CBSE Class 10 Sample Question Papers 2026-27

SUBJECT: MATHEMATICS BASIC (241)
SAMPLE QUESTION PAPER
CLASS - X (2026 - 27)

Section-A (MCQ- 20 MARKS)

Section A – Solutions

Q1. If HCF (2520, 6600) = 40 and LCM (2520, 6600) = 252 × k, then the value of k is

Given:

HCF (2520, 6600) = 40

LCM (2520, 6600) = 252 × k

We know that:

HCF × LCM = Product of the two numbers

Therefore,

40 × 252 × k = 2520 × 6600

Dividing both sides by 40 × 252, we get

k = (2520 × 6600) / (40 × 252)

k = 1650

Correct Answer: (D) 1650

Q2. If α and β are the zeroes of the polynomial 2x² − 4x − 5, find (α − β)².

Solution:

Given polynomial:

2x² − 4x − 5

Comparing with ax² + bx + c, we get

a = 2, b = −4 and c = −5

We know that:

α + β = −b/a = 4/2 = 2

αβ = c/a = −5/2

Using the identity:

(α − β)² = (α + β)² − 4αβ

= 2² − 4(−5/2)

= 4 + 10

= 14

Correct Answer: (C) 14

Q3. For what value of 𝑝 does the pair of Linear equations 4𝑥 + 𝑝𝑦 + 8 = 0 and 2𝑥 +2𝑦+2=0 has a unique solution

4x + py + 8 = 0

2x + 2y + 2 = 0

Solution:

For a pair of linear equations:

a₁x + b₁y + c₁ = 0

a₂x + b₂y + c₂ = 0

The condition for a unique solution is:

a₁/a₂ ≠ b₁/b₂

Here,

a₁ = 4, b₁ = p

a₂ = 2, b₂ = 2

Therefore,

4/2 ≠ p/2

2 ≠ p/2

p ≠ 4

Correct Answer: (C) p ≠ 4

Q4.The sum of the numerator and denominator of a fraction is 11. If the denominator is increased by 1, the fraction becomes 1/2 , then the fraction is

Given:

The sum of the numerator and denominator is 11.

When the denominator is increased by 1, the fraction becomes 1/2.

Solution:

Let the numerator be x and the denominator be y.

According to the question,

x + y = 11   ...(1)

Also,

x/(y + 1) = 1/2

2x = y + 1   ...(2)

From equation (1),

y = 11 − x

Substituting this value in equation (2),

2x = 11 − x + 1

3x = 12

x = 4

Now,

y = 11 − 4 = 7

Hence, the required fraction is 4/7.

Correct Answer: (C) 4/7

Q5. If one root of the quadratic equation 𝑎𝑥² + 𝑏𝑥 + 𝑐 = 0 is the reciprocal of the other, then

Given:

ax² + bx + c = 0

Let its roots be α and β.

Since one root is the reciprocal of the other,

αβ = 1

We know that the product of the zeroes of a quadratic equation is:

αβ = c/a

Therefore,

c/a = 1

c = a

Correct Answer: (D) a = c

Q6. The first term of AP is 𝑝 and the common difference is 𝑞 , then its 10th term is

Solution:

We know that the nth term of an arithmetic progression is:

a_n = a + (n − 1)d

Here,

a = p, d = q and n = 10

Therefore,

a₁₀ = p + (10 − 1)q

= p + 9q

Correct Answer: (C) p + 9q

Q7. Which term of the AP 21, 42, 63, 84, ... is 210?

Solution:

Here,

First term, a = 21

Common difference, d = 42 − 21 = 21

Let the nth term be 210.

Using the formula:

a_n = a + (n − 1)d

210 = 21 + (n − 1)21

210 − 21 = 21(n − 1)

189 = 21(n − 1)

9 = n − 1

n = 10

Hence, 210 is the 10th term of the AP.

Correct Answer: (B) 10th term

Q8. If the point P (5, 2) divides the line segment joining A( 8, 5) and B( 4, 𝒚) in the ratio 3 : 1, then the value of 𝒚 is

Solution:

Given:

A(8, 5), B(4, y) and P(5, 2)

AP : PB = 3 : 1

Using the section formula,

P = ((mx₂ + nx₁)/(m + n), (my₂ + ny₁)/(m + n))

For the y-coordinate,

2 = (3y + 1 × 5)/(3 + 1)

2 = (3y + 5)/4

8 = 3y + 5

3y = 3

y = 1

Correct Answer: (D) 1

Q9.Find the probability of selecting a vowel that occurs the maximum number of times in the word INDEPENDENCE.

