Samacheer Kalvi Class 9 Maths Solutions Chapter 2 Real Numbers Exercise 2.9

Question 1. If n is a natural number then \( \sqrt{n} \) is..........
(a) always a natural number
(b) always an irrational number
(c) always a rational number
(d) may be rational or irrational
Answer: (d) may be rational or irrational
In simple words: The square root of a natural number can be rational (like \( \sqrt{4}=2 \)) or irrational (like \( \sqrt{2} \)). It depends on whether the number inside the square root is a perfect square or not.

🎯 Exam Tip: Remember that rational numbers can be written as a fraction \( \frac{p}{q} \), while irrational numbers cannot.

Question 2. Which of the following is not true?
(a) Every rational number is a real number
(b) Every integer is a rational number
(c) Every real number is an irrational number
(d) Every natural number is a whole number
Answer: (c) Every real number is an irrational number
In simple words: This statement is false because real numbers include both rational and irrational numbers. Not every real number is irrational; many are rational. For example, 2 is a real number, but it is rational, not irrational.

🎯 Exam Tip: Understand the hierarchy of number systems: Natural numbers \( \subset \) Whole numbers \( \subset \) Integers \( \subset \) Rational numbers \( \subset \) Real numbers. Irrational numbers are also part of real numbers but are separate from rational numbers.

Question 3. Which one of the following, regarding sum of two irrational numbers, is true?
(a) always an irrational number
(b) may be a rational or irrational number
(c) always a rational number
(d) always an integer
Answer: (b) may be a rational or irrational number
In simple words: When you add two irrational numbers, the answer can sometimes be irrational (like \( \sqrt{2} + \sqrt{3} \)) or sometimes rational (like \( \sqrt{2} + (-\sqrt{2}) = 0 \)). It is not always one or the other.

🎯 Exam Tip: Give examples to test properties of number types. For this question, consider \( \sqrt{2} + \sqrt{3} \) (irrational) and \( (2 + \sqrt{3}) + (2 - \sqrt{3}) = 4 \) (rational).

Question 4. Which one of the following has a terminating decimal expansion?
(a) \( \frac{5}{64} \)
(b) \( \frac{8}{9} \)
(c) \( \frac{14}{15} \)
(d) \( \frac{1}{12} \)
Answer: (a) \( \frac{5}{64} \)
\( \frac{5}{64} = \frac{5}{2^6} \). A fraction has a terminating decimal expansion if its denominator, after being simplified to its lowest terms, has only 2 or 5 as prime factors. Here, 64 is \( 2^6 \), so it has only 2 as a prime factor.
In simple words: A fraction will stop after some decimal places if you can write its bottom number (denominator) using only the prime factors 2 and 5. For \( \frac{5}{64} \), the denominator 64 is just 2 multiplied by itself six times, so it will terminate.

🎯 Exam Tip: To check for terminating decimals, reduce the fraction to its simplest form and then find the prime factors of the denominator. If the prime factors are only 2 and/or 5, it terminates.

Question 5. Which one of the following is an irrational number?
(a) \( \sqrt{25} \)
(b) \( \sqrt{\frac{9}{4}} \)
(c) \( \frac{7}{11} \)
(d) \( \pi \)
Answer: (d) \( \pi \)
\( \sqrt{25} = 5 \) (rational). \( \sqrt{\frac{9}{4}} = \frac{3}{2} \) (rational). \( \frac{7}{11} \) is a rational number. \( \pi \) is an irrational number because its decimal expansion is non-terminating and non-repeating. Although we often use \( \frac{22}{7} \) as an approximation for \( \pi \), this is not its exact value.
In simple words: An irrational number cannot be written as a simple fraction. From the given options, \( \pi \) is the only number that cannot be expressed as a fraction of two integers, making it irrational.

🎯 Exam Tip: Remember that \( \pi \) is always irrational, even though approximations like \( \frac{22}{7} \) or 3.14 are rational numbers.

