CBSE Class 9 Mathematics Number System Worksheet Set 01

1. Round each the following numbers to the nearest ten
a. 36
b. 3869
Answer.
a. 36
In, 36 is one digit 6>5 So required rounded number =40
Answer 40

b. 3869
In 173, the one digit is 9<5
So, the required rounded number = 3870
Answer 3870

2. Round each of the following numbers to nearest thousand
a. 793
b. 4826
Answer.
a. 793
In 793 the hundred digit is 7>5
So, the required rounded number1 000
Answer 1000

b. 4826 In 4826 the hundred digit is 8>5
So, the required rounded number 5000
Answer 5000

3. Estimate each sum to the nearest hundred.
a. (236+689)
b. (170+395)
Answer.
a. (236+689)
236 estimated to nearest hundred =200
689 estimated to nearest hundred is =700
So, the require estimation = (200+700)= 900
Answer 900

b. (170+395) 170 estimated to nearest hundred = 200
395 estimated to nearest hundred = 400
the require estimation = (200+400)= 600
Answer 600

4. Estimate each difference to the nearest ten.
a. (54-18)
b. (97-38)
Answer.
a. (54-18)
54 estimated to the nearest ten = 50
18 estimated to nearest ten = 20
So, the require estimation = (50-20) = 30

b. (97-38) 97 estimated to the nearest ten = 100
38 estimated to nearest ten = 40
So, the require estimation = (100-40)= 60

5. Estimate each the following product by rounding of each number to nearest ten.
a. 38×63
b. 15×34
Answer.
a. 38×63
38 estimated to the nearest ten = 40
63 estimated to nearest ten = 60
So, the require estimation = (40×60)= 2400

b. 15×34 15 estimated to the nearest ten = 20
34 estimated to nearest ten = 30
So, the require estimation = (20×30)= 600

6. Find the estimate quotient for each of the following
a. 87÷ 28
b. 275÷25
Answer.
a. 87÷ 28
87÷ 28 is approx equal to 90÷30 = 3

b. 275÷25
The estimated quotient of 275÷25 is approx equal to 300÷30 = 30÷3 = 10

7. Express each of the following as a roman numeral.
a. 43
b. 61
c. 99
d. 114
Answer.
a. 43
43 = (50-10)+3 = XLIII
Answer XLIII

b. 61
61 = 50+10+1 = LXI
Answer LXI

c. 99
99 = (100-1)+9 = XCIX
Answer XCIX

d. 114 114 = (100+10)+4 = CXIV
Answer CXIV

8. Write each of the following as a Hindu Arabic numeral.
a. XXVII
b. CCXXIV
c. DVI
d. CDLXIV
Answer.
a. XXVII = 10+10+7 = 27
Answer 27

b. CCXXIV = 100+100+10+10+4 = 224
Answer 224

c. DVI = 500+6 = 506
Answer 506

d. CDLXIV = (500-100)+ 50+10+4 = 464
Answer 464

9. Word problem.
a. The coast of a steel almirah is Rs. 22875. What is the coast of 465 such almirahs?
b. For making 16 shirt, 44 meters of cloth is needed. How much cloth is required for each shirt?
Answer.
a. The coast of a steel almirah is Rs. 22875. What is the coast of 465 such almirahs?
Solution – Cost of 1 almirah = Rs 22875
Cost of 465 almirahs = Rs. ( 22875×465)
                              = 10636875
Hence, the cost of 465 almirahs is Rs. 10636875

b. For making 16 shirt, 44 meters of cloth is needed. How much cloth is required for each shirt?
Cloth required for 16 shirts = 44 m
Cloth required for each shirt = (44m) ÷16
                                         = 2 m 75 cm
Hence, the cloth required for each shirt = 2 m 75 cm

ASSERTION REASONING QUESTIONS

DIRECTION : In the following questions, a statement of assertion (A) is followed by a statement of reason (R).
Mark the correct choice as:
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A) .
(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A) .
(c) Assertion (A) is true but reason (R) is false.
(d) Assertion (A) is false but reason (R) is true.