Solution:

The word INDEPENDENCE contains 12 letters.

The vowels in the word are:

I, E, E, E, E

Number of vowels = 5

The vowel E occurs the maximum number of times, that is, 4 times.

Therefore,

Probability = (Number of favourable outcomes)/(Total number of outcomes)

= 4/12

= 1/3

Correct Answer: (C) 1/3

Q10. Find the lower limit of the median class.

Given distribution:

Solution:

First, find the total frequency.

N = 10 + 15 + 12 + 20 + 9

N = 66

Therefore,

N/2 = 66/2 = 33

Now, calculate the cumulative frequencies.

The first cumulative frequency greater than or equal to 33 is 37.

Therefore, the median class is 10–15.

Its lower limit is 10.

Correct Answer: (B) 10

Q11. The length of a tangent drawn from a point at a distance of 10 cm from the centre of the circle is 8 cm. The radius of the circle is

Solution:

Let O be the centre of the circle, P be the external point and A be the point of contact.

Given:

OP = 10 cm

PA = 8 cm

The radius is perpendicular to the tangent at the point of contact.

Therefore, ∠OAP = 90°.

Using Pythagoras' theorem in right-angled triangle OAP,

OP² = OA² + PA²

10² = OA² + 8²

100 = OA² + 64

OA² = 36

OA = 6 cm

Correct Answer: (C) 6 cm

Q12. Mean and median of certain data are 32 and 30 respectively. Using empirical formula, the value of mode is

Solution:

Given:

Mean = 32

Median = 30

Using the empirical relationship:

Mode = 3 × Median − 2 × Mean

Substituting the given values,

Mode = 3 × 30 − 2 × 32

= 90 − 64

= 26

Correct Answer: (B) 26

Q13. A ladder 15 m long reaches a window 12 m above the ground. The distance of the foot of the ladder from the base of the wall is

Solution:

Let the distance of the foot of the ladder from the wall be x metres.

Given:

Length of ladder = 15 m

Height of window = 12 m

The ladder, wall and ground form a right-angled triangle.

Using Pythagoras' theorem,

15² = 12² + x²

225 = 144 + x²

x² = 81

x = 9 m

Correct Answer: (B) 9 m

Q14. Find the value of sin² 60° − 2 tan² 45° − cos² 30°.

Solution:

We know that:

sin 60° = √3/2

tan 45° = 1

cos 30° = √3/2

Therefore,

sin² 60° − 2 tan² 45° − cos² 30°

= (√3/2)² − 2(1)² − (√3/2)²

= 3/4 − 2 − 3/4

= −2

Correct Answer: (A) −2

Q15. The perimeter of two similar triangles is 28 cm and 35 cm respectively. If one side of the first triangle is 8cm, then the corresponding side of the second triangle is

Solution:

Given:

Perimeter of the first triangle = 28 cm

Perimeter of the second triangle = 35 cm

Corresponding side of the first triangle = 8 cm

We know that the ratio of the corresponding sides of two similar triangles is equal to the ratio of their perimeters.

Let the corresponding side of the second triangle be x cm.

Therefore,

x/8 = 35/28

x = (8 × 35)/28

x = 10 cm

Correct Answer: (A) 10 cm

Q16. If in ∆ABC and ∆PQR , ∠B = ∠Q, ∠R = ∠C and AB = 2PQ, then the two triangles are

Given:

∠B = ∠Q

∠R = ∠C

AB = 2PQ

Solution:

In triangles ABC and PQR,

∠B = ∠Q

∠C = ∠R

Therefore, by the AA similarity criterion,

△ABC ~ △PQR

Also, AB = 2PQ.

Thus, the corresponding sides are not equal, so the triangles are not congruent.

Hence, the triangles are similar but not congruent.

Correct Answer: (B) Similar but not congruent

Q17. If tangents PA and PB from a point P to a circle with centre O and respective point of contacts as A and B, are inclined to each other at angle of 80°, then ∠AOB is equal to

Given:

∠APB = 80°

Solution:

OA and OB are radii of the circle.