Question 6. An irrational number between 2 and 2.5 is.........
(a) \( \sqrt{11} \)
(b) \( \sqrt{5} \)
(c) \( \sqrt{2.5} \)
(d) \( \sqrt{8} \)
Answer: (b) \( \sqrt{5} \)
Let's check the approximate values: \( \sqrt{11} \approx 3.31 \), \( \sqrt{5} \approx 2.236 \), \( \sqrt{2.5} \approx 1.58 \), \( \sqrt{8} \approx 2.82 \). We can see that \( \sqrt{5} \approx 2.236 \) lies between 2 and 2.5. Also, \( \sqrt{5} \) is irrational because 5 is not a perfect square.
In simple words: To find an irrational number between 2 and 2.5, we can look for numbers that are not perfect squares and whose square roots fall in that range. \( \sqrt{5} \) is about 2.236, which fits perfectly.

🎯 Exam Tip: To find a number between two values, it's often helpful to estimate the values of the given options. You can also square the bounds (2 and 2.5) to get 4 and 6.25, and then look for non-perfect square options between 4 and 6.25.

Question 7. The smallest rational number by which \( \frac{1}{3} \) should be multiplied so that its decimal expansion terminates with one place of decimal is .........
(a) \( \frac{1}{10} \)
(b) \( \frac{3}{10} \)
(c) 3
(d) 30
Answer: (b) \( \frac{3}{10} \)
We want the product to terminate with one decimal place. This means the denominator of the product, when simplified, should be \( 10^1 = 10 \), or a factor of 10 like \( 2 \times 5 \).
If we multiply \( \frac{1}{3} \) by \( \frac{3}{10} \), we get \( \frac{1}{3} \times \frac{3}{10} = \frac{1}{10} \).
The decimal expansion of \( \frac{1}{10} \) is 0.1, which terminates with one decimal place. This is the smallest such rational number.
In simple words: To make the fraction \( \frac{1}{3} \) turn into a decimal that stops after one number (like 0.1), you need to multiply it by a fraction that cancels out the '3' at the bottom and leaves a '10' or a factor of '10'. Multiplying by \( \frac{3}{10} \) works because \( \frac{1}{3} \times \frac{3}{10} = \frac{1}{10} \), which is 0.1.

🎯 Exam Tip: A terminating decimal with one place implies a denominator of 10. For the product to have a denominator of 10, the '3' in \( \frac{1}{3} \) must be cancelled out.

Question 8. If \( \frac{1}{7} = 0.\overline {142857} \) then the value of \( \frac{5}{7} \) is...........
(a) \( 0.\overline {142857} \)
(b) \( 0.\overline {714285} \)
(c) \( 0.\overline {571428} \)
(d) 0.714285
Answer: (b) \( 0.\overline {714285} \)
We are given \( \frac{1}{7} = 0.142857142857... \)
To find \( \frac{5}{7} \), we simply multiply the decimal representation of \( \frac{1}{7} \) by 5:
\( \frac{5}{7} = 5 \times \frac{1}{7} = 5 \times 0.\overline{142857} \)
\( \implies 0.714285714285... = 0.\overline{714285} \)
In simple words: If you know what \( \frac{1}{7} \) looks like as a repeating decimal, you can find \( \frac{5}{7} \) by simply multiplying that decimal pattern by 5. The new number will have a repeating block that starts with 714285.

🎯 Exam Tip: When multiplying a repeating decimal by an integer, multiply the repeating block by the integer. The new product will have its own repeating block starting from where the original product's digits begin to repeat.