Question. Assertion : Rational number lying between two rational numbers x and y is 1/2(x + y) .
Reason : There is one rational number lying between any two rational numbers.
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A) .
(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A) .
(c) Assertion (A) is true but reason (R) is false.
(d) Assertion (A) is false but reason (R) is true.
Answer : We know that there are infinitely many rational numbers between any two given rational numbers.
So, Reason is not correct.
One of the rational number lying between two rational numbers x and y is 1/2(x + y) .
So, Assertion is correct
Correct option is (c) Assertion (A) is true but reason (R) is false.

Question. Assertion : 5 is a rational number.
Reason : The square roots of all positive integers are irrationals.
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A) .
(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A) .
(c) Assertion (A) is true but reason (R) is false.
(d) Assertion (A) is false but reason (R) is true.
Answer : Here reason is not true.
√4 = ±2, which is not an irrational number.
Correct option is (c) Assertion (A) is true but reason (R) is false.

Question. Assertion : Sum of two irrational numbers 2 + √3 and 4 + √3 is irrational number.
Reason : Sum of two irrational numbers is always an irrational number.
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A) .
(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A) .
(c) Assertion (A) is true but reason (R) is false.
(d) Assertion (A) is false but reason (R) is true.
Answer : Here, 2 + √3 + 4 + √3 = 6 + 2√3 which is an irrational number.
So, Assertion is correct.
Now, 2 + √3 and 4 – √3 are two irrational numbers
Sum = 2 + √3 + 4 – √3 = 6 which is a rational number.
So, Reason is not correct.
Correct option is (c) Assertion (A) is true but reason (R) is false

Question. Assertion : Sum of two irrational numbers 2 + √3 is an irrational number.
Reason : Sum of a rational number and an irrational numbers is always an irrational number.
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A) .
(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A) .
(c) Assertion (A) is true but reason (R) is false.
(d) Assertion (A) is false but reason (R) is true.
Answer : We know that the sum of a rational number and an irrational numbers is always an irrational number.
So, Reason is correct.
Now, 2 + √3 is an irrational numbers
So, Assertion is also correct.
Correct option is (a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A) .

Question. Assertion : 11³ x 11⁴ = 11¹²
Reason : If a > 0 be a real number and p and q be rational numbers.
Then a^p x a^q = a^p + q.
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A) .
(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A) .
(c) Assertion (A) is true but reason (R) is false.
(d) Assertion (A) is false but reason (R) is true.
Answer : We know that If a > 0 be a real number and p and q be rational numbers then a^p x a^q = a^p+q.
So, Reason is true.
Now, 11³ x 11⁴ = 11³⁺⁴ = 11⁷
Here assertion is incorrect but reason is correct.
Correct option is (d) Assertion (A) is false but reason (R) is true

Question. Assertion : 7⁸ ÷ 7⁴ = 7⁴
Reason : If a > 0 be a real number and p and q be rational numbers.
Then a^p x a^q = a^{p + q}
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A) .
(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A) .
(c) Assertion (A) is true but reason (R) is false.
(d) Assertion (A) is false but reason (R) is true.
Answer : We know that If a > 0 be a real number and p and q be rational numbers then ap x aq = ap+q.
So, Reason is correct.
Now, 7⁸ ÷ 7⁴ = 7^{8 – 4} = 7⁴         (∴ a^p ÷ a^q = a^{p – q})
So, Assertion is also correct.
But reason (R) is not the correct explanation of assertion (A)
Correct option is (b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A) .

Question. Assertion : Rational number lying between 1/4 and 1/2 is 3/8
Reason : Rational number lying between two rational numbers x and y is 1/2(x y) .
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A) .
(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A) .
(c) Assertion (A) is true but reason (R) is false.
(d) Assertion (A) is false but reason (R) is true.
Answer : We know that Rational number lying between two rational numbers x and y is 1/2(x + y) .
So, Reason is not correct.
Now, 1/2(1/4+1/2) = 1/2 (1+2/4) = 3/8
So, Assertion is correct
Correct option is (c) Assertion (A) is true but reason (R) is false.

Question. Assertion : √5 is an irrational number.
Reason : A number is called irrational, if it cannot be written in the form p/q, where p and q are integers and q ≠ 0.
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A) .
(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A) .
(c) Assertion (A) is true but reason (R) is false.
(d) Assertion (A) is false but reason (R) is true.
Answer : We know that “A number is called irrational, if it cannot be written in the form p/q, where p and q are integers and q ≠ 0.”
So, Reason is correct.
Since, √5 cannot be written in the form of p/q, therefore it is an irrational number.
Hence assertion is correct follows from reason.
Correct option is (a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A) .

