CBSE Class 9 Mathematics Number System Worksheet Set 07

Question. If 20% of your Maths score is equal to 30% of my Maths score on a test, which of the following best describes the relation between our scores?
A. Your score is 1(1/2) times mine.
B. My score is 1/2 times yours
C. You scored 10 marks more than me.
D. I scored 10% more than you.

Answer : A

Question. Which of the following statements can be true for a natural number N?
A. When N is divided by 3, the remainder is 3 and when N is divided by 6, the remainder is 0.
B. When N is divided by 3, the remainder is 0 and when N is divided by 6, the remainder is 3.
C. When N is divided by 3, the remainder is 1 and when N is divided by 6, the remainder is 0.
D. All the three statements can be true.

Answer : B

Question. The numbers in the (n + 1)th row of the Pascal's triangle also represent the coefficients of the terms in the expansion of  (x + y)ⁿ. For instance: (x + y)² = 1x² + 2xy + 1y², (x+y)³ = 1x³ + 3x²y + 3xy² + 1y³ .The 6th row of the triangle would represent the coefficients of which of these expressions?
A. (x + y)⁵
B. (x + 5y)⁶
C. (x + y)⁶
D. (x + y)⁴

Answer : A

Question. Which of the following is closest to 8.413/0.19 ?
A. 0.44
B. 4.4
C. 44
D. 440

Answer : C

Question. What decimal fraction of the figure below is shaded

A. 0.45
B. 0.045
C. 0.09
D. 0.18

Answer : D

Question. Scientific Notation is a concise way of writing any number in the form a x 10b where a is a number between 1 and 10 and b is an integer. For example, 652 would be written as 6.52 x 102 in this notation. Which of the following is most likely to contain a negative exponent of 10 when represented in scientific notation?
A. Number of cells in a honeycomb
B. Speed of light in kilometres per second
C. Diameter of a blood cell in centimetres
D. Mass of an airplane in kilograms

Answer : D

Question. In which of the following lists are ALL three numbers irrational?
A. 0, -3, 3.14
B. 22/7 , 0.0141414....., √3
C. √2,√3,√4
D. 3.919119111911119....,√5   , 

Answer : D

Question. If x is such that it satisfies BOTH the inequalities: -2 <x and x≤ 2      which of the following number lines correctly depicts the values of x?

Answer : A

Question. Which of the following numbers will have a '6' in the unit's place?
A. 2⁶
B. 2¹⁶
C. 2²⁶
D. 2⁴⁶  

Answer : B

Question. The set of Natural numbers is said to be 'CLOSED' under the operation '+' because for any Natural numbers m and n, the number m + n is also a Natural number. Under which operations is the set of Integers 'CLOSED'?
A. +, -, x and &divide;
B. +, - and x
C. only + and x
D. only + and -

Answer : B

Question. If p is a non-zero Rational number, which of the following might NOT be Rational?
A. √3p
B. p^-3
C. -p³
D. 3/p

Answer : A

Question. Which of the following is equal to 1?
A. 17^-2 + 17²
B. 17² - 17^-2
C. 17^-2 x 17²
D. 17²/17^-2

Answer : C

Question. Which of these numbers can be written as a non-terminating, repeating decimal?
A. √3p
B. √3
C. 3/24
D. 1/12

Answer : A

Question. If the square of  ‘x’ is 5, then square of ‘4x’ will be
A. 400
B. 80
C. 60
D. 20

Answer : B
 

Work Sheet - Number System

Question 1. Find 2 rational nos. and 2 irrational nos. between
a) 1/3 & 1/4
b) 1.3 & 1.4
c) 0.2 & 0.2̄
d) 1 & √2
Answer:
a) Between \( \frac{1}{4} = 0.25 \) and \( \frac{1}{3} = 0.3333\dots \):
Two rational numbers: \( 0.26 = \frac{13}{50} \) and \( 0.30 = \frac{3}{10} \).
Two irrational numbers: \( 0.26010010001\dots \) and \( 0.27010010001\dots \)

b) Between \( 1.3 \) and \( 1.4 \):
Two rational numbers: \( 1.31 = \frac{131}{100} \) and \( 1.35 = \frac{27}{20} \).
Two irrational numbers: \( 1.31010010001\dots \) and \( 1.32010010001\dots \)

