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CBSE Class 10 Maths Standard 2026 Question Paper 30/5/1 with Solutions

All 44 questions from the CBSE Class 10 Mathematics Standard board paper, Set 30/5/1 (2026), with answers and step-by-step solutions. Total 80 marks. Tap “Show answer & solution” under any question.

Set 30/1/1 Set 30/1/2 Set 30/1/3 Set 30/2/1 Set 30/2/2 Set 30/2/3 Set 30/3/1 Set 30/3/2 Set 30/3/3 Set 30/4/1 Set 30/4/2 Set 30/4/3 Set 30/5/1 Set 30/5/2 Set 30/5/3

Q11 markMCQQuadratic Equations

The value of k for which the equation $k x^{2} - 6 x - 4 = 0$ has real and equal roots, is

(A) $\frac{9}{4}$
(B) $- 4$
(C) $- \frac{9}{4}$
(D) $- 2$

Show answer & solution

Answer: (C)

- \frac{9}{4}

For real and equal roots, $D = b^{2} - 4 a c = 0$ .
$(- 6)^{2} - 4 (k) (- 4) = 0$
$36 + 16 k = 0$
$k = - \frac{9}{4}$

Also asked in: 2026 Standard 30/5/2

Q21 markMCQCoordinate Geometry

The line segment joining the points $P (- 4, - 2)$ and $Q (10, 4)$ is divided by y-axis in the ratio

(A)2 : 5
(B)1 : 2
(C)2 : 1
(D)5 : 2

Show answer & solution

Answer: (A) 2 : 5

Let the y-axis divide PQ in the ratio $k : 1$ .
The point of division has x-coordinate $\frac{10 k - 4}{k + 1}$ , which is 0 on the y-axis.
$10 k - 4 = 0$ , so $k = \frac{2}{5}$ .
Ratio = 2 : 5

Also asked in: 2026 Standard 30/5/2, 2026 Standard 30/5/3

Q31 markMCQPolynomials

If the zeroes of a polynomial $p (x)$ are $- 3$ and 8, then $p (x)$ equals

(A) $x^{2} + 5 x - 4$
(B) $(x + 3) (- x + 8)$
(C) $a (x^{2} + 5 x - 24)$
(D) $x^{2} - 24$

Show answer & solution

Answer: (B)

(x + 3) (- x + 8)

A polynomial with zeroes $- 3$ and 8 is $k (x + 3) (x - 8)$ for some non-zero constant k.
With $k = - 1$ : $- (x + 3) (x - 8) = (x + 3) (- x + 8)$ , whose zeroes are $- 3$ and 8.
Option (C) $a (x^{2} + 5 x - 24) = a (x + 8) (x - 3)$ has zeroes $- 8$ and 3, so it is wrong; (A) and (D) do not have these zeroes either.

Also asked in: 2026 Standard 30/5/3

Q41 markMCQCircles

In the given figure, PQ is tangent to the circle with centre O. S is a point on the circle such that $∠ S QT = 5 5^{\circ}$ . The m $∠ QP S$ is

Diagram for CBSE 2026 Class 10 Maths question 4

(A) $5 5^{\circ}$
(B) $2 0^{\circ}$
(C) $3 5^{\circ}$
(D) $7 0^{\circ}$

Show answer & solution

Answer: (B)

2 0^{\circ}

$O Q ⊥ P T$ (radius is perpendicular to tangent), so $∠ O QS = 9 0^{\circ} - 5 5^{\circ} = 3 5^{\circ}$ .
$O Q = O S$ (radii), so $∠ O S Q = ∠ O QS = 3 5^{\circ}$ .
$∠ P QS = 18 0^{\circ} - ∠ S QT = 18 0^{\circ} - 5 5^{\circ} = 12 5^{\circ}$ .
In $△ P QS$ : $∠ QP S = 18 0^{\circ} - 12 5^{\circ} - 3 5^{\circ} = 2 0^{\circ}$ .

Also asked in: 2026 Standard 30/5/2

Q51 markMCQTriangles

Devansh proved that $△ A B C \sim △ P QR$ using SAS similarity criteria. If he found $∠ C = ∠ R$ , then which of the following was proved true ?

(A) $\frac{A C}{A B} = \frac{P R}{P Q}$
(B) $\frac{B C}{A C} = \frac{P R}{QR}$
(C) $\frac{A C}{B C} = \frac{P R}{P Q}$
(D) $\frac{A C}{B C} = \frac{P R}{QR}$

Show answer & solution

Answer: (D)

\frac{A C}{B C} = \frac{P R}{QR}

In SAS similarity, the sides including the equal angles must be proportional.
The sides including $∠ C$ are AC and BC; the sides including $∠ R$ are PR and QR.
So $\frac{A C}{P R} = \frac{B C}{QR}$ , i.e. $\frac{A C}{B C} = \frac{P R}{QR}$ .