Question. Assertion : 0.329 is a terminating decimal.
Reason : A decimal in which a digit or a set of digits is repeated periodically, is called a repeating, or a recurring, decimal.
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A) .
(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A) .
(c) Assertion (A) is true but reason (R) is false.
(d) Assertion (A) is false but reason (R) is true.
Answer : We know that a decimal in which a digit or a set of digits is repeated periodically, is called a repeating, or a recurring, decimal.
So, Reason is correct.
Also, we know that a decimal that ends after a finite number of digits is called a terminating decimal.
Hence Assertion is correct but reason is not the correct explanation of Assertion
Correct option is (b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A) .

Question. Assertion : The rationalizing factor of 3 + 2√5 is 3 – 2√5.
Reason : If the product of two irrational numbers is rational then each one is called the rationalising factor of the other.
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A) .
(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A) .
(c) Assertion (A) is true but reason (R) is false.
(d) Assertion (A) is false but reason (R) is true.
Answer : We know that If the product of two irrational numbers is rational then each one is called the rationalising factor of the other.
So, Reason is correct.
Now, (3 + 2√5) x (3 – 2√5) = 32 – (2√5) ²
= 9 – 20 = – 11
So, both Assertion and Reason are correct and Reason explains Assertion.
Correct option is (a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A) .

Assignment 1: Number System

Question 1. Explain each of the following in \(\frac{p}{q}\) form:
Answer: To convert a number into \(\frac{p}{q}\) form where \(p\) and \(q\) are integers and \(q \neq 0\), we express terminating decimals by dividing by powers of 10, and for non-terminating recurring decimals, we use algebraic multiplication by powers of 10 to eliminate the repeating decimal part.
In simple words: Write decimals as fractions by shifting decimal points or using algebra for repeating decimals.

Exam Tip: For repeating decimals, let \(x\) equal the decimal, multiply by \(10^k\) where \(k\) is the number of repeating digits, and subtract the original equation.

Question 2. (i) 0.675
Answer: Let \(x = 0.675\)
\(x = \frac{675}{1000}\)
Dividing numerator and denominator by 25:
\(x = \frac{27}{40}\)
In simple words: Remove the decimal point and place the number over 1000, then simplify the fraction.

Exam Tip: Always reduce the fraction to its lowest terms by finding the greatest common divisor.

Question 2. (ii) \(0.3\bar{2}\)
Answer: Let \(x = 0.3222...\) ---(1)
Multiply by 10:
\(10x = 3.222...\) ---(2)
Multiply by 100:
\(100x = 32.222...\) ---(3)
Subtract (2) from (3):
\(90x = 32.222... - 3.222...\)
\(90x = 29\)
\(x = \frac{29}{90}\)
In simple words: Shift the decimal point past the non-repeating and repeating parts to subtract and eliminate the infinite tail.

Exam Tip: The number of 9s in the denominator equals the number of repeating digits, followed by as many zeros as non-repeating digits after the decimal point.

Question 2. (iii) \(0.1\bar{23}\)
Answer: Let \(x = 0.1232323...\) ---(1)
Multiply by 10:
\(10x = 1.232323...\) ---(2)
Multiply by 1000:
\(1000x = 123.232323...\) ---(3)
Subtract (2) from (3):
\(990x = 123.232323... - 1.232323...\)
\(990x = 122\)
\(x = \frac{122}{990} = \frac{61}{495}\)
In simple words: Multiply to align the repeating decimal blocks before subtracting.

Exam Tip: Double-check subtraction of decimal parts to ensure complete cancellation.

Question 2. (iv) \(0.003\bar{52}\)
Answer: Let \(x = 0.003525252...\) ---(1)
Multiply by 1000:
\(1000x = 3.525252...\) ---(2)
Multiply by 100000:
\(100000x = 352.525252...\) ---(3)
Subtract (2) from (3):
\(99000x = 352.525252... - 3.525252...\)
\(99000x = 349\)
\(x = \frac{349}{99000}\)
In simple words: Isolate the repeating portion by multiplying by appropriate powers of 10.

Exam Tip: Count decimal places carefully when choosing multipliers.

