CBSE Class 10 Maths Important Questions 2026: Chapter-Wise with Solutions

Problem: Find the HCF and LCM of 306 and 657 by prime factorisation. Also verify that HCF $\times$ LCM $=$ Product of the two numbers.

Solution:
$306 = 2 \times 3^2 \times 17$
$657 = 3 \times 3 \times 73 = 3^2 \times 73$

HCF $= 3^2 = 9$ (product of common prime factors with lowest powers)

LCM $= 2 \times 3^2 \times 17 \times 73 = 22338$ (product of all prime factors with highest powers)

Verification: HCF $\times$ LCM $= 9 \times 22338 = 201042$
Product $= 306 \times 657 = 201042$ $\checkmark$

Marking tip: Always show the prime factorisation step and the verification. The verification alone can earn you 1 mark.

Problem: Prove that $\sqrt{3}$ is irrational.

Solution:
Assume $\sqrt{3}$ is rational. Then $\sqrt{3} = \frac{p}{q}$ where $p$ and $q$ are co-prime integers ($q \neq 0$).

Squaring: $3 = \frac{p^2}{q^2} \implies p^2 = 3q^2$.

This means $3$ divides $p^2$, so $3$ divides $p$ (since 3 is prime).

Let $p = 3k$. Then $p^2 = 9k^2$, so $3q^2 = 9k^2 \implies q^2 = 3k^2$.

This means $3$ divides $q^2$, so $3$ divides $q$.

But if $3$ divides both $p$ and $q$, they are not co-prime — a contradiction.

Therefore, $\sqrt{3}$ is irrational. $\square$

Marking tip: State the assumption clearly, show the contradiction, and write the conclusion. Each of these three parts carries marks.

Problem: Find the zeroes of $p(x) = 6x^2 - 3 - 7x$ and verify the relationship between zeroes and coefficients.

Solution:
Rearranging: $6x^2 - 7x - 3 = 0$.

Factorising: $(2x - 3)(3x + 1) = 0$.

Zeroes: $x = \frac{3}{2}$ and $x = -\frac{1}{3}$.

Verification:
Sum $= \frac{3}{2} + (-\frac{1}{3}) = \frac{9 - 2}{6} = \frac{7}{6}$ and $-\frac{b}{a} = -\frac{(-7)}{6} = \frac{7}{6}$ $\checkmark$

Product $= \frac{3}{2} \times (-\frac{1}{3}) = -\frac{1}{2}$ and $\frac{c}{a} = \frac{-3}{6} = -\frac{1}{2}$ $\checkmark$

Problem: Find all zeroes of $p(x) = x^4 - 6x^3 - 26x^2 + 138x - 35$, given that $2 + \sqrt{3}$ and $2 - \sqrt{3}$ are two of its zeroes.

Solution:
Since $2 + \sqrt{3}$ and $2 - \sqrt{3}$ are zeroes:
$[x - (2+\sqrt{3})][x - (2-\sqrt{3})] = (x-2)^2 - 3 = x^2 - 4x + 1$ is a factor.

Divide $x^4 - 6x^3 - 26x^2 + 138x - 35$ by $x^2 - 4x + 1$:

Quotient $= x^2 - 2x - 35$.

Factorise: $x^2 - 2x - 35 = (x - 7)(x + 5)$.

Remaining zeroes: $x = 7$ and $x = -5$.

All zeroes: $2 + \sqrt{3}, 2 - \sqrt{3}, 7, -5$.

Marking tip: Show the long division clearly. Each step of the division and each factor earn marks.

Problem: A fraction becomes $\frac{1}{3}$ when 1 is subtracted from the numerator and it becomes $\frac{1}{4}$ when 8 is added to its denominator. Find the fraction.

Solution:
Let the fraction be $\frac{x}{y}$.

Subtract (1) from (2): $x = 5$.

From (1): $y = 3(5) - 3 = 12$.

Answer: The fraction is $\frac{5}{12}$.

Problem: Solve: $\frac{5}{x-1} + \frac{1}{y-2} = 2$ and $\frac{6}{x-1} - \frac{3}{y-2} = 1$.

Solution:
Let $u = \frac{1}{x-1}$ and $v = \frac{1}{y-2}$:

Multiply (1) by 3: $15u + 3v = 6 \quad \cdots (1')$

Add $(1')$ and $(2)$: $21u = 7 \implies u = \frac{1}{3}$.

From (1): $v = 2 - 5 \times \frac{1}{3} = \frac{1}{3}$.

So $x - 1 = 3 \implies x = 4$ and $y - 2 = 3 \implies y = 5$.

Answer: $x = 4$, $y = 5$.

Problem: Find the value of $k$ for which $2x^2 + kx + 3 = 0$ has two equal real roots.

Solution:
For equal roots, discriminant $= 0$:

Answer: $k = 2\sqrt{6}$ or $k = -2\sqrt{6}$.

Problem: A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, it would have taken 1 hour less. Find the speed of the train.

Solution:
Let speed $= x$ km/h. Time $= \frac{360}{x}$ hours.

New speed $= (x + 5)$ km/h. New time $= \frac{360}{x + 5}$ hours.

Answer: Speed of the train $= 40$ km/h.

Marking tip: Show the equation formation (2 marks), solving (2 marks), and stating the answer with reason for rejection (1 mark).

Problem: The $17^{\text{th}}$ term of an AP exceeds its $10^{\text{th}}$ term by 7. Find the common difference.

Solution:
$a_{17} = a + 16d$ and $a_{10} = a + 9d$.

Given: $a_{17} - a_{10} = 7$:

Answer: Common difference $d = 1$.

Problem: Find the sum of the first 15 terms of the AP: $7, 13, 19, 25, \ldots$

Solution:
Here $a = 7$, $d = 13 - 7 = 6$, $n = 15$.

Answer: $S_{15} = 735$.

Marking tip: Write the formula, substitute values, and simplify step by step. Even if you make an arithmetic error, showing the correct formula earns partial marks.

Problem: In $\triangle ABC$, $DE \parallel BC$ with $D$ on $AB$ and $E$ on $AC$. If $\frac{AD}{DB} = \frac{3}{4}$ and $AC = 15$ cm, find $AE$.

Solution:
By BPT: $\frac{AD}{DB} = \frac{AE}{EC} = \frac{3}{4}$.

So $\frac{AE}{AC - AE} = \frac{3}{4}$:

Answer: $AE = \frac{45}{7}$ cm.

Problem: Prove that in a right triangle, the square of the hypotenuse equals the sum of the squares of the other two sides.

Solution:
Given: $\triangle ABC$ with $\angle B = 90^\circ$.
To prove: $AC^2 = AB^2 + BC^2$.
Construction: Draw $BD \perp AC$.

In $\triangle ADB$ and $\triangle ABC$:
$\angle ADB = \angle ABC = 90^\circ$ and $\angle A$ is common.
By AA similarity: $\triangle ADB \sim \triangle ABC$.

In $\triangle BDC$ and $\triangle ABC$:
$\angle BDC = \angle ABC = 90^\circ$ and $\angle C$ is common.
By AA similarity: $\triangle BDC \sim \triangle ABC$.

Adding (1) and (2):

Marking tip: The construction, both similarity proofs, and the final addition step each carry marks. Draw a clear diagram.

Problem: Find the coordinates of the point which divides the line segment joining $A(4, -3)$ and $B(8, 5)$ in the ratio $3:1$ internally.

Solution:
Using the section formula:

Answer: The point is $(7, 3)$.

Problem: Find the area of the triangle with vertices $A(2, 3)$, $B(-1, 0)$, and $C(2, -4)$.

Solution:

Answer: Area $= 10.5$ sq units.

Problem: Prove that $\frac{\sin\theta}{1 + \cos\theta} + \frac{1 + \cos\theta}{\sin\theta} = 2\csc\theta$.

Solution:

Problem: From the top of a 7 m high building, the angle of elevation of the top of a tower is $60^\circ$ and the angle of depression of its foot is $45^\circ$. Find the height of the tower.

Solution:
Let the tower be $AB$ and the building be $CD = 7$ m. Let $CE$ be drawn horizontal from $C$ to the tower at $E$, so $AE = AB - BE = AB - 7$.

From $\angle$ of depression of foot $B$: $\tan 45^\circ = \frac{CD}{BD} = \frac{7}{BD}$.

From $\angle$ of elevation of top $A$: $\tan 60^\circ = \frac{AE}{CE} = \frac{AB - 7}{7}$.

Answer: Height of tower $= 7(\sqrt{3} + 1) \approx 19.12$ m.

Marking tip: Draw the diagram (1 mark), set up both equations (2 marks), solve (2 marks). The diagram is essential for this type of question.

Problem: Prove that the tangents drawn from an external point to a circle are equal in length.

Solution:
Given: A circle with centre $O$. $PA$ and $PB$ are tangents from external point $P$ to the circle at points $A$ and $B$.
To prove: $PA = PB$.
Proof:
In $\triangle OAP$ and $\triangle OBP$:
1. $OA = OB$ (radii of same circle)
2. $OP = OP$ (common side)
3. $\angle OAP = \angle OBP = 90^\circ$ (radius $\perp$ tangent, Theorem 10.1)

By RHS congruence: $\triangle OAP \cong \triangle OBP$.

By CPCTC: $PA = PB$. $\square$

Marking tip: State Given, To Prove, draw diagram, and provide reasons for each step.

Problem: Two tangents $TP$ and $TQ$ are drawn from an external point $T$ to a circle with centre $O$. If $\angle PTQ = 70^\circ$, find $\angle POQ$.

Solution:
$\angle OPT = \angle OQT = 90^\circ$ (radius $\perp$ tangent).

In quadrilateral $OPTQ$:

Answer: $\angle POQ = 110^\circ$.

Problem: Find the area of the minor segment of a circle of radius 21 cm, if the angle of the sector is $120^\circ$. (Use $\pi = \frac{22}{7}$ and $\sqrt{3} = 1.73$.)

Solution:
Area of sector:

**Area of $\triangle OAB$:** With $\theta = 120^\circ$, using $\frac{1}{2}r^2 \sin\theta$:

Area of segment $= 462 - 190.73 = 271.27$ cm$^2$.

Answer: Area of minor segment $\approx 271.27$ cm$^2$.

Problem: In a square of side 14 cm, four equal circles are drawn. Find the area of the shaded region (the region inside the square but outside all four circles).

Solution:
Since four equal circles fit in a square of side 14 cm, each circle has diameter $= 14/2 = 7$ cm, so radius $= 3.5$ cm.

Answer: Shaded area $= 42$ cm$^2$.

Problem: A metallic sphere of radius 4.2 cm is melted and recast into the shape of a cylinder of radius 6 cm. Find the height of the cylinder.

Solution:
Volume of sphere $=$ Volume of cylinder:

Answer: Height of cylinder $= 2.744$ cm.

Problem: A solid is in the shape of a cone surmounted on a hemisphere. The radius of each is 3.5 cm and the total height of the solid is 9.5 cm. Find the volume of the solid. (Use $\pi = \frac{22}{7}$.)

Solution:
Radius $r = 3.5$ cm. Height of cone $= 9.5 - 3.5 = 6$ cm.

Volume of hemisphere:

Volume of cone:

Total volume $= 89.83 + 77 = 166.83$ cm$^3$.

Answer: Volume $\approx 166.83$ cm$^3$.

Problem: Find the mean of the following data using the assumed mean method:

Solution:
Let assumed mean $a = 35$, class size $h = 10$.

Answer: Mean $= 35.5$.

Problem: Find the median of the following data:

Solution:
$N = 40$, so $\frac{N}{2} = 20$.

Cumulative frequencies: 6, 15, 27, 35, 40.

The median class is $20-30$ (since cumulative frequency first exceeds 20 at this class).

$l = 20$, $f = 12$, $cf = 15$ (cumulative frequency of the class before median class), $h = 10$.

Answer: Median $\approx 24.17$.

Marking tip: Show the cumulative frequency column, identify the median class correctly, and write the formula before substituting.

Problem: Two dice are thrown simultaneously. Find the probability of getting a sum of 7.

Solution:
Total outcomes $= 6 \times 6 = 36$.

Favourable outcomes (sum $= 7$): $(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)$ $= 6$ outcomes.

Answer: $P = \frac{1}{6}$.

Problem: A card is drawn at random from a well-shuffled deck of 52 cards. Find the probability that the card drawn is (i) a king or a queen, (ii) neither a king nor a queen.

Solution:
(i) Kings $= 4$, Queens $= 4$. Total favourable $= 8$.

(ii) Using complementary events:

Answer: (i) $\frac{2}{13}$, (ii) $\frac{11}{13}$.

Marking tip: For multiple parts, clearly label (i) and (ii). Use complementary probability to save time in part (ii).