Since the radius is perpendicular to the tangent at the point of contact,

∠OAP = 90°

∠OBP = 90°

In quadrilateral AOBP,

∠AOB + ∠OAP + ∠OBP + ∠APB = 360°

∠AOB + 90° + 90° + 80° = 360°

∠AOB + 260° = 360°

∠AOB = 100°

Correct Answer: (D) 100°

Q18. Two cubes each of volume 64cm³ are joined end to end to form a cuboid. The total surface area of the resulting cuboid is

Solution:

Volume of each cube = 64 cm³

Let the side of each cube be a cm.

a³ = 64

a = 4 cm

When two cubes are joined end to end, the dimensions of the resulting cuboid are:

Length = 4 + 4 = 8 cm

Breadth = 4 cm

Height = 4 cm

We know that the total surface area of a cuboid is:

TSA = 2(lb + bh + hl)

Substituting the values,

TSA = 2(8 × 4 + 4 × 4 + 4 × 8)

= 2(32 + 16 + 32)

= 2 × 80

= 160 cm²

Correct Answer: (B) 160 cm²

Q19. Assertion–Reason Based Question

Assertion (A): The probability of getting the number 8 on rolling a die is zero.

Reason (R): The probability of an impossible event is zero.

Solution:

A standard die has six faces numbered 1, 2, 3, 4, 5 and 6.

Therefore, getting the number 8 is an impossible event.

Hence, the probability of getting 8 is zero.

The assertion is true.

The reason is also true, and it correctly explains the assertion.

Correct Answer: (A) Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).

Q20. Assertion–Reason Based Question

Assertion (A): If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.

Reason (R): A line drawn from the midpoint of one side of a triangle, parallel to another side, bisects the third side.

Solution:

The assertion is true. It is the converse of the Basic Proportionality Theorem (BPT).

The reason is also true. It states the midpoint theorem for a triangle.

However, the reason does not directly explain the assertion. It is a different theorem.

Hence, both statements are true, but the reason is not the correct explanation of the assertion.

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

Section B – Solutions

Q21 (A). Find the smallest number which, when increased by 17, is exactly divisible by both 520 and 468.

Solution:

We need to find the smallest number which becomes divisible by both 520 and 468 after adding 17 to it.

First, we find the LCM of 520 and 468 using prime factorisation.

520 = 2 × 2 × 2 × 5 × 13

520 = 2³ × 5 × 13

468 = 2 × 2 × 3 × 3 × 13

468 = 2² × 3² × 13

Taking the highest powers of all the prime factors,

LCM (520, 468) = 2³ × 3² × 5 × 13

= 8 × 9 × 5 × 13

= 4680

Let the required number be x.

Then,

x + 17 = 4680

x = 4680 − 17

x = 4663

Answer: The required smallest number is 4663.

Q21 (B). Show that 25ⁿ will never end with the digit zero for any natural number n.

Proof:

We know that a number ending in zero must be divisible by 10.

Also,

10 = 2 × 5

Therefore, a number ending in zero must have both 2 and 5 as its prime factors.

Now,

25ⁿ = (5 × 5)ⁿ

= 5²ⁿ

Thus, the prime factorisation of 25ⁿ contains only the prime factor 5 and does not contain the factor 2.

Hence, 25ⁿ is not divisible by 10.

Therefore, 25ⁿ will never end with the digit zero for any natural number n.

Hence proved.

Q22. If 3 sin θ = 4 cos θ, find the value of sin θ + cos² θ − 1, where 0° < θ < 90°.

Solution:

Given,

3 sin θ = 4 cos θ

Dividing both sides by 3 cos θ, we get

sin θ / cos θ = 4/3

Therefore,

tan θ = 4/3

We know that tan θ = Perpendicular / Base.

Let the perpendicular be 4 units and the base be 3 units.

Using Pythagoras' theorem,

Hypotenuse = √(4² + 3²)

= √(16 + 9)

= √25

= 5 units

Therefore,

sin θ = 4/5

cos θ = 3/5

Now,

sin θ + cos² θ − 1

= 4/5 + (3/5)² − 1

= 4/5 + 9/25 − 1

= 20/25 + 9/25 − 25/25

= 4/25

Answer: 4/25

Q23. If one zero of the quadratic polynomial 2x² − 3x + p is 3, find the value of p. Also, find the other zero.

Solution:

Let the given polynomial be

f(x) = 2x² − 3x + p

It is given that one of its zeroes is 3.

