Bihar Board Class 10 Higher Mathematics Question Paper 2026 PDF with Solutions - Available

Concept:
To compare exponents, express the number as a product of its prime factors.

Step 1: Prime factorization of 540 \[ 540 = 54 \times 10 \]
\[ 54 = 2 \times 27 = 2 \times 3^3 \] \[ 10 = 2 \times 5 \]

So, \[ 540 = (2 \times 3^3)(2 \times 5) = 2^2 \times 3^3 \times 5^1 \]

Step 2: Compare with given form \[ 540 = 2^x \times 3^y \times 5^z \]

Thus, \[ x = 2, \quad y = 3, \quad z = 1 \]

Step 3: Find required value \[ x + y - z = 2 + 3 - 1 = 4 \]

Final Answer: \[ \boxed{4} \] Quick Tip: To compare exponents: Always do prime factorization first. Match bases and compare powers. This method is common in factorization problems.

Concept:

Rational numbers can be written in the form \(\frac{p}{q}\).
Irrational numbers cannot be written as fractions.
Square root of a perfect square is rational.


Option (A): \(2.3\) \[ 2.3 = \frac{23}{10} \]
This is a fraction, so it is rational.

Option (B): \(\sqrt{13} \times \sqrt{13}\) \[ \sqrt{13} \times \sqrt{13} = 13 \]
Since 13 is an integer, it is rational.

Option (C): \(\sqrt{441}\) \[ 441 = 21^2 \] \[ \sqrt{441} = 21 \]
This is also rational.

Conclusion:
All given options are rational numbers.
Hence, none of them is irrational.
\[ \boxed{None of these} \] Quick Tip: Quick checks: Terminating decimals → rational \(\sqrt{a} \times \sqrt{a} = a\) Square root of perfect square → rational Only roots of non-perfect squares are irrational.

Concept:

A prime number has only two factors: \(1\) and itself.
HCF (Highest Common Factor) is the greatest common factor of two numbers.


Case 1: \(m\) and \(n\) are different prime numbers
Factors of \(m\): \(1, m\)
Factors of \(n\): \(1, n\)

Common factor = \(1\)

So, \[ HCF(m, n) = 1 \]

Case 2: If \(m = n\) (same prime)
Then, \[ HCF(m, n) = m \]

Conclusion:
Generally, for two distinct prime numbers: \[ \boxed{1} \] Quick Tip: Remember: HCF of two different primes = 1 Same primes → HCF is the number itself This is a common objective question.

Concept:
Standard trigonometric values: \[ \sin 90^\circ = 1, \quad \cos 0^\circ = 1 \]

Step 1: Substitute known values \[ \sin 90^\circ + \cos 0^\circ = 1 + 1 \]

Step 2: Simplify \[ = 2 \]

Final Answer: \[ \boxed{2} \] Quick Tip: Important values to memorize: \[ \sin 0^\circ = 0, \quad \sin 90^\circ = 1 \] \[ \cos 0^\circ = 1, \quad \cos 90^\circ = 0 \] These are frequently used in basic trigonometry.

Concept:
Distance of a point \((x, y)\) from the origin \((0,0)\) is given by: \[ Distance = \sqrt{x^2 + y^2} \]

Step 1: Substitute coordinates \[ x = 3, \quad y = 4 \]
\[ Distance = \sqrt{3^2 + 4^2} \]

Step 2: Simplify \[ = \sqrt{9 + 16} = \sqrt{25} \]
\[ = 5 \]

Final Answer: \[ \boxed{5} \] Quick Tip: Distance from origin formula: \[ \sqrt{x^2 + y^2} \] Recognize Pythagorean triples like \((3,4,5)\) to solve quickly.

Concept:
The degree of a polynomial is the highest power of the variable in its expanded form.

Step 1: Expand the polynomial \[ (x + 3)(x - 8) \]

Using distributive property: \[ x(x - 8) + 3(x - 8) \]
\[ = x^2 - 8x + 3x - 24 \]
\[ = x^2 - 5x - 24 \]

Step 2: Identify highest power
The highest power of \(x\) is \(2\).

Final Answer: \[ \boxed{2} \] Quick Tip: Degree rules: Product of two linear polynomials → degree 2 Add degrees when multiplying polynomials Example: \((x)(x)\) gives degree \(2\).

Concept:
To find the HCF using prime factorization:

Express each number as a product of prime factors.
Take only the common prime factors with the smallest powers.


Step 1: Prime factorization

36: \[ 36 = 2^2 \times 3^2 \]

54: \[ 54 = 2 \times 27 = 2 \times 3^3 \]

90: \[ 90 = 9 \times 10 = 3^2 \times 2 \times 5 = 2 \times 3^2 \times 5 \]

Step 2: Identify common prime factors
Common primes in all three: \[ 2 \quad and \quad 3 \]

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

Step 3: Find HCF \[ HCF = 2 \times 3^2 = 2 \times 9 = 18 \]

Final Answer: \[ \boxed{18} \] Quick Tip: Steps for HCF using prime factors: Write prime factorization of all numbers. Pick common primes only. Take smallest powers. This method works for any number of values.

