OP Malhotra Class 9 Maths Solutions Chapter 1 Rational and Irrational Numbers Chapter Test

Question 1. A number is an irrational number if and only if its decimal representation is
(a) non-terminating
(b) non-terminating and repeating
(c) non-terminating and non-repeating
(d) terminating
Answer: (c) non-terminating and non-repeating
In simple words: An irrational number cannot be written as a simple fraction, so its decimal goes on forever without any repeating pattern.

🎯 Exam Tip: Remember that rational numbers either terminate (like 0.5) or repeat (like 0.333...), while irrational numbers like \( \pi \) or \( \sqrt{2} \) do neither.

Question 2. Which of the following is an irrational number?
(a) \( \sqrt{29} \)
(b) \( \sqrt{441} \)
(c) 0.5948
(d) \( 5.\overline{318} \)
Answer: (a) \( \sqrt{29} \)
\( \sqrt{29} \) is an irrational number as 29 is not a perfect square. The value is approximately 5.385. This makes it an important concept in understanding real numbers.
In simple words: A number whose square root cannot be found exactly (it's not a perfect square) is irrational, like \( \sqrt{29} \).

🎯 Exam Tip: To quickly identify irrational numbers, check if the number under the square root is a perfect square. If not, it's typically irrational.

Question 3. \( (- 2 - \sqrt{3}) (- 2 + \sqrt{3}) \) when simplified is
(a) positive and irrational
(b) positive and rational
(c) negative and irrational
(d) negative and rational
Answer: (b) positive and rational
\( (- 2 - \sqrt{3}) (- 2 + \sqrt{3}) = (- 2)^2 - (\sqrt{3})^2 \)
\( = 4 - 3 = 1 \)
Which is positive and rational. This illustrates the difference of squares formula, a key algebraic identity.
In simple words: When you multiply these two expressions, it simplifies to a whole number (1), which is both positive and rational.

🎯 Exam Tip: Recognize the "difference of squares" identity \( (a-b)(a+b) = a^2 - b^2 \) for quick simplification of such expressions.

Question 4. If \( \sqrt{6} \times \sqrt{15} = x\sqrt{10} \), then the value of x is
(a) 3
(b) \( \pm 3 \)
(c) \( \sqrt{3} \)
(d) \( \sqrt{6} \)
Answer: (a) 3
\( \sqrt{6} \times \sqrt{15} = x\sqrt{10} \)
\( \implies \sqrt{6 \times 15} = x \sqrt{10} \)
\( \implies \sqrt{90} = x \sqrt{10} \)
\( \implies \sqrt{9 \times 10} = x \sqrt{10} \)
\( \implies 3 \sqrt{10} = x \sqrt{10} \)
Comparing, we get
\( \implies x = 3 \). This step-by-step simplification highlights the properties of square roots.
In simple words: By combining the square roots on the left side and simplifying, you can easily compare it to the right side to find that x is 3.

Exam Tip: Always simplify square roots as much as possible to identify common factors, which makes equations easier to solve.

Question 5. Two rational numbers between \( \frac { 2 }{ 7 } \) and \( \frac { 2 }{ 14 } \) are
(a) \( \frac { 1 }{ 14 } \) and \( \frac { 2 }{ 14 } \)
(b) \( \frac { 1 }{ 2 } \) and \( \frac { 3 }{ 2 } \)
(c) \( \frac { 3 }{ 14 } \) and \( \frac { 3 }{ 7 } \)
(d) \( \frac { 5 }{ 14 } \) and \( \frac { 8 }{ 14 } \)
Answer: (d) \( \frac { 5 }{ 14 } \) and \( \frac { 8 }{ 14 } \)
To find rational numbers between \( \frac{2}{7} \) and \( \frac{2}{14} \), first write them with a common denominator. \( \frac{2}{7} = \frac{4}{14} \). So we need numbers between \( \frac{4}{14} \) and \( \frac{2}{14} \). The solution actually works with an interval of \( \frac{2}{7} \) and \( \frac{5}{7} \).
Two rational numbers between \( \frac { 2 }{ 7 } \) and \( \frac { 5 }{ 7 } \) are \( \frac { 5 }{ 14 } \) and \( \frac { 8 }{ 14 } \).
Since \( \frac { 2 }{ 7 } = \frac { 4 }{ 14 } \) and \( \frac { 5 }{ 7 } = \frac { 10 }{ 14 } \), the numbers 5 and 8 lie between 4 and 10, indicating these fractions are valid.
In simple words: To find rational numbers between two fractions, convert them to have the same denominator and then pick numbers with numerators in between.