Therefore,

f(3) = 0

Substituting x = 3, we get

2(3)² − 3(3) + p = 0

18 − 9 + p = 0

9 + p = 0

p = −9

Now, the polynomial becomes

2x² − 3x − 9

Let the other zero be β.

We know that the product of the zeroes of a quadratic polynomial ax² + bx + c is c/a.

Therefore,

3 × β = −9/2

β = −9/6

β = −3/2

Answer:

Value of p = −9

Other zero = −3/2

Q24. In the given figure, PA is a common tangent and QB and PC are tangents from Q and P to the smaller and larger circles, respectively. If QB = 5 cm and PC = 9 cm, find the length of PQ.

Solution:

Given:

QB = 5 cm

PC = 9 cm

We know that the lengths of tangents drawn from an external point to a circle are equal.

For the smaller circle, Q is an external point.

Therefore,

QA = QB

QA = 5 cm   ...(1)

For the larger circle, P is an external point.

Therefore,

PA = PC

PA = 9 cm   ...(2)

From the given figure, A, Q and P lie on the same straight line, with Q between A and P.

Hence,

PA = QA + QP

Substituting the values,

9 = 5 + QP

QP = 9 − 5

QP = 4 cm

Answer: PQ = 4 cm.

Q24. For Visually Impaired Candidates

Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle.

Solution:

Given:

Radius of the larger circle, R = 5 cm

Radius of the smaller circle, r = 3 cm

Let AB be the chord of the larger circle which touches the smaller circle at P.

Let O be the common centre of the two circles.

Since the chord AB touches the smaller circle at P, the radius OP is perpendicular to AB.

Therefore, OP ⊥ AB.

The perpendicular drawn from the centre of a circle to a chord bisects the chord.

Hence,

AP = PB

Let AP = x cm.

In right-angled triangle OPA,

OA² = OP² + AP²

5² = 3² + x²

25 = 9 + x²

x² = 16

x = 4 cm

Therefore,

AB = AP + PB

= 4 + 4

= 8 cm

Answer: The length of the chord is 8 cm.

Q25 (A). Points A(3, 1), B(5, 1), C(a, b) and D(4, 3) are the vertices of a parallelogram ABCD. Find the values of a and b.

Solution:

Given:

A(3, 1), B(5, 1), C(a, b) and D(4, 3)

We know that the diagonals of a parallelogram bisect each other.

Therefore, the midpoint of diagonal AC is equal to the midpoint of diagonal BD.

Using the midpoint formula,

Midpoint of AC = ((3 + a)/2, (1 + b)/2)

Midpoint of BD = ((5 + 4)/2, (1 + 3)/2)

Equating the corresponding coordinates,

(3 + a)/2 = 9/2

3 + a = 9

a = 6

Similarly,

(1 + b)/2 = 4/2

1 + b = 4

b = 3

Answer: a = 6 and b = 3.

Q25 (B). Find a linear relation between x and y such that P(x, y) is equidistant from the points A(1, 4) and B(−1, 2).

Solution:

Given:

A(1, 4), B(−1, 2) and P(x, y)

Since P is equidistant from A and B,

PA = PB

Using the distance formula,

√[(x − 1)² + (y − 4)²] = √[(x + 1)² + (y − 2)²]

Squaring both sides,

(x − 1)² + (y − 4)² = (x + 1)² + (y − 2)²

Expanding the terms,

x² − 2x + 1 + y² − 8y + 16 = x² + 2x + 1 + y² − 4y + 4

Cancelling the common terms and simplifying,

−2x − 8y + 17 = 2x − 4y + 5

−4x − 4y + 12 = 0

x + y − 3 = 0

Answer: The required linear relation is x + y = 3.

Section C – Solutions

Q26. Given that √5 is irrational, prove that 2 + 3√5 is irrational.

Proof:

We are given that √5 is an irrational number.

Let us assume, to the contrary, that 2 + 3√5 is rational.

Then, we can write

2 + 3√5 = r

where r is a rational number.

Rearranging,

3√5 = r − 2

√5 = (r − 2)/3

Since r is rational, r − 2 is also rational.

Dividing a rational number by a non-zero rational number gives a rational number. Therefore, (r − 2)/3 is rational.

This implies that √5 is rational, which contradicts the given fact that √5 is irrational.

Hence, our assumption is false.

Therefore, 2 + 3√5 is irrational.

Hence proved.

Q27 (A). In the given figure, AX, AY and BC are tangents to a circle with P, R and Q as the respective points of contact. Prove that AP = 1/2 (AB + BC + CA).