Also asked in: 2026 Standard 30/5/2, 2026 Standard 30/5/3

Q61 markMCQStatistics

While calculating mean of a grouped frequency distribution, step deviation method was used $(\frac{x - a}{h} = u)$ . It was found that $x ˉ = 64$ , h = 5 and a = 62.5. The value of $u ˉ$ is

(A)0.5
(B)1.5
(C)0.3
(D)7.5

Show answer & solution

Answer: (C) 0.3

$x ˉ = a + h u ˉ$
$64 = 62.5 + 5 u ˉ$
$5 u ˉ = 1.5$ , so $u ˉ = 0.3$

Also asked in: 2026 Standard 30/5/2, 2026 Standard 30/5/3

Q71 markMCQSurface Areas and Volumes

A conical cavity of maximum volume is carved out from a wooden solid hemisphere of radius 10 cm. Curved surface area of the cavity carved out is (use $π = 3.14$ )

Diagram for CBSE 2026 Class 10 Maths question 7

(A) $3142$ $c m^{2}$
(B)314 $c m^{2}$
(C) $\frac{3140}{3}$ $c m^{2}$
(D) $31402$ $c m^{2}$

Show answer & solution

Answer: (A)

3142

c m^{2}

The largest cone has radius $r = 10$ cm and height $h = 10$ cm (radius of hemisphere).
Slant height $l = 1 0^{2} + 1 0^{2} = 102$ cm
CSA $= π r l = 3.14 \times 10 \times 102 = 3142$ $c m^{2}$

Also asked in: 2026 Standard 30/5/2, 2026 Standard 30/5/3

Q81 markMCQArithmetic Progressions

If $a_{n}$ represents $n^{t h}$ term of the A.P. $- \frac{15}{4}, - \frac{10}{4}, - \frac{5}{4}, \dots$ then value of $a_{16} - a_{12}$ is

(A)4
(B) $\frac{5}{4}$
(C)5
(D) $\frac{25}{4}$

Show answer & solution

Answer: (C) 5

$d = - \frac{10}{4} - (- \frac{15}{4}) = \frac{5}{4}$
$a_{16} - a_{12} = (a + 15 d) - (a + 11 d) = 4 d = 4 \times \frac{5}{4} = 5$

Also asked in: 2026 Standard 30/5/3

Q91 markMCQQuadratic Equations

The value of p for which roots of the quadratic equation $x^{2} - p x + 6 = 0$ are rational, is

(A)1
(B) $- 5$
(C)25
(D) $5$

Show answer & solution

Answer: (B)

- 5

Roots are rational when $D = p^{2} - 24$ is a perfect square (and p is rational).
$p = 1$ : $D = - 23$ (no real roots).
$p = - 5$ : $D = 25 - 24 = 1 = 1^{2}$ , a perfect square; $x^{2} + 5 x + 6 = 0$ gives $x = - 2, - 3$ , which are rational.
$p = 25$ : $D = 601$ , not a perfect square. $p = 5$ : $D = - 19$ .
So $p = - 5$ .

Also asked in: 2026 Standard 30/5/2, 2026 Standard 30/5/3

Q101 markMCQProbability

Which of the following can not be the probability of an event ?

(A) $\frac{39}{100}$
(B) $\frac{0.001}{20}$
(C) $\frac{10}{0.2}$
(D)10%

Show answer & solution

Answer: (C)

\frac{10}{0.2}

Probability of an event lies between 0 and 1.
$\frac{10}{0.2} = 50 > 1$ , so it cannot be a probability.
The others are 0.39, 0.00005 and 0.1, all between 0 and 1.

Also asked in: 2026 Standard 30/5/2, 2026 Standard 30/5/3

Q111 markMCQProbability

A card is drawn at random from a well shuffled deck of 52 playing cards. The probability that it is either a ten or a king is

(A) $\frac{1}{26}$
(B) $\frac{2}{13}$
(C) $\frac{1}{13}$
(D) $\frac{8}{26}$

Show answer & solution

Answer: (B)

\frac{2}{13}

Number of tens = 4, number of kings = 4, so favourable outcomes = 8.
P(ten or king) $= \frac{8}{52} = \frac{2}{13}$

Also asked in: 2026 Standard 30/5/2

Q121 markMCQSurface Areas and Volumes

A camping tent in hemispherical shape of radius 1.4 m, has a door opening of area 0.50 $m^{2}$ . Outer surface area of the tent is

(A)11.78 $m^{2}$
(B)12.32 $m^{2}$
(C)11.82 $m^{2}$
(D)12.86 $m^{2}$

Show answer & solution

Answer: (C) 11.82

m^{2}

CSA of hemisphere $= 2 π r^{2} = 2 \times \frac{22}{7} \times 1.4 \times 1.4 = 12.32$ $m^{2}$
Outer surface area $= 12.32 - 0.50 = 11.82$ $m^{2}$

Also asked in: 2026 Standard 30/5/2, 2026 Standard 30/5/3

Q131 markMCQAreas Related to Circles

An arc of length 2.2 cm subtends an angle $θ$ at the centre of the circle with radius 2.8 cm. The value of $θ$ is

(A) $5 0^{\circ}$
(B) $6 0^{\circ}$
(C) $4 5^{\circ}$
(D) $3 0^{\circ}$

Show answer & solution

Answer: (C)

