MP Board Class 10 Mathematics Basic Question Paper 2026 PDF with Solutions - Available

Concept:

A rational number can be written in the form \(\frac{p}{q}\), where \(p, q\) are integers and \(q \ne 0\).
An irrational number cannot be written in fractional form (e.g., \(\sqrt{2}, \pi\)).


Explanation:
Let: \[ r = rational number, \quad x = irrational number \]

Assume their sum is rational: \[ r + x = y \quad (rational) \]

Then, \[ x = y - r \]

Since the difference of two rational numbers is rational, this would imply \(x\) is rational.

But this contradicts the fact that \(x\) is irrational.

Conclusion:
Hence, the sum of a rational and an irrational number is always: \[ \boxed{Irrational} \] Quick Tip: Key results to remember: Rational + Irrational = Irrational Rational \(\times\) Irrational = Irrational (generally) Irrational + Irrational may be rational or irrational These are common exam fill-in-the-blanks.

Concept:
In an Arithmetic Progression (A.P.), the common difference is the difference between any two consecutive terms: \[ d = a_2 - a_1 \]

Step 1: Identify first two terms \[ a_1 = 4, \quad a_2 = 10 \]

Step 2: Find the common difference \[ d = 10 - 4 = 6 \]

Verification: \[ 16 - 10 = 6, \quad 22 - 16 = 6 \]
So, the difference is constant.

Conclusion: \[ \boxed{6} \] Quick Tip: To find common difference quickly: \[ d = second term - first term \] Always check one more pair to confirm it is an A.P.

Concept:
Two triangles are similar if:

Their corresponding angles are equal (AAA similarity), or
Their corresponding sides are proportional.


Explanation:
In an equilateral triangle:

All three sides are equal.
All three angles are \(60^\circ\).


Consider any two equilateral triangles: \[ \triangle ABC \quad and \quad \triangle PQR \]

Then, \[ \angle A = \angle B = \angle C = 60^\circ \] \[ \angle P = \angle Q = \angle R = 60^\circ \]

So, corresponding angles are equal.

Using AAA similarity criterion:
If all corresponding angles of two triangles are equal, then the triangles are similar.

Hence, \[ \triangle ABC \sim \triangle PQR \]

Conclusion:
All equilateral triangles have the same angles, so they are always similar.
\[ \boxed{All equilateral triangles are similar.} \] Quick Tip: Remember: Equilateral triangle → all angles \(60^\circ\). Same angles ⇒ AAA similarity ⇒ triangles are similar. But they may not be congruent (sizes can differ).

Concept:
Prime factorization means expressing a number as a product of prime numbers only.

Step 1: Divide by smallest prime number \[ 140 \div 2 = 70 \]

Step 2: Divide again by 2 \[ 70 \div 2 = 35 \]

Step 3: Divide by next prime \[ 35 \div 5 = 7 \]

Step 4: Final prime \[ 7 \div 7 = 1 \]

So, prime factors are: \[ 2, 2, 5, 7 \]

Step 5: Write in exponential form \[ 140 = 2^2 \times 5 \times 7 \]

Final Answer: \[ \boxed{2^2 \times 5 \times 7} \] Quick Tip: For prime factorization: Start dividing by 2, then 3, 5, 7, etc. Stop when quotient becomes 1. Write repeated primes using powers. This is useful for HCF and LCM problems.

Concept:
The zeroes of a quadratic polynomial are the values of \(x\) that make the polynomial equal to zero.
We solve: \[ x^2 + 7x + 10 = 0 \]

Step 1: Factorize the quadratic
We need two numbers whose:

Sum = 7
Product = 10


These numbers are \(5\) and \(2\).

Step 2: Split the middle term \[ x^2 + 5x + 2x + 10 = 0 \]

Step 3: Factor by grouping \[ x(x + 5) + 2(x + 5) = 0 \]
\[ (x + 5)(x + 2) = 0 \]

Step 4: Find zeroes \[ x + 5 = 0 \Rightarrow x = -5 \] \[ x + 2 = 0 \Rightarrow x = -2 \]

Final Answer: \[ \boxed{x = -5, \; -2} \] Quick Tip: For quick factorization: Find two numbers whose sum = middle coefficient. Product = constant term. If factorization is difficult, use quadratic formula.

Concept:
The \(n^{th}\) term of an Arithmetic Progression (A.P.) is given by: \[ a_n = a + (n-1)d \]
where:

\(a\) = first term
\(d\) = common difference


Step 1: Identify values \[ a = 2, \quad d = 7 - 2 = 5, \quad n = 10 \]

Step 2: Apply formula \[ a_{10} = 2 + (10 - 1)\times 5 \]
\[ = 2 + 9 \times 5 \]
\[ = 2 + 45 = 47 \]

Final Answer: \[ \boxed{47} \] Quick Tip: To find any term of an A.P.: \[ a_n = a + (n-1)d \] This is one of the most frequently used formulas in A.P. problems.

Concept:
Probability is defined as: \[ Probability = \frac{Number of favourable outcomes}{Total number of possible outcomes} \]

A standard die has 6 faces numbered: \[ 1, 2, 3, 4, 5, 6 \]

Step 1: Identify prime numbers on a die
Prime numbers are numbers greater than 1 that have only two factors (1 and itself).

Prime numbers between 1 and 6: \[ 2, 3, 5 \]

Step 2: Count favourable outcomes
Number of prime numbers = 3

Step 3: Total outcomes
Total outcomes when rolling a die = 6

Step 4: Find probability \[ P(prime number) = \frac{3}{6} = \frac{1}{2} \]

Final Answer: \[ \boxed{\frac{1}{2}} \] Quick Tip: Steps for probability: List all possible outcomes. Identify favourable outcomes. Use \(\frac{favourable}{total}\). Always simplify the fraction to lowest form.

Concept:
A tangent to a circle is perpendicular to the radius at the point of contact.
We prove the result using congruent triangles.

Given:
Let \(P\) be an external point, and \(PA\) and \(PB\) be tangents to a circle with center \(O\), touching the circle at \(A\) and \(B\) respectively.

To Prove: \[ PA = PB \]

Construction:
Join \(OA\), \(OB\), and \(OP\).

Proof:

Step 1: Radius is perpendicular to tangent \[ OA \perp PA \quad and \quad OB \perp PB \]
So, \[ \angle OAP = \angle OBP = 90^\circ \]

Step 2: Consider triangles \(\triangle OAP\) and \(\triangle OBP\)
We have:

\(OA = OB\) (radii of the same circle)
\(OP = OP\) (common side)
\(\angle OAP = \angle OBP = 90^\circ\)


Step 3: Apply RHS congruence \[ \triangle OAP \cong \triangle OBP \quad (RHS) \]

Step 4: Corresponding parts are equal \[ PA = PB \]

Conclusion:
Hence, tangents drawn from an external point to a circle are equal in length. Quick Tip: To prove tangent properties: Join center to points of contact. Use radius ⟂ tangent property. Apply RHS congruence. This is a standard geometry proof.