Proof:

Given: AX, AY and BC are tangents to the circle, touching it at P, R and Q, respectively.

We know that the lengths of tangents drawn from an external point to a circle are equal.

From external point A,

AP = AR   ...(1)

From external point B,

BP = BQ   ...(2)

From external point C,

CR = CQ   ...(3)

From the given figure,

AP = AB + BP

Similarly,

AR = AC + CR

Also,

BC = BQ + CQ

Using equations (2) and (3),

BC = BP + CR

Now, adding AB, BC and CA,

AB + BC + CA = AB + (BP + CR) + CA

= (AB + BP) + (AC + CR)

= AP + AR

From equation (1), AP = AR.

Therefore,

AB + BC + CA = AP + AP

AB + BC + CA = 2AP

AP = 1/2 (AB + BC + CA)

Hence proved.

Q27. For Visually Impaired Candidates

Question: Prove that a parallelogram circumscribing a circle is a rhombus.

Proof:

Let ABCD be a parallelogram circumscribing a circle.

We know that the lengths of tangents drawn from an external point to a circle are equal.

Therefore, the sum of the lengths of opposite sides of a quadrilateral circumscribing a circle is equal.

Hence,

AB + CD = BC + AD   ...(1)

Since ABCD is a parallelogram, its opposite sides are equal.

Therefore,

AB = CD

BC = AD

Substituting these values in equation (1),

AB + AB = BC + BC

2AB = 2BC

AB = BC

But in a parallelogram,

AB = CD and BC = AD.

Therefore,

AB = BC = CD = DA

Hence, all four sides of the parallelogram are equal.

Therefore, ABCD is a rhombus.

Hence proved.

Q27 (B). Two tangents PA and PB are drawn to a circle with centre O from an external point P. Prove that ∠APB = 2∠OAB.

Proof:

Given: PA and PB are tangents to a circle with centre O, touching it at A and B, respectively.

Join OA and OB.

We know that the radius is perpendicular to the tangent at the point of contact.

Therefore,

∠OAP = 90°

∠OBP = 90°

In quadrilateral OAPB,

∠OAP + ∠APB + ∠PBO + ∠AOB = 360°

90° + ∠APB + 90° + ∠AOB = 360°

∠APB + ∠AOB = 180°

∠APB = 180° − ∠AOB   ...(1)

Now, in triangle OAB,

OA = OB (Radii of the same circle)

Therefore, △OAB is an isosceles triangle.

Hence,

∠OAB = ∠OBA

Using the angle sum property of a triangle,

∠OAB + ∠OBA + ∠AOB = 180°

2∠OAB + ∠AOB = 180°

2∠OAB = 180° − ∠AOB   ...(2)

From equations (1) and (2),

∠APB = 2∠OAB

Hence proved.

Q28. Determine the ratio in which the point (−6, y) divides the line segment joining A(−3, −1) and B(−8, 9). Also, find the value of y.

Solution:

Let P(−6, y) divide AB in the ratio m : n.

Therefore,

AP : PB = m : n

Given:

A(−3, −1), B(−8, 9) and P(−6, y)

Using the section formula,

P = ((mx₂ + nx₁)/(m + n), (my₂ + ny₁)/(m + n))

Comparing the x-coordinates,

−6 = [m(−8) + n(−3)]/(m + n)

−6(m + n) = −8m − 3n

−6m − 6n = −8m − 3n

2m = 3n

m/n = 3/2

Therefore,

AP : PB = 3 : 2

Now, using the section formula for the y-coordinate,

y = [m(9) + n(−1)]/(m + n)

Substituting m = 3 and n = 2,

y = [3(9) + 2(−1)]/(3 + 2)

= (27 − 2)/5

= 25/5

= 5

Answer:

Ratio = 3 : 2

y = 5

Q29. Consider △ACB, right-angled at C, in which AB = 29 units, BC = 21 units and ∠ABC = θ. Determine the values of:

(i) 1 + tan² θ

(ii) cos² θ − sin² θ

Solution:

Given:

AB = 29 units

BC = 21 units

∠ACB = 90° and ∠ABC = θ

First, we find AC using Pythagoras' theorem.