4 5^{\circ}

Arc length $= \frac{θ}{36 0 ^{\circ}} \times 2 π r$
$2.2 = \frac{θ}{36 0 ^{\circ}} \times 2 \times \frac{22}{7} \times 2.8 = \frac{θ}{36 0 ^{\circ}} \times 17.6$
$θ = \frac{2.2}{17.6} \times 36 0^{\circ} = 4 5^{\circ}$

Also asked in: 2026 Standard 30/5/3

Q141 markMCQIntroduction to Trigonometry

For an acute angle $θ$ , if $sin θ = \frac{1}{9}$ , then value of $\frac{9 cosec θ + 1}{9 cosec θ - 1}$ is

(A)0
(B) $\frac{80}{81}$
(C)1
(D) $\frac{82}{80}$

Show answer & solution

Answer: (D)

\frac{82}{80}

$cosec θ = \frac{1}{s i n θ} = 9$
$\frac{9 \times 9 + 1}{9 \times 9 - 1} = \frac{82}{80}$

Also asked in: 2026 Standard 30/5/3

Q151 markMCQTriangles

In the given figure, $P Q ∥ Y Z$ such that XP : PY = 2 : 3. If PQ = 5 cm, then YZ equals

Diagram for CBSE 2026 Class 10 Maths question 15

(A)12.5 cm
(B)10 cm
(C)15 cm
(D)7.5 cm

Show answer & solution

Answer: (A) 12.5 cm

$P Q ∥ Y Z$ , so $△ X P Q \sim △ X Y Z$ (AA).
$\frac{X P}{X Y} = \frac{2}{2 + 3} = \frac{2}{5}$
$\frac{P Q}{Y Z} = \frac{2}{5}$ , so $Y Z = \frac{5 \times 5}{2} = 12.5$ cm

Also asked in: 2026 Standard 30/5/2

Q161 markMCQProbability

Meena calculates that the probability of her winning the first prize in a lottery is 0.08. If total 800 tickets were sold, the number of tickets bought by her, is

(A)64
(B)640
(C)100
(D)10

Show answer & solution

Answer: (A) 64

P(winning) $= \frac{tickets bought by her}{800} = 0.08$
Tickets bought $= 0.08 \times 800 = 64$

Also asked in: 2026 Standard 30/5/2, 2026 Standard 30/5/3

Q171 markMCQSome Applications of Trigonometry

A wire is attached from a point A on the ground to the top of a pole BC, making an angle of elevation as $6 0^{\circ}$ . If AB = $53$ m, then length of the wire is

(A)10 m
(B) $103$ m
(C)15 m
(D) $\frac{5}{2} 3$ m

Show answer & solution

Answer: (B)

103

In right $△ A B C$ , $cos 6 0^{\circ} = \frac{A B}{A C}$
$\frac{1}{2} = \frac{5 3}{A C}$
$A C = 103$ m

Also asked in: 2026 Standard 30/5/2, 2026 Standard 30/5/3

Q181 markMCQIntroduction to Trigonometry

Simplest form of $\frac{s e c A}{s e c ^{2} A - 1}$ is

(A) $sin A$
(B) $tan A$
(C) $cosec A$
(D) $cos A$

Show answer & solution

Answer: (C)

cosec A

$sec^{2} A - 1 = tan A$ (for acute A)
$\frac{s e c A}{t a n A} = \frac{1/ c o s A}{s i n A / c o s A} = \frac{1}{s i n A} = cosec A$

Also asked in: 2026 Standard 30/5/2, 2026 Standard 30/5/3

Q191 markAssertion–ReasonPair of Linear Equations in Two Variables

Assertion (A): The system of linear equations $3 x - 5 y + 7 = 0$ and $- 6 x + 10 y + 14 = 0$ is inconsistent.
Reason (R): When two linear equations don’t have unique solution, they always represent parallel lines.

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

Show answer & solution

Answer: (C) Assertion (A) is true, but Reason (R) is false.

$\frac{a _{1}}{a _{2}} = \frac{3}{- 6} = - \frac{1}{2}$ , $\frac{b _{1}}{b _{2}} = \frac{- 5}{10} = - \frac{1}{2}$ , $\frac{c _{1}}{c _{2}} = \frac{7}{14} = \frac{1}{2}$
$\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} \neq = \frac{c _{1}}{c _{2}}$ , so the lines are parallel and the system is inconsistent. A is true.
If there is no unique solution, the lines may be parallel or coincident (infinitely many solutions). So R is false.

Also asked in: 2026 Standard 30/5/2, 2026 Standard 30/5/3

Q201 markAssertion–ReasonReal Numbers

Assertion (A): H.C.F. $(36 m^{2}, 18 m) = 18 m$ , where m is a prime number.
Reason (R): H.C.F. of two numbers is always less than or equal to the smaller number.

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

Show answer & solution

Answer: (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).