c) Between \( 0.2 = 0.2000\dots \) and \( 0.\overline{2} = 0.2222\dots \):
Two rational numbers: \( 0.21 = \frac{21}{100} \) and \( 0.215 = \frac{43}{200} \).
Two irrational numbers: \( 0.21010010001\dots \) and \( 0.215010010001\dots \)

d) Between \( 1 \) and \( \sqrt{2} \approx 1.4142\dots \):
Two rational numbers: \( 1.1 = \frac{11}{10} \) and \( 1.2 = \frac{6}{5} \).
Two irrational numbers: \( 1.2010010001\dots \) and \( 1.3010010001\dots \)
In simple words: Rational numbers have decimals that terminate or repeat, while irrational numbers have non-terminating and non-repeating decimal expansions.

Exam Tip: To create an irrational number easily, write a non-repeating pattern of zeros such as \( 0.1010010001\dots \)

Question 2. Express the following nos. in the form p/q where p & q are integers and q ≠ 0.
a) 0.6̄
b) 1.307̄
c) 0.023̄
d) 3.37̄9̄
Answer:
a) Let \( x = 0.\overline{6} = 0.666\dots \)
Multiplying by 10:
\( 10x = 6.666\dots \)
Subtracting the first equation from the second:
\( 9x = 6 \)
\( \implies x = \frac{6}{9} = \frac{2}{3} \).

b) Let \( x = 1.\overline{307} = 1.307307\dots \)
Multiplying by 1000:
\( 1000x = 1307.307\dots \)
Subtracting:
\( 999x = 1306 \)
\( \implies x = \frac{1306}{999} \).

c) Let \( x = 0.02\overline{3} = 0.02333\dots \)
Multiply by 100 to shift non-repeating digits:
\( 100x = 2.\overline{3} \)
Multiply by 1000:
\( 1000x = 23.\overline{3} \)
Subtracting gives:
\( 900x = 21 \)
\( \implies x = \frac{21}{900} = \frac{7}{300} \).

d) Let \( x = 3.3\overline{79} = 3.37979\dots \)
Multiply by 10:
\( 10x = 33.\overline{79} \)
Multiply by 1000:
\( 1000x = 3379.\overline{79} \)
Subtracting gives:
\( 990x = 3346 \)
\( \implies x = \frac{3346}{990} = \frac{1673}{495} \).
In simple words: Multiply by powers of 10 to line up the repeating parts, subtract to cancel the infinite decimal, and simplify the fraction.

Exam Tip: Always reduce the final fraction \( \frac{p}{q} \) to its lowest terms to ensure full credit.

Question 3. Represent on the number line geometrically.
a) √7
b) √17
c) √6.5
Answer:
a) Representation of \( \sqrt{7} \):
By Pythagoras theorem, \( \sqrt{7} = \sqrt{(\sqrt{6})^2 + 1^2} \).
First, construct \( \sqrt{4} = 2 \) units along the line, erect a perpendicular of 1 unit to get \( \sqrt{5} \). Repeat successively to obtain \( \sqrt{6} \) and then \( \sqrt{7} \).

b) Representation of \( \sqrt{17} \):
Express 17 as the sum of two squares: \( 17 = 4^2 + 1^2 \).
Take 4 units along the baseline from 0 to point A. Draw a perpendicular line segment AB of length 1 unit at A. By Pythagoras theorem:
\( \text{OB} = \sqrt{4^2 + 1^2} = \sqrt{17} \).
With O as centre and radius OB, sweep an arc down to meet the number line at \( \sqrt{17} \approx 4.12 \).

c) Representation of \( \sqrt{6.5} \):
1. Mark segment AB = 6.5 units on a straight line.
2. Extend AB by 1 unit to point C such that BC = 1 unit.
3. Find the midpoint M of AC, where \( \text{AC} = 7.5 \) units and \( \text{MC} = 3.75 \) units.
4. Draw a semicircle with centre M and radius MA.
5. Erect a perpendicular at B intersecting the semicircle at D. The length \( \text{BD} = \sqrt{6.5} \).
6. With B as origin (0) and radius BD, draw an arc cutting the number line at point E, which represents \( \sqrt{6.5} \). In simple words: Mark the length plus 1 unit, draw a semicircle over it, and raise a perpendicular at the join to get the square root.