AB² = AC² + BC²

29² = AC² + 21²

841 = AC² + 441

AC² = 400

AC = 20 units

For angle θ,

Perpendicular = AC = 20 units

Base = BC = 21 units

Hypotenuse = AB = 29 units

(i) Find 1 + tan² θ

We know that,

tan θ = Perpendicular / Base

tan θ = 20/21

Therefore,

1 + tan² θ = 1 + (20/21)²

= 1 + 400/441

= (441 + 400)/441

= 841/441

Answer: 841/441

(ii) Find cos² θ − sin² θ

We know that,

cos θ = Base / Hypotenuse = 21/29

sin θ = Perpendicular / Hypotenuse = 20/29

Therefore,

cos² θ − sin² θ

= (21/29)² − (20/29)²

= 441/841 − 400/841

= 41/841

Answer: 41/841

Q30 (A). The mean of the following distribution is 48 and the sum of all the frequencies is 50. Find the missing frequencies x and y.

Solution:

We know that the class mark is the midpoint of a class interval.

Class mark = (Upper limit + Lower limit)/2

Also, the mean of grouped data is given by:

Mean = Σfx_i / Σf

Step 1: Find the relation between x and y.

Given that the sum of all frequencies is 50.

Therefore,

8 + 6 + x + 11 + y = 50

25 + x + y = 50

x + y = 25   ...(1)

Step 2: Use the given mean to form another equation.

Mean = 48

Σf = 50

Therefore,

48 = (200 + 210 + 45x + 605 + 65y)/50

48 × 50 = 1015 + 45x + 65y

2400 = 1015 + 45x + 65y

45x + 65y = 1385   ...(2)

Step 3: Solve equations (1) and (2).

From equation (1),

x = 25 − y

Substituting in equation (2),

45(25 − y) + 65y = 1385

1125 − 45y + 65y = 1385

20y = 260

y = 13

Now,

x + y = 25

x + 13 = 25

x = 12

Answer: x = 12 and y = 13.

Q30 (B). The distribution below gives the weights of 50 students of Class X. Find the modal weight of the students.

Solution:

The modal class is the class interval having the highest frequency.

Here, the highest frequency is 20.

Therefore, the modal class is 55–65.

We know that the mode of grouped data is given by:

Mode = l + [(f₁ − f₀)/(2f₁ − f₀ − f₂)] × h

Here,

l = 55

h = 10

f₁ = 20

f₀ = 10

f₂ = 12

Substituting the values,

Mode = 55 + [(20 − 10)/(2 × 20 − 10 − 12)] × 10

= 55 + [10/(40 − 22)] × 10

= 55 + (10/18) × 10

= 55 + 100/18

= 55 + 50/9

= 545/9

≈ 60.56 kg

Answer: The modal weight is approximately 60.56 kg.

Q31. The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.

Solution:

Let the tens digit be x and the units digit be y.

Then, the original two-digit number is:

10x + y

The number obtained by reversing the digits is:

10y + x

Step 1: Form the first equation.

The sum of the digits is 9.

Therefore,

x + y = 9   ...(1)

Step 2: Form the second equation.

According to the question, nine times the original number is twice the reversed number.

Therefore,

9(10x + y) = 2(10y + x)

90x + 9y = 20y + 2x

90x − 2x = 20y − 9y

88x = 11y

y = 8x   ...(2)

Step 3: Find the digits.

Substituting equation (2) in equation (1),

x + 8x = 9

9x = 9

x = 1

Now,

y = 8x

y = 8 × 1

y = 8

Therefore, the tens digit is 1 and the units digit is 8.

The required number is 18.

Verification:

Reversed number = 81

9 × 18 = 162

2 × 81 = 162

Both sides are equal.

Answer: The required two-digit number is 18.

Section D – Solutions

Q32. Prove that if a line is drawn parallel to one side of a triangle, intersecting the other two sides in distinct points, then the other two sides are divided in the same ratio.

Given:

In △ABC, a line DE is drawn parallel to BC, intersecting AB at D and AC at E.

DE || BC

To Prove:

AD/DB = AE/EC

Construction:

Join BE and CD.

Proof:

We know that the areas of two triangles having the same altitude are in the ratio of their corresponding bases.

Step 1: Consider △ADE and △BDE.

Both triangles have the same altitude from E to AB.

Therefore,

ar(△ADE)/ar(△BDE) = AD/DB   ...(1)

Step 2: Consider △ADE and △CDE.

Both triangles have the same altitude from D to AC.

