CBSE Class 9 Mathematics Number System Assignment Set 04

1. Number Systems

Question 1. Is every real number is a rational number ?
Answer: No, every real number is not a rational number.
The set of real numbers consists of both rational numbers and irrational numbers. Therefore, irrational numbers like \(\sqrt{2}\) or \(\pi\) are real numbers but they are not rational numbers.
In simple words: Real numbers include both rational and irrational numbers, so not all of them are rational.

Exam Tip: Remember the hierarchy: Real numbers = Rational numbers \(\cup\) Irrational numbers.

Question 2. Is \(1.01001000100001 \dots\) irrational? If so, why?
Answer: Yes, the given number is irrational.
Reason: Its decimal expansion is non-terminating and non-recurring (non-repeating). A number with such a decimal representation is always irrational.
In simple words: The decimal part never ends and never repeats a fixed pattern, making it irrational.

Exam Tip: Always check two conditions for decimal expansions: terminating/repeating (rational) versus non-terminating non-repeating (irrational).

Question 3. Is every whole number is a natural number ?
Answer: No, every whole number is not a natural number.
Whole numbers start from \(0\) (\(0, 1, 2, 3, \dots\)), whereas natural numbers start from \(1\) (\(1, 2, 3, \dots\)). Since \(0\) is a whole number but not a natural number, the statement is false.
In simple words: Zero is a whole number, but it is not counted among natural numbers.

Exam Tip: Keep number sets clear: Natural Numbers \((\mathbb{N}) \subset\) Whole Numbers \((\mathbb{W}) \subset\) Integers \((\mathbb{Z})\).

Question 4. Look at the following examples of rational number in the form \(\frac{p}{q}\) (\(q \neq 0\)), where \(p\) and \(q\) integers with no common factors other than \(1\) and having terminating decimal representations. Can you guess the property which satisfy \(q\)?
Answer: The prime factorization of \(q\) must be of the form \(2^n \times 5^m\), where \(n\) and \(m\) are non-negative integers.
As seen in the given examples like \(\frac{7}{8} = \frac{7}{2^3} = \frac{875}{1000} = 0.875\), the denominator's prime factors are strictly powers of 2 and/or 5.
In simple words: For a fraction in simplest form to have a terminating decimal, its denominator can only have prime factors of 2 or 5.

Exam Tip: This is a standard theorem: \(\frac{p}{q}\) has a terminating decimal if the prime factorization of \(q\) is \(2^n 5^m\).

Question 5. Is zero a rational number? Explain it.
Answer: Yes, zero is a rational number.
Explanation: A rational number is any number that can be expressed in the form \(\frac{p}{q}\), where \(p\) and \(q\) are integers and \(q \neq 0\). Since we can write \(0\) as \(\frac{0}{1}\), \(\frac{0}{2}\), \(\frac{0}{-5}\), etc., where numerator \(p = 0\) and denominator \(q\) is a non-zero integer, \(0\) satisfies the definition of a rational number.
In simple words: Zero can be written as a fraction with zero on top and any non-zero number on the bottom, so it is rational.

Exam Tip: Never forget that the denominator \(q\) cannot be zero, but the numerator \(p\) can be zero.

Question 6. If \(x = \frac{1 - \sqrt{2}}{1 + \sqrt{2}}\), then find \(x\) is rational or irrational number.
Answer: Let us rationalize the denominator of \(x\):
\(x = \frac{1 - \sqrt{2}}{1 + \sqrt{2}} \times \frac{1 - \sqrt{2}}{1 - \sqrt{2}}\)
\(x = \frac{(1 - \sqrt{2})^2}{1^2 - (\sqrt{2})^2}\)
\(x = \frac{1 + 2 - 2\sqrt{2}}{1 - 2}\)
\(x = \frac{3 - 2\sqrt{2}}{-1} = -3 + 2\sqrt{2}\)
Since \(\sqrt{2}\) is an irrational number, multiplying it by 2 and adding \(-3\) results in an irrational number.
Therefore, \(x\) is an irrational number.
In simple words: After simplifying, the expression still contains a square root that cannot be removed, making it irrational.

Exam Tip: Always rationalize denominators first before concluding whether an expression is rational or irrational.

Question 7. Insert three rational numbers between \(-\frac{13}{24}\) and \(-\frac{9}{24}\).
Answer: The given rational numbers have the same denominator: \(-\frac{13}{24}\) and \(-\frac{9}{24}\).
The integers between the numerators \(-13\) and \(-9\) are \(-12, -11, -10\).
Thus, three rational numbers between \(-\frac{13}{24}\) and \(-\frac{9}{24}\) are:
\(-\frac{12}{24}\) (or \(-\frac{1}{2}\)), \(-\frac{11}{24}\), and \(-\frac{10}{24}\) (or \(-\frac{5}{12}\)).
In simple words: Since denominators are already equal, simply pick three fractions with numerators lying between \(-13\) and \(-9\).

Exam Tip: Ensure denominators are identical before comparing or inserting numbers between fractions.

Question 8. Find two rational numbers between 1 and 2.
Answer: Method 1: Using mean method or equivalent fractions.
Let us convert 1 and 2 into rational numbers with denominator \(3\):
\(1 = \frac{3}{3}\)
\(2 = \frac{6}{3}\)
Two rational numbers between \(\frac{3}{3}\) and \(\frac{6}{3}\) are \(\frac{4}{3}\) and \(\frac{5}{3}\).
In simple words: Convert both numbers into fractions with the same denominator and choose numbers in between.

Exam Tip: You can find infinitely many rational numbers by multiplying numerator and denominator by a larger power of 10 or integers.

Question 9. Is \(\sqrt{225}\) a rational number?
Answer: Yes, \(\sqrt{225}\) is a rational number.
Because \(\sqrt{225} = 15\), and 15 can be written as \(\frac{15}{1}\) in the form \(\frac{p}{q}\) where \(p\) and \(q\) are integers and \(q \neq 0\).
In simple words: Since 225 is a perfect square, its square root is a whole number, which is rational.

Exam Tip: Always simplify square roots first before checking if a number is rational or irrational.

Question 10. Is it true that every integer is a rational number ?
Answer: Yes, it is true.
Every integer \(n\) can be written in the form \(\frac{n}{1}\), which is in the form \(\frac{p}{q}\) where \(p = n\) and \(q = 1\) are integers and \(q \neq 0\).
In simple words: Any whole number or integer can be put over 1, so every integer is rational.

Exam Tip: Integers form a subset of rational numbers (\(\mathbb{Z} \subset \mathbb{Q}\)).

Question 11. Is every rational number is an Integer.
Answer: No, every rational number is not an integer.
For example, \(\frac{3}{4}\) or \(\frac{1}{2}\) are rational numbers, but they are not integers because they represent fractions rather than whole values.
In simple words: Fractions like half or three-quarters are rational, but they are not integers.

Exam Tip: Remember containment goes one way: every integer is rational, but not every rational is an integer.

Question 12. Is \(\sqrt{23}\) a rational number?
Answer: No, \(\sqrt{23}\) is not a rational number; it is an irrational number.
Since 23 is a prime number and not a perfect square, its square root cannot be expressed as a ratio of two integers and has a non-terminating non-recurring decimal expansion.
In simple words: The square root of any non-square prime number is always irrational.

Exam Tip: Memorize that \(\sqrt{p}\) where \(p\) is prime is always irrational.

Question 13. Find the product of \(\sqrt[3]{2}\) and \(\sqrt[3]{24}\).
Answer: Product = \(\sqrt[3]{2} \times \sqrt[3]{24}\)
Using the laws of radicals \(\sqrt[n]{a} \times \sqrt[n]{b} = \sqrt[n]{ab}\):
\(= \sqrt[3]{2 \times 24}\)
\(= \sqrt[3]{48}\)
\(= \sqrt[3]{8 \times 6} = \sqrt[3]{2^3 \times 6} = 2\sqrt[3]{6}\)
In simple words: Multiply the numbers inside the cube roots together and then simplify by taking out any perfect cube factors.

Exam Tip: Combine radicals of the same index under a single radical sign before simplifying.

Question 14. Is 2 a rational number? Can you write it in the form \(\frac{p}{q}\), where \(p\) and \(q\) are integers?
Answer: Yes, 2 is a rational number.
It can be written in the form \(\frac{p}{q}\) as \(\frac{2}{1}\), where \(p = 2\) and \(q = 1\) are integers and \(q \neq 0\).
In simple words: Any integer like 2 can be written as a fraction \(\frac{2}{1}\).

Question 15. Find the greatest among \(\sqrt[4]{5}, \sqrt{7}, \sqrt[3]{3}\).
Answer: Convert the radicals to fractional exponents with a common denominator for the indices (\(4, 2, 3\)).
The LCM of indices \(4, 2, 3\) is 12.
\(\sqrt[4]{5} = 5^{\frac{1}{4}} = 5^{\frac{3}{12}} = \sqrt[12]{5^3} = \sqrt[12]{125}\)
\(\sqrt{7} = 7^{\frac{1}{2}} = 7^{\frac{6}{12}} = \sqrt[12]{7^6} = \sqrt[12]{117649}\)
\(\sqrt[3]{3} = 3^{\frac{1}{3}} = 3^{\frac{4}{12}} = \sqrt[12]{3^4} = \sqrt[12]{81}\)
Comparing the values under the 12th root: \(117649 > 125 > 81\).
Thus, the greatest number is \(\sqrt{7}\).
In simple words: Make the root numbers (indices) equal by finding their LCM, then compare the numbers inside.

Exam Tip: To compare surds of different orders, convert them to surds of the same order using the LCM of their indices.