$36 m^{2} = 2^{2} \times 3^{2} \times m^{2}$ and $18 m = 2 \times 3^{2} \times m$ .
HCF $= 2 \times 3^{2} \times m = 18 m$ (18m divides $36 m^{2}$ ). A is true.
The HCF divides both numbers, so it cannot exceed the smaller one. R is true.
But R does not explain why the HCF here equals 18m; that follows from 18m being a factor of $36 m^{2}$ . So the answer is (B).

Also asked in: 2026 Standard 30/5/2, 2026 Standard 30/5/3

Q212 marksVery Short AnswerReal Numbers

Prove that $2 - 53$ is an irrational number given that $3$ is irrational.

Show answer & solution

Answer: Proved.

Suppose $2 - 53$ is rational. Let $2 - 53 = r$ , where r is rational.
Then $53 = 2 - r$ , so $3 = \frac{2 - r}{5}$ .
Since r is rational, $\frac{2 - r}{5}$ is rational, so $3$ would be rational.
This contradicts the fact that $3$ is irrational.
Hence $2 - 53$ is irrational.

Q222 marksVery Short AnswerCoordinate Geometry

Vertices of a right triangle ABC with $∠ B = 9 0^{\circ}$ are $A (3, 4)$ , $B (1, 1)$ and $C (- 8, 7)$ . Find the value of tan A.

Show answer & solution

Answer:

tan A = 3

$A B = (3 - 1)^{2} + (4 - 1)^{2} = 4 + 9 = 13$
$B C = (1 + 8)^{2} + (1 - 7)^{2} = 81 + 36 = 117 = 313$
In right $△ A B C$ with $∠ B = 9 0^{\circ}$ : $tan A = \frac{B C}{A B} = \frac{3 13}{13} = 3$

Also asked in: 2026 Standard 30/5/2

Q22 (OR) (OR)2 marksVery Short AnswerCoordinate Geometry

Using distance formula, prove that the points $A (2, 3)$ , $B (- 7, 0)$ and $C (- 1, 2)$ are collinear.

Show answer & solution

Answer: Proved.

$A B = (2 + 7)^{2} + (3 - 0)^{2} = 81 + 9 = 90 = 310$
$B C = (- 1 + 7)^{2} + (2 - 0)^{2} = 36 + 4 = 40 = 210$
$A C = (2 + 1)^{2} + (3 - 2)^{2} = 9 + 1 = 10$
$B C + C A = 210 + 10 = 310 = A B$
Hence A, B, C are collinear (C lies between A and B).

Also asked in: 2026 Standard 30/5/2

Q232 marksVery Short AnswerTriangles

In the given figure, $A B ∥ D E$ and $A C ∥ D F$ . Show that $△ A B C \sim △ D E F$ . If BC = 10 cm, EB = CF = 5 cm and AB = 7 cm, then find the length DE.

Diagram for CBSE 2026 Class 10 Maths question 23

Show answer & solution

Answer: DE = 14 cm

$A B ∥ D E$ , so $∠ A B C = ∠ D E F$ (corresponding angles, with transversal EF).
$A C ∥ D F$ , so $∠ A C B = ∠ D F E$ (corresponding angles).
Hence $△ A B C \sim △ D E F$ (AA similarity).
$E F = E B + B C + C F = 5 + 10 + 5 = 20$ cm
$\frac{D E}{A B} = \frac{E F}{B C} = \frac{20}{10} = 2$ , so $D E = 2 \times 7 = 14$ cm

Also asked in: 2026 Standard 30/5/2, 2026 Standard 30/5/3

Q242 marksVery Short AnswerIntroduction to Trigonometry

Evaluate : $\frac{s i n ^{3} 6 0 ^{\circ} - t a n 3 0 ^{\circ}}{c o s ^{2} 4 5 ^{\circ}}$

Show answer & solution

Answer:

\frac{3}{12}

$sin^{3} 6 0^{\circ} = (\frac{3}{2})^{3} = \frac{3 3}{8}$ , $tan 3 0^{\circ} = \frac{1}{3} = \frac{3}{3}$ , $cos^{2} 4 5^{\circ} = \frac{1}{2}$
Numerator $= \frac{3 3}{8} - \frac{3}{3} = \frac{9 3 - 8 3}{24} = \frac{3}{24}$
Value $= \frac{3}{24} \div \frac{1}{2} = \frac{3}{12}$

Also asked in: 2026 Standard 30/5/3

Q24 (OR) (OR)2 marksVery Short AnswerIntroduction to Trigonometry

For acute angles A and B and A + 2B and 2A + B are acute if $tan (A + 2 B) = 3$ and $sin (2 A + B) = \frac{1}{2}$ , then find the measures of angles A and B.

Show answer & solution

Answer:

A = 1 0^{\circ}

B = 2 5^{\circ}

$tan (A + 2 B) = 3 = tan 6 0^{\circ}$ , so $A + 2 B = 6 0^{\circ}$ ... (1)
$sin (2 A + B) = \frac{1}{2} = sin 4 5^{\circ}$ , so $2 A + B = 4 5^{\circ}$ ... (2)
Adding: $3 A + 3 B = 10 5^{\circ}$ , so $A + B = 3 5^{\circ}$ ... (3)
(1) − (3): $B = 2 5^{\circ}$ ; then $A = 1 0^{\circ}$ .