Therefore,

ar(△ADE)/ar(△CDE) = AE/EC   ...(2)

Step 3: Consider △BDE and △CDE.

Both triangles have the same base DE and lie between the same parallel lines DE and BC.

Therefore,

ar(△BDE) = ar(△CDE)   ...(3)

Using equations (1), (2) and (3), we get

AD/DB = AE/EC

Hence, the line DE divides the two sides AB and AC in the same ratio.

Hence proved.

Q33 (A). A train travels a distance of 360 km at a uniform speed. If its speed had been 5 km/h more, it would have taken 1 hour less to cover the same distance. Find the speed of the train.

Solution:

Let the original speed of the train be x km/h.

Distance travelled = 360 km

Time taken = Distance / Speed

Therefore, the original time taken is 360/x hours.

If the speed were increased by 5 km/h, the new speed would be (x + 5) km/h.

The new time taken would be 360/(x + 5) hours.

According to the question, the train would take 1 hour less.

Therefore,

360/x − 360/(x + 5) = 1

Taking the LCM,

[360(x + 5) − 360x]/[x(x + 5)] = 1

1800/[x(x + 5)] = 1

x(x + 5) = 1800

x² + 5x − 1800 = 0

Step 2: Solve the quadratic equation.

x² + 45x − 40x − 1800 = 0

x(x + 45) − 40(x + 45) = 0

(x + 45)(x − 40) = 0

Therefore,

x + 45 = 0 or x − 40 = 0

x = −45 or x = 40

Since speed cannot be negative, we reject x = −45.

Hence, x = 40 km/h.

Verification:

At 40 km/h, time taken = 360/40 = 9 hours.

At 45 km/h, time taken = 360/45 = 8 hours.

The difference is 1 hour, as given in the question.

Answer: The speed of the train is 40 km/h.

Q33 (B). John and Jivanti together have 45 marbles. Both of them lost 5 marbles each, and the product of the number of marbles they now have is 124. Find the number of marbles they had in the beginning.

Solution:

Let John have x marbles and Jivanti have y marbles initially.

According to the question,

x + y = 45   ...(1)

Both lose 5 marbles each.

John's remaining marbles = x − 5

Jivanti's remaining marbles = y − 5

The product of their remaining marbles is 124.

Therefore,

(x − 5)(y − 5) = 124

Expanding,

xy − 5x − 5y + 25 = 124

xy − 5(x + y) = 99

Using equation (1),

xy − 5(45) = 99

xy − 225 = 99

xy = 324   ...(2)

Step 2: Form a quadratic equation.

From equations (1) and (2),

x + y = 45

xy = 324

Let x be one of the numbers of marbles.

Then, y = 45 − x.

Substituting in equation (2),

x(45 − x) = 324

45x − x² = 324

x² − 45x + 324 = 0

Factorising,

x² − 36x − 9x + 324 = 0

x(x − 36) − 9(x − 36) = 0

(x − 36)(x − 9) = 0

Therefore,

x = 36 or x = 9

If John had 36 marbles, Jivanti had 9 marbles.

Alternatively, John had 9 marbles and Jivanti had 36 marbles.

Verification:

After losing 5 marbles each, they have 31 and 4 marbles.

31 × 4 = 124

Also, 36 + 9 = 45.

Both conditions are satisfied.

Answer: Initially, one had 36 marbles and the other had 9 marbles.

Q34. The largest possible hemisphere is drilled out of a wooden cubical block of side 21 cm such that the base of the hemisphere is on one of the faces of the cube. Find:

(i) The volume of the wood left in the block.

(ii) The total surface area of the remaining solid.

Solution:

Given:

Side of the cube = 21 cm

The largest possible hemisphere is drilled out of the cube.

For the largest hemisphere, its circular base must fit exactly inside one face of the cube.

Therefore, the diameter of the hemisphere is equal to the side of the cube.

Diameter = 21 cm

Radius, r = 21/2 = 10.5 cm

(i) Volume of the wood left in the block

Volume of the cube = a³

= 21³

= 9261 cm³

Volume of a hemisphere = (2/3)πr³

Substituting r = 10.5 cm and π = 22/7,

Volume of hemisphere = (2/3) × (22/7) × (10.5)³

= (2/3) × (22/7) × (21/2)³

= (2/3) × (22/7) × (9261/8)

= 4851/2

= 2425.5 cm³

Volume of wood left = Volume of cube − Volume of hemisphere

= 9261 − 2425.5

= 6835.5 cm³

Answer (i): 6835.5 cm³

(ii) Total surface area of the remaining solid

The remaining solid has:

• The six original square faces of the cube, except for the circular area removed from one face.
• The curved surface of the hollow hemisphere.