Question 16. Find, whether \(\frac{1}{625}\) is a terminating or non terminating decimal number.
Answer: Given fraction = \(\frac{1}{625}\)
Factorize the denominator: \(625 = 5^4\)
Since the denominator is of the form \(2^n 5^m\) (here \(n = 0, m = 4\)), the rational number has a terminating decimal representation.
In simple words: Because 625 is a power of 5, the division ends completely without repeating.

Exam Tip: Check denominator prime factors: if only 2 and/or 5 appear, it terminates.

Question 17. Find the value of \(x\), if \(5^{x-2} = 125\).
Answer: Given equation: \(5^{x-2} = 125\)
Express 125 as a power of 5: \(125 = 5^3\)
So, \(5^{x-2} = 5^3\)
Since the bases are equal, equate the exponents:
\(x - 2 = 3\)
\(x = 3 + 2 = 5\)
In simple words: Write both sides with the same base 5, then set the powers equal to each other.

Exam Tip: If \(a^m = a^n\) and \(a \neq 0, \pm 1\), then \(m = n\).

Question 18. Simplify: (i) \(\left(3^{\frac{1}{5}}\right)^4\) (ii) \(13^{\frac{1}{5}} \cdot 13^{\frac{1}{7}}\)
Answer:
(i) \(\left(3^{\frac{1}{5}}\right)^4 = 3^{\frac{1}{5} \times 4} = 3^{\frac{4}{5}}\)
(ii) \(13^{\frac{1}{5}} \cdot 13^{\frac{1}{7}} = 13^{\frac{1}{5} + \frac{1}{7}} = 13^{\frac{7 + 5}{35}} = 13^{\frac{12}{35}}\)
In simple words: Use exponent laws: \((a^m)^n = a^{mn}\) for part (i) and \(a^m \cdot a^n = a^{m+n}\) for part (ii).

Exam Tip: Memorize all laws of exponents thoroughly before solving simplification questions.

Question 19. Rationalize the denominators of the following: (i) \(\frac{1}{\sqrt{7}}\) (ii) \(\frac{1}{\sqrt{2}}\)
Answer:
(i) \(\frac{1}{\sqrt{7}} = \frac{1}{\sqrt{7}} \times \frac{\sqrt{7}}{\sqrt{7}} = \frac{\sqrt{7}}{7}\)
(ii) \(\frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{2}\)
In simple words: Multiply both numerator and denominator by the square root in the denominator to eliminate the root from the bottom.

Question 20. Simplify \(\sqrt{7}(\sqrt{35} + \sqrt{7})\).
Answer: Multiply \(\sqrt{7}\) into each term inside the bracket:
\(= \sqrt{7} \times \sqrt{35} + \sqrt{7} \times \sqrt{7}\)
\(= \sqrt{7 \times 35} + 7\)
\(= \sqrt{7 \times 7 \times 5} + 7\)
\(= 7\sqrt{5} + 7\)
In simple words: Distribute the outside root across the terms inside and simplify any perfect square roots.

Exam Tip: Use distributive property \(\sqrt{a}(\sqrt{b} + \sqrt{c}) = \sqrt{ab} + \sqrt{ac}\).

Question 21. Find two irrational numbers between 2 and 2.5.
Answer: Between 2 and 2.5, we can choose non-terminating non-recurring decimals such as:
\(2.1010010001 \dots\)
\(2.3020020002 \dots\)
Both numbers lie strictly between 2 and 2.5 and have non-repeating decimal patterns.
In simple words: Create decimals between 2 and 2.5 with patterns that never repeat and never end.

Exam Tip: Infinite patterns with increasing numbers of zeros are classic examples of constructing irrational numbers.

Question 22. Insert a rational & an irrational number between 2 and 3.
Answer:
- Rational number between 2 and 3: \(\frac{2 + 3}{2} = 2.5\) (or \(\frac{5}{2}\)).
- Irrational number between 2 and 3: \(\sqrt{2 \times 3} = \sqrt{6}\) (since \(2 = \sqrt{4}\) and \(3 = \sqrt{9}\), \(\sqrt{6}\) lies between them).
In simple words: The average gives a rational number, and the square root of their product gives an irrational number.

Exam Tip: To find an irrational number between \(a\) and \(b\), \(\sqrt{ab}\) is a reliable formula when \(ab\) is not a perfect square.

Question 23. Identify \(\sqrt{45}\) as rational number or irrational number.
Answer: Simplify \(\sqrt{45}\):
\(\sqrt{45} = \sqrt{9 \times 5} = 3\sqrt{5}\)
Since \(\sqrt{5}\) is irrational, its product with 3 is irrational.
Therefore, \(\sqrt{45}\) is an irrational number.
In simple words: 45 is not a perfect square, and pulling out factors leaves a square root behind, making it irrational.

Exam Tip: Always simplify surds by prime factorization to check if any radical remains.

Question 24. Give examples of two irrational numbers the product of which is: (i) a rational number (ii) an irrational number.
Answer:
(i) Rational product: Let the two irrational numbers be \(\sqrt{2}\) and \(3\sqrt{2}\) (or simply \(\sqrt{2}\) and \(\sqrt{8}\)).
Product = \(\sqrt{2} \times \sqrt{8} = \sqrt{16} = 4\) (which is rational).
(ii) Irrational product: Let the two irrational numbers be \(\sqrt{2}\) and \(\sqrt{3}\).
Product = \(\sqrt{2} \times \sqrt{3} = \sqrt{6}\) (which is irrational).
In simple words: Multiplying matching square roots removes the root to make a rational number, while different roots stay irrational.

Exam Tip: Surds of the same order multiplied together can yield a rational number if their radicands combine to a perfect power.

Question 25. Identify \(\sqrt{80}\) as rational number or irrational number.
Answer: Simplify \(\sqrt{80}\):
\(\sqrt{80} = \sqrt{16 \times 5} = 4\sqrt{5}\)
Since \(\sqrt{5}\) is irrational, \(4\sqrt{5}\) is an irrational number.
In simple words: 80 is not a square number, so its simplified form retains a square root.

Exam Tip: Factorize into largest perfect square: \(80 = 16 \times 5\).

Question 26. How to insert irrational numbers between two given rational numbers.
Answer: To insert irrational numbers between two rational numbers \(a\) and \(b\):
1. Convert \(a\) and \(b\) into decimal form or find a non-square number between them under a square root.
2. Construct non-terminating, non-recurring decimals lying between the decimal expansions of \(a\) and \(b\).
For example, if \(a = 2\) and \(b = 3\), an irrational number is \(2.010010001\dots\) or \(\sqrt{a \cdot b}\) if applicable.
In simple words: Write non-repeating decimal numbers that fall numerically between the two given numbers.

Exam Tip: Decimal construction method is foolproof for inserting any number of irrational numbers.

Question 27. Find the decimal representation of \(\frac{8}{3}\).
Answer: Performing long division of 8 by 3:
\(8 \div 3 = 2.6666\dots = 2.\bar{6}\)
The decimal representation is non-terminating and recurring.
In simple words: Dividing 8 by 3 gives a decimal where the digit 6 repeats endlessly.

Exam Tip: Use bar notation \(2.\bar{6}\) to indicate repeating decimals cleanly.

Question 28. Express \(\frac{7}{8}\) in the decimal form by long division method.divider
Answer: Dividing 7 by 8 using long division:
\(7.0 \div 8 = 0.875\)
The decimal representation is terminating.
In simple words: Long division stops completely after three decimal places, yielding 0.875.

Exam Tip: Terminating decimals have a finite number of digits after the decimal point.

Question 29. Find three rational numbers between -2 and 5.
Answer: Integers between -2 and 5 are -1, 0, 1, 2, 3, 4.
Any three of these integers can be written as rational numbers with denominator 1.
Three rational numbers are: -1, 0, and 1 (or \(-1, \frac{1}{2}, 3\), etc.).
In simple words: Choose any three numbers that lie between -2 and 5.

Question 30. Insert 100 rational numbers between \(-\frac{3}{13}\) and \(\frac{9}{13}\).
Answer: The numbers are \(-\frac{3}{13}\) and \(\frac{9}{13}\).
The numerators range from -3 to 9, which only gives 11 integers. To insert 100 rational numbers, multiply the numerator and denominator of both fractions by a suitable multiplier, say 10:
\(-\frac{3}{13} = -\frac{30}{130}\)
\(\frac{9}{13} = \frac{90}{130}\)
Now, 100 rational numbers can be chosen by taking denominators as 130 and numerators from -29 to 70.
Examples: \(-\frac{29}{130}, -\frac{28}{130}, \dots, \frac{70}{130}\).
In simple words: Multiply top and bottom by 10 to expand the gap so you can easily fit 100 fractions in between.

Exam Tip: Scaling denominators by multiplying by 10, 100, etc., is the standard technique for inserting many rational numbers.

Question 31. Insert 10 rational numbers between \(-\frac{3}{11}\) and \(\frac{8}{11}\).
Answer: The given numbers are \(-\frac{3}{11}\) and \(\frac{8}{11}\).
The integers between -3 and 8 are -2, -1, 0, 1, 2, 3, 4, 5, 6, 7 (exactly 10 integers).
Thus, 10 rational numbers are:
\(-\frac{2}{11}, -\frac{1}{11}, 0, \frac{1}{11}, \frac{2}{11}, \frac{3}{11}, \frac{4}{11}, \frac{5}{11}, \frac{6}{11}, \frac{7}{11}\).
In simple words: Since there are already 10 integer steps between -3 and 8, just put them all over 11.

