Real Numbers Class 10 Mathematics — Competency Based Questions

20 Questions | 60 Marks

A baker is creating a special fruitcake. He uses 36 almonds and 48 cherries as toppings. He wants to arrange them in separate layers such that each layer has the same number of fruits, and the number of fruits in each layer is maximized. What is the maximum number of fruits he can put in each layer? 3marks

Assertion (A): The number

6^{n}

, where

n

is a natural number, can end with the digit 0.Reason (R): For a number to end with the digit 0, its prime factorization must include both 2 and 5.(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).(C) Assertion (A) is true, but Reason (R) is false.(D) Assertion (A) is false, but Reason (R) is true. 2marks

A scientist is studying the decay of a radioactive substance. The initial amount of the substance is 100 grams. After 1 hour, the amount remaining is 80 grams. If the decay process follows an exponential model, can the amount of the substance remaining after

n

hours be represented by

100 \times (4 / 5)^{n}

? Justify your answer using the concept of prime factorization. 3marks

Consider two positive integers

a

and

b

. If HCF

(a, b) = 12

and LCM

(a, b) = 72

, and one of the numbers is 24, find the other number. Explain your reasoning using the relationship between HCF, LCM, and the numbers. 3marks

A group of students is preparing for a science fair. They have two different colored threads: red threads are 60 cm long, and blue threads are 84 cm long. They want to cut these threads into smaller pieces of equal length, such that the length of each piece is maximized, and no thread is wasted. What is the maximum possible length of each piece? 3marks

Prove that

7 + 2 \sqrt{3}

is an irrational number, given that

\sqrt{3}

is an irrational number. 3marks

A circular track has a length of 1200 meters. Three runners start running from the same point at the same time. Runner A completes a round in 4 minutes, Runner B in 5 minutes, and Runner C in 6 minutes. After how much time will they all meet again at the starting point? 3marks

Assertion (A): The number

2^{n} \times 5^{m}

, where

n

and

m

are natural numbers, will always end with the digit 0.Reason (R): A number ends with the digit 0 if and only if its prime factorization contains both 2 and 5.(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).(C) Assertion (A) is true, but Reason (R) is false.(D) Assertion (A) is false, but Reason (R) is true. 2marks

A rectangular garden measures 15 meters by 20 meters. The owner wants to pave a square-shaped path of uniform width around the garden. If the area of the path is equal to the area of the garden, find the width of the path. Explain the steps to solve this problem, relating it to the properties of numbers. 5marks

Prove that

\sqrt{7}

is an irrational number. 3marks

A school organizes a quiz competition. There are 48 questions in the first round and 72 questions in the second round. The students are asked to divide the questions into sections such that each section has the same number of questions, and this number is maximized. Find the maximum number of questions in each section. 3marks

Case Study: A school is organizing a sports day. The physical education teacher wants to arrange the students in rows for a drill. There are 360 students in Class 9 and 480 students in Class 10. The teacher wants to arrange them in rows such that each row has the same number of students, and the number of students in each row is maximized.(a) What is the maximum number of students that can be placed in each row? [2 marks](b) How many rows will be needed for Class 9 students? [1 mark](c) How many rows will be needed for Class 10 students? [1 mark](d) If the teacher wanted to arrange the students in groups of 10, would it be possible to do so without any remainder for both classes? Justify. 1mark

Assertion (A): The sum of a rational number and an irrational number is always irrational.Reason (R): If

x

is rational and

y

is irrational, then

x + y

is irrational.(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).(C) Assertion (A) is true, but Reason (R) is false.(D) Assertion (A) is false, but Reason (R) is true. 2marks

Prove that

3 \sqrt{2} - 1

is an irrational number, given that

\sqrt{2}

is an irrational number. 3marks

Two positive integers

a

and

b

satisfy the property that HCF

(a, b) \times

LCM

(a, b) = a \times b

. If HCF

(a, b) = 10

and

a = 20

, find the value of

b

. 2marks

A chocolate company is producing two types of chocolates, A and B. Type A has 64 peanuts, and Type B has 80 peanuts. They want to pack them in boxes such that each box has the same number of peanuts, and this number is maximized. What is the maximum number of peanuts that can be in each box? 3marks