Exam Tip: For decimals like \( \sqrt{x} \), always use the geometric semicircle method where diameter is \( x + 1 \).

Question 4. Evaluate
a) (1³ + 2³ + 3³)^{-1/2}
b) \frac{(25)^{5/2} \times (729)^{1/2} \times (8)^{-2/3}}{(125)^{2/3} \times (27)^{2/3} \times (225)^{3/2}}
c) \frac{1}{1+\sqrt{2}} + \frac{1}{\sqrt{2}+\sqrt{3}} + \frac{1}{\sqrt{3}+\sqrt{4}} + \dots + \frac{1}{\sqrt{8}+\sqrt{9}}
Answer:
a) Evaluate the sum of cubes inside the parentheses:
\( 1^3 + 2^3 + 3^3 = 1 + 8 + 27 = 36 \)
\( (36)^{-1/2} = \frac{1}{\sqrt{36}} = \frac{1}{6} \).

b) Express all bases in prime factor form:
\( (25)^{5/2} = (5^2)^{5/2} = 5^5 \)
\( (729)^{1/2} = (3^6)^{1/2} = 3^3 \)
\( (8)^{-2/3} = (2^3)^{-2/3} = 2^{-2} = \frac{1}{2^2} \)
Denominator:
\( (125)^{2/3} = (5^3)^{2/3} = 5^2 \)
\( (27)^{2/3} = (3^3)^{2/3} = 3^2 \)
\( (225)^{3/2} = (15^2)^{3/2} = 15^3 = (3 \times 5)^3 = 3^3 \times 5^3 \)
Combine powers:
Numerator = \( \frac{5^5 \times 3^3}{4} \)
Denominator = \( 5^2 \times 3^2 \times 3^3 \times 5^3 = 5^5 \times 3^5 \)
Dividing numerator by denominator:
\( = \frac{5^5 \times 3^3}{4 \times 5^5 \times 3^5} = \frac{1}{4 \times 3^2} = \frac{1}{36} \).

c) Rationalize each term using \( \frac{1}{\sqrt{n} + \sqrt{n+1}} = \sqrt{n+1} - \sqrt{n} \):
\( \frac{1}{1+\sqrt{2}} = \sqrt{2} - 1 \)
\( \frac{1}{\sqrt{2}+\sqrt{3}} = \sqrt{3} - \sqrt{2} \)
\( \dots \)
\( \frac{1}{\sqrt{8}+\sqrt{9}} = \sqrt{9} - \sqrt{8} \)
Adding all rationalized terms, intermediate terms cancel out in a telescoping sum:
\( = (\sqrt{2} - 1) + (\sqrt{3} - \sqrt{2}) + \dots + (\sqrt{9} - \sqrt{8}) \)
\( = \sqrt{9} - 1 = 3 - 1 = 2 \).
In simple words: Change bases to prime powers to cancel common factors, and rationalize terms in long sums to watch intermediate values cancel.

Exam Tip: In telescoping radical sums, only the very first and very last terms survive after cancellation.

Question 5. Simplify by rationalising the denominator.
a) \frac{30}{5\sqrt{3} - 3\sqrt{5}}
b) \frac{1}{\sqrt{3} - \sqrt{2} - 1}
c) \frac{3\sqrt{2}}{\sqrt{6} - \sqrt{3}} - \frac{4\sqrt{3}}{\sqrt{6} - \sqrt{2}} + \frac{2\sqrt{3}}{2 + \sqrt{6}}
d) \frac{3\sqrt{6}}{5\sqrt{3} - \sqrt{32} + 2\sqrt{12} - \sqrt{50}}
Answer:
a) Multiply numerator and denominator by the conjugate \( 5\sqrt{3} + 3\sqrt{5} \):
\( \frac{30(5\sqrt{3} + 3\sqrt{5})}{(5\sqrt{3})^2 - (3\sqrt{5})^2} = \frac{30(5\sqrt{3} + 3\sqrt{5})}{75 - 45} = \frac{30(5\sqrt{3} + 3\sqrt{5})}{30} = 5\sqrt{3} + 3\sqrt{5} \).