Question 9. Find the odd one out of the following.
(a) \( \sqrt{32} \times \sqrt{2} \)
(b) \( \frac{\sqrt{27}}{\sqrt{3}} \)
(c) \( \sqrt{72} \times \sqrt{8} \)
(d) \( \frac{\sqrt{54}}{\sqrt{18}} \)
Answer: (b) \( \frac{\sqrt{27}}{\sqrt{3}} \)
Let's evaluate each option:
(a) \( \sqrt{32} \times \sqrt{2} = \sqrt{32 \times 2} = \sqrt{64} = 8 \)
(b) \( \frac{\sqrt{27}}{\sqrt{3}} = \sqrt{\frac{27}{3}} = \sqrt{9} = 3 \)
(c) \( \sqrt{72} \times \sqrt{8} = \sqrt{72 \times 8} = \sqrt{576} = 24 \)
(d) \( \frac{\sqrt{54}}{\sqrt{18}} = \sqrt{\frac{54}{18}} = \sqrt{3} \)
The values are 8, 3, 24, and \( \sqrt{3} \). All options (a), (b), and (c) simplify to rational numbers (integers), while option (d) simplifies to an irrational number (\( \sqrt{3} \)). Wait, the provided hint states: " \( \frac{\sqrt{27}}{\sqrt{3}} = \frac{\sqrt{27}}{\sqrt{3}} = \sqrt{9} = 3 \). It is an odd number". This suggests that (b) is the 'odd one out' because it results in '3', which is an odd number. The other options (a) 8, (c) 24, (d) \( \sqrt{3} \) (approx 1.732). If the question is about an 'odd number' in the sense of 'not even', then 3 is the only one. If it's about rational vs. irrational, then \( \sqrt{3} \) would be the odd one out. Given the hint explicitly calls 3 an "odd number", it seems the question implies the arithmetic property of 'odd'. Let's re-evaluate based on the provided values after simplification.
(a) 8 (even)
(b) 3 (odd)
(c) 24 (even)
(d) \( \sqrt{3} \) (irrational, not an integer, so not odd or even in the traditional sense).
Based on the hint, 3 is the only odd *integer* among the simplified values of the *integer-result* options. So (b) is the odd one out in that specific interpretation.
In simple words: We need to simplify each expression. After simplifying, most of them give even numbers, but option (b) simplifies to 3, which is an odd number. This makes it different from the others.

🎯 Exam Tip: Always simplify all options completely before comparing them. Sometimes, the "odd one out" could refer to an integer property (like odd/even), while other times it might refer to a number type (like rational/irrational).

Question 10. \( 0.\overline {34} + 0.3\overline {4} = \) ..........
(a) \( 0.6\overline {87} \)
(b) \( 0.\overline {68} \)
(c) \( 0.6\overline {8} \)
(d) \( 0.68\overline {7} \)
Answer: (a) \( 0.6\overline {87} \)
Let's expand the decimals:
\( 0.\overline{34} = 0.34343434... \)
\( 0.3\overline{4} = 0.34444444... \)
Now, let's add them:
\( 0.34343434 \)
\( + 0.34444444 \)
\( = 0.68787878... \)
This can be written as \( 0.6\overline{87} \).
In simple words: When you add two repeating decimals, you write them out for several decimal places and then add them column by column. The repeating pattern in the sum will then become clear. In this case, the sum shows a pattern of '87' repeating after the first '6'.

🎯 Exam Tip: For adding or subtracting repeating decimals, write out enough digits to see the full repeating pattern of the sum. Align them carefully according to their decimal points.

Question 11. Which of the following statement is false?
(a) The square root of 25 is 5 or -5
(b) \( -\sqrt{25} = -5 \)
(c) \( \sqrt{25} = 5 \)
(d) \( \sqrt{25} = \pm 5 \)
Answer: (d) \( \sqrt{25} = \pm 5 \)
The symbol \( \sqrt{} \) (radical sign) by definition denotes the principal (non-negative) square root. So, \( \sqrt{25} \) is always 5, not \( \pm 5 \). If we want to represent both positive and negative roots, we use \( \pm\sqrt{} \) or state "the square roots of 25 are \( \pm 5 \)". Statement (a) correctly describes "the square root", implying both roots. Statement (b) is true because \( -\sqrt{25} = -(5) = -5 \). Statement (c) is also true, as \( \sqrt{25} = 5 \). The only false statement is (d) which incorrectly assigns both positive and negative values directly to the principal square root symbol.
In simple words: The symbol \( \sqrt{} \) always means "the positive square root". So, \( \sqrt{25} \) is only 5, never -5. If you want to show both positive and negative roots, you must write \( \pm \sqrt{25} \).

🎯 Exam Tip: Distinguish between "the square root of x" (which refers to the principal, non-negative root) and "the solutions to \( y^2=x \)" (which are \( y = \pm\sqrt{x} \)).