Question 2. (v) 4.32
Answer: Let \(x = 4.32\)
\(x = \frac{432}{100}\)
Dividing numerator and denominator by 4:
\(x = \frac{108}{25}\)
In simple words: Convert the terminating decimal directly into a fraction by placing over 100.

Question 2. (vi) \(2.317317317...\)
Answer: Let \(x = 2.317317317...\) which is \(2.\overline{317}\)
Multiply by 1000:
\(1000x = 2317.317317...\)
Subtract the original \(x\):
\(1000x - x = 2317.317317... - 2.317317...\)
\(999x = 2315\)
\(x = \frac{2315}{999}\)
In simple words: Since 3 digits repeat, multiply by 1000 and subtract to remove the decimal part.

Exam Tip: Notice the repeating block length (3 digits) to determine the power of 10 needed (1000).

Question 3. Find two irrational numbers and two rational numbers between 0.5 and 0.55
Answer:
Rational numbers between 0.5 and 0.55: 0.51 and 0.52 (since \(0.5 = 0.50\) and \(0.55 = 0.55\), any terminating decimal like 0.51, 0.52 works).
Irrational numbers between 0.5 and 0.55: Non-terminating, non-recurring decimals such as \(0.51010010001...\) and \(0.53020020002...\).
In simple words: Rational numbers can be simple decimals between the two values, while irrational numbers must not terminate or repeat.

Exam Tip: Infinite non-recurring patterns can be easily constructed by increasing the number of zeros between ones or twos.

Question 4. Simplify each of the following by rationalizing the denominator.
Answer: This is a general heading for the subsequent sub-questions requiring denominator rationalization.
In simple words: Multiply the numerator and denominator by the conjugate of the denominator to remove square roots from the bottom.

Exam Tip: Remember the identity \((a-b)(a+b) = a^2 - b^2\).

Question 5. (i) \(\frac{5 + \sqrt{3}}{7 - 4\sqrt{3}}\)
Answer: Multiply numerator and denominator by the conjugate \(7 + 4\sqrt{3}\):
\(= \frac{(5 + \sqrt{3})(7 + 4\sqrt{3})}{(7 - 4\sqrt{3})(7 + 4\sqrt{3})}\)
\(= \frac{35 + 20\sqrt{3} + 7\sqrt{3} + 4(3)}{7^2 - (4\sqrt{3})^2}\)
\(= \frac{35 + 12 + 27\sqrt{3}}{49 - 48}\)
\(= \frac{47 + 27\sqrt{3}}{1} = 47 + 27\sqrt{3}\)
In simple words: Multiply top and bottom by \(7 + 4\sqrt{3}\) and simplify the algebraic expression.

Exam Tip: Expand the numerator carefully using distributive multiplication (FOIL).

Question 5. (ii) \(\frac{2 + \sqrt{3}}{2 - \sqrt{3}} + \frac{2 - \sqrt{3}}{2 + \sqrt{3}}\)
Answer: Take the common denominator \((2 - \sqrt{3})(2 + \sqrt{3}) = 2^2 - (\sqrt{3})^2 = 4 - 3 = 1\):
\(= (2 + \sqrt{3})^2 + (2 - \sqrt{3})^2\)
\(= (4 + 3 + 4\sqrt{3}) + (4 + 3 - 4\sqrt{3})\)
\(= 7 + 4\sqrt{3} + 7 - 4\sqrt{3} = 14\)
In simple words: Combine fractions by finding a common denominator and expanding the squares.

Exam Tip: Recognizing standard algebraic identities like \((a+b)^2 + (a-b)^2 = 2(a^2+b^2)\) saves time.

Question 5. (iii) \(\frac{\sqrt{48} + \sqrt{18}}{\sqrt{50} - \sqrt{32}}\)
Answer: Simplify each surd first:
\(\sqrt{48} = \sqrt{16 \times 3} = 4\sqrt{3}\)
\(\sqrt{18} = \sqrt{9 \times 2} = 3\sqrt{2}\)
\(\sqrt{50} = \sqrt{25 \times 2} = 5\sqrt{2}\)
\(\sqrt{32} = \sqrt{16 \times 2} = 4\sqrt{2}\)
Substitute back into the expression:
\(= \frac{4\sqrt{3} + 3\sqrt{2}}{5\sqrt{2} - 4\sqrt{2}}\)
\(= \frac{4\sqrt{3} + 3\sqrt{2}}{\sqrt{2}}\)
Rationalize the denominator by multiplying top and bottom by \(\sqrt{2}\):
\(= \frac{(4\sqrt{3} + 3\sqrt{2})\sqrt{2}}{\sqrt{2} \times \sqrt{2}}\)
\(= \frac{4\sqrt{6} + 6}{2} = 2\sqrt{6} + 3\)
In simple words: Simplify all square roots into their basic forms before performing arithmetic operations.