b) Group as \( (\sqrt{3} - 1) - \sqrt{2} \) and multiply by \( (\sqrt{3} - 1) + \sqrt{2} \):
\( \frac{(\sqrt{3} - 1) + \sqrt{2}}{[(\sqrt{3} - 1) - \sqrt{2}][(\sqrt{3} - 1) + \sqrt{2}]} = \frac{\sqrt{3} - 1 + \sqrt{2}}{(\sqrt{3} - 1)^2 - 2} = \frac{\sqrt{3} + \sqrt{2} - 1}{4 - 2\sqrt{3} - 2} = \frac{\sqrt{3} + \sqrt{2} - 1}{2 - 2\sqrt{3}} \)
Alternatively, grouping as \( \sqrt{3} - (\sqrt{2} + 1) \):
Multiply numerator and denominator by \( \sqrt{3} + (\sqrt{2} + 1) \):
Denominator: \( 3 - (\sqrt{2} + 1)^2 = 3 - (3 + 2\sqrt{2}) = -2\sqrt{2} \).
Multiplying numerator and denominator by \( \sqrt{2} \):
\( = \frac{(\sqrt{3} + \sqrt{2} + 1)\sqrt{2}}{-2\sqrt{2} \times \sqrt{2}} = -\frac{1}{4}(\sqrt{6} + 2 + \sqrt{2}) \).

c) Rationalize each term individually:
Term 1: \( \frac{3\sqrt{2}(\sqrt{6} + \sqrt{3})}{6 - 3} = \frac{3\sqrt{2}(\sqrt{6} + \sqrt{3})}{3} = \sqrt{12} + \sqrt{6} = 2\sqrt{3} + \sqrt{6} \)
Term 2: \( \frac{4\sqrt{3}(\sqrt{6} + \sqrt{2})}{6 - 2} = \frac{4\sqrt{3}(\sqrt{6} + \sqrt{2})}{4} = \sqrt{18} + \sqrt{6} = 3\sqrt{2} + \sqrt{6} \)
Term 3: \( \frac{2\sqrt{3}(\sqrt{6} - 2)}{6 - 4} = \frac{2\sqrt{3}(\sqrt{6} - 2)}{2} = \sqrt{18} - 2\sqrt{3} = 3\sqrt{2} - 2\sqrt{3} \)
Combining:
\( (2\sqrt{3} + \sqrt{6}) - (3\sqrt{2} + \sqrt{6}) + (3\sqrt{2} - 2\sqrt{3}) = 0 \).

d) Simplify the terms in the denominator first:
\( \sqrt{32} = 4\sqrt{2} \), \( 2\sqrt{12} = 4\sqrt{3} \), \( \sqrt{50} = 5\sqrt{2} \)
Denominator = \( 5\sqrt{3} - 4\sqrt{2} + 4\sqrt{3} - 5\sqrt{2} = 9\sqrt{3} - 9\sqrt{2} = 9(\sqrt{3} - \sqrt{2}) \).
Expression = \( \frac{3\sqrt{6}}{9(\sqrt{3} - \sqrt{2})} = \frac{\sqrt{6}}{3(\sqrt{3} - \sqrt{2})} \)
Multiply numerator and denominator by \( \sqrt{3} + \sqrt{2} \):
\( = \frac{\sqrt{6}(\sqrt{3} + \sqrt{2})}{3(3 - 2)} = \frac{\sqrt{18} + \sqrt{12}}{3} = \frac{3\sqrt{2} + 2\sqrt{3}}{3} = \sqrt{2} + \frac{2\sqrt{3}}{3} \).
In simple words: Multiply top and bottom by the conjugate expression to eliminate square roots from denominators.

Exam Tip: Simplify individual surds such as \( \sqrt{32} = 4\sqrt{2} \) and \( \sqrt{50} = 5\sqrt{2} \) before attempting to rationalize a multi-term denominator.

