Their product is (4√2)(5√2) = 20×2 = 40. Thus option A is correct. Option B doubles the answer, option C is an incomplete or incorrect simplification because the two √2 factors must also be multiplied, and option D adds radicands instead of multiplying them.

Direct answer: Option B, 3. Rewrite the expression as \(\frac{\sqrt{180}}{\sqrt{20}}\). Since both radicands are positive, \(\frac{\sqrt a}{\sqrt b}=\sqrt{\frac ab}\). Therefore \(\frac{\sqrt{180}}{\sqrt{20}}=\sqrt{\frac{180}{20}}=\sqrt9=3\). Another check is to simplify first: \(180=9\times20\), so \(\sqrt{180}=\sqrt{9\times20}=3\sqrt{20}efore dividing by \(\sqrt{20}\), giving 3. Option A, 9, is the radicand quotient before taking the square root; it stops one step too early. Option B is correct because \(\sqrt9=3\). Option C, \(\sqrt{160}\), incorrectly subtracts or combines the radicands instead of dividing them. Option D, 6, has no valid calculation and is not the result. The denominator is non-zero, so the division is defined. Remember: when dividing square roots of positive numbers, divide the radicands and then take the square root.

\(\frac{\sqrt{200}}{\sqrt{2}}=\sqrt{\frac{200}{2}}=\sqrt{100}=10\), and \(\sqrt{16}=4\). Therefore, the value is \(10+4=14\). Option B incorrectly reflects a subtraction-based calculation rather than evaluating the quotient and then adding \(\sqrt{16}\). In the exam, remember that for positive radicands, \(\frac{\sqrt{a}}{\sqrt{b}}=\sqrt{\frac{a}{b}}\).

Since \(4.84=(2.2)^2\), we have \(\sqrt{4.84}=2.2\). Similarly, \(1.44=(1.2)^2\), so \(\sqrt{1.44}=1.2\). Therefore, the required value is \(2.2-1.2=1\). Options B and D are only the values of the second and first square roots respectively, while option C is not the correct difference. Exam tip: Recognise decimal numbers as squares of terminating decimals before subtracting.

The first term is ( \frac{4}{5} ) and the second is ( \frac{3}{8} ). The sum is ( \frac{32}{40}+\frac{15}{40}=\frac{47}{40} ).

( \sqrt{\frac{81}{121}}=\frac{9}{11} ) and ( \sqrt{\frac{4}{49}}=\frac{2}{7} ). The difference is ( \frac{63}{77}-\frac{22}{77}=\frac{41}{77} ).

Since (5^2<31<6^2), ( \sqrt{31} ) lies between (5) and (6). Since (31) is not a perfect square, it is irrational.

Compare the squares of the two positive numbers: \((\sqrt{45})^2=45\), whereas \(\left(\frac{13}{2}\right)^2=\frac{169}{4}=42.25\). Since \(45>42.25\), we have \(\sqrt{45}>\frac{13}{2}\). Therefore, option A is correct. Exam tip: For positive square roots or positive numbers, comparing their squares is often quicker than using decimal approximations.

(8^2=64) and (9^2=81), so ( \sqrt{68} ) lies between (8) and (9). Compare nearby perfect squares.

The governing concept is locating a principal square root by comparing the number with consecutive perfect squares. The nearby perfect squares are 9² = 81 and 10² = 100. Because 81 < 86 < 100, taking the positive square root preserves the order and gives 9 < √86 < 10. Therefore √86 lies between 9 and 10, making option C correct. Its approximate value is about 9.27, which provides an additional check. The interval 7 to 8 is too low because 8² is only 64. The interval 8 to 9 is also too low because 9² is 81. The interval 10 to 11 is too high because √86 is less than 10.

( \sqrt{50}\approx7.07 ), so ( -\sqrt{50}\approx-7.07 ). It is greater than ( -7.2 ) because it is closer to zero.

A non-terminating decimal without regular repetition is irrational. Only fixed repeating decimals are rational.

A rational number has a terminating decimal when, after reducing the fraction, the denominator has no prime factors other than 2 and 5. This happens because powers of 2 and 5 can be multiplied to make a power of 10. Therefore the decimal form in this question ends after a finite number of digits, so option A is correct. It is not irrational or undefined.

Here, the denominator is 80, and the factorisation is \(80=2^4\times5\). The numerator 21 has no common factor with 80, so the fraction is already in simplest form. Since its denominator contains only 2 and 5, \(21/80=0.2625\), which terminates. Thus the correct statement is that it is a terminating decimal.

The diagonal of the square is \(\sqrt{1^2+1^2}=\sqrt{2}\) cm. Since 2 is not a perfect square, \(\sqrt{2}\) cannot be expressed exactly as a ratio of two integers; therefore, it is irrational. It is neither an integer nor a natural number, so options C and D are incorrect. Exam tip: the square root of a non-perfect square is generally irrational.

The diagonal of a square is given by side × \(\sqrt{2}\), so its length is \(5\sqrt{2}\) cm. Since \(\sqrt{2}\) is irrational and 5 is a non-zero rational number, \(5\sqrt{2}\) is also irrational. Therefore, option B is correct. Exam tip: multiplying an irrational number by a non-zero rational number keeps the result irrational.

In the decimal expansion \(0.101001000100001\ldots\), the number of zeros between successive 1s keeps increasing, so no fixed block repeats regularly. Hence it is a non-terminating, non-recurring decimal and therefore irrational. The use of only the digits 0 and 1 does not make a number rational. Exam tip: a decimal is rational only if it terminates or eventually repeats a fixed pattern.

The decimal expansion \(0.101001000100001\ldots\) is non-terminating, and the number of zeros between successive 1s keeps increasing. Hence, no fixed block of digits repeats regularly. A rational number has either a terminating or an infinitely repeating decimal expansion, so this number is irrational. Exam tip: an infinite decimal is not automatically irrational; an infinite recurring decimal, such as \(0.333\ldots\), is rational.

The two expressions are conjugates, so use \((a+b)(a-b)=a^2-b^2\): \((3+\sqrt{14})(3-\sqrt{14})=3^2-(\sqrt{14})^2=9-14=-5\). Option A results from missing the negative sign. Exam tip: whenever conjugate pairs appear, apply the difference-of-squares identity directly.

Multiplying by the conjugate ( \sqrt{7}+2 ) makes the denominator (7-4=3). Use difference of squares in the conjugate method.