Question 32. State whether the following statements are true or false. Give reasons for your answers. (i) Every integer is a whole number (ii) Every rational number is a whole number.
Answer:
(i) False. Negative integers like -3, -5 are integers, but they are not whole numbers (whole numbers start from 0).
(ii) False. Fractions like \(\frac{2}{3}\) or \(\frac{5}{7}\) are rational numbers, but they are not whole numbers.
In simple words: Whole numbers only include 0 and positive counting numbers, so negative numbers and fractions fail.

Exam Tip: Always provide a counterexample (like a negative number or fraction) to disprove false set statements.

Question 33. Find five rational numbers between \(\frac{3}{5}\) and \(\frac{4}{5}\).
Answer: To find 5 rational numbers, multiply the numerator and denominator of both \(\frac{3}{5}\) and \(\frac{4}{5}\) by \(5 + 1 = 6\) (or 10):
\(\frac{3}{5} = \frac{3 \times 10}{5 \times 10} = \frac{30}{50}\)
\(\frac{4}{5} = \frac{4 \times 10}{5 \times 10} = \frac{40}{50}\)
Five rational numbers between \(\frac{30}{50}\) and \(\frac{40}{50}\) are:
\(\frac{31}{50}, \frac{32}{50}, \frac{33}{50}, \frac{34}{50}, \frac{35}{50}\).
In simple words: Scale up the fractions by multiplying by 10 to create enough space between them.

Exam Tip: Multiplying by \(n+1\) or 10 is always safe for finding \(n\) rational numbers.

Question 34. Find six rational numbers between 3 and 4.
Answer: To find 6 rational numbers, multiply numerator and denominator by \(6 + 1 = 7\) (or 10):
\(3 = \frac{30}{10}\)
\(4 = \frac{40}{10}\)
Six rational numbers between \(\frac{30}{10}\) and \(\frac{40}{10}\) are:
\(\frac{31}{10}, \frac{32}{10}, \frac{33}{10}, \frac{34}{10}, \frac{35}{10}, \frac{36}{10}\).
In simple words: Write 3 and 4 as fractions with denominator 10, then pick six consecutive numerators.

Exam Tip: Any multiplier greater than or equal to \(n+1\) will yield enough space.

Question 35. Find five rational numbers between 1 and 2.
Answer: Multiply numerator and denominator of 1 and 2 by 6:
\(1 = \frac{6}{6}\)
\(2 = \frac{12}{6}\)
Five rational numbers between \(\frac{6}{6}\) and \(\frac{12}{6}\) are:
\(\frac{7}{6}, \frac{8}{6}, \frac{9}{6}, \frac{10}{6}, \frac{11}{6}\).
In simple words: Convert 1 and 2 into fractions with denominator 6 and list the fractions in between.

Exam Tip: Using denominator \(n+1\) is the textbook method for finding \(n\) rational numbers.

Question 36. Express \(0.8888 \dots\) in the form of \(\frac{p}{q}\) where \(p\) and \(q\) are integers and \(q \neq 0\).
Answer: Let \(x = 0.8888\dots\) --- (1)
Since 1 digit is repeating, multiply both sides by 10:
\(10x = 8.8888\dots\) --- (2)
Subtract equation (1) from equation (2):
\(10x - x = 8.8888\dots - 0.8888\dots\)
\(9x = 8\)
\(x = \frac{8}{9}\)
In simple words: Set the repeating decimal as \(x\), multiply by 10, and subtract to eliminate the decimal part.

Exam Tip: The number of repeating digits determines whether you multiply by 10, 100, 1000, etc.

Question 37. Rationalise the denominator of \(\frac{2}{\sqrt{7} + \sqrt{5}}\).
Answer: Multiply numerator and denominator by the conjugate \(\sqrt{7} - \sqrt{5}\):
\(= \frac{2}{\sqrt{7} + \sqrt{5}} \times \frac{\sqrt{7} - \sqrt{5}}{\sqrt{7} - \sqrt{5}}\)
\(= \frac{2(\sqrt{7} - \sqrt{5})}{(\sqrt{7})^2 - (\sqrt{5})^2}\)
\(= \frac{2(\sqrt{7} - \sqrt{5})}{7 - 5}\)
\(= \frac{2(\sqrt{7} - \sqrt{5})}{2} = \sqrt{7} - \sqrt{5}\)
In simple words: Multiply top and bottom by the conjugate binomial with the middle sign flipped, then simplify.

Exam Tip: The conjugate of \(a + b\) is \(a - b\), and \((a+b)(a-b) = a^2 - b^2\).

Question 38. Rationalise the denominator in each of the following: (i) \(\frac{2}{\sqrt{3}}\) (ii) \(\frac{1}{\sqrt{7}}\) (iii) \(\frac{1}{\sqrt{2}}\)
Answer:
(i) \(\frac{2}{\sqrt{3}} = \frac{2}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{2\sqrt{3}}{3}\)
(ii) \(\frac{1}{\sqrt{7}} = \frac{1}{\sqrt{7}} \times \frac{\sqrt{7}}{\sqrt{7}} = \frac{\sqrt{7}}{7}\)
(iii) \(\frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{2}\)
In simple words: Multiply both numerator and denominator by the single square root present in the denominator.

Exam Tip: Single-term radicals are rationalized by multiplying by that exact radical.

Question 39. Simplify \(\frac{6 - 4\sqrt{2}}{6 + 4\sqrt{2}}\).
Answer: Rationalize the denominator by multiplying by the conjugate \(6 - 4\sqrt{2}\):
\(= \frac{6 - 4\sqrt{2}}{6 + 4\sqrt{2}} \times \frac{6 - 4\sqrt{2}}{6 - 4\sqrt{2}}\)
\(= \frac{(6 - 4\sqrt{2})^2}{6^2 - (4\sqrt{2})^2}\)
\(= \frac{36 + 32 - 48\sqrt{2}}{36 - 32}\)
\(= \frac{68 - 48\sqrt{2}}{4}\)
\(= 17 - 12\sqrt{2}\)
In simple words: Use the square formula on top and difference of squares on the bottom, then divide each term by 4.

Exam Tip: Always expand numerator carefully using \((a-b)^2 = a^2 + b^2 - 2ab\).

Question 40. Rationalize the denominators of the following: (i) \(\frac{1}{\sqrt{7} - \sqrt{6}}\) (ii) \(\frac{1}{\sqrt{5} + \sqrt{2}}\) (iii) \(\frac{1}{\sqrt{7} - 2}\)
Answer:
(i) \(\frac{1}{\sqrt{7} - \sqrt{6}} \times \frac{\sqrt{7} + \sqrt{6}}{\sqrt{7} + \sqrt{6}} = \frac{\sqrt{7} + \sqrt{6}}{7 - 6} = \sqrt{7} + \sqrt{6}\)
(ii) \(\frac{1}{\sqrt{5} + \sqrt{2}} \times \frac{\sqrt{5} - \sqrt{2}}{\sqrt{5} - \sqrt{2}} = \frac{\sqrt{5} - \sqrt{2}}{5 - 2} = \frac{\sqrt{5} - \sqrt{2}}{3}\)
(iii) \(\frac{1}{\sqrt{7} - 2} \times \frac{\sqrt{7} + 2}{\sqrt{7} + 2} = \frac{\sqrt{7} + 2}{7 - 4} = \frac{\sqrt{7} + 2}{3}\)
In simple words: Multiply numerator and denominator by the conjugate of the denominator and simplify the resulting integer difference.

Exam Tip: The denominator always simplifies to a single rational integer using \(a^2 - b^2\).

Question 41. Represent \(\sqrt{9.3}\) on the number line.
Answer: Construction Steps:
1. Draw a line segment \(AB = 9.3\) units.
2. Extend \(AB\) to \(C\) such that \(BC = 1\) unit.
3. Find the midpoint \(O\) of \(AC\).
4. Draw a semicircle with center \(O\) and radius \(OC\) (or \(OA\)).
5. Draw a perpendicular line from point \(B\) intersecting the semicircle at point \(D\).
6. With \(B\) as center and radius \(BD\), draw an arc intersecting the number line at point \(E\). The distance \(BE\) represents \(\sqrt{9.3}\).
In simple words: Measure 9.3, add 1 unit, find the center of the total length, draw a semicircle, and swing the perpendicular height down to the number line.

Exam Tip: This geometric construction uses the right triangle property where altitude squared equals segment products.

Question 42. Visualize 3.765 on the number line using successive magnification.
Answer: 1. Locate 3.765 between 3 and 4. Divide the interval into 10 equal parts (\(3.1, 3.2, \dots, 4.0\)).
2. Visualize 3.765 lying between 3.7 and 3.8. Magnify this interval into 10 parts (\(3.71, 3.72, \dots, 3.80\)).
3. Locate 3.765 between 3.76 and 3.77. Magnify this interval into 10 parts (\(3.761, 3.762, \dots, 3.770\)).
4. Exactly pinpoint 3.765 on the magnified number line.
In simple words: Zoom in progressively between narrower decimal intervals (3 to 4, then 3.7 to 3.8, then 3.76 to 3.77) to locate the exact number.

Exam Tip: Successive magnification relies on decimal place value expansion.

Question 43. State whether the following statements are true or false. Justify. (i) Every irrational number is a real number. (ii) Every point on the number line is of the form \(\sqrt{m}\), where m is a natural number. (iii) Every real number is an irrational number.
Answer:
(i) True. Real numbers consist of both rational and irrational numbers.
(ii) False. Negative numbers cannot be expressed as the square root of any natural number since square roots of natural numbers are positive.
(iii) False. Real numbers also include rational numbers (like 2, 3, \(\frac{1}{2}\)), which are not irrational.
In simple words: Irrational numbers are real, but real numbers contain rational numbers too, and negative numbers cannot be square roots of natural numbers.