Question 6. If x = 3+√8 find the value of x² + 1/x² and x³ - 1/x³
Answer: Given \( x = 3 + \sqrt{8} \).
Calculate \( \frac{1}{x} \):
\( \frac{1}{x} = \frac{1}{3 + \sqrt{8}} = \frac{3 - \sqrt{8}}{3^2 - (\sqrt{8})^2} = \frac{3 - \sqrt{8}}{9 - 8} = 3 - \sqrt{8} \).
Then:
\( x + \frac{1}{x} = (3 + \sqrt{8}) + (3 - \sqrt{8}) = 6 \)
\( x - \frac{1}{x} = (3 + \sqrt{8}) - (3 - \sqrt{8}) = 2\sqrt{8} = 4\sqrt{2} \).

Now, calculate \( x^2 + \frac{1}{x^2} \):
\( x^2 + \frac{1}{x^2} = \left(x + \frac{1}{x}\right)^2 - 2 = 6^2 - 2 = 36 - 2 = 34 \).

Next, calculate \( x^3 - \frac{1}{x^3} \):
\( x^3 - \frac{1}{x^3} = \left(x - \frac{1}{x}\right)^3 + 3\left(x - \frac{1}{x}\right) \)
\( = (2\sqrt{8})^3 + 3(2\sqrt{8}) = 8(8\sqrt{8}) + 6\sqrt{8} = 70\sqrt{8} = 70(2\sqrt{2}) = 140\sqrt{2} \).
In simple words: Find \( 1/x \) by rationalizing, then use standard algebraic identities with \( (x + 1/x) \) and \( (x - 1/x) \).

Exam Tip: Notice that \( 3^2 - (\sqrt{8})^2 = 1 \), so \( \frac{1}{x} \) is simply the conjugate \( 3 - \sqrt{8} \).

Question 7. If x = \frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}} and y = \frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}} find x³ + y³
Answer: Rationalize \( x \) and \( y \):
\( x = \frac{(\sqrt{3} - \sqrt{2})^2}{3 - 2} = 3 + 2 - 2\sqrt{6} = 5 - 2\sqrt{6} \)
\( y = \frac{(\sqrt{3} + \sqrt{2})^2}{3 - 2} = 3 + 2 + 2\sqrt{6} = 5 + 2\sqrt{6} \)
Now find the sum and product:
\( x + y = (5 - 2\sqrt{6}) + (5 + 2\sqrt{6}) = 10 \)
\( xy = (5 - 2\sqrt{6})(5 + 2\sqrt{6}) = 25 - 24 = 1 \)
Using the identity \( x^3 + y^3 = (x + y)^3 - 3xy(x + y) \):
\( x^3 + y^3 = (10)^3 - 3(1)(10) = 1000 - 30 = 970 \).
In simple words: Simplify both fractions first to find their sum and product, then evaluate using the sum of cubes formula.

Exam Tip: Do not cube the radical expressions directly; always use the identity \( (x+y)^3 - 3xy(x+y) \).

Question 8. If a^x = b^y = c^z and b² = ac, then S.T. y = \frac{2xz}{x+z}
Answer: Let \( a^x = b^y = c^z = k \).
Express \( a, b, c \) in terms of \( k \):
\( a = k^{1/x}, \quad b = k^{1/y}, \quad c = k^{1/z} \).
Substitute into the given equation \( b^2 = ac \):
\( (k^{1/y})^2 = (k^{1/x})(k^{1/z}) \)
\( k^{2/y} = k^{1/x + 1/z} \)
Equating the exponents since bases are equal:
\( \frac{2}{y} = \frac{1}{x} + \frac{1}{z} \)
\( \frac{2}{y} = \frac{x + z}{xz} \)
Taking the reciprocal on both sides:
\( \implies y = \frac{2xz}{x + z} \).
Hence proved.
In simple words: Equate all powers to a single constant k, express the bases as fractional powers, and equate exponents.

Exam Tip: State clearly that since the base \( k \neq 0, 1 \), the exponents on both sides must be equal.