Concept:
Zeroes of a polynomial are values of \(x\) that make it equal to zero.

Step 1: Set the polynomial equal to zero \[ 2x^2 - 6 = 0 \]

Step 2: Solve for \(x\) \[ 2x^2 = 6 \]
\[ x^2 = 3 \]

Step 3: Take square root \[ x = \pm \sqrt{3} \]

Final Answer: \[ \boxed{x = \pm \sqrt{3}} \] Quick Tip: For equations of type \(ax^2 - b = 0\): Move constant term to RHS. Divide by coefficient of \(x^2\). Take square root. Always include both \(+\) and \(-\) roots.

Concept:
In the graphical method:

Each linear equation represents a straight line.
The solution is the point of intersection of the two lines.


Step 1: Convert equations into convenient form

Equation (1): \[ 4x - 5y = 20 \]

Find intercepts:

If \(x = 0\), then \(-5y = 20 \Rightarrow y = -4\)
If \(y = 0\), then \(4x = 20 \Rightarrow x = 5\)


So points: \((0, -4)\) and \((5, 0)\)

Equation (2): \[ 3x + 5y = 15 \]

Find intercepts:

If \(x = 0\), then \(5y = 15 \Rightarrow y = 3\)
If \(y = 0\), then \(3x = 15 \Rightarrow x = 5\)


So points: \((0, 3)\) and \((5, 0)\)

Step 2: Graph the lines
Plot both pairs of points and draw the lines.

Step 3: Identify intersection point
Both lines intersect at: \[ (5, 0) \]

Conclusion:
The graphical solution of the equations is: \[ \boxed{(5, 0)} \] Quick Tip: For graphical method: Find two points for each line. Plot and draw lines. Intersection point = solution. If lines intersect at one point → unique solution.

Statement (Pythagoras Theorem):
In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Given:
In \(\triangle ABC\), \(\angle B = 90^\circ\).
So, \(AC\) is the hypotenuse.

To Prove: \[ AC^2 = AB^2 + BC^2 \]

Proof (Using similarity):

Step 1: Draw altitude
Draw \(BD \perp AC\) from the right angle \(B\) to the hypotenuse \(AC\).

Step 2: Consider similar triangles
Triangles \(\triangle ABC\), \(\triangle ADB\), and \(\triangle BDC\) are similar because they share common angles.

From similarity of \(\triangle ABC \sim \triangle ADB\): \[ \frac{AB}{AC} = \frac{AD}{AB} \]
\[ AB^2 = AC \cdot AD \quad \cdots (1) \]

From similarity of \(\triangle ABC \sim \triangle BDC\): \[ \frac{BC}{AC} = \frac{DC}{BC} \]
\[ BC^2 = AC \cdot DC \quad \cdots (2) \]

Step 3: Add equations (1) and (2) \[ AB^2 + BC^2 = AC \cdot AD + AC \cdot DC \]
\[ AB^2 + BC^2 = AC(AD + DC) \]

But, \[ AD + DC = AC \]

So, \[ AB^2 + BC^2 = AC^2 \]

Conclusion:
Hence proved, \[ \boxed{AC^2 = AB^2 + BC^2} \] Quick Tip: Pythagoras Theorem applies only to right-angled triangles. Remember the formula: \[ (Hypotenuse)^2 = (Base)^2 + (Perpendicular)^2 \] Useful in distance formula and coordinate geometry.

Concept:
Translate the given statements into equations and solve simultaneously.

Let: \[ Smaller number = x, \quad Larger number = y \]

Step 1: Form equations

Difference of squares: \[ y^2 - x^2 = 180 \quad \cdots (1) \]

Square of smaller number is 8 times the larger: \[ x^2 = 8y \quad \cdots (2) \]

Step 2: Substitute equation (2) into (1) \[ y^2 - 8y = 180 \]
\[ y^2 - 8y - 180 = 0 \]

Step 3: Solve quadratic \[ y^2 - 8y - 180 = 0 \]

Factorization: \[ (y - 18)(y + 10) = 0 \]
\[ y = 18 \quad or \quad y = -10 \]

Since larger number must be positive: \[ y = 18 \]

Step 4: Find \(x\)
Using \(x^2 = 8y\): \[ x^2 = 8 \times 18 = 144 \]
\[ x = 12 \quad (taking positive value) \]

Final Answer: \[ \boxed{Smaller number = 12, \quad Larger number = 18} \] Quick Tip: Word problems → form equations first. Steps: Assign variables clearly. Convert statements into equations. Substitute to reduce variables. Always check which root is valid.