Prove that

5 \sqrt{2}

is an irrational number. 2marks

Assertion (A): The decimal expansion of

\frac{23}{2^{2} \times 5}

is terminating.Reason (R): A rational number

\frac{p}{q}

(where

q \neq 0

) terminates if and only if the prime factorization of

q

is of the form

2^{n} \times 5^{m}

, where

n

and

m

are non-negative integers.(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).(C) Assertion (A) is true, but Reason (R) is false.(D) Assertion (A) is false, but Reason (R) is true. 2marks

A gardener wants to plant trees in a rectangular field of size 48 meters by 60 meters. He wants to plant them in rows such that the distance between any two trees in the same row and the distance between any two rows are equal. This distance should be maximized. Find the maximum possible distance between the trees. 3marks

Prove that

4 - \sqrt{5}

is an irrational number, given that

\sqrt{5}

is an irrational number. 3marks

Answers

The problem requires finding the HCF of 36 and 48.Prime factorization of 36:

2^{2} \times 3^{2}

Prime factorization of 48:

2^{4} \times 3

HCF(36, 48) =

2^{2} \times 3 = 4 \times 3 = 12

.Therefore, the maximum number of fruits he can put in each layer is 12. 3marks

(D) Assertion (A) is false, but Reason (R) is true.Reasoning: For a number to end with the digit 0, its prime factorization must include both 2 and 5. The prime factorization of

6^{n}

(2 \times 3)^{n} = 2^{n} \times 3^{n}

. This factorization only contains the prime factors 2 and 3, not 5. Therefore,

6^{n}

can never end with the digit 0 for any natural number

n

. The reason provided is correct. 2marks

The amount of substance remaining after

n

hours can be modeled as

A (n) = A_{0} \times r^{n}

, where

A_{0}

is the initial amount and

r

is the decay factor.Given

A_{0} = 100

grams and

A (1) = 80

grams.So,

80 = 100 \times r^{1} \Rightarrow r = \frac{80}{100} = \frac{4}{5}

.Thus, the amount remaining after

n

hours is

A (n) = 100 \times (\frac{4}{5})^{n}

.The question asks if this can be represented using prime factorization. The base of the exponentiation is

\frac{4}{5}

.The prime factorization of 4 is

2^{2}

.The prime factorization of 5 is

5

.So, the decay factor is

\frac{2^{2}}{5^{1}}

.The formula

100 \times (\frac{4}{5})^{n}

can be written as

100 \times (\frac{2^{2}}{5})^{n}

. This form uses prime factors in the base. So, yes, it can be represented using prime factorization of the decay factor. 3marks

We know that for any two positive integers

a

and

b

, HCF

(a, b) \times

LCM

(a, b) = a \times b

.Given HCF

(a, b) = 12

, LCM

(a, b) = 72

, and let

a = 24

.So,

12 \times 72 = 24 \times b

b = \frac{12 \times 72}{24} = \frac{12 \times 72}{2 \times 12} = \frac{72}{2} = 36

.The other number is 36. This property is derived from the unique prime factorization of numbers, where HCF accounts for the minimum powers of common prime factors and LCM accounts for the maximum powers of all prime factors. 3marks

The problem requires finding the HCF of 60 and 84.Prime factorization of 60:

2^{2} \times 3 \times 5

Prime factorization of 84:

2^{2} \times 3 \times 7

HCF(60, 84) =

2^{2} \times 3 = 4 \times 3 = 12

.Therefore, the maximum possible length of each piece is 12 cm. 3marks

Let's assume, for the sake of contradiction, that

7 + 2 \sqrt{3}

is a rational number.Then,

7 + 2 \sqrt{3} = \frac{a}{b}

, where

a

and

b

are integers,

b \neq 0

, and

a

and

b

are coprime.Rearranging the equation:

2 \sqrt{3} = \frac{a}{b} - 7

2 \sqrt{3} = \frac{a - 7 b}{b}

\sqrt{3} = \frac{a - 7 b}{2 b}

Since $a$ , $b$ , 7, and 2 are integers, $\frac{a - 7 b}{2 b}$ is a rational number.
This implies that $\sqrt{3}$ is a rational number.
However, we are given that $\sqrt{3}$ is an irrational number. This is a contradiction.
Therefore, our initial assumption that $7 + 2 \sqrt{3}$ is rational must be false.
Hence, $7 + 2 \sqrt{3}$ is an irrational number. [3 marks]

The problem requires finding the LCM of 4, 5, and 6 minutes.Prime factorization of 4:

2^{2}

Prime factorization of 5:

5^{1}

Prime factorization of 6:

2 \times 3

LCM(4, 5, 6) =

2^{2} \times 3 \times 5 = 4 \times 3 \times 5 = 60

.They will all meet again at the starting point after 60 minutes. 3marks

(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).Reasoning: The number

2^{n} \times 5^{m}

contains both prime factors 2 and 5 in its prime factorization. According to Reason (R), a number terminates if and only if its prime factorization contains only powers of 2 and 5. Therefore, any number of the form

2^{n} \times 5^{m}

will end with the digit 0. 2marks

Let the width of the square-shaped path be

x

meters.The dimensions of the rectangular garden are 15 m by 20 m.Area of the garden =

15 \times 20 = 300

^{2}

.The dimensions of the garden including the path will be

(15 + 2 x)

m by

(20 + 2 x)

m.Area of the garden with the path =

(15 + 2 x) (20 + 2 x)

^{2}

.Area of the path = Area of the garden with path - Area of the gardenArea of the path =

(15 + 2 x) (20 + 2 x) - 300

Area of the path =

300 + 30 x + 40 x + 4 x^{2} - 300

Area of the path =

4 x^{2} + 70 x

^{2}

.According to the problem, the area of the path is equal to the area of the garden.So,

4 x^{2} + 70 x = 300

4 x^{2} + 70 x - 300 = 0

.Divide by 2:

2 x^{2} + 35 x - 150 = 0

.We can solve this quadratic equation using the quadratic formula

x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}

x = \frac{- 35 \pm \sqrt{35^{2} - 4 (2) (- 150)}}{2 (2)}

x = \frac{- 35 \pm \sqrt{1225 + 1200}}{4}

x = \frac{- 35 \pm \sqrt{2425}}{4}

x = \frac{- 35 \pm 5 \sqrt{97}}{4}

Since the width $x$ must be positive, we take the positive root:
$x = \frac{- 35 + 5 \sqrt{97}}{4}$ .
This problem involves solving a quadratic equation, which can be related to number properties through coefficients and discriminant, but the core is algebraic. The justification relates to setting up the problem based on geometric areas. [5 marks]

Let's assume, for the sake of contradiction, that

\sqrt{7}

is a rational number.Then,

\sqrt{7} = \frac{a}{b}

, where

a

and

b

are integers,

b \neq 0

, and

a

and

b

are coprime.Squaring both sides:

7 = \frac{a^{2}}{b^{2}} \Rightarrow 7 b^{2} = a^{2}

.This implies that

a^{2}

is divisible by 7.According to Theorem 1.2 from NCERT, if a prime

p

divides

a^{2}

, then

p

divides

a

. Here,

p = 7

.So,

a

is divisible by 7. We can write

a = 7 k

for some integer

k

.Substitute this into the equation

7 b^{2} = a^{2}

7 b^{2} = (7 k)^{2}

7 b^{2} = 49 k^{2}

$b^{2} = 7 k^{2}$ .
This implies that $b^{2}$ is divisible by 7.
Therefore, $b$ is also divisible by 7.
We have shown that both $a$ and $b$ are divisible by 7. This contradicts our assumption that $a$ and $b$ are coprime.
Therefore, our initial assumption that $\sqrt{7}$ is rational must be false.
Hence, $\sqrt{7}$ is an irrational number. [3 marks]

The problem requires finding the HCF of 48 and 72.Prime factorization of 48:

2^{4} \times 3

Prime factorization of 72:

2^{3} \times 3^{2}

HCF(48, 72) =

2^{3} \times 3 = 8 \times 3 = 24

.Therefore, the maximum number of questions in each section is 24. 3marks

Case Study:(a) We need to find the HCF of 360 and 480.Prime factorization of 360:

2^{3} \times 3^{2} \times 5

Prime factorization of 480:

2^{5} \times 3 \times 5

HCF(360, 480) =

2^{3} \times 3 \times 5 = 8 \times 3 \times 5 = 120

.The maximum number of students that can be placed in each row is 120. [2 marks](b) Number of rows for Class 9 = Total students in Class 9 / Students per row =

360 / 120 = 3

. [1 mark](c) Number of rows for Class 10 = Total students in Class 10 / Students per row =

480 / 120 = 4

. [1 mark](d) For Class 9:

360 / 10 = 36

(no remainder).For Class 10:

480 / 10 = 48

(no remainder).Yes, it would be possible to arrange them in groups of 10 without any remainder for both classes. This is because both 360 and 480 are divisible by 10. Their prime factorization (

360 = 2^{3} \times 3^{2} \times 5

and

480 = 2^{5} \times 3 \times 5

) both contain the prime factor 5 and at least one factor of 2, making them divisible by

2 \times 5 = 10

. 1mark

(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).Reasoning: Let

x

be a rational number and

y

be an irrational number. Assume

x + y

is rational. Then

x + y = r

for some rational number

r

. This means

y = r - x

. Since

r

and

x

are rational,

r - x

is also rational. This implies

y

is rational, which contradicts the fact that

y

is irrational. Therefore,

x + y

must be irrational. The reason provided accurately explains the assertion. 2marks

Let's assume, for the sake of contradiction, that

3 \sqrt{2} - 1

is a rational number.Then,

3 \sqrt{2} - 1 = \frac{a}{b}

, where

a

and

b

are integers,

b \neq 0

, and

a

and

b

are coprime.Rearranging the equation:

3 \sqrt{2} = \frac{a}{b} + 1

3 \sqrt{2} = \frac{a + b}{b}

\sqrt{2} = \frac{a + b}{3 b}

Since $a$ , $b$ , 1, and 3 are integers, $\frac{a + b}{3 b}$ is a rational number.
This implies that $\sqrt{2}$ is a rational number.
However, we are given that $\sqrt{2}$ is an irrational number. This is a contradiction.
Therefore, our initial assumption that $3 \sqrt{2} - 1$ is rational must be false.
Hence, $3 \sqrt{2} - 1$ is an irrational number. [3 marks]

We know that for any two positive integers

a

and

b

, HCF

(a, b) \times

LCM

(a, b) = a \times b

.Given HCF

(a, b) = 10

and

a = 20

.So,

10 \times

LCM

(a, b) = 20 \times b

.We also know that

a

and

b

are multiples of HCF. So,

a = 10 k_{1}

and

b = 10 k_{2}

where

k_{1}, k_{2}

are coprime.Since

a = 20

20 = 10 k_{1} \Rightarrow k_{1} = 2

.So,

a = 10 \times 2

.Since

k_{1} = 2

, and

k_{1}, k_{2}

are coprime,

k_{2}

cannot be an even number.The LCM of

a

and

b

10 \times k_{1} \times k_{2} = 10 \times 2 \times k_{2} = 20 k_{2}

.Substitute into the formula:

10 \times (20 k_{2}) = 20 \times (10 k_{2})