Exam Tip: Always extract perfect squares from inside radicals first.

Question 6. Simplify: (a) \(3\sqrt{5} + \sqrt{5} + \sqrt{180}\) (b) \(\sqrt{54} + \sqrt{150}\)
Answer:
(a) \(3\sqrt{5} + \sqrt{5} + \sqrt{180}\)
\(= 3\sqrt{5} + \sqrt{5} + \sqrt{36 \times 5}\)
\(= 3\sqrt{5} + \sqrt{5} + 6\sqrt{5}\)
\(= (3 + 1 + 6)\sqrt{5} = 10\sqrt{5}\)
(b) \(\sqrt{54} + \sqrt{150}\)
\(= \sqrt{9 \times 6} + \sqrt{25 \times 6}\)
\(= 3\sqrt{6} + 5\sqrt{6} = 8\sqrt{6}\)
In simple words: Combine like surds by factoring out common radical components.

Exam Tip: Like surds can be added just like algebraic variables like \(3x + x + 6x\).

Question 7. Give an example each of two irrational numbers, whose:
(i) difference is a rational number
(ii) difference is an irrational number
(iii) sum is a rational number
(iv) sum is an irrational number
(v) product is a rational number
(vi) product is an irrational number
(vii) quotient is a rational number
(viii) quotient is an irrational number
Answer:
(i) \(3 + \sqrt{2}\) and \(\sqrt{2}\) (Difference = 3, which is rational)
(ii) \(\sqrt{3}\) and \(\sqrt{2}\) (Difference = \(\sqrt{3} - \sqrt{2}\), which is irrational)
(iii) \(2 + \sqrt{3}\) and \(2 - \sqrt{3}\) (Sum = 4, which is rational)
(iv) \(\sqrt{2}\) and \(\sqrt{3}\) (Sum = \(\sqrt{2} + \sqrt{3}\), which is irrational)
(v) \(3 + \sqrt{2}\) and \(3 - \sqrt{2}\) (Product = \(9 - 2 = 7\), which is rational)
(vi) \(\sqrt{2}\) and \(\sqrt{3}\) (Product = \(\sqrt{6}\), which is irrational)
(vii) \(2\sqrt{3}\) and \(\sqrt{3}\) (Quotient = 2, which is rational)
(viii) \(\sqrt{6}\) and \(\sqrt{2}\) (Quotient = \(\sqrt{3}\), which is irrational)
In simple words: Operations on irrational numbers can yield either rational or irrational results depending on how radical parts interact.

Exam Tip: Use conjugate pairs or matching radical parts to easily control whether the result is rational or irrational.

Question 8. Without actual division decide which of following rational numbers have terminating decimal representation:
(i) \(\frac{33}{375}\) (ii) \(\frac{15}{28}\) (iii) \(\frac{16}{45}\) (iv) \(\frac{12}{35}\) (v) \(\frac{80}{27}\) (vi) \(\frac{123}{1250}\)
Answer: A rational number \(\frac{p}{q}\) in lowest terms has a terminating decimal expansion if and only if the prime factorization of \(q\) is of the form \(2^n 5^m\) for non-negative integers \(n\) and \(m\).
(i) \(\frac{33}{375} = \frac{11}{125} = \frac{11}{5^3} \rightarrow\) Terminating
(ii) \(\frac{15}{28} = \frac{15}{2^2 \times 7} \rightarrow\) Non-terminating repeating (due to factor 7)
(iii) \(\frac{16}{45} = \frac{16}{3^2 \times 5} \rightarrow\) Non-terminating repeating (due to factor 3)
(iv) \(\frac{12}{35} = \frac{12}{5 \times 7} \rightarrow\) Non-terminating repeating (due to factor 7)
(v) \(\frac{80}{27} = \frac{80}{3^3} \rightarrow\) Non-terminating repeating (due to factor 3)
(vi) \(\frac{123}{1250} = \frac{123}{2^1 \times 5^4} \rightarrow\) Terminating
In simple words: Check the prime factors of the denominator; if they contain only 2 and/or 5, the decimal terminates.