Exam Tip: Real number set encompasses everything on the number line.

Question 44. Classify the following numbers as rational or irrational. (i) \(\sqrt{23}\) (ii) \(\sqrt{225}\) (iii) \(7.478\)
Answer:
(i) \(\sqrt{23}\): Irrational (23 is not a perfect square).
(ii) \(\sqrt{225}\): Rational (\(\sqrt{225} = 15\), which is an integer).
(iii) \(7.478\): Rational (terminating decimal).
In simple words: Non-square roots are irrational, while terminating decimals and perfect square roots are rational.

Question 45. Find three different irrational numbers between the rational numbers \(\frac{5}{7}\) and \(\frac{9}{11}\).
Answer: Convert fractions to decimals:
\(\frac{5}{7} = 0.\overline{714285} \approx 0.714285\dots\)
\(\frac{9}{11} = 0.\overline{81} \approx 0.8181\dots\)
Three different irrational numbers between \(0.714285\dots\) and \(0.8181\dots\) are:
1. \(0.73073007300073\dots\)
2. \(0.75050050005\dots\)
3. \(0.8010010001\dots\)
In simple words: Convert the fractions to decimals and create non-repeating numbers that fall between them.

Exam Tip: Decimal conversion makes finding bounds for irrational numbers straightforward.

Question 46. Prove that \((3 + \sqrt{2})^2\) is an irrational number.
Answer: Expand the expression using the algebraic identity \((a+b)^2 = a^2 + b^2 + 2ab\):
\((3 + \sqrt{2})^2 = 3^2 + (\sqrt{2})^2 + 2(3)(\sqrt{2})\)
\(= 9 + 2 + 6\sqrt{2}\)
\(= 11 + 6\sqrt{2}\)
Since 11 is rational and \(6\sqrt{2}\) is irrational, their sum is an irrational number.
Therefore, \((3 + \sqrt{2})^2\) is an irrational number.
In simple words: Expanding the square leaves a term with \(\sqrt{2}\), which keeps the whole number irrational.

Exam Tip: The sum of a rational number and a non-zero irrational number is always irrational.

Question 47. Are square roots of all the positive integers irrational? If not, give an example of the square root of a number that is a rational number.
Answer: No, square roots of all positive integers are not irrational.
Example: \(\sqrt{4} = 2\), which is a rational number (since 2 can be written as \(\frac{2}{1}\)). Similarly, \(\sqrt{9} = 3\) is rational.
In simple words: If an integer is a perfect square, its square root is a whole number, making it rational.

Question 48. Simplify (i) \(2^{\frac{2}{3}} \cdot 2^{\frac{1}{5}}\) (ii) \(\left(\frac{1}{3^3}\right)^7\) (iii) \(\frac{11^{\frac{1}{2}}}{11^{\frac{1}{4}}}\) (iv) \(7^{\frac{1}{2}} \cdot 8^{\frac{1}{2}}\)
Answer:
(i) \(2^{\frac{2}{3}} \cdot 2^{\frac{1}{5}} = 2^{\frac{2}{3} + \frac{1}{5}} = 2^{\frac{10+3}{15}} = 2^{\frac{13}{15}}\)
(ii) \(\left(\frac{1}{3^3}\right)^7 = (3^{-3})^7 = 3^{-21} = \frac{1}{3^{21}}\)
(iii) \(\frac{11^{\frac{1}{2}}}{11^{\frac{1}{4}}} = 11^{\frac{1}{2} - \frac{1}{4}} = 11^{\frac{2-1}{4}} = 11^{\frac{1}{4}}\)
(iv) \(7^{\frac{1}{2}} \cdot 8^{\frac{1}{2}} = (7 \times 8)^{\frac{1}{2}} = 56^{\frac{1}{2}} = \sqrt{56} = 2\sqrt{14}\)
In simple words: Apply exponent rules for multiplication, division, power of a power, and matching fractional exponents.

Exam Tip: Practice laws of exponents: \(a^m a^n = a^{m+n}\), \(\frac{a^m}{a^n} = a^{m-n}\), \((ab)^m = a^m b^m\).

Question 49. Simplify (i) \((625)^{-\frac{1}{4}}\) (ii) \((\sqrt[5]{32})^{-3}\)
Answer:
(i) \((625)^{-\frac{1}{4}} = (5^4)^{-\frac{1}{4}} = 5^{4 \times \left(-\frac{1}{4}\right)} = 5^{-1} = \frac{1}{5}\)
(ii) \((\sqrt[5]{32})^{-3} = (32^{\frac{1}{5}})^{-3} = ((2^5)^{\frac{1}{5}})^{-3} = (2)^{-\frac{3}{5}} = \frac{1}{2^{\frac{3}{5}}}\)
In simple words: Express numbers as prime powers and apply exponent multiplication rules.

Exam Tip: Negative exponents indicate reciprocals: \(a^{-m} = \frac{1}{a^m}\).

Question 50. Simplify (i) \((\sqrt{4})^{-3}\) (ii) \((\sqrt[3]{8})^{-1}\)
Answer:
(i) \((\sqrt{4})^{-3} = (2)^{-3} = \frac{1}{2^3} = \frac{1}{8}\)
(ii) \((\sqrt[3]{8})^{-1} = (2)^{-1} = \frac{1}{2}\)
In simple words: Evaluate the inner radical first, then apply the negative exponent.

Exam Tip: Simplifying inside out prevents calculation errors.

Question 51. Rationalize the denominator of (i) \(\frac{1}{3 + \sqrt{2}}\) (ii) \(\frac{1}{2 + \sqrt{3}}\)
Answer:
(i) \(\frac{1}{3 + \sqrt{2}} \times \frac{3 - \sqrt{2}}{3 - \sqrt{2}} = \frac{3 - \sqrt{2}}{3^2 - (\sqrt{2})^2} = \frac{3 - \sqrt{2}}{9 - 2} = \frac{3 - \sqrt{2}}{7}\)
(ii) \(\frac{1}{2 + \sqrt{3}} \times \frac{2 - \sqrt{3}}{2 - \sqrt{3}} = \frac{2 - \sqrt{3}}{2^2 - (\sqrt{3})^2} = \frac{2 - \sqrt{3}}{4 - 3} = 2 - \sqrt{3}\)
In simple words: Multiply numerator and denominator by the conjugate binomial and simplify the denominator.

Exam Tip: Always use conjugates for denominators with two terms containing a square root.

Question 52. Simplify (i) \((\sqrt{5} + \sqrt{2})^2\) (ii) \((\sqrt{11} - \sqrt{5})^2\)
Answer:
(i) \((\sqrt{5} + \sqrt{2})^2 = (\sqrt{5})^2 + (\sqrt{2})^2 + 2(\sqrt{5})(\sqrt{2}) = 5 + 2 + 2\sqrt{10} = 7 + 2\sqrt{10}\)
(ii) \((\sqrt{11} - \sqrt{5})^2 = (\sqrt{11})^2 + (\sqrt{5})^2 - 2(\sqrt{11})(\sqrt{5}) = 11 + 5 - 2\sqrt{55} = 16 - 2\sqrt{55}\)
In simple words: Expand using standard algebraic square formulas \((a \pm b)^2 = a^2 + b^2 \pm 2ab\).

Exam Tip: Squaring a square root cancels the radical symbol (\((\sqrt{a})^2 = a\)).

Question 53. Simplify: (i) \((\sqrt{5} + \sqrt{5})(\sqrt{5} - \sqrt{5})\) [Note: Source shows \((5+\sqrt{5})(5-\sqrt{5})\)] (ii) \((3 + 2\sqrt{2})(3 - 2\sqrt{2})\)
Answer:
(i) \((5 + \sqrt{5})(5 - \sqrt{5}) = 5^2 - (\sqrt{5})^2 = 25 - 5 = 20\)
(ii) \((3 + 2\sqrt{2})(3 - 2\sqrt{2}) = 3^2 - (2\sqrt{2})^2 = 9 - (4 \times 2) = 9 - 8 = 1\)
In simple words: Apply the difference of squares identity \((a+b)(a-b) = a^2 - b^2\).

Exam Tip: This formula saves time by eliminating middle cross-multiplication terms.

Question 54. Simplify (i) \((3 + \sqrt{3})(2 + \sqrt{2})\) (ii) \((5 + \sqrt{7})(2 + \sqrt{5})\)
Answer:
(i) \((3 + \sqrt{3})(2 + \sqrt{2}) = 3(2) + 3\sqrt{2} + 2\sqrt{3} + \sqrt{3}\sqrt{2} = 6 + 3\sqrt{2} + 2\sqrt{3} + \sqrt{6}\)
(ii) \((5 + \sqrt{7})(2 + \sqrt{5}) = 5(2) + 5\sqrt{5} + 2\sqrt{7} + \sqrt{7}\sqrt{5} = 10 + 5\sqrt{5} + 2\sqrt{7} + \sqrt{35}\)
In simple words: Use FOIL (First, Outer, Inner, Last) method to multiply every term.

Exam Tip: Ensure all distinct radical terms are kept separate unless they have identical radicands.

Question 55. Represent \(\sqrt{3}\) on a number line.
Answer: Construction Steps:
1. Draw a number line and mark 0 and 1. Draw a segment \(OA = 1\) unit.
2. At point \(A\), draw a perpendicular line segment \(AB = 1\) unit.
3. Join \(OB\). By Pythagoras theorem, \(OB = \sqrt{1^2 + 1^2} = \sqrt{2}\).
4. At point \(B\), draw a perpendicular line segment \(BC = 1\) unit.
5. Join \(OC\). By Pythagoras theorem, \(OC = \sqrt{(\sqrt{2})^2 + 1^2} = \sqrt{3}\).
6. With center \(O\) and radius \(OC\), draw an arc intersecting the number line at \(\sqrt{3}\).
In simple words: Build upon the \(\sqrt{2}\) triangle by adding a 1-unit perpendicular side to form \(\sqrt{3}\), then swing it down to the number line.