Question 9. S.T. (x^{a-b})^{a+b} × (x^{b-c})^{b+c} × (x^{c-a})^{c+a} = 1
Answer: Applying the law of exponents \( (x^m)^n = x^{mn} \):
\( \text{LHS} = x^{(a-b)(a+b)} \times x^{(b-c)(b+c)} \times x^{(c-a)(c+a)} \)
Using the identity \( (u - v)(u + v) = u^2 - v^2 \):
\( = x^{a^2 - b^2} \times x^{b^2 - c^2} \times x^{c^2 - a^2} \)
Using the product rule \( x^p \times x^q \times x^r = x^{p+q+r} \):
\( = x^{(a^2 - b^2) + (b^2 - c^2) + (c^2 - a^2)} \)
\( = x^0 = 1 = \text{RHS} \).
Hence proved.
In simple words: Multiply powers using the difference of squares, then add exponents together to get power zero, which equals 1.

Exam Tip: Recall that any non-zero base raised to the power zero equals 1: \( x^0 = 1 \).

Question 10. S.T. \frac{a^{-1}}{a^{-1} + b^{-1}} - \frac{a^{-1}}{a^{-1} - b^{-1}} = \frac{2ab}{a^2 - b^2}
Answer: Rewrite negative exponents as reciprocals:
\( a^{-1} = \frac{1}{a} \) and \( b^{-1} = \frac{1}{b} \).
Then:
\( \frac{a^{-1}}{a^{-1} + b^{-1}} = \frac{\frac{1}{a}}{\frac{1}{a} + \frac{1}{b}} = \frac{\frac{1}{a}}{\frac{b + a}{ab}} = \frac{b}{a + b} \)
Similarly:
\( \frac{a^{-1}}{a^{-1} - b^{-1}} = \frac{\frac{1}{a}}{\frac{1}{a} - \frac{1}{b}} = \frac{\frac{1}{a}}{\frac{b - a}{ab}} = \frac{b}{b - a} = -\frac{b}{a - b} \)
Substitute into LHS:
\( \text{LHS} = \frac{b}{a + b} - \left(-\frac{b}{a - b}\right) = \frac{b}{a + b} + \frac{b}{a - b} \)
Find a common denominator:
\( = \frac{b(a - b) + b(a + b)}{(a + b)(a - b)} = \frac{ab - b^2 + ab + b^2}{a^2 - b^2} = \frac{2ab}{a^2 - b^2} = \text{RHS} \).
Hence proved.
In simple words: Replace negative powers with fractions, simplify the compound fractions, and add them with a common denominator.

Exam Tip: Watch the minus sign in the denominator: \( \frac{b}{b - a} = -\frac{b}{a - b} \).

Question 11. If \frac{7+\sqrt{5}}{7-\sqrt{5}} - \frac{7-\sqrt{5}}{7+\sqrt{5}} = a + b\sqrt{5} find the value of a & b.
Answer: Simplify the left-hand side by taking a common denominator:
\( \text{LHS} = \frac{(7 + \sqrt{5})^2 - (7 - \sqrt{5})^2}{(7 - \sqrt{5})(7 + \sqrt{5})} \)
Using the identity \( (x + y)^2 - (x - y)^2 = 4xy \):
Numerator = \( 4(7)(\sqrt{5}) = 28\sqrt{5} \)
Denominator = \( 7^2 - (\sqrt{5})^2 = 49 - 5 = 44 \)
\( \text{LHS} = \frac{28\sqrt{5}}{44} = \frac{7\sqrt{5}}{11} = 0 + \frac{7}{11}\sqrt{5} \)
Comparing with \( a + b\sqrt{5} \):
\( a = 0 \) and \( b = \frac{7}{11} \).
In simple words: Combine fractions over a common denominator, simplify the numerator, and equate corresponding rational and irrational parts.

Exam Tip: When there is no rational term on the simplified left side, \( a \) is simply 0.

Question 12. Solve for x. 2^{3x-6} × 3^{2x-3} = 216
Answer: Express 216 as a product of its prime factors:
\( 216 = 2^3 \times 3^3 \)
The given equation becomes:
\( 2^{3x-6} \times 3^{2x-3} = 2^3 \times 3^3 \)
Equating the exponents of corresponding prime bases:
For base 2:
\( 3x - 6 = 3 \)
\( \implies 3x = 9 \)
\( \implies x = 3 \)
For base 3:
\( 2x - 3 = 3 \)
\( \implies 2x = 6 \)
\( \implies x = 3 \)
Both give the consistent solution \( x = 3 \).
In simple words: Factor 216 into prime powers of 2 and 3, then set the matching powers equal to each other.