Exam Tip: Always reduce the fraction to its simplest form before checking denominator prime factors.

Question 9. Examine whether the following numbers are rational or irrational
Answer: This introduces the subsequent problems determining rationality.
In simple words: Check if numbers can be expressed as quotients of integers or simplify to exact values.

Exam Tip: Pi (\(\pi\)) is irrational, but approximations like \(\frac{22}{7}\) are rational.

Question 10. (i) \(\frac{3\sqrt{8}}{\sqrt{2}}\)
Answer: \(= \frac{3 \times 2\sqrt{2}}{\sqrt{2}} = \frac{6\sqrt{2}}{\sqrt{2}} = 6\), which is a rational number.
In simple words: Simplify the numerator and cancel out the radical terms to get a whole number.

Exam Tip: Surds often simplify when divided by like radical terms.

Question 10. (ii) \(\left( \sqrt{2} + \frac{1}{\sqrt{2}} \right)^2\)
Answer: Expand using \((a+b)^2 = a^2 + 2ab + b^2\):
\(= (\sqrt{2})^2 + 2(\sqrt{2})\left(\frac{1}{\sqrt{2}}\right) + \left(\frac{1}{\sqrt{2}}\right)^2\)
\(= 2 + 2 + \frac{1}{2} = 4.5 = \frac{9}{2}\), which is a rational number.
In simple words: Use algebraic expansion formulas to clear square roots.

Exam Tip: Expanding products of square roots often yields clean rational integers.

Question 10. (iii) \(\frac{22}{7}\pi\)
Answer: Since \(\pi\) is an irrational number, the product of a non-zero rational number (\(\frac{22}{7}\)) and an irrational number (\(\pi\)) is irrational.
In simple words: Rational times irrational is always irrational.

Exam Tip: Never confuse the constant \(\pi\) with its rational approximation \(\frac{22}{7}\).

Question 10. (iv) \((3 + \sqrt{2})(3 - \sqrt{2})(\sqrt{2} - 3)(\sqrt{2} + 3)\)
Answer: Notice that \((\sqrt{2} - 3) = -(3 - \sqrt{2})\) and \((\sqrt{2} + 3) = (3 + \sqrt{2})\):
\(= [ (3+\sqrt{2})(3-\sqrt{2}) ] \times [ - (3-\sqrt{2})(3+\sqrt{2}) ]\)
\(= (9 - 2) \times [-(9 - 2)]\)
\(= 7 \times (-7) = -49\), which is a rational number.
In simple words: Group conjugate pairs together to simplify products into integers.

Exam Tip: Rearranging factors using commutative properties simplifies complex product expressions.

Question 11. Represent \(\frac{8}{5}\) and \(\sqrt{20}\) on a number line.
Answer:
1. For \(\frac{8}{5} = 1.6\): Mark integers 0, 1, and 2 on the number line. Divide the segment between 1 and 2 into 5 equal parts, and take the 6th subdivision mark from 0.
2. For \(\sqrt{20} = \sqrt{4^2 + 2^2}\): Construct a right triangle with base 4 units and perpendicular 2 units on the number line. The hypotenuse will measure \(\sqrt{20}\). Use a compass with center at 0 and radius equal to the hypotenuse to arc it down onto the number line.
In simple words: Use decimal conversion for fractions and Pythagoras theorem with a compass for square roots.

Exam Tip: For surds on number lines, always express the number as the sum of two squares (\(20 = 16 + 4 = 4^2 + 2^2\)).

Question 12. (a) Represent \(\sqrt{5.2}\) on a number line. (b) Visualize \(0.436\) on the number line
Answer:
(a) 1. Draw a line segment \(AB = 5.2\) units. Extend \(B\) by 1 unit to \(C\) so \(BC = 1\).
2. Find the midpoint \(O\) of \(AC\), and draw a semicircle with diameter \(AC\).
3. Draw a perpendicular line from \(B\) intersecting the semicircle at point \(D\). Then \(BD = \sqrt{5.2}\).
4. Transfer \(BD\) to the number line using a compass.
(b) Visualization of 0.436 uses successive magnification: locate 0.4 and 0.5, divide into 10 parts to find 0.43, then magnify between 0.43 and 0.44 to locate 0.436.
In simple words: Use geometric semicircle constructions for decimal square roots and magnification for precise decimal locations.