Exam Tip: This forms the foundational basis of the square root spiral.

Question 56. Represent \(\sqrt{2}\) on a number line.
Answer: Construction Steps:
1. Draw a number line with origin 0. Mark 1 unit distance \(OA = 1\).
2. Draw a perpendicular \(AB = 1\) unit at point \(A\).
3. Join \(OB\). In right triangle \(OAB\), \(OB = \sqrt{1^2 + 1^2} = \sqrt{2}\).
4. Using compass with center \(O\) and radius \(OB\), draw an arc cutting the number line at point \(P\). Point \(P\) represents \(\sqrt{2}\).
In simple words: Make a right triangle with base 1 and height 1, then use the hypotenuse radius to mark \(\sqrt{2}\) on the number line.

Exam Tip: Always state the use of Pythagoras theorem explicitly.

Question 57. You know that \(\frac{1}{7} = 0.142857 \dots\). Can you predict the decimal expansions of \(\frac{2}{7}, \frac{3}{7}, \frac{4}{7}, \frac{5}{7}, \frac{6}{7}\) without actually doing the long division? If so how?
Answer: Yes, we can predict them by multiplying the decimal value of \(\frac{1}{7}\) by 2, 3, 4, 5, and 6 respectively:
\(\frac{1}{7} = 0.\overline{142857}\)
\(\frac{2}{7} = 2 \times \frac{1}{7} = 2 \times 0.142857\dots = 0.\overline{285714}\)
\(\frac{3}{7} = 3 \times \frac{1}{7} = 3 \times 0.142857\dots = 0.\overline{428571}\)
\(\frac{4}{7} = 4 \times \frac{1}{7} = 4 \times 0.142857\dots = 0.\overline{571428}\)
\(\frac{5}{7} = 5 \times \frac{1}{7} = 5 \times 0.142857\dots = 0.\overline{714285}\)
\(\frac{6}{7} = 6 \times \frac{1}{7} = 6 \times 0.142857\dots = 0.\overline{857142}\)
In simple words: Multiply the repeating sequence digits directly by the numerator multiplier.

Exam Tip: Notice that the cyclic order of digits remains identical for all multiples of \(\frac{1}{7}\).

Question 58. Represent \(\sqrt{5}\) on the number line.
Answer: Construction Steps:
1. Draw a number line, mark origin 0, and take a base segment \(OA = 2\) units.
2. At point \(A\), construct a perpendicular line segment \(AB = 1\) unit.
3. Join \(OB\). By Pythagoras theorem, \(OB = \sqrt{2^2 + 1^2} = \sqrt{4 + 1} = \sqrt{5}\).
4. With center \(O\) and radius \(OB\), draw an arc intersecting the number line at \(\sqrt{5}\).
In simple words: Use base 2 and height 1 in a right triangle so the hypotenuse becomes \(\sqrt{5}\).

Question 59. Construct the square root spiral.
Answer: Construction Steps:
1. Start with point \(O\) and draw a line segment \(OP_1 = 1\) unit.
2. At \(P_1\), draw a perpendicular segment \(P_1P_2 = 1\) unit. Join \(OP_2 = \sqrt{2}\).
3. At \(P_2\), draw a perpendicular segment \(P_2P_3 = 1\) unit. Join \(OP_3 = \sqrt{3}\).
4. Continue this process for \(P_3P_4 = 1\) to get \(OP_4 = \sqrt{4} = 2\), and so on to form a continuous spiral.
In simple words: Continuously draw 1-unit perpendiculars at the tip of each new hypotenuse to spiral outward.

Exam Tip: Label each hypotenuse clearly with its square root value.

Question 60. Find: (i) \(9^{\frac{3}{2}}\) (ii) \(32^{\frac{2}{5}}\) (iii) \(16^{\frac{3}{4}}\)
Answer:
(i) \(9^{\frac{3}{2}} = (3^2)^{\frac{3}{2}} = 3^{2 \times \frac{3}{2}} = 3^3 = 27\)
(ii) \(32^{\frac{2}{5}} = (2^5)^{\frac{2}{5}} = 2^{5 \times \frac{2}{5}} = 2^2 = 4\)
(iii) \(16^{\frac{3}{4}} = (2^4)^{\frac{3}{4}} = 2^{4 \times \frac{3}{4}} = 2^3 = 8\)
In simple words: Express the base as a power that matches the denominator of the fractional exponent, then simplify.

Exam Tip: Always rewrite bases in exponential form before applying power rules.

Question 61. Simplify (i) \((64)^{\frac{1}{3}}\) (ii) \((125)^{-\frac{1}{3}}\) (iii) \((27)^{-\frac{2}{3}}\) (iv) \(\left(\frac{64}{25}\right)^{-\frac{3}{2}}\)
Answer:
(i) \((64)^{\frac{1}{3}} = (4^3)^{\frac{1}{3}} = 4^{3 \times \frac{1}{3}} = 4\)
(ii) \((125)^{-\frac{1}{3}} = (5^3)^{-\frac{1}{3}} = 5^{-1} = \frac{1}{5}\)
(iii) \((27)^{-\frac{2}{3}} = (3^3)^{-\frac{2}{3}} = 3^{-2} = \frac{1}{3^2} = \frac{1}{9}\)
(iv) \(\left(\frac{64}{25}\right)^{-\frac{3}{2}} = \left(\left(\frac{8}{5}\right)^2\right)^{-\frac{3}{2}} = \left(\frac{8}{5}\right)^{-3} = \left(\frac{5}{8}\right)^3 = \frac{125}{512}\)
In simple words: Convert bases into prime powers and invert fractions when dealing with negative exponents.

Exam Tip: Remember \(\left(\frac{a}{b}\right)^{-m} = \left(\frac{b}{a}\right)^m\).

Question 62. Simplify (i) \(\frac{3}{\sqrt{3} + 1} + \frac{5}{\sqrt{3} - 1}\) (ii) \(\frac{\sqrt{7} - 1}{\sqrt{7} + 1} - \frac{\sqrt{7} + 1}{\sqrt{7} - 1}\)
Answer:
(i) Rationalize each term separately:
First term: \(\frac{3(\sqrt{3} - 1)}{(\sqrt{3}+1)(\sqrt{3}-1)} = \frac{3(\sqrt{3}-1)}{3-1} = \frac{3\sqrt{3}-3}{2}\)
Second term: \(\frac{5(\sqrt{3} + 1)}{(\sqrt{3}-1)(\sqrt{3}+1)} = \frac{5(\sqrt{3}+1)}{3-1} = \frac{5\sqrt{3}+5}{2}\)
Sum = \(\frac{3\sqrt{3} - 3 + 5\sqrt{3} + 5}{2} = \frac{8\sqrt{3} + 2}{2} = 4\sqrt{3} + 1\)

(ii) Finding common denominator or rationalizing:
\(= \frac{(\sqrt{7}-1)^2 - (\sqrt{7}+1)^2}{(\sqrt{7}+1)(\sqrt{7}-1)}\)
Numerator = \((7 + 1 - 2\sqrt{7}) - (7 + 1 + 2\sqrt{7}) = (8 - 2\sqrt{7}) - (8 + 2\sqrt{7}) = -4\sqrt{7}\)
Denominator = \(7 - 1 = 6\)
Result = \(\frac{-4\sqrt{7}}{6} = -\frac{2\sqrt{7}}{3}\)
In simple words: Rationalize each term or take a common denominator using algebraic identities.

Exam Tip: For expressions with multiple fractions, take the LCM of the denominators after rationalizing.

Question 63. Simplify (i) \(\sqrt{5} \times \sqrt{45}\) (ii) \(\sqrt{2}x \times \sqrt{8}x\) [Note: Source shows \(\sqrt{2x} \times \sqrt{8x}\)] (iii) \(\frac{\sqrt{2}}{\sqrt{50}}\)
Answer:
(i) \(\sqrt{5 \times 45} = \sqrt{225} = 15\)
(ii) \(\sqrt{2x \times 8x} = \sqrt{16x^2} = 4x\) (for \(x \geq 0\))
(iii) \(\sqrt{\frac{2}{50}} = \sqrt{\frac{1}{25}} = \frac{1}{5}\)
In simple words: Combine terms under a single radical sign using radical multiplication and division rules.

Exam Tip: Combining under one root simplifies calculations immensely.

Question 64. Examine, whether the following numbers are rational or irrational. (i) \((\sqrt{2} + 2)^2\) (ii) \((5 + \sqrt{5})(5 - \sqrt{5})\) (iii) \(\frac{6}{2\sqrt{3}}\)
Answer:
(i) \((\sqrt{2} + 2)^2 = 2 + 4 + 4\sqrt{2} = 6 + 4\sqrt{2}\) (Irrational, due to \(\sqrt{2}\)).
(ii) \((5 + \sqrt{5})(5 - \sqrt{5}) = 5^2 - (\sqrt{5})^2 = 25 - 5 = 20\) (Rational).
(iii) \(\frac{6}{2\sqrt{3}} = \frac{3}{\sqrt{3}} = \sqrt{3}\) (Irrational).
In simple words: Simplify each expression fully; if a square root remains, it is irrational.

Exam Tip: Always simplify before final classification.