Question 12. Which one of the following is not a rational number?
(a) \( \sqrt{0.0004} \)
(b) \( \frac{7}{3} \)
(c) \( \sqrt{0.01} \)
(d) \( \sqrt{13} \)
Answer: (d) \( \sqrt{13} \)
Let's evaluate each option:
(a) \( \sqrt{0.0004} = \sqrt{\frac{4}{10000}} = \frac{2}{100} = \frac{1}{50} \) (rational)
(b) \( \frac{7}{3} \) (rational, already in \( \frac{p}{q} \) form)
(c) \( \sqrt{0.01} = \sqrt{\frac{1}{100}} = \frac{1}{10} \) (rational)
(d) \( \sqrt{13} \) (irrational, because 13 is not a perfect square)
In simple words: A rational number can be written as a simple fraction, while an irrational number cannot. From the choices, \( \sqrt{13} \) is the only one that cannot be turned into a fraction of two whole numbers, as 13 is not a perfect square.

🎯 Exam Tip: Perfect squares are numbers like 1, 4, 9, 16, 25... The square root of any non-perfect square is an irrational number.

Question 13. \( \sqrt{27} + \sqrt{12} = \) ..........
(a) \( \sqrt{39} \)
(b) \( 5\sqrt{6} \)
(c) \( 5\sqrt{3} \)
(d) \( 3\sqrt{5} \)
Answer: (c) \( 5\sqrt{3} \)
To add these square roots, we first simplify each term by finding perfect square factors:
\( \sqrt{27} = \sqrt{9 \times 3} = \sqrt{9} \times \sqrt{3} = 3\sqrt{3} \)
\( \sqrt{12} = \sqrt{4 \times 3} = \sqrt{4} \times \sqrt{3} = 2\sqrt{3} \)
Now, we add the simplified terms:
\( 3\sqrt{3} + 2\sqrt{3} = (3+2)\sqrt{3} = 5\sqrt{3} \)
In simple words: To add numbers with square roots, first try to simplify each square root. Here, \( \sqrt{27} \) becomes \( 3\sqrt{3} \) and \( \sqrt{12} \) becomes \( 2\sqrt{3} \). Once the 'root part' is the same (\( \sqrt{3} \) in this case), you can simply add the numbers in front of them, like adding \( 3x + 2x \) to get \( 5x \).

🎯 Exam Tip: You can only add or subtract square roots if the number inside the square root symbol (the radicand) is the same after simplification.

Question 14. If \( \sqrt{80} = k\sqrt{5} \), then k = .........
(a) 2
(b) 4
(c) 8
(d) 16
Answer: (b) 4
We are given the equation \( \sqrt{80} = k\sqrt{5} \).
First, simplify \( \sqrt{80} \) by finding its largest perfect square factor:
\( \sqrt{80} = \sqrt{16 \times 5} = \sqrt{16} \times \sqrt{5} = 4\sqrt{5} \)
Now, compare this with the given expression: \( 4\sqrt{5} = k\sqrt{5} \).
By comparing both sides, we can see that \( k = 4 \).
In simple words: To find 'k', first break down \( \sqrt{80} \) into simpler parts. Since 80 is \( 16 \times 5 \), \( \sqrt{80} \) becomes \( \sqrt{16} \times \sqrt{5} \), which is \( 4\sqrt{5} \). So, \( 4\sqrt{5} \) is equal to \( k\sqrt{5} \), meaning k must be 4.

🎯 Exam Tip: When simplifying square roots, always look for the largest perfect square that is a factor of the number inside the root. This makes the simplification easier and quicker.

Question 15. \( 4\sqrt{7} \times 2\sqrt{3} = \) ..........
(a) \( 6\sqrt{10} \)
(b) \( 8\sqrt{21} \)
(c) \( 8\sqrt{10} \)
(d) \( 6\sqrt{21} \)
Answer: (b) \( 8\sqrt{21} \)
To multiply expressions with square roots, we multiply the numbers outside the root together and the numbers inside the root together:
\( 4\sqrt{7} \times 2\sqrt{3} = (4 \times 2) \times (\sqrt{7} \times \sqrt{3}) \)
\( \implies 8 \times \sqrt{7 \times 3} \)
\( \implies 8\sqrt{21} \)
In simple words: When multiplying terms with square roots, you multiply the normal numbers outside the root sign, and then you multiply the numbers inside the root sign. So, 4 times 2 gives 8, and \( \sqrt{7} \) times \( \sqrt{3} \) gives \( \sqrt{21} \).