Exam Tip: Check that the value of \( x \) satisfies the exponent equations for both prime bases.

Question 13. If x = 9 + 4√5 find the value of √x + 1/√x
Answer: Express \( x \) as a perfect square:
\( x = 9 + 4\sqrt{5} = 5 + 4 + 2(2)(\sqrt{5}) = (\sqrt{5})^2 + 2^2 + 2(2)(\sqrt{5}) = (\sqrt{5} + 2)^2 \)
Taking the positive square root:
\( \sqrt{x} = \sqrt{5} + 2 \)
Now find \( \frac{1}{\sqrt{x}} \):
\( \frac{1}{\sqrt{x}} = \frac{1}{\sqrt{5} + 2} = \frac{\sqrt{5} - 2}{(\sqrt{5})^2 - 2^2} = \frac{\sqrt{5} - 2}{5 - 4} = \sqrt{5} - 2 \)
Adding both:
\( \sqrt{x} + \frac{1}{\sqrt{x}} = (\sqrt{5} + 2) + (\sqrt{5} - 2) = 2\sqrt{5} \).
In simple words: Rewrite x as the square of \( (\sqrt{5} + 2) \), take its square root, and add it to its rationalized reciprocal.

Exam Tip: Decompose the constant term 9 into \( a^2 + b^2 = 5 + 4 \) to spot the square pattern \( (a+b)^2 \) easily.

Question 14. Choose the most suitable option from the bracket and give eg. to support your answer. "Product of a rational number and an irrational number is _______" [rational / irrational / not a real number / rational or irrational]
Answer: The most suitable option is rational or irrational.
Supporting Examples:
1. If the rational number is zero:
\( 0 \times \sqrt{3} = 0 \), which is a rational number.
2. If the rational number is non-zero (e.g., 2):
\( 2 \times \sqrt{3} = 2\sqrt{3} \), which is an irrational number.
Because zero makes the product rational while any non-zero rational keeps it irrational, the product can be rational or irrational.
In simple words: The result is usually irrational, but multiplying by zero produces a rational number, so both outcomes are possible.

Exam Tip: Always remember that zero is a rational number that acts as an exception in multiplication.

Question 15. Find the value of 0.6 + 0.7̄ + 0.47̄ in the p/q form where p & q are integers and q ≠ 0
Answer: Convert each decimal to fractional form:
1. \( 0.6 = \frac{6}{10} = \frac{3}{5} \)
2. Let \( x = 0.\overline{7} \implies 10x = 7.\overline{7} \implies 9x = 7 \implies x = \frac{7}{9} \)
3. Let \( y = 0.4\overline{7} \implies 10y = 4.\overline{7} \) and \( 100y = 47.\overline{7} \)
\( 90y = 43 \implies y = \frac{43}{90} \)
Now add the three fractions:
\( \frac{6}{10} + \frac{7}{9} + \frac{43}{90} = \frac{6 \times 9 + 7 \times 10 + 43}{90} \)
\( = \frac{54 + 70 + 43}{90} = \frac{167}{90} \).
In simple words: Convert the terminating and repeating decimals into fractions with common denominator 90, then add them up.

Exam Tip: Notice carefully whether a digit has a bar: 0.6 has no bar, while 0.7 and the 7 in 0.47 have bars.

Question 16(i). 0.123̄ =
(a) 111/990
(b) 123/990
(c) 120/900
(d) 111/900
Answer: (d) 111/900
In simple words: Subtract the non-repeating part 12 from 123 to get 111, placed over 900 for one repeating digit and two non-repeating digits.

Exam Tip: The shortcut formula is: \( \frac{\text{Complete Number} - \text{Non-repeating part}}{900} = \frac{123 - 12}{900} = \frac{111}{900} \).

Question 16(ii). which of the following is irrational?
(a) √4/9
(b) √12/√3
(c) √7 × √2
(d) √2025
Answer: (c) √7 × √2
In simple words: \( \sqrt{7} \times \sqrt{2} = \sqrt{14} \), which cannot be simplified to a whole number or simple fraction.