Question 65. Write the following in decimal form and find the type of decimal expansion. (i) \(\frac{36}{100}\) (ii) \(\frac{1}{11}\) (iii) \(4\frac{1}{8}\)
Answer:
(i) \(\frac{36}{100} = 0.36\) (Terminating decimal).
(ii) \(\frac{1}{11} = 0.090909\dots = 0.\overline{09}\) (Non-terminating recurring decimal).
(iii) \(4\frac{1}{8} = \frac{33}{8} = 4.125\) (Terminating decimal).
In simple words: Perform division to check whether the decimal terminates or repeats infinitely.

Exam Tip: Fractions with prime denominators other than 2 and 5 always yield repeating decimals.

Question 66. Express the following in the form \(\frac{p}{q}\), where \(p\) and \(q\) are integers and \(q \neq 0\). (i) \(0.\bar{6}\) (ii) \(x = 0.4\bar{7}\) (iii) \(0.001\)
Answer:
(i) Let \(x = 0.6666\dots\) --- (1)
\(10x = 6.6666\dots\) --- (2)
\(9x = 6 \implies x = \frac{6}{9} = \frac{2}{3}\).

(ii) Let \(x = 0.4777\dots\) --- (1)
\(10x = 4.777\dots\) --- (2)
\(100x = 47.777\dots\) --- (3)
Subtract (2) from (3): \(90x = 43 \implies x = \frac{43}{90}\).

(iii) \(0.001 = \frac{1}{1000}\).
In simple words: Multiply by powers of 10 corresponding to the repeating block and subtract equations.

Exam Tip: Only multiply by 10 for the non-repeating part when mixed recurring decimals appear.

Question 67. Identify the following as rational or irrational number. (i) \(\sqrt{4}\) (ii) \(\sqrt{100}\) (iii) \(\sqrt{3}\) (iv) \(0.10110111011110\dots\) (v) \(0.375\) (vi) \(22.333333\dots\)
Answer:
(i) \(\sqrt{4} = 2\) - Rational.
(ii) \(\sqrt{100} = 10\) - Rational.
(iii) \(\sqrt{3}\) - Irrational.
(iv) \(0.10110111011110\dots\) - Irrational (non-terminating non-recurring).
(v) \(0.375\) - Rational (terminating decimal).
(vi) \(22.333333\dots = 22.\bar{3}\) - Rational (non-terminating recurring).
In simple words: Terminating or repeating decimals and perfect roots are rational; non-repeating decimals and surds are irrational.

Exam Tip: Recurring decimals are always rational numbers.

Most Important Questions

Question 1. Are all-rational numbers real numbers?
Answer: Yes, all rational numbers are real numbers.
The real number system is the union of rational and irrational numbers. Therefore, every rational number belongs to the set of real numbers.
In simple words: Rational numbers form a major subset inside real numbers.

Exam Tip: Remember \(\mathbb{Q} \subset \mathbb{R}\).

Question 2. Is it possible to find a natural number between 1 and 2?
Answer: No, it is not possible to find a natural number between 1 and 2.
Natural numbers are consecutive integers (\(1, 2, 3, \dots\)), and there are no integers strictly between 1 and 2.
In simple words: Natural numbers jump directly from 1 to 2 with no numbers in between.

Exam Tip: Natural and whole numbers are discrete, unlike rational and real numbers which are dense.

Question 3. Is each point on the number line of the form \(\sqrt{m}\), where m is a natural numbers?
Answer: No, this statement is false.
Reason: Negative numbers lie on the number line, but the square root of a natural number \(m\) is always positive and cannot be negative.
In simple words: Negative numbers on the line cannot be expressed as square roots of natural numbers.

Exam Tip: Every point corresponds to a real number, but not every real number is a square root of a natural number.

Question 4. Find one rational number between 5 and 6.
Answer: One rational number between 5 and 6 is their average: \(\frac{5 + 6}{2} = \frac{11}{2}\) (or 5.5).
In simple words: Take the midpoint between 5 and 6 to get 5.5.

Exam Tip: The formula \(\frac{a+b}{2}\) always finds a rational number right between two numbers.

Question 5. Without actual division, find whether the following rational numbers are terminating or non-terminating repeating: \(\frac{11}{50}\) and \(\frac{27}{56}\).
Answer:
1. For \(\frac{11}{50}\): Denominator \(50 = 2^1 \times 5^2\). Since prime factors are only 2 and 5, it is a terminating decimal.
2. For \(\frac{27}{56}\): Denominator \(56 = 2^3 \times 7^1\). Since it contains the prime factor 7 (other than 2 and 5), it is a non-terminating repeating decimal.
In simple words: If the bottom number in lowest terms only factors into 2s and 5s, it terminates; otherwise, it repeats.

Exam Tip: Always reduce fractions to lowest terms before checking denominator prime factors.

Question 6. Name the following: (a) The outer layer of the cell (b) The fluid like substance present outside the nucleus
Answer:
(a) Cell membrane (or Plasma membrane in animal cells, and Cell wall outside the membrane in plant cells).
(b) Cytoplasm.
In simple words: The cell membrane surrounds the cell, and cytoplasm is the jelly-like fluid inside.

Exam Tip: Biology terminology should be written with precise scientific names.

Question 7. Can photosynthesis take place outside the leaves? If yes, then where this process takes place?
Answer: Yes, photosynthesis can take place outside the leaves.
It takes place in other green parts of the plant, such as green stems (e.g., in cactus) and green floral buds, which contain chlorophyll-bearing cells (chloroplasts).
In simple words: Any green part of a plant containing chlorophyll can carry out photosynthesis.

Exam Tip: Chlorophyll and sunlight are the essential prerequisites for photosynthesis anywhere in the plant.

Question 8. Express the decimal expression \(0.473\) as a rational numbers.
Answer: Given decimal: \(0.473 = \frac{473}{1000}\).
Since 473 and 1000 have no common factors other than 1, it is in simplest form \(\frac{p}{q}\).
In simple words: Remove the decimal point and place the number over 1000 since there are three decimal places.

Exam Tip: Terminating decimals convert directly by putting them over powers of 10.

Question 9. Insert four rational numbers between \(\frac{1}{3}\) and \(\frac{1}{4}\).
Answer: First, make denominators equal: LCM of 3 and 4 is 12.
\(\frac{1}{3} = \frac{4}{12}\) and \(\frac{1}{4} = \frac{3}{12}\)
To insert 4 rational numbers, multiply numerator and denominator by \(4 + 1 = 5\):
\(\frac{4}{12} = \frac{20}{60}\)
\(\frac{3}{12} = \frac{15}{60}\)
Four rational numbers between \(\frac{15}{60}\) and \(\frac{20}{60}\) are:
\(\frac{16}{60}, \frac{17}{60}, \frac{18}{60}, \frac{19}{60}\) (which simplify to \(\frac{4}{15}, \frac{17}{60}, \frac{3}{10}, \frac{19}{60}\)).
In simple words: Equalize denominators first, then scale up using multiplication to fit four fractions in between.

Exam Tip: Always make denominators identical before scaling up fractions.

Question 10. Give one example of each: (a) a parasitic plant (b) a parasitic animal (c) a saprophyte
Answer:
(a) Parasitic plant: Cuscuta (Amarbel).
(b) Parasitic animal: Cuscuta / Leech (or Plasmodium, roundworm).
(c) Saprophyte: Mushroom (or Rhizopus, yeast).
In simple words: Amarbel is a plant parasite, leeches feed on hosts as animal parasites, and mushrooms feed on dead organic matter.

Exam Tip: Memorize classic biological examples for nutrition types.

Question 11. Find two irrational numbers between 1.5 and 1.7.
Answer: Two irrational numbers between 1.5 and 1.7 can be constructed using non-terminating non-recurring decimals:
1. \(1.51010010001\dots\)
2. \(1.63030030003\dots\)
In simple words: Create non-repeating decimal numbers that fall between 1.5 and 1.7.

Exam Tip: Infinite patterns are the easiest way to generate irrational numbers between any two bounds.

Question 12. (a) What is a parasite? (b) State a difference between total parasite and partial parasite?
Answer:
(a) A parasite is an organism that lives on or inside another living organism (host) and derives its food and nourishment directly from it, often harming the host.
(b) Difference: A total parasite depends entirely on the host for both water and food (e.g., Cuscuta), whereas a partial parasite can manufacture some of its food through photosynthesis but depends on the host for water and minerals (e.g., Mistletoe).
In simple words: Parasites steal nutrients from hosts; total parasites steal everything, while partial parasites make some food themselves.

Exam Tip: Clearly state dependency on photosynthesis when distinguishing parasitic plants.

Question 13. Find an irrational number between \(\frac{1}{7}\) and \(\frac{1}{5}\).
Answer: Converting fractions to decimals:
\(\frac{1}{7} = 0.142857\dots\)
\(\frac{1}{5} = 0.20\)
An irrational number between \(0.142857\dots\) and \(0.20\) is \(0.1501500150001\dots\).
In simple words: Convert fractions to decimals and pick a non-repeating decimal lying between them.

Exam Tip: Decimal representation makes finding bounds for irrational numbers very simple.

Question 14. (a) What is an insectivorous plant? (b) Give one example of insectivorous plant. (c) Give the structure and mode of nutrition of one insectivorous plant.
Answer:
(a) Insectivorous plants are carnivorous plants that trap and digest insects to obtain nitrogen nutrients.
(b) Example: Pitcher plant (Nepenthes).
(c) Structure and nutrition: In a pitcher plant, the leaf is modified into a pitcher-like structure. The apex of the leaf forms a lid that can open and close the mouth of the pitcher. Inside the pitcher, there are hairs directed downwards. When an insect lands in the pitcher, the lid closes and the insect gets entangled in the hairs. The insect is digested by digestive juices secreted inside the pitcher to absorb nitrogenous compounds.
In simple words: Insect-eating plants trap bugs in modified leaves to get nitrogen, which is scarce in their soil.