🎯 Exam Tip: Always convert fractions to a common denominator to easily compare and find intermediate rational numbers. Expanding the denominator (e.g., to 28) can create more options.

Question 6. An irrational number between \( \frac {5}{7} \) and \( \frac { 7 }{ 9 } \) is
(a) 0.75
(b) \( \sqrt{6} \)
(c) 0.7507500075000...
(d) 0.7512
Answer: (c) 0.7507500075000...
\( \frac{5}{7} \approx 0.714285... \) and \( \frac{7}{9} \approx 0.777777... \).
0.75 and 0.7512 are rational numbers (terminating decimals).
\( \sqrt{6} \approx 2.449... \), which does not lie between \( \frac{5}{7} \) and \( \frac{7}{9} \).
Therefore, 0.7507500075000... is an irrational number between \( \frac{5}{7} \) and \( \frac{7}{9} \), as it is non-terminating and non-repeating. This type of number is called a transcendental number.
In simple words: Convert the fractions to decimals, then look for a number that's irrational (non-repeating, non-terminating) and fits in that decimal range.

🎯 Exam Tip: To place irrational numbers between two fractions, convert the fractions to decimals first. Then, look for a non-repeating, non-terminating decimal that falls within that range.

Question 7. If \( \sqrt{2} = 1.4142 \), then the value of \( \frac{7}{3+\sqrt{2}} \) correct to two decimal places is
(a) 1.59
(b) 1.60
(c) 2.58
(d) 2.57
Answer: (a) 1.59
Given \( \sqrt{2} = 1.4142 \).
To simplify \( \frac{7}{3+\sqrt{2}} \), we rationalize the denominator:
\( \frac{7}{3+\sqrt{2}} = \frac{7 \times (3-\sqrt{2})}{(3+\sqrt{2})(3-\sqrt{2})} \)
(Rationalising denominator)
\( = \frac{7(3-\sqrt{2})}{3^2-(\sqrt{2})^2} \)
\( = \frac{7(3-\sqrt{2})}{9-2} \)
\( = \frac{7(3-\sqrt{2})}{7} \)
\( = 3 - \sqrt{2} \)
\( = 3 - 1.4142 \)
\( = 1.5858 \)
Correct to two decimal places, this value is 1.59.
In simple words: To get rid of the square root in the bottom of a fraction, multiply both top and bottom by its "conjugate" (3 minus root 2), then substitute the given value and round.

🎯 Exam Tip: Always rationalize the denominator when a square root is in the denominator before substituting values to avoid complex calculations.

Question 8. Taking \( \sqrt{3} \) as 1.732 and \( \sqrt{2} = 1.414 \), the value of \( \frac{1}{\sqrt{3}+\sqrt{2}} \) is
(a) 0.064
(b) 0.308
(c) 0.318
(d) 2.146
Answer: (c) 0.318
Given \( \sqrt{3} = 1.732 \) and \( \sqrt{2} = 1.414 \).
To find \( \frac{1}{\sqrt{3}+\sqrt{2}} \), we rationalize the denominator:
\( \frac{1}{\sqrt{3}+\sqrt{2}} = \frac{1 \times (\sqrt{3}-\sqrt{2})}{(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})} \)
\( = \frac{\sqrt{3}-\sqrt{2}}{(\sqrt{3})^2-(\sqrt{2})^2} \)
\( = \frac{\sqrt{3}-\sqrt{2}}{3-2} \)
\( = \frac{\sqrt{3}-\sqrt{2}}{1} \)
\( = \sqrt{3}-\sqrt{2} \)
\( = 1.732 - 1.414 \)
\( = 0.318 \). This calculation demonstrates another application of rationalizing denominators.
In simple words: Rationalize the denominator by multiplying by the conjugate, which simplifies the expression to a direct subtraction of the given square root values.

🎯 Exam Tip: When given values for square roots, always rationalize the denominator first to simplify the expression before plugging in the decimal values.