🎯 Exam Tip: Remember the rule for multiplying radicals: \( a\sqrt{x} \times b\sqrt{y} = (a \times b)\sqrt{x \times y} \).

Question 16. When written with a rational denominator, the expression \( \frac{2\sqrt{3}}{3\sqrt{2}} \) can be simplified as...........
(a) \( \frac{\sqrt{2}}{3} \)
(b) \( \frac{\sqrt{3}}{2} \)
(c) \( \frac{\sqrt{6}}{3} \)
(d) \( \frac{\sqrt{2}}{3} \)
Answer: (c) \( \frac{\sqrt{6}}{3} \)
To rationalize the denominator, we multiply the numerator and the denominator by \( \sqrt{2} \):
\( \frac{2\sqrt{3}}{3\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} \)
\( \implies \frac{2\sqrt{3 \times 2}}{3 \times \sqrt{2 \times 2}} \)
\( \implies \frac{2\sqrt{6}}{3 \times 2} \)
\( \implies \frac{2\sqrt{6}}{6} \)
Now, simplify the fraction by dividing the numerator and denominator by 2:
\( \implies \frac{\sqrt{6}}{3} \)
In simple words: To remove the square root from the bottom of a fraction, you multiply both the top and bottom of the fraction by that square root. This gets rid of the root at the bottom. Then, you simplify the fraction if possible.

🎯 Exam Tip: Always multiply by the square root itself, not the number in front of it, to rationalize the denominator. Simplify the resulting fraction carefully.

Question 17. When \( (2\sqrt{5} - \sqrt{2})^2 \) is simplified, we get.........
(a) \( 4\sqrt{5} + 2\sqrt{2} \)
(b) \( 22 - 4\sqrt{10} \)
(c) \( 22 - 2\sqrt{10} \)
(d) \( 2\sqrt{10} - 2 \)
Answer: (b) \( 22 - 4\sqrt{10} \)
We use the algebraic identity \( (a-b)^2 = a^2 - 2ab + b^2 \).
Here, \( a = 2\sqrt{5} \) and \( b = \sqrt{2} \).
\( (2\sqrt{5} - \sqrt{2})^2 = (2\sqrt{5})^2 - 2(2\sqrt{5})(\sqrt{2}) + (\sqrt{2})^2 \)
\( \implies (2^2 \times (\sqrt{5})^2) - (2 \times 2 \times \sqrt{5 \times 2}) + 2 \)
\( \implies (4 \times 5) - (4\sqrt{10}) + 2 \)
\( \implies 20 - 4\sqrt{10} + 2 \)
\( \implies 22 - 4\sqrt{10} \)
In simple words: To simplify this expression, we use the formula for squaring a subtraction: \( (a-b)^2 = a^2 - 2ab + b^2 \). We square the first term, subtract two times the product of both terms, and then add the square of the second term. Finally, we combine the normal numbers.

🎯 Exam Tip: Be careful with signs and with squaring terms involving both a number and a square root. Remember that \( (a\sqrt{b})^2 = a^2 \times b \).

Question 18. \( (0.000729)^{-\frac{3}{4}} \times (0.09)^{-\frac{3}{4}} = \) ..........
(a) \( \frac{10^3}{3^3} \)
(b) \( \frac{10^5}{3^5} \)
(c) \( \frac{10^2}{3^2} \)
(d) \( \frac{10^6}{3^6} \)
Answer: (d) \( \frac{10^6}{3^6} \)
We can combine the terms since they have the same exponent:
\( (0.000729)^{-\frac{3}{4}} \times (0.09)^{-\frac{3}{4}} = (0.000729 \times 0.09)^{-\frac{3}{4}} \)
First, convert the decimals to scientific notation:
\( 0.000729 = 729 \times 10^{-6} \)
\( 0.09 = 9 \times 10^{-2} \)
Now, multiply them:
\( 729 \times 10^{-6} \times 9 \times 10^{-2} = (729 \times 9) \times (10^{-6} \times 10^{-2}) \)
\( \implies 6561 \times 10^{-8} \)
We know that \( 6561 = 3^8 \). So, \( 6561 \times 10^{-8} = 3^8 \times 10^{-8} = (3 \times 10^{-1})^8 = (0.3)^8 \).
So, the expression becomes \( ((0.3)^8)^{-\frac{3}{4}} \).
Using the power rule \( (a^m)^n = a^{m \times n} \):
\( (0.3)^{8 \times (-\frac{3}{4})} = (0.3)^{-6} \)
\( \implies \left(\frac{3}{10}\right)^{-6} = \left(\frac{10}{3}\right)^6 = \frac{10^6}{3^6} \)
In simple words: First, multiply the two decimal numbers inside the parenthesis. Then, write the result in a simpler form using powers of 3 and 10. Finally, apply the negative fractional power, which means flipping the base and raising it to the positive power.

🎯 Exam Tip: When dealing with negative fractional exponents, first simplify the base, then apply the power. Remember that \( a^{-n} = \frac{1}{a^n} \) and \( (a^m)^n = a^{mn} \).

Question 19. If \( \sqrt{9^x} = \sqrt[3]{9^2} \), then x = ..........
(a) \( \frac{2}{3} \)
(b) \( \frac{4}{3} \)
(c) \( \frac{1}{3} \)
(d) \( \frac{5}{3} \)
Answer: (b) \( \frac{4}{3} \)
We can rewrite the square root and cube root as fractional exponents:
\( \sqrt{9^x} = 9^{\frac{x}{2}} \)
\( \sqrt[3]{9^2} = 9^{\frac{2}{3}} \)
So, the equation becomes:
\( 9^{\frac{x}{2}} = 9^{\frac{2}{3}} \)
Since the bases are the same (9), we can equate the exponents:
\( \frac{x}{2} = \frac{2}{3} \)
Now, solve for x:
\( x = \frac{2}{3} \times 2 \)
\( x = \frac{4}{3} \)
In simple words: Change the square root and cube root signs into powers with fractions. Once both sides of the equation have the same base number (here, 9), you can just set their fractional powers equal to each other. Then, solve that simple equation to find x.

🎯 Exam Tip: Always convert roots to fractional exponents \( \sqrt[n]{a^m} = a^{\frac{m}{n}} \) to solve equations involving powers and roots. This makes the algebraic manipulation much simpler.

Question 20. The length and breadth of a rectangular plot are \( 5 \times 10^5 \) and \( 4 \times 10^4 \) metres respectively. Its area is ..........
(a) \( 9 \times 10^1 \) m²
(b) \( 9 \times 10^9 \) m²
(c) \( 2 \times 10^{10} \) m²
(d) \( 20 \times 10^{20} \) m²
Answer: (c) \( 2 \times 10^{10} \) m²
The area of a rectangle is given by the formula: Area = Length \( \times \) Breadth.
Given: Length \( L = 5 \times 10^5 \) m, Breadth \( B = 4 \times 10^4 \) m.
Area \( = (5 \times 10^5) \times (4 \times 10^4) \)
We multiply the numerical parts and the powers of 10 separately:
Area \( = (5 \times 4) \times (10^5 \times 10^4) \)
Area \( = 20 \times 10^{(5+4)} \) (using the rule \( a^m \times a^n = a^{m+n} \))
Area \( = 20 \times 10^9 \)
To express this in standard scientific notation, we convert 20 to \( 2 \times 10^1 \):
Area \( = (2 \times 10^1) \times 10^9 \)
Area \( = 2 \times 10^{(1+9)} \)
Area \( = 2 \times 10^{10} \) m²
In simple words: To find the area of the rectangle, multiply its length and breadth. First, multiply the regular numbers (5 times 4 gives 20). Then, add the powers of 10 together (5 plus 4 gives 9). So you get \( 20 \times 10^9 \). To make it a standard scientific number, change 20 to \( 2 \times 10^1 \), which then combines with \( 10^9 \) to make \( 2 \times 10^{10} \).

🎯 Exam Tip: When multiplying numbers in scientific notation, multiply the coefficients and add the exponents of 10. Always ensure the final answer is in standard scientific notation (coefficient between 1 and 10).