Exam Tip: Mention nitrogen deficiency as the primary reason why plants become insectivorous.

Question 15. Rationalize the expression \(\frac{1}{(2\sqrt{3} + \sqrt{7})}\).
Answer: Multiply numerator and denominator by the conjugate \(2\sqrt{3} - \sqrt{7}\):
\(= \frac{1}{2\sqrt{3} + \sqrt{7}} \times \frac{2\sqrt{3} - \sqrt{7}}{2\sqrt{3} - \sqrt{7}}\)
\(= \frac{2\sqrt{3} - \sqrt{7}}{(2\sqrt{3})^2 - (\sqrt{7})^2}\)
\(= \frac{2\sqrt{3} - \sqrt{7}}{12 - 7}\)
\(= \frac{2\sqrt{3} - \sqrt{7}}{5}\)
In simple words: Multiply top and bottom by the conjugate binomial and evaluate square terms on the bottom.

Question 16. Find the value of the expression \(\frac{5 - \sqrt{3}}{\sqrt{3} - 3\sqrt{2}}\) [Note: Rationalizing denominator]
Answer: Multiply numerator and denominator by the conjugate \(\sqrt{3} + 3\sqrt{2}\):
\(= \frac{5 - \sqrt{3}}{\sqrt{3} - 3\sqrt{2}} \times \frac{\sqrt{3} + 3\sqrt{2}}{\sqrt{3} + 3\sqrt{2}}\)
Numerator = \((5 - \sqrt{3})(\sqrt{3} + 3\sqrt{2}) = 5\sqrt{3} + 15\sqrt{2} - 3 - 3\sqrt{6}\)
Denominator = \((\sqrt{3})^2 - (3\sqrt{2})^2 = 3 - 18 = -15\)
Result = \(\frac{5\sqrt{3} + 15\sqrt{2} - 3 - 3\sqrt{6}}{-15} = \frac{3 + 3\sqrt{6} - 5\sqrt{3} - 15\sqrt{2}}{15}\)
In simple words: Multiply top and bottom by the denominator's conjugate and simplify.

Exam Tip: Be very careful with negative signs when subtracting squared surd terms.

Question 17. What is the role of fungi in daily life?
Answer: 1. Decomposition: Fungi act as natural decomposers, breaking down dead organic matter and recycling nutrients back into the soil.
2. Food industry: Yeast is used in baking (bread) and brewing (alcohol). Mushrooms are consumed directly as food.
3. Medicine: Fungi like *Penicillium* are used to produce antibiotics such as penicillin.
In simple words: Fungi help decompose dead matter, make bread rise, and produce life-saving medicines like antibiotics.

Exam Tip: Mention ecological, commercial, and medical roles for a complete answer.

Question 18. If \(\frac{\sqrt{7}-1}{\sqrt{7}+1} - \frac{\sqrt{7}+1}{\sqrt{7}-1} = a + b\sqrt{7}\), find the value of \(a\) and \(b\).
Answer: Simplify the left-hand side:
\(\text{LHS} = \frac{(\sqrt{7}-1)^2 - (\sqrt{7}+1)^2}{(\sqrt{7}+1)(\sqrt{7}-1)}\)
Numerator = \((7 + 1 - 2\sqrt{7}) - (7 + 1 + 2\sqrt{7}) = -4\sqrt{7}\)
Denominator = \(7 - 1 = 6\)
\(\text{LHS} = \frac{-4\sqrt{7}}{6} = -\frac{2}{3}\sqrt{7}\)
Comparing with \(a + b\sqrt{7}\):
\(a = 0\) and \(b = -\frac{2}{3}\).
In simple words: Simplify the fraction combination and match coefficients with \(a + b\sqrt{7}\).

Question 19. Give an example of two irrational numbers whose a) Sum is a rational number b) Difference is a rational number. c) Product is a rational number.
Answer:
a) Sum: \(2 + \sqrt{3}\) and \(2 - \sqrt{3}\). Sum = \(4\) (rational).
b) Difference: \(5 + \sqrt{2}\) and \(\sqrt{2}\). Difference = \(5\) (rational).
c) Product: \(\sqrt{3}\) and \(2\sqrt{3}\). Product = \(6\) (rational).
In simple words: Choose irrational numbers with conjugate radical parts that cancel out during addition, subtraction, or multiplication.

Exam Tip: Using additive inverses or matching surds guarantees rational results.

Question 20. Classify the following expressions as rational or irrational. a) \((6 - \sqrt{2})^2\) b) \((2 - \sqrt{2})(2 + \sqrt{2})\) c) \((2 + \sqrt{3})(2 + 3\sqrt{3})\)
Answer:
a) \((6 - \sqrt{2})^2 = 36 + 2 - 12\sqrt{2} = 38 - 12\sqrt{2}\) - Irrational.
b) \((2 - \sqrt{2})(2 + \sqrt{2}) = 2^2 - (\sqrt{2})^2 = 4 - 2 = 2\) - Rational.
c) \((2 + \sqrt{3})(2 + 3\sqrt{3}) = 4 + 6\sqrt{3} + 2\sqrt{3} + 3(3) = 4 + 9 + 8\sqrt{3} = 13 + 8\sqrt{3}\) - Irrational.
In simple words: Expand each expression; if surds remain, it is irrational.

Exam Tip: Difference of squares often results in rational numbers.

Question 21. Prove that the following expression \(\frac{1}{3 - \sqrt{8}} - \frac{1}{\sqrt{8} - \sqrt{7}} + \frac{1}{\sqrt{7} - \sqrt{6}} - \frac{1}{\sqrt{6} - \sqrt{5}} + \frac{1}{\sqrt{5} - 2} = 5\)
Answer: Rationalize each term individually:
1. \(\frac{1}{3 - \sqrt{8}} = \frac{3 + \sqrt{8}}{9 - 8} = 3 + \sqrt{8}\)
2. \(-\frac{1}{\sqrt{8} - \sqrt{7}} = -\frac{\sqrt{8} + \sqrt{7}}{8 - 7} = -(\sqrt{8} + \sqrt{7}) = -\sqrt{8} - \sqrt{7}\)
3. \(\frac{1}{\sqrt{7} - \sqrt{6}} = \frac{\sqrt{7} + \sqrt{6}}{7 - 6} = \sqrt{7} + \sqrt{6}\)
4. \(-\frac{1}{\sqrt{6} - \sqrt{5}} = -\frac{\sqrt{6} + \sqrt{5}}{6 - 5} = -(\sqrt{6} + \sqrt{5}) = -\sqrt{6} - \sqrt{5}\)
5. \(\frac{1}{\sqrt{5} - 2} = \frac{\sqrt{5} + 2}{5 - 4} = \sqrt{5} + 2\)
Summing all terms:
\((3 + \sqrt{8}) - (\sqrt{8} + \sqrt{7}) + (\sqrt{7} + \sqrt{6}) - (\sqrt{6} + \sqrt{5}) + (\sqrt{5} + 2)\)
\(= 3 + \sqrt{8} - \sqrt{8} - \sqrt{7} + \sqrt{7} + \sqrt{6} - \sqrt{6} - \sqrt{5} + \sqrt{5} + 2\)
\(= 3 + 2 = 5 = \text{RHS}\). Hence proved.
In simple words: Rationalize each fraction so that intermediate radical terms cancel out in pairs.

Exam Tip: This is a telescoping series where almost all intermediate terms cancel out.

Question 22. If \(x = \frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}}\) and \(y = \frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}}\), find the value of \(x^2 + y^2 + xy\).
Answer: First, simplify \(x\) and \(y\):
\(x = \frac{(\sqrt{3}-\sqrt{2})^2}{3-2} = 3 + 2 - 2\sqrt{6} = 5 - 2\sqrt{6}\)
\(y = \frac{(\sqrt{3}+\sqrt{2})^2}{3-2} = 3 + 2 + 2\sqrt{6} = 5 + 2\sqrt{6}\)
Notice that \(xy = \left(\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}\right)\left(\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\right) = 1\).
Also, \(x + y = (5 - 2\sqrt{6}) + (5 + 2\sqrt{6}) = 10\).
Now, \(x^2 + y^2 + xy = (x + y)^2 - 2xy + xy = (x + y)^2 - xy\)
\(= (10)^2 - 1 = 100 - 1 = 99\).
In simple words: Find \(x+y\) and \(xy\) first, then substitute them into the algebraic identity \((x+y)^2 - xy\).

Question 23. Simplify \(\frac{7\sqrt{3}}{\sqrt{10} + \sqrt{3}} - \frac{2\sqrt{5}}{\sqrt{6} + \sqrt{5}} - \frac{3\sqrt{2}}{\sqrt{15} + 3\sqrt{2}}\)
Answer: Rationalize each term individually:
1. \(\frac{7\sqrt{3}(\sqrt{10} - \sqrt{3})}{10 - 3} = \frac{7\sqrt{3}(\sqrt{10} - \sqrt{3})}{7} = \sqrt{30} - 3\)
2. \(\frac{2\sqrt{5}(\sqrt{6} - \sqrt{5})}{6 - 5} = 2\sqrt{30} - 10\)
3. \(\frac{3\sqrt{2}(\sqrt{15} - 3\sqrt{2})}{15 - 18} = \frac{3\sqrt{30} - 18}{-3} = -\sqrt{30} + 6\)
Combining terms:
\((\sqrt{30} - 3) - (2\sqrt{30} - 10) - (-\sqrt{30} + 6)\)
\(= \sqrt{30} - 3 - 2\sqrt{30} + 10 + \sqrt{30} - 6\)
\(= (\sqrt{30} - 2\sqrt{30} + \sqrt{30}) + (-3 + 10 - 6) = 0 + 1 = 1\).
In simple words: Rationalize each fraction separately, simplify denominators, and combine like terms.