200 k_{2} = 200 k_{2}

. This equation is always true and does not help find

b

Let's re-evaluate. The relationship $H C F (a, b) \times L C M (a, b) = a \times b$ is key.
Given HCF $(a, b) = 10$ and $a = 20$ .
We know that $b$ must be a multiple of HCF(a, b), so $b$ must be a multiple of 10. Let $b = 10 k$ .
Also, $a = 20$ , which is $10 \times 2$ .
The HCF of $a$ and $b$ is 10. This means that the common prime factors with their lowest powers must result in 10. $10 = 2 \times 5$ .
$a = 20 = 2^{2} \times 5$ .
For HCF $(20, b) = 10$ , $b$ must have $2^{1}$ and $5^{1}$ in its prime factorization.
Let $b = 2^{1} \times 5^{1} \times m = 10 m$ .
For HCF to be exactly 10, $m$ should not have any factors of 2 or 5 that would increase the HCF. Also, $a$ has $2^{2}$ , so $b$ cannot have $2^{2}$ or higher powers of 2 if HCF is 10. Thus, the highest power of 2 in $b$ must be $2^{1}$ .
The number $b$ could be $10 \times 1 = 10$ . HCF(20, 10) = 10. LCM(20, 10) = 20. $10 \times 20 = 200$ . $20 \times 10 = 200$ . This works. So $b = 10$ is a possible value.
Another possibility: let $m = 3$ . Then $b = 10 \times 3 = 30$ . HCF(20, 30) = 10. LCM(20, 30) = 60. $10 \times 60 = 600$ . $20 \times 30 = 600$ . This works. So $b = 30$ is a possible value.
The question implies a unique value for $b$ . Let's re-read the question. "find the value of b". This suggests a unique answer.

Let's use the formula directly: $10 \times LCM (20, b) = 20 \times b$ .
We know $b$ must be a multiple of 10. Let $b = 10 k$ .
HCF $(20, 10 k) = 10$ .
LCM $(20, 10 k) = \frac{20 \times 10 k}{10} = 20 k$ .
So, $10 \times (20 k) = 20 \times (10 k)$ . This simplifies to $200 k = 200 k$ , which is always true.

Let's consider the properties of HCF and LCM more deeply.
$a = 20 = 2^{2} \times 5^{1}$ .
HCF $(a, b) = 10 = 2^{1} \times 5^{1}$ .
This implies that in the prime factorization of $b$ :
The power of 2 must be exactly 1 (since $a$ has $2^{2}$ , and HCF has $2^{1}$ ).
The power of 5 must be at least 1 (since HCF has $5^{1}$ ).
Let $b = 2^{1} \times 5^{x} \times (other prime factors)$ .
Since HCF is 10, $x \geq 1$ .
Let's take the simplest case: $b = 2^{1} \times 5^{1} = 10$ .
HCF(20, 10) = 10. LCM(20, 10) = 20. $10 \times 20 = 200$ . $20 \times 10 = 200$ . This is correct. So $b = 10$ .

What if $b = 2^{1} \times 5^{2} = 50$ ?
HCF(20, 50) = HCF( $2^{2} \times 5$ , $2 \times 5^{2}$ ) = $2^{1} \times 5^{1} = 10$ . This is correct.
LCM(20, 50) = LCM( $2^{2} \times 5$ , $2 \times 5^{2}$ ) = $2^{2} \times 5^{2} = 4 \times 25 = 100$ .
Check: HCF $\times$ LCM = $10 \times 100 = 1000$ .
$a \times b = 20 \times 50 = 1000$ . This is also correct. So $b = 50$ is a possible value.

The question must be interpreted such that there's a single answer. Usually, in such cases, $b$ is assumed to be different from $a$ . If $b$ is different from $a$ , $b = 10$ is a valid answer.
Let's check the wording. "Two positive integers a and b". It doesn't state they are different.

Let's go back to $a = 20 = 2^{2} \times 5^{1}$ . HCF = $10 = 2^{1} \times 5^{1}$ .
For HCF to be $2^{1} \times 5^{1}$ , $b$ must have $2^{1}$ as the power of 2. The power of 5 must be at least 1. Any other prime factors of $b$ must not be common with $a$ .
Let $b = 2^{1} \times 5^{k} \times \dots$ , where $k \geq 1$ .
If $k = 1$ , $b = 2^{1} \times 5^{1} = 10$ . HCF(20, 10) = 10.
If $k = 2$ , $b = 2^{1} \times 5^{2} = 50$ . HCF(20, 50) = 10.
If $k = 1$ and $b$ has another factor, say 3. $b = 2^{1} \times 5^{1} \times 3^{1} = 30$ . HCF(20, 30) = 10.

It seems there are multiple possible values for $b$ . However, standard textbook problems of this type usually intend for the simplest or smallest positive integer value when not specified. The smallest positive integer $b$ such that HCF(20, b) = 10 is $b = 10$ .

Let's try to use the LCM formula: LCM $(a, b) = \frac{a \times b}{H C F (a, b)}$ .
LCM $(20, b) = \frac{20 \times b}{10} = 2 b$ .
We also know that LCM $(a, b)$ must be a multiple of $a$ . So $2 b$ must be a multiple of 20.
This means $b$ must be a multiple of 10. Let $b = 10 k$ .
So, LCM $(20, 10 k) = 2 (10 k) = 20 k$ .
And HCF $(20, 10 k) = 10$ .
We need HCF $(2^{2} \times 5, 2 \times 5 \times k) = 2^{1} \times 5^{1}$ .
This implies that the power of 2 in $b$ must be 1. So, $10 k = 2^{1} \times 5^{1} \times k$ .
For HCF to be 10, $k$ cannot have a factor of 2. So $k$ must be odd.
The smallest odd positive integer $k$ is 1.
If $k = 1$ , then $b = 10 \times 1 = 10$ .
If $k = 3$ , then $b = 10 \times 3 = 30$ .
If $k = 5$ , then $b = 10 \times 5 = 50$ .

Given the wording "find the value of b", and that this is a typical school problem, it's highly probable that the intended answer is the smallest such positive integer, which is 10. [2 marks]

The problem requires finding the HCF of 64 and 80.Prime factorization of 64:

2^{6}

Prime factorization of 80:

2^{4} \times 5

HCF(64, 80) =

2^{4} = 16

.Therefore, the maximum number of peanuts that can be in each box is 16. 3marks

Let's assume, for the sake of contradiction, that

5 \sqrt{2}

is a rational number.Then,

5 \sqrt{2} = \frac{a}{b}

, where

a

and

b

are integers,

b \neq 0

, and

a

and

b

are coprime.Rearranging the equation:

\sqrt{2} = \frac{a}{5 b}

Since $a$ , $b$ , and 5 are integers, $\frac{a}{5 b}$ is a rational number.
This implies that $\sqrt{2}$ is a rational number.
However, it is a known fact that $\sqrt{2}$ is an irrational number. This is a contradiction.
Therefore, our initial assumption that $5 \sqrt{2}$ is rational must be false.
Hence, $5 \sqrt{2}$ is an irrational number. [2 marks]

(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).Reasoning: The rational number is

\frac{23}{2^{2} \times 5}

. The denominator is

2^{2} \times 5^{1}

. This is in the form

2^{n} \times 5^{m}

where

n = 2

and

m = 1

are non-negative integers. Therefore, the decimal expansion terminates, supporting Assertion (A). Reason (R) correctly states the condition for a terminating decimal expansion. 2marks

The number is

8^{n}

, where

n

is a natural number.

8^{n} = (2^{3})^{n} = 2^{3 n}

.The prime factorization of

8^{n}

only contains the prime factor 2.For a number to end with the digit 0, its prime factorization must contain both 2 and 5.Since the prime factorization of

8^{n}

does not contain the prime factor 5, it cannot end with the digit 0.Therefore, the number

8^{n}

cannot end with the digit 0 for any natural number

n

. 2marks

Let's assume, for the sake of contradiction, that

\frac{2 - \sqrt{3}}{5}

is a rational number.Then,

\frac{2 - \sqrt{3}}{5} = \frac{a}{b}

, where

a

and

b

are integers,

b \neq 0

, and

a

and

b

are coprime.Rearranging the equation:

b (2 - \sqrt{3}) = 5 a

2 b - b \sqrt{3} = 5 a

2 b - 5 a = b \sqrt{3}

\frac{2 b - 5 a}{b} = \sqrt{3}

Since $a$ , $b$ , 2, and 5 are integers, $\frac{2 b - 5 a}{b}$ is a rational number.
This implies that $\sqrt{3}$ is a rational number.
However, we are given that $\sqrt{3}$ is an irrational number. This is a contradiction.
Therefore, our initial assumption that $\frac{2 - \sqrt{3}}{5}$ is rational must be false.
Hence, $\frac{2 - \sqrt{3}}{5}$ is an irrational number. [3 marks]

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Real Numbers — Competency Based Questions

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