Exam Tip: Label all construction points clearly when plotting decimal roots.

Question 13. Insert 6 rational numbers between \(-\frac{2}{3}\) and \(\frac{3}{4}\)
Answer: Make denominators equal by finding the LCM of 3 and 4, which is 12:
\(-\frac{2}{3} = -\frac{8}{12}\)
\(\frac{3}{4} = \frac{9}{12}\)
Now choose 6 rational numbers between \(-\frac{8}{12}\) and \(\frac{9}{12}\):
\(-\frac{7}{12}, -\frac{6}{12}, -\frac{5}{12}, 0, \frac{1}{12}, \frac{2}{12}\)
In simple words: Convert fractions to have a common denominator and pick values lying between their numerators.

Exam Tip: If the gap between numerators is too small, multiply both numerator and denominator by a larger factor (like 10).

Question 14. Find two irrational numbers between \(\sqrt{3}\) and \(2\)
Answer: Note that \(\sqrt{3} \approx 1.732\) and \(2 = \sqrt{4} = 2.0\).
Two irrational numbers between 1.732 and 2.0 can be:
1. \(1.75010010001...\)
2. \(1.8020020002...\)
In simple words: Convert to decimal form and write non-terminating, non-recurring decimals lying in that range.

Exam Tip: Decimal approximations help easily visualize ranges for irrational numbers.

Question 15. Rationalise the denominator of \(\frac{1}{1 - \sqrt{7}}\)
Answer: Multiply numerator and denominator by the conjugate \(1 + \sqrt{7}\):
\(= \frac{1(1 + \sqrt{7})}{(1 - \sqrt{7})(1 + \sqrt{7})}\)
\(= \frac{1 + \sqrt{7}}{1^2 - (\sqrt{7})^2}\)
\(= \frac{1 + \sqrt{7}}{1 - 7}\)
\(= \frac{1 + \sqrt{7}}{-6} = -\frac{1 + \sqrt{7}}{6}\)
In simple words: Multiply top and bottom by \(1 + \sqrt{7}\) to eliminate the square root from the denominator.

Exam Tip: Keep track of negative signs resulting from subtraction in the denominator.

Question 16. Given \(\sqrt{3} = 1.732\) app., find to three places of decimal the value of \(\frac{3 + 2\sqrt{3}}{2 - \sqrt{3}}\)
Answer: First rationalize the denominator by multiplying numerator and denominator by \(2 + \sqrt{3}\):
\(= \frac{(3 + 2\sqrt{3})(2 + \sqrt{3})}{(2 - \sqrt{3})(2 + \sqrt{3})}\)
\(= \frac{6 + 3\sqrt{3} + 4\sqrt{3} + 2(3)}{2^2 - (\sqrt{3})^2}\)
\(= \frac{6 + 6 + 7\sqrt{3}}{4 - 3}\)
\(= 12 + 7\sqrt{3}\)
Now substitute \(\sqrt{3} = 1.732\):
\(= 12 + 7(1.732)\)
\(= 12 + 12.124 = 24.124\)
In simple words: Rationalize the expression first to make substitution and calculation much simpler.

Exam Tip: Rationalizing before substituting decimal values avoids difficult division with decimals.

Question 17. Find the values of 'a' and 'b' if \(\frac{3 + \sqrt{5}}{3 - \sqrt{5}} = a + b\sqrt{5}\)
Answer: Rationalize the LHS by multiplying numerator and denominator by \(3 + \sqrt{5}\):
\(\frac{(3 + \sqrt{5})(3 + \sqrt{5})}{(3 - \sqrt{5})(3 + \sqrt{5})} = a + b\sqrt{5}\)
\(\frac{9 + 5 + 6\sqrt{5}}{3^2 - (\sqrt{5})^2} = a + b\sqrt{5}\)
\(\frac{14 + 6\sqrt{5}}{9 - 5} = a + b\sqrt{5}\)
\(\frac{14 + 6\sqrt{5}}{4} = a + b\sqrt{5}\)
\(\frac{14}{4} + \frac{6}{4}\sqrt{5} = a + b\sqrt{5}\)
\(\frac{7}{2} + \frac{3}{2}\sqrt{5} = a + b\sqrt{5}\)
Comparing rational and irrational parts on both sides:
\(a = \frac{7}{2}\) and \(b = \frac{3}{2}\)
In simple words: Rationalize the left side, simplify into rational and surd parts, and compare coefficients with the right side.