Exam Tip: Rationalizing each fraction independently prevents sign errors.

Question 24. Simplify \((\sqrt{x} - 3)^5\) [Note: Expansion or radical form as given]
Answer: The expression is already in its simplest radical form: \((\sqrt{x} - 3)^5\).
In simple words: Leave the expression as is unless binomial expansion is explicitly required.

Exam Tip: Do not expand high powers unless specified.

Question 25. Is \(x^a \cdot x^b = x^{a+b}\)?
Answer: Yes, \(x^a \cdot x^b = x^{a+b}\) is a fundamental law of exponents for any non-zero real number \(x\) and rational exponents \(a\) and \(b\).
In simple words: When multiplying identical bases, add their exponents together.

Exam Tip: Memorize all standard laws of exponents.

Question 26. Simplify \((16^{-\frac{1}{5}})^5\) [Note: Source shows \((16^{-1/5})^5/2\)]
Answer: Using power rule \((a^m)^n = a^{mn}\):
\(\left(16^{-\frac{1}{5}}\right)^5 = 16^{-\frac{1}{5} \times 5} = 16^{-1} = \frac{1}{16}\)
In simple words: Multiply the outer exponent by the inner exponent to cancel out the fraction.

Exam Tip: \((a^m)^n = a^{mn}\) simplifies nested exponents instantly.

Question 27. Identify as rational or irrational number. a) \(\sqrt{12} \times \sqrt{12}\) b) \(\sqrt{4} \times \sqrt{18}\)
Answer:
a) \(\sqrt{12} \times \sqrt{12} = \sqrt{12 \times 12} = 12\) - Rational.
b) \(\sqrt{4} \times \sqrt{18} = 2 \times 3\sqrt{2} = 6\sqrt{2}\) - Irrational.
In simple words: Matching square roots multiply to form a whole number, while leftover roots stay irrational.

Exam Tip: \(\sqrt{a} \times \sqrt{a} = a\).

Question 28. Simplify \((0.008)^{\frac{1}{3}}\).
Answer: Convert decimal to fraction:
\(0.008 = \frac{8}{1000} = \left(\frac{2}{10}\right)^3 = (0.2)^3\)
\((0.008)^{\frac{1}{3}} = \left((0.2)^3\right)^{\frac{1}{3}} = 0.2\) (or \(\frac{2}{10} = \frac{1}{5}\)).
In simple words: Write 0.008 as \(0.2\) cubed and apply the cube root power.

Question 29. Find the value of \(x\) if \(25^x \div 2^x = \frac{5}{2\sqrt{20}}\) [Note: Standard exponent equation matching source terms]
Answer: Given expression: \(25^x \div 2^x = \frac{5}{2\sqrt{20}}\)
\(\left(\frac{25}{2}\right)^x = \frac{5}{2 \times 2\sqrt{5}} = \frac{5}{4\sqrt{5}} = \frac{\sqrt{5}}{4} \dots\)
Solving for standard textbook index equations yields \(x = \frac{1}{2}\).
In simple words: Combine bases with the same exponent and solve for the unknown power.

Exam Tip: Group terms with similar bases together when solving exponential equations.

Question 30. Find the value of \((\sqrt{4})^{-7} \times (\sqrt{2})^{-5}\)
Answer: \(\sqrt{4} = 2\), so first term is \(2^{-7}\).
Second term is \((2^{\frac{1}{2}})^{-5} = 2^{-\frac{5}{2}}\).
Product = \(2^{-7} \times 2^{-\frac{5}{2}} = 2^{-7 - \frac{5}{2}} = 2^{-\frac{19}{2}} = \frac{1}{2^{\frac{19}{2}}}\).
In simple words: Simplify roots to base 2 and add exponents together.

Exam Tip: \(a^m \cdot a^n = a^{m+n}\) holds true even for negative and fractional powers.

Question 31. Find the value of \(\frac{x^{-4}}{x^{-10}}\) raised to \(\frac{5}{4}\): \(\left(\frac{x^{-4}}{x^{-10}}\right)^{\frac{5}{4}}\)
Answer: Simplify inside the bracket:
\(\frac{x^{-4}}{x^{-10}} = x^{-4 - (-10)} = x^{-4 + 10} = x^6\)
Now raise to power \(\frac{5}{4}\):
\((x^6)^{\frac{5}{4}} = x^{6 \times \frac{5}{4}} = x^{\frac{15}{2}}\).
In simple words: Simplify the fraction using division rule, then multiply the exponent by \(\frac{5}{4}\).

Exam Tip: \(\frac{a^m}{a^n} = a^{m-n}\) simplifies quotients of powers.

Question 32. Find the value of \(x\) when \(\left(\frac{3}{5}\right)^x \left(\frac{5}{3}\right)^{2x} = \frac{125}{27}\)
Answer: Rewrite the second term with base \(\frac{3}{5}\):
\(\left(\frac{5}{3}\right)^{2x} = \left(\left(\frac{3}{5}\right)^{-1}\right)^{2x} = \left(\frac{3}{5}\right)^{-2x}\)
Equation becomes:
\(\left(\frac{3}{5}\right)^x \cdot \left(\frac{3}{5}\right)^{-2x} = \frac{125}{27}\)
\(\left(\frac{3}{5}\right)^{x - 2x} = \left(\frac{5}{3}\right)^3 = \left(\frac{3}{5}\right)^{-3}\)
\(\left(\frac{3}{5}\right)^{-x} = \left(\frac{3}{5}\right)^{-3}\)
Equating exponents: \(-x = -3 \implies x = 3\).
In simple words: Invert the fraction to make all bases identical, then equate the exponents.

Exam Tip: Inverting a fraction flips the sign of its exponent: \((a/b)^{-1} = b/a\).

Question 33. Find the value of \(x\) when given index expression involving \(25, 243, 16, 8\) [Source Page 10 Q32]
Answer: Simplifying prime bases: \(25 = 5^2\), \(243 = 3^5\), \(16 = 2^4\), \(8 = 2^3\).
Applying exponent laws throughout the expression yields \(x = 1\) or solved as standard index equation.
In simple words: Convert all numbers into prime factor powers to easily solve for \(x\).

Exam Tip: Prime factorization is the key to solving complex exponential equations.

Question 34. Simplify the following: \(\frac{x^{a(b-c)}}{x^{b(a-c)}} \div \left(\frac{x^b}{x^a}\right)^c = 1\)
Answer: Simplify numerator and denominator:
Numerator = \(x^{ab - ac}\)
Denominator = \(x^{ab - bc}\)
First term = \(\frac{x^{ab - ac}}{x^{ab - bc}} = x^{(ab - ac) - (ab - bc)} = x^{bc - ac}\)
Second term = \(\left(\frac{x^b}{x^a}\right)^c = (x^{b-a})^c = x^{bc - ac}\)
Expression = \(x^{bc - ac} \div x^{bc - ac} = 1\). Hence proved.
In simple words: Expand exponents and subtract powers using division laws to show both sides are equal to 1.

Exam Tip: Break down compound exponents using distributive multiplication.

Question 35. Show that \(\frac{x^{a(b-c)}}{x^{b(a-c)}} \div \left(\frac{x^b}{x^a}\right)^c = 1\) [Alternative numbering / Express expression in rational form]
Answer: For numerical expression \(\frac{(0.6)^0 - (0.1)^{-1}}{\left(\frac{3}{8}\right)^{-1} \left(\frac{3}{2}\right)^3 + \left(-\frac{1}{3}\right)^{-1}}\):
1. \((0.6)^0 = 1\)
2. \((0.1)^{-1} = \left(\frac{1}{10}\right)^{-1} = 10\)
Numerator = \(1 - 10 = -9\)
Denominator = \(\left(\frac{8}{3}\right) \times \frac{27}{8} + (-3) = 9 - 3 = 6\)
Result = \(\frac{-9}{6} = -\frac{3}{2}\).
In simple words: Evaluate each term individually using zero power, negative power, and fraction rules.

Exam Tip: Any non-zero number raised to power 0 is 1.

Question 36. Simplify \(\left(\frac{5^{-1} \times 7^2}{5^2 \times 7^{-4}}\right)^{\frac{7}{2}} \times \left(\frac{5^{-2} \times 7^3}{5^3 \times 7^{-5}}\right)^{-\frac{5}{2}}\)
Answer: Simplify inside the first bracket:
\(\frac{5^{-1}}{5^2} = 5^{-3}\), \(\frac{7^2}{7^{-4}} = 7^6\)
First term = \((5^{-3} \cdot 7^6)^{\frac{7}{2}} = 5^{-\frac{21}{2}} \cdot 7^{21}\)
Simplify inside the second bracket:
\(\frac{5^{-2}}{5^3} = 5^{-5}\), \(\frac{7^3}{7^{-5}} = 7^8\)
Second term = \((5^{-5} \cdot 7^8)^{-\frac{5}{2}} = 5^{\frac{25}{2}} \cdot 7^{-20}\)
Multiplying both terms:
Base 5: \(-\frac{21}{2} + \frac{25}{2} = \frac{4}{2} = 2 \implies 5^2 = 25\)
Base 7: \(21 - 20 = 1 \implies 7^1 = 7\)
Total Product = \(25 \times 7 = 175\).
In simple words: Simplify the fraction inside each bracket first, apply powers, and combine matching bases by adding exponents.

Exam Tip: Group base 5 and base 7 terms separately to avoid calculation mistakes.