Exam Tip: Evaluate options fully: \( \sqrt{4/9} = 2/3 \), \( \sqrt{12}/\sqrt{3} = 2 \), and \( \sqrt{2025} = 45 \) are all rational.

Question 16(iii). When you represent √3 on the number line, it lies in between
(a) 0 & 1
(b) 1 & 2
(c) 2 & 3
(d) 3 & 4
Answer: (b) 1 & 2
In simple words: Since 3 is between 1 and 4, its square root lies between 1 and 2 (approximately 1.732).

Exam Tip: Use perfect squares: \( 1 < 3 < 4 \implies \sqrt{1} < \sqrt{3} < \sqrt{4} \implies 1 < \sqrt{3} < 2 \).

Question 16(iv). If (27)^x = 9 / 3^x then x =
(a) 3
(b) 1/2
(c) 2
(d) 1
Answer: (b) 1/2
In simple words: Write in base 3 to get \( 3^{3x} = 3^{2-x} \), leading to \( 4x = 2 \), so \( x = 1/2 \).

Exam Tip: Convert all powers to the common base 3: \( (3^3)^x = 3^2 / 3^x \implies 3^{3x} = 3^{2-x} \).

Question 16(v). Which of the following is equal to x?
(a) x^{12/7 + 5/7}
(b) x^{12/7} × x^{7/12}
(c) (√(x^3))^{3/2}
(d) \sqrt[12]{(x^4)^3}
Answer: (d) \sqrt[12]{(x^4)^3}
In simple words: In option (d), the power is \( 4 \times 3 = 12 \), and the 12th root of \( x^{12} \) is simply x.

Exam Tip: Multiply powers using the law \( (x^a)^b = x^{ab} \): \( \sqrt[12]{x^{12}} = x^{12/12} = x^1 \).

Question 16(vi). If 4^{2y} = 256 then 4^{-y} =
(a) 1/16
(b) -1/16
(c) 1/4
(d) 16
Answer: (a) 1/16
In simple words: Since \( 256 = 4^4 \), \( 2y = 4 \) giving \( y = 2 \). Then \( 4^{-2} = 1/16 \).

Exam Tip: A negative exponent means reciprocal: \( 4^{-y} = \frac{1}{4^y} = \frac{1}{4^2} = \frac{1}{16} \).

Question 16(vii). \sqrt[4]{\sqrt[3]{2^2}} = _______
(a) 2^{1/6}
(b) 2^{-6}
(c) 2^{2/3}
(d) 2^6
Answer: (a) 2^{1/6}
In simple words: Multiply the radical indices: \( \sqrt[4]{\sqrt[3]{2^2}} = (2^2)^{1/12} = 2^{2/12} = 2^{1/6} \).

Exam Tip: Combine radical roots: \( \sqrt[m]{\sqrt[n]{a}} = \sqrt[mn]{a} \).

Question 16(viii). (√7 - √5)² ÷ (6 - √35) = _______
(a) 1
(b) -1
(c) 2
(d) -2
Answer: (c) 2
In simple words: Expanding the numerator gives \( 12 - 2\sqrt{35} \), which is exactly twice \( 6 - \sqrt{35} \).

Exam Tip: Factor out 2 from \( 12 - 2\sqrt{35} = 2(6 - \sqrt{35}) \) to cancel the entire denominator cleanly.

Question 16(ix). An irrational no. between √2 & √3 is
(a) (√2+√3)/2
(b) (√3-√2)/2
(c) (√2×√3)/2
(d) √3/√2
Answer: (a) (√2+√3)/2
In simple words: The average of two distinct irrational numbers lies strictly between them and is itself irrational here.

Exam Tip: Check decimal estimates: \( \frac{1.414 + 1.732}{2} \approx 1.573 \), which falls right between \( 1.414 \) and \( 1.732 \).

Question 16(x). A rational no. equivalent to 7/19 is
(a) 17/119
(b) 14/57
(c) 21/57
(d) 49/361
Answer: (c) 21/57
In simple words: Multiply numerator and denominator by 3 to get \( \frac{21}{57} \).

Exam Tip: Equivalent rational numbers must maintain the exact same ratio: \( \frac{7 \times 3}{19 \times 3} = \frac{21}{57} \).