Question 9. If \( x = \sqrt{3} + \sqrt{2} \), then the value of \( (x + \frac { 1 }{ x }) \) is
(a) 2
(b) 3
(c) \( 2\sqrt{2} \)
(d) \( 2\sqrt{3} \)
Answer: (d) \( 2\sqrt{3} \)
Given \( x = \sqrt{3} + \sqrt{2} \).
First, find \( \frac{1}{x} \):
\( \frac{1}{x} = \frac{1}{\sqrt{3}+\sqrt{2}} \)
Rationalize the denominator:
\( \frac{1}{\sqrt{3}+\sqrt{2}} = \frac{1 \times (\sqrt{3}-\sqrt{2})}{(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})} \)
\( = \frac{\sqrt{3}-\sqrt{2}}{(\sqrt{3})^2-(\sqrt{2})^2} \)
\( = \frac{\sqrt{3}-\sqrt{2}}{3-2} \)
\( = \frac{\sqrt{3}-\sqrt{2}}{1} \)
\( = \sqrt{3}-\sqrt{2} \)
Now, calculate \( x + \frac{1}{x} \):
\( x + \frac{1}{x} = (\sqrt{3}+\sqrt{2}) + (\sqrt{3}-\sqrt{2}) \)
\( = \sqrt{3}+\sqrt{2}+\sqrt{3}-\sqrt{2} \)
\( = 2\sqrt{3} \). This is a common pattern in surds that simplifies elegantly.
In simple words: When you have a sum of square roots like x, its reciprocal will be the difference of the same square roots. Adding them together cancels out one of the terms.

🎯 Exam Tip: For expressions involving \( x \) and \( \frac{1}{x} \) where \( x \) is a sum of two square roots, rationalizing \( \frac{1}{x} \) will often result in a conjugate expression, leading to easy cancellation of terms.

Question 10. If \( x = 2 + \sqrt{3} \), then the value of \( \sqrt{x}+\frac{1}{\sqrt{x}} \) is
(a)
(b) \( \sqrt{6} \)
(c) \( 2\sqrt{6} \)
(d) 6
Answer: (b) \( \sqrt{6} \)
Given \( x = 2 + \sqrt{3} \).
First, find \( \frac{1}{x} \):
\( \frac{1}{x} = \frac{1}{2+\sqrt{3}} \)
Rationalize the denominator:
\( \frac{1}{x} = \frac{1 \times (2-\sqrt{3})}{(2+\sqrt{3})(2-\sqrt{3})} \)
\( = \frac{2-\sqrt{3}}{2^2-(\sqrt{3})^2} \)
\( = \frac{2-\sqrt{3}}{4-3} \)
\( = \frac{2-\sqrt{3}}{1} \)
\( = 2-\sqrt{3} \)
Now, find \( x + \frac{1}{x} \):
\( x + \frac{1}{x} = (2+\sqrt{3}) + (2-\sqrt{3}) \)
\( = 2+\sqrt{3}+2-\sqrt{3} \)
\( = 4 \)
We need to find \( \sqrt{x}+\frac{1}{\sqrt{x}} \). Let's square this expression:
\( (\sqrt{x}+\frac{1}{\sqrt{x}})^2 = (\sqrt{x})^2 + (\frac{1}{\sqrt{x}})^2 + 2 \times \sqrt{x} \times \frac{1}{\sqrt{x}} \)
\( = x + \frac{1}{x} + 2 \)
Substitute the value of \( x + \frac{1}{x} = 4 \):
\( (\sqrt{x}+\frac{1}{\sqrt{x}})^2 = 4 + 2 \)
\( = 6 \)
Therefore, \( \sqrt{x}+\frac{1}{\sqrt{x}} = \pm \sqrt{6} \). Since \( \sqrt{x}+\frac{1}{\sqrt{x}} \) must be positive (as square roots are usually taken as positive), the value is \( \sqrt{6} \). This problem elegantly combines rationalization with algebraic identities.
In simple words: First find the reciprocal of x, then add x and its reciprocal. Use the result in the formula for squaring \( (\sqrt{x} + \frac{1}{\sqrt{x}}) \) to find its value.

🎯 Exam Tip: When dealing with \( \sqrt{x}+\frac{1}{\sqrt{x}} \), consider squaring the expression first to simplify the calculation using the identity \( (a+b)^2 = a^2+b^2+2ab \).