JKBOSE Class 10 Mathematics Question Paper with Solution PDF

Concept:
A number is irrational if it cannot be expressed as a ratio of two integers.
A common method to prove irrationality is proof by contradiction:

Assume the number is rational.
Express it in lowest terms \( \frac{p}{q} \).
Derive a contradiction (both numerator and denominator becoming divisible by the same number).


Step 1: Prove that \( \sqrt{3} \) is irrational.

Assume \( \sqrt{3} \) is rational.
Then it can be written as: \[ \sqrt{3} = \frac{p}{q}, \quad where p, q \in \mathbb{Z}, \; q \neq 0, and \gcd(p,q)=1. \]

Squaring both sides: \[ 3 = \frac{p^2}{q^2} \quad \Rightarrow \quad p^2 = 3q^2. \]

Thus, \( p^2 \) is divisible by 3, so \( p \) must also be divisible by 3.
Let \( p = 3k \).

Substitute back: \[ (3k)^2 = 3q^2 \Rightarrow 9k^2 = 3q^2 \Rightarrow q^2 = 3k^2. \]

So \( q^2 \) is divisible by 3, hence \( q \) is divisible by 3.

This means both \( p \) and \( q \) are divisible by 3, contradicting the assumption that \( \frac{p}{q} \) is in lowest terms.
\[ \therefore \sqrt{3} is irrational. \]



Step 2: Prove that \( \sqrt{5} \) is irrational.

Assume \( \sqrt{5} = \frac{p}{q} \) in lowest terms.

Squaring: \[ 5 = \frac{p^2}{q^2} \Rightarrow p^2 = 5q^2. \]

Thus, \( p \) is divisible by 5. Let \( p = 5k \).

Substitute: \[ 25k^2 = 5q^2 \Rightarrow q^2 = 5k^2. \]

Hence \( q \) is also divisible by 5, contradicting lowest terms assumption.
\[ \therefore \sqrt{5} is irrational. \]



Conclusion:
Both \( \sqrt{3} \) and \( \sqrt{5} \) are irrational numbers. Quick Tip: To prove square roots of non-perfect squares are irrational, assume the number is rational and use contradiction by showing numerator and denominator share a common factor.

Concept:
Using the prime factorisation method:

HCF (Highest Common Factor): Product of the lowest powers of common prime factors.
LCM (Least Common Multiple): Product of the highest powers of all prime factors.



(i) For 56 and 72

Step 1: Prime factorisation
\[ 56 = 2 \times 2 \times 2 \times 7 = 2^3 \times 7 \] \[ 72 = 2 \times 2 \times 2 \times 3 \times 3 = 2^3 \times 3^2 \]

Step 2: HCF

Common prime factors: only \( 2 \)
Lowest power of \( 2 \) is \( 2^3 \)
\[ HCF = 2^3 = 8 \]

Step 3: LCM

Take highest powers of all primes: \[ 2^3, \quad 3^2, \quad 7 \]
\[ LCM = 2^3 \times 3^2 \times 7 = 8 \times 9 \times 7 = 504 \]
\[ \therefore HCF = 8, \quad LCM = 504 \]


(ii) For 6, 72, and 120

Step 1: Prime factorisation
\[ 6 = 2 \times 3 = 2^1 \times 3^1 \] \[ 72 = 2^3 \times 3^2 \] \[ 120 = 2^3 \times 3 \times 5 = 2^3 \times 3^1 \times 5^1 \]

Step 2: HCF

Common prime factors in all numbers: \( 2 \) and \( 3 \)

Lowest powers: \[ 2^1, \quad 3^1 \]
\[ HCF = 2 \times 3 = 6 \]

Step 3: LCM

Take highest powers of all primes: \[ 2^3, \quad 3^2, \quad 5^1 \]
\[ LCM = 2^3 \times 3^2 \times 5 = 8 \times 9 \times 5 = 360 \]
\[ \therefore HCF = 6, \quad LCM = 360 \] Quick Tip: For HCF, take the smallest powers of common primes. For LCM, take the largest powers of all primes present in any number.

Concept:
For a quadratic polynomial \( ax^2 + bx + c \) with zeroes \( \alpha \) and \( \beta \):
\[ \alpha + \beta = -\frac{b}{a}, \quad \alpha\beta = \frac{c}{a} \]

We find the zeroes and verify these relations.


Example 1: \( x^2 - 3x - 10 \)

Step 1: Factorisation
\[ x^2 - 3x - 10 = (x - 5)(x + 2) \]

Step 2: Zeroes
\[ x - 5 = 0 \Rightarrow x = 5 \] \[ x + 2 = 0 \Rightarrow x = -2 \]

So, zeroes are \( 5 \) and \( -2 \).