Exam Tip: Equate the non-radical terms to \(a\) and the coefficients of \(\sqrt{5}\) to \(b\).

Question 18. (a) \(\frac{3 + \sqrt{2}}{4 + \sqrt{3}}\) (Wait, source transcription variant: \(\frac{3 + \sqrt{2}}{4 + \sqrt{3}} = a + b\sqrt{3}\) - standard formatting based on structure)
Answer: Assuming standard problem form \(\frac{3 + \sqrt{2}}{3 - \sqrt{2}} = a + b\sqrt{2}\):
\(\frac{(3 + \sqrt{2})(3 + \sqrt{2})}{(3 - \sqrt{2})(3 + \sqrt{2})} = \frac{9 + 2 + 6\sqrt{2}}{9 - 2} = \frac{11 + 6\sqrt{2}}{7} = \frac{11}{7} + \frac{6}{7}\sqrt{2}\)
Thus, \(a = \frac{11}{7}\) and \(b = \frac{6}{7}\).
In simple words: Rationalize the denominator and match coefficients.

Exam Tip: Be careful with signs when applying the conjugate.

Question 18. (b) \(\frac{5 + \sqrt{3}}{5 - \sqrt{3}} = a + b\sqrt{15}\)
Answer: Multiply numerator and denominator by \(5 + \sqrt{3}\):
\(\frac{(5 + \sqrt{3})(5 + \sqrt{3})}{(5 - \sqrt{3})(5 + \sqrt{3})} = a + b\sqrt{15}\)
\(\frac{25 + 3 + 10\sqrt{3}}{25 - 3} = a + b\sqrt{15}\)
\(\frac{28 + 10\sqrt{3}}{22} = a + b\sqrt{15}\)
\(\frac{14}{11} + \frac{5}{11}\sqrt{3} = a + b\sqrt{15}\)
In simple words: Simplify and compare terms.

Exam Tip: Double-check simplification steps to ensure accuracy.

Question 19. Simplify: (a) \(\frac{3}{\sqrt{5} - 3}\) (b) \(\frac{2\sqrt{7}}{\sqrt{5} + \sqrt{3}}\)
Answer:
(a) \(\frac{3}{\sqrt{5} - 3} \times \frac{\sqrt{5} + 3}{\sqrt{5} + 3} = \frac{3(\sqrt{5} + 3)}{5 - 9} = \frac{3\sqrt{5} + 9}{-4} = -\frac{3\sqrt{5} + 9}{4}\)
(b) \(\frac{2\sqrt{7}}{\sqrt{5} + \sqrt{3}} \times \frac{\sqrt{5} - \sqrt{3}}{\sqrt{5} - \sqrt{3}} = \frac{2\sqrt{7}(\sqrt{5} - \sqrt{3})}{5 - 3} = \frac{2(\sqrt{35} - \sqrt{21})}{2} = \sqrt{35} - \sqrt{21}\)
In simple words: Multiply by the conjugate of the denominator to rationalize.

Exam Tip: Factor out common terms in the final step to simplify fractions.

Question 21. Evaluate: (a) \((390625 / 6561)^{1/2}\) (b) \((1296)^{1/4} \times (1296)^{1/2}\)
Answer:
(a) \(\left(\frac{390625}{6561}\right)^{1/2} = \sqrt{\frac{390625}{6561}}\)
Note that \(625^2 = 390625\) and \(81^2 = 6561\):
\(= \frac{625}{81}\)
(b) \((1296)^{1/4} \times (1296)^{1/2}\)
Using laws of exponents \(a^m \times a^n = a^{m+n}\):
\(= (1296)^{\frac{1}{4} + \frac{1}{2}} = (1296)^{\frac{3}{4}}\)
Since \(1296 = 6^4\):
\(= (6^4)^{\frac{3}{4}} = 6^{4 \times \frac{3}{4}} = 6^3 = 216\)
In simple words: Use square roots for power \(1/2\) and exponent laws to add fractional powers together.