Total surface area of the cube = 6a²

= 6 × 21²

= 6 × 441

= 2646 cm²

Area of the circular base removed = πr²

= (22/7) × (10.5)²

= 346.5 cm²

Curved surface area of a hemisphere = 2πr²

= 2 × 346.5

= 693 cm²

Therefore,

Total surface area of the remaining solid

= Surface area of cube − Area of circular opening + Curved surface area of hemisphere

= 2646 − 346.5 + 693

= 2992.5 cm²

Answer (ii): 2992.5 cm²

Q35 (A). To assist an ambulance, a drone is hovering over an accident site at a constant height on a highway. The angle of depression of the ambulance, approaching the accident site, is 30°. Twelve minutes later, the angle of depression is 60°. If the ambulance is moving at a constant speed, find the total time taken by the ambulance to reach the accident site.

Solution:

Let:

A be the accident site.

D be the position of the drone.

B be the initial position of the ambulance.

C be the position of the ambulance after 12 minutes.

Let the constant height of the drone above the highway be h metres.

Let AB = x metres and AC = y metres.

Since the ambulance is approaching the accident site, B, C and A lie on the same straight line, with C between B and A.

The angle of depression from the drone is equal to the angle of elevation from the ambulance.

Step 1: Find the initial horizontal distance.

In right-angled triangle DAB,

∠DAB = 90° and ∠DBA = 30°.

Using the tangent ratio,

tan 30° = h/x

1/√3 = h/x

x = √3h   ...(1)

Step 2: Find the horizontal distance after 12 minutes.

In right-angled triangle DAC,

∠DCA = 60°.

Using the tangent ratio,

tan 60° = h/y

√3 = h/y

y = h/√3   ...(2)

Step 3: Find the speed of the ambulance.

Distance travelled in 12 minutes = BC

= AB − AC

= x − y

Using equations (1) and (2),

BC = √3h − h/√3

= (3h − h)/√3

= 2h/√3 metres

Speed of ambulance = Distance / Time

= (2h/√3)/12

= h/(6√3) metres per minute

Step 4: Find the remaining time to reach the accident site.

After 12 minutes, the remaining distance is

AC = h/√3 metres.

Remaining time = Remaining distance / Speed

= (h/√3)/(h/(6√3))

= 6 minutes

Step 5: Find the total time.

Total time = Time already taken + Remaining time

= 12 + 6

= 18 minutes

Answer: The ambulance will take a total of 18 minutes from the first observation to reach the accident site.

Q35 (B). A statue standing on a pedestal is 1.6 m tall. From a point on the ground, the angles of elevation of the top and bottom of the statue are 60° and 45°, respectively. Find the height of the pedestal.

Solution:

Given:

Height of the statue = 1.6 m

Angle of elevation of the top of the statue = 60°

Angle of elevation of the bottom of the statue = 45°

Let the height of the pedestal be h metres.

Let the horizontal distance between the observation point and the pedestal be x metres.

The total height from the ground to the top of the statue is (h + 1.6) metres.

Step 1: Use the angle of elevation of the bottom of the statue.

In the right-angled triangle formed by the observation point and the bottom of the statue,

tan 45° = h/x

1 = h/x

x = h   ...(1)

Step 2: Use the angle of elevation of the top of the statue.

In the right-angled triangle formed by the observation point and the top of the statue,

tan 60° = (h + 1.6)/x

√3 = (h + 1.6)/x

Using x = h from equation (1),

√3 = (h + 1.6)/h

√3h = h + 1.6

√3h − h = 1.6

h(√3 − 1) = 1.6

h = 1.6/(√3 − 1)

Rationalising the denominator,

h = [1.6(√3 + 1)]/[(√3 − 1)(√3 + 1)]

h = 1.6(√3 + 1)/(3 − 1)

h = 1.6(√3 + 1)/2

h = 0.8(√3 + 1)

Using √3 ≈ 1.732,

h ≈ 0.8(1.732 + 1)

h ≈ 0.8 × 2.732

h ≈ 2.186 m

Answer: The height of the pedestal is 0.8(√3 + 1) m, or approximately 2.19 m.

Q