Step 3: Verify relations

Here \( a = 1, b = -3, c = -10 \)
\[ \alpha + \beta = 5 + (-2) = 3 \] \[ -\frac{b}{a} = -\frac{-3}{1} = 3 \quad \checkmark \]
\[ \alpha\beta = 5 \times (-2) = -10 \] \[ \frac{c}{a} = \frac{-10}{1} = -10 \quad \checkmark \]

Relations verified.


Example 2: \( 4u^2 + 8u \)

Step 1: Factorisation
\[ 4u^2 + 8u = 4u(u + 2) \]

Step 2: Zeroes
\[ 4u = 0 \Rightarrow u = 0 \] \[ u + 2 = 0 \Rightarrow u = -2 \]

So, zeroes are \( 0 \) and \( -2 \).

Step 3: Verify relations

Here \( a = 4, b = 8, c = 0 \)
\[ \alpha + \beta = 0 + (-2) = -2 \] \[ -\frac{b}{a} = -\frac{8}{4} = -2 \quad \checkmark \]
\[ \alpha\beta = 0 \times (-2) = 0 \] \[ \frac{c}{a} = \frac{0}{4} = 0 \quad \checkmark \]

Relations verified.


Conclusion:
For both polynomials, the sum and product of zeroes satisfy: \[ \alpha + \beta = -\frac{b}{a}, \quad \alpha\beta = \frac{c}{a} \] Quick Tip: For any quadratic \( ax^2 + bx + c \): Sum of zeroes = \( -\frac{b}{a} \), Product of zeroes = \( \frac{c}{a} \). Always verify after finding roots.

Concept:
The HCF (Highest Common Factor) is the greatest number that divides both numbers exactly.
We use the prime factorisation method:

Write each number as a product of prime factors.
Take the common primes with the smallest powers.


Step 1: Prime factorisation
\[ 18 = 2 \times 3 \times 3 = 2 \times 3^2 \] \[ 24 = 2 \times 2 \times 2 \times 3 = 2^3 \times 3 \]

Step 2: Identify common prime factors

Common primes: \( 2 \) and \( 3 \)

Lowest powers: \[ 2^1, \quad 3^1 \]

Step 3: Find HCF
\[ HCF = 2 \times 3 = 6 \]
\[ \therefore HCF of 18 and 24 is 6. \] Quick Tip: HCF = product of common primes with smallest powers. LCM uses highest powers instead.

Concept:
A rational number can be expressed in the form \( \frac{p}{q} \), where \( p \) and \( q \) are integers and \( q \neq 0 \).
The square root of a number is rational only if the number is a perfect square.

Step 1: Examine the number

Since \( 7 \) is not a perfect square, its square root cannot be written as a ratio of two integers.

Step 2: Conclusion
\[ \sqrt{7} is an irrational number. \]

Therefore, the given statement is False. Quick Tip: The square root of any non-perfect square number is always irrational.

Concept:
Probability measures the chance of an event occurring and lies between 0 and 1.

Probability = 0 \quad \(\Rightarrow\) Impossible event
Probability = 1 \quad \(\Rightarrow\) Certain event


Step 1: Definition of an impossible event

An impossible event is one that can never happen.
Example: Getting 7 on a standard die.

Step 2: Probability value

Since the event cannot occur at all, the number of favourable outcomes is 0.
\[ Probability = \frac{0}{Total outcomes} = 0 \]

Final Answer: \[ \boxed{0} \] Quick Tip: Probability always lies between 0 and 1. Impossible event \(\rightarrow\) 0, Certain event \(\rightarrow\) 1.

Concept:
An Arithmetic Progression (AP) is a sequence of numbers in which the difference between consecutive terms is constant.
This constant difference is called the common difference \( d \).

Let:

First term = \( a \)
Common difference = \( d \)


Step 1: General form of an AP
\[ a,\; a+d,\; a+2d,\; a+3d,\; \dots \]

Step 2: Pattern of terms

\begin{align*
1st term &= a

\text{2nd term &= a + d

\text{3rd term &= a + 2d

\text{4th term &= a + 3d
\end{align*

We observe that the coefficient of \( d \) is always one less than the term number.

Step 3: Formula for the \( n^{\text{th} \) term
\[ a_n = a + (n-1)d \]

Final Answer: \[ \boxed{a_n = a + (n-1)d} \] Quick Tip: To find any term in an AP, use: \( Term = First term + (n-1) \times Common difference \).