Also asked in: 2026 Standard 30/5/3

Q252 marksVery Short AnswerProbability

A bag contains 25 balls. Some of them are yellow and others are green. One ball is drawn at random. If probability of getting a green ball is 3/5, then find the number of yellow balls.

Show answer & solution

Answer: 10 yellow balls

Number of green balls $= \frac{3}{5} \times 25 = 15$
Number of yellow balls $= 25 - 15 = 10$

Also asked in: 2026 Standard 30/5/2, 2026 Standard 30/5/3

Q263 marksShort AnswerAreas Related to Circles

A circle of diameter 20 cm is equally divided into five sectors. Find the area and perimeter of one of the sectors.

Show answer & solution

Answer: Area

= \frac{440}{7}

c m^{2}

(≈ 62.86

c m^{2}

); perimeter

= \frac{228}{7}

cm (≈ 32.57 cm)

Radius $r = 10$ cm; each sector has angle $\frac{36 0 ^{\circ}}{5} = 7 2^{\circ}$ .
Area of one sector $= \frac{1}{5} \times π r^{2} = \frac{1}{5} \times \frac{22}{7} \times 100 = \frac{440}{7}$ $c m^{2}$ ≈ 62.86 $c m^{2}$
Arc length $= \frac{1}{5} \times 2 π r = \frac{1}{5} \times 2 \times \frac{22}{7} \times 10 = \frac{88}{7}$ cm
Perimeter $= 2 r + arc = 20 + \frac{88}{7} = \frac{228}{7}$ cm ≈ 32.57 cm

Also asked in: 2026 Standard 30/5/2

Q273 marksShort AnswerArithmetic Progressions

In an A.P., $1 5^{t h}$ term exceeds the $8^{t h}$ term by 21. If sum of first 10 terms is 55, then form the A.P.

Show answer & solution

Answer:

- 8, - 5, - 2, 1, \dots

$a_{15} - a_{8} = 7 d = 21$ , so $d = 3$ .
$S_{10} = \frac{10}{2} [2 a + 9 d] = 5 (2 a + 27) = 55$
$2 a + 27 = 11$ , so $a = - 8$ .
A.P.: $- 8, - 5, - 2, 1, \dots$

Also asked in: 2026 Standard 30/5/2, 2026 Standard 30/5/3

Q27 (OR) (OR)3 marksShort AnswerArithmetic Progressions

The sum of first n terms of an A.P. is $2 n^{2} + 13 n$ . Find its $n^{t h}$ term and hence $1 0^{t h}$ term.

Show answer & solution

Answer:

a_{n} = 4 n + 11

;

a_{10} = 51

$S_{n} = 2 n^{2} + 13 n$
$S_{n - 1} = 2 (n - 1)^{2} + 13 (n - 1) = 2 n^{2} + 9 n - 11$
$a_{n} = S_{n} - S_{n - 1} = 4 n + 11$
$a_{10} = 4 (10) + 11 = 51$

Also asked in: 2026 Standard 30/5/2, 2026 Standard 30/5/3

Q283 marksShort AnswerIntroduction to Trigonometry

Prove that : $\frac{t a n θ}{1 - c o t θ} + \frac{c o t θ}{1 - t a n θ} = 1 + tan θ + cot θ$ .

Show answer & solution

Answer: Proved.

Let $t = tan θ$ , so $cot θ = \frac{1}{t}$ .
$\frac{t}{1 - \frac{1}{t}} = \frac{t ^{2}}{t - 1}$ and $\frac{1/ t}{1 - t} = - \frac{1}{t ( t - 1 )}$
LHS $= \frac{t ^{2}}{t - 1} - \frac{1}{t ( t - 1 )} = \frac{t ^{3} - 1}{t ( t - 1 )}$
$= \frac{( t - 1 ) ( t ^{2} + t + 1 )}{t ( t - 1 )} = \frac{t ^{2} + t + 1}{t} = t + 1 + \frac{1}{t}$
$= 1 + tan θ + cot θ$ = RHS

Also asked in: 2026 Standard 30/5/3

Q293 marksShort AnswerCoordinate Geometry

A circle centered at $(2, 1)$ passes through the points $A (5, 6)$ and $B (- 3, K)$ . Find the value(s) of K. Hence find length of chord AB.

Show answer & solution

Answer: K = 4 or K = −2; AB =

217

units (K = 4) or

82

units (K = −2)

Let centre $O (2, 1)$ . $O A^{2} = (5 - 2)^{2} + (6 - 1)^{2} = 9 + 25 = 34$
$O B^{2} = O A^{2}$ : $(- 3 - 2)^{2} + (K - 1)^{2} = 34$
$25 + (K - 1)^{2} = 34$ , so $(K - 1)^{2} = 9$ and $K - 1 = \pm 3$
$K = 4$ or $K = - 2$
For $K = 4$ : $A B = (5 + 3)^{2} + (6 - 4)^{2} = 68 = 217$
For $K = - 2$ : $A B = (5 + 3)^{2} + (6 + 2)^{2} = 128 = 82$

Also asked in: 2026 Standard 30/5/2, 2026 Standard 30/5/3

Q29 (OR) (OR)3 marksShort AnswerCoordinate Geometry

Prove that the point P dividing the line segment joining the points $A (- 1, 7)$ and $B (4, - 3)$ in the ratio 3 : 2, lies on the line $x - 3 y = - 1$ . Also find length of PA and PB.

Show answer & solution

Answer: P(2, 1) lies on the line; PA =

35

units, PB =

25

units

$P = (\frac{3 ( 4 ) + 2 ( - 1 )}{3 + 2}, \frac{3 ( - 3 ) + 2 ( 7 )}{3 + 2}) = (2, 1)$
Check: $x - 3 y = 2 - 3 = - 1$ . So P lies on the line $x - 3 y = - 1$ .
$P A = (2 + 1)^{2} + (1 - 7)^{2} = 9 + 36 = 35$
$P B = (4 - 2)^{2} + (- 3 - 1)^{2} = 4 + 16 = 25$

Also asked in: 2026 Standard 30/5/2, 2026 Standard 30/5/3

Q303 marksShort AnswerPair of Linear Equations in Two Variables

Solve the system of linear equations : $x = 4$ and $3 x - 2 y = 6$ graphically.

Show answer & solution

Answer:

x = 4

y = 3

$x = 4$ is a vertical line through $(4, 0)$ .
For $3 x - 2 y = 6$ : $y = \frac{3 x - 6}{2}$ . Points: $(0, - 3)$ , $(2, 0)$ , $(4, 3)$ .
Plot both lines on the same axes.
The lines intersect at $(4, 3)$ .
Solution: $x = 4$ , $y = 3$

Q313 marksShort AnswerReal Numbers

The dimensions of a window are 156 cm × 216 cm. Arjun wants to put grill on the window creating complete squares of maximum size. Determine the side length of the square and hence find the number of squares formed.

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Answer: Side = 12 cm; number of squares = 234

$156 = 2^{2} \times 3 \times 13$ , $216 = 2^{3} \times 3^{3}$
HCF $= 2^{2} \times 3 = 12$ , so the side of each square is 12 cm.
Number of squares $= \frac{156}{12} \times \frac{216}{12} = 13 \times 18 = 234$

Also asked in: 2026 Standard 30/5/2, 2026 Standard 30/5/3

Q325 marksLong AnswerQuadratic Equations

Venkat can row a boat in still water at the speed of 12 km/h. He ferries tourists 15 km upstream and 18 km downstream in 3 hours. Find the speed of the stream.

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Answer: 4 km/h

Let the speed of the stream be x km/h. Upstream speed $= 12 - x$ , downstream speed $= 12 + x$ .
$\frac{15}{12 - x} + \frac{18}{12 + x} = 3$
$15 (12 + x) + 18 (12 - x) = 3 (144 - x^{2})$
$396 - 3 x = 432 - 3 x^{2}$
$3 x^{2} - 3 x - 36 = 0$ , i.e. $x^{2} - x - 12 = 0$
$(x - 4) (x + 3) = 0$ , so $x = 4$ (speed cannot be negative).
Speed of the stream = 4 km/h

Q335 marksLong AnswerTriangles

D is the mid-point of side BC of $△ A B C$ . CE and BF intersect at O, a point on AD. AD is produced to G such that OD = DG. Prove that
(i) OBGC is a parallelogram.
(ii) $E F ∥ B C$
(iii) $△ A E F \sim △ A B C$

Diagram for CBSE 2026 Class 10 Maths question 33

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Answer: Proved.

(i) BD = DC (D is the mid-point of BC) and OD = DG (given). So the diagonals BC and OG of quadrilateral OBGC bisect each other. Hence OBGC is a parallelogram.
(ii) As OBGC is a parallelogram, $O C ∥ B G$ and $O B ∥ C G$ , i.e. $O E ∥ B G$ and $O F ∥ C G$ .
In $△ A B G$ , $O E ∥ B G$ , so by BPT $\frac{A E}{E B} = \frac{A O}{O G}$ .
In $△ A C G$ , $O F ∥ C G$ , so by BPT $\frac{A F}{F C} = \frac{A O}{O G}$ .
Hence $\frac{A E}{E B} = \frac{A F}{F C}$ , and by the converse of BPT in $△ A B C$ , $E F ∥ B C$ .
(iii) In $△ A E F$ and $△ A B C$ : $∠ A$ is common, and $∠ A E F = ∠ A B C$ (corresponding angles, $E F ∥ B C$ ).
Hence $△ A E F \sim △ A B C$ (AA similarity).

Also asked in: 2026 Standard 30/5/2, 2026 Standard 30/5/3

Q33 (OR) (OR)5 marksLong AnswerTriangles

Through the mid-point Q of side CD of a parallelogram ABCD, the line AR is drawn which intersects BD at P and produced BC at R. Prove that
(i) AQ = QR
(ii) AP = 2PQ
(iii) PR = 2AP

Diagram for CBSE 2026 Class 10 Maths question 33 (OR)

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Answer: Proved.

(i) In $△ A D Q$ and $△ R C Q$ : DQ = CQ (Q is the mid-point of CD), $∠ A Q D = ∠ R QC$ (vertically opposite), $∠ A D Q = ∠ R C Q$ (alternate angles, $A D ∥ B R$ ).
So $△ A D Q ≅ △ R C Q$ (ASA), hence AQ = QR.
(ii) In $△ A P B$ and $△ QP D$ : $∠ P A B = ∠ P Q D$ and $∠ P B A = ∠ P D Q$ (alternate angles, $A B ∥ D C$ ).
So $△ A P B \sim △ QP D$ (AA), and $\frac{A P}{P Q} = \frac{A B}{D Q}$ .
$D Q = \frac{1}{2} C D = \frac{1}{2} A B$ , so $\frac{A P}{P Q} = 2$ , i.e. AP = 2PQ.
(iii) PR = PQ + QR = PQ + AQ (from (i)) = PQ + (AP + PQ) = AP + 2PQ = AP + AP = 2AP (using (ii)).

Also asked in: 2026 Standard 30/5/2, 2026 Standard 30/5/3

Q345 marksLong AnswerStatistics

The mean of the following frequency distribution is 28. If sum of all frequencies is 100, then find the values of p and q :

Class Interval	0 – 10	10 – 20	20 – 30	30 – 40	40 – 50	50 – 60
Frequency	12	p	27	20	q	6

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Answer: p = 18, q = 17

Sum of frequencies: $12 + p + 27 + 20 + q + 6 = 100$ , so $p + q = 35$ ... (1)
Class marks: 5, 15, 25, 35, 45, 55
$\sum f_{i} x_{i} = 60 + 15 p + 675 + 700 + 45 q + 330 = 1765 + 15 p + 45 q$
Mean $= \frac{1765 + 15 p + 45 q}{100} = 28$ , so $15 p + 45 q = 1035$ , i.e. $p + 3 q = 69$ ... (2)
(2) − (1): $2 q = 34$ , so $q = 17$ and $p = 18$ .

Q34 (OR) (OR)5 marksLong AnswerStatistics

Find median and mode of the following distribution :

Class Interval	0 – 15	15 – 30	30 – 45	45 – 60	60 – 75	75 – 90	90 – 105
Frequency	15	10	12	9	8	10	6

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Answer: Median = 42.5; mode = 11.25

N = 15 + 10 + 12 + 9 + 8 + 10 + 6 = 70, $\frac{N}{2} = 35$
Cumulative frequencies: 15, 25, 37, 46, 54, 64, 70
Median class 30 – 45: l = 30, cf = 25, f = 12, h = 15
Median $= 30 + \frac{35 - 25}{12} \times 15 = 30 + 12.5 = 42.5$
Modal class 0 – 15 (highest frequency 15): l = 0, $f_{1} = 15$ , $f_{0} = 0$ , $f_{2} = 10$ , h = 15
Mode $= 0 + \frac{15 - 0}{2 ( 15 ) - 0 - 10} \times 15 = \frac{15}{20} \times 15 = 11.25$

Q355 marksLong AnswerCircles

PQ and PR are two tangents to a circle with centre O and radius 5 cm. AB is another tangent to the circle at C which lies on OP. If OP = 13 cm, then find the length AB and PA.

Diagram for CBSE 2026 Class 10 Maths question 35

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Answer: AB =

\frac{20}{3}

cm, PA =

\frac{26}{3}

$O Q ⊥ P Q$ , so $P Q = O P^{2} - O Q^{2} = 169 - 25 = 12$ cm.
C lies on OP and OC = 5 cm, so PC = 13 − 5 = 8 cm. $A B ⊥ O P$ at C.
Let AC = x. Tangents from A are equal, so AQ = AC = x and PA = 12 − x.
In right $△ A C P$ : $P A^{2} = A C^{2} + P C^{2}$
$(12 - x)^{2} = x^{2} + 64$ , so $144 - 24 x = 64$ and $x = \frac{10}{3}$ cm.
PA $= 12 - \frac{10}{3} = \frac{26}{3}$ cm
By symmetry BC = AC, so AB = 2x $= \frac{20}{3}$ cm.

Also asked in: 2026 Standard 30/5/2, 2026 Standard 30/5/3

Q364 marksCase StudyPolynomials

An arch of a railway bridge, built on Chenab riverbed, is shown in the above diagram. It is a parabolic arch connecting two hills at P and Q. If the parabolic curve is represented by the polynomial $p (x) = - 0.0025 x^{2} - 0.025 x + 136$ .
Observe the diagram and based on above information, answer the following questions :
(i) Write the co-ordinates of point A. (1)
(ii) Find the span of the arch. (1)
(iii) (a) Write the zeroes of the polynomial using diagram and verify the relationship between sum of zeroes and polynomials. (2)
OR (iii) (b) Find the values of $p (x)$ at $x = 100$ and $x = - 100$ . Are they same ? (2)

Diagram for CBSE 2026 Class 10 Maths question 36

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Answer: (i) A(0, 136) (ii) 467 m (iii) (a) Zeroes 228.5 and −238.5; sum = −10 =

- \frac{b}{a}

OR (iii) (b) p(100) = 108.5, p(−100) = 113.5; not the same

(i) A lies on the y-axis, so x = 0 and $p (0) = 136$ . A = (0, 136).
(ii) Span = PQ = 228.5 − (−238.5) = 467 m.
(iii) (a) From the diagram the zeroes are 228.5 and −238.5.
Sum of zeroes = 228.5 + (−238.5) = −10.
$- \frac{b}{a} = - \frac{- 0.025}{- 0.0025} = - 10$ . Hence sum of zeroes $= - \frac{b}{a}$ , verified.
(iii) (b) $p (100) = - 0.0025 (10000) - 0.025 (100) + 136 = - 25 - 2.5 + 136 = 108.5$
$p (- 100) = - 25 + 2.5 + 136 = 113.5$
The values are not the same.

Also asked in: 2026 Standard 30/5/2, 2026 Standard 30/5/3

Q374 marksCase StudySurface Areas and Volumes

A wall mounted lamp, made of fabric, is shown below. Lamp has cuboidal shape, open from top and bottom. A spherical bulb of diameter 7 cm is latched with a very thin rod. (Ignore the rod while making calculations.)
Dimensions of the cuboid are 24 cm × 12 cm × 17 cm.
(i) Find the surface area of the bulb. (1)
(ii) What could be the maximum diameter of the bulb if at least 1 cm space is left from each side ? (1)
(iii) (a) Find the area of the fabric used if there is a fold of 2 cm on top and bottom edges. (2)
OR (iii) (b) Find the space available inside the lamp. (2)

Diagram for CBSE 2026 Class 10 Maths question 37

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Answer: (i) 154

c m^{2}

(ii) 10 cm (iii) (a) 1512

c m^{2}

OR (iii) (b) 4896

c m^{3}

(lamp volume); 4716.33

c m^{3}

after excluding the bulb

(i) r = 3.5 cm. Surface area $= 4 π r^{2} = 4 \times \frac{22}{7} \times 3.5 \times 3.5 = 154$ $c m^{2}$
(ii) The smallest dimension of the lamp is 12 cm. Leaving 1 cm on each side, maximum diameter = 12 − 2 = 10 cm.
(iii) (a) The fabric covers the four vertical faces (open top and bottom). Height with folds = 17 + 2 + 2 = 21 cm.
Area $= 2 (24 + 12) \times 21 = 72 \times 21 = 1512$ $c m^{2}$
(iii) (b) Volume of the lamp $= 24 \times 12 \times 17 = 4896$ $c m^{3}$
Volume of bulb $= \frac{4}{3} \times \frac{22}{7} \times (3.5)^{3} = \frac{539}{3} \approx 179.67$ $c m^{3}$
Space available (excluding the bulb) $= 4896 - 179.67 = 4716.33$ $c m^{3}$

Also asked in: 2026 Standard 30/5/2, 2026 Standard 30/5/3

Q384 marksCase StudySome Applications of Trigonometry

Elevated water storage tanks are built to store and supply water to nearby colonies. In the diagram given above, AB is an elevated water tank and CD is a nearby multistorey building. The building is 54 metres away from the water tank.
From a window (W) of the building, the angle of elevation of top of the tank is $4 5^{\circ}$ and angle of depression of its foot is $3 0^{\circ}$ .
(i) Write a relation between d (the height of window) and y. (1)
(ii) Determine the value of h. (1)
(iii) (a) Determine height of the water tank. (2)
OR (iii) (b) Find the value of $x$ and height of the window above ground level. (2)

Diagram for CBSE 2026 Class 10 Maths question 38

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Answer: (i) y = 2d (ii) h = 54 m (iii) (a)

54 + 183

m ≈ 85.18 m OR (iii) (b)

x = 542

m, window height

= 183

m ≈ 31.18 m

XW = AC = 54 m, $∠ B W X = 4 5^{\circ}$ , $∠ X W A = 3 0^{\circ}$ , AX = WC = d.
(i) In right $△ A X W$ : $sin 3 0^{\circ} = \frac{A X}{A W} = \frac{d}{y}$ , so $y = 2 d$ .
(ii) In right $△ B X W$ : $tan 4 5^{\circ} = \frac{h}{54}$ , so h = 54 m.
(iii) (a) $tan 3 0^{\circ} = \frac{d}{54}$ , so $d = \frac{54}{3} = 183$ m.
Height of tank AB $= h + d = 54 + 183 \approx 54 + 31.18 = 85.18$ m
(iii) (b) $cos 4 5^{\circ} = \frac{54}{x}$ , so $x = 542$ m (≈ 76.37 m).
Height of window $= d = 183 \approx 31.18$ m

Also asked in: 2026 Standard 30/5/2, 2026 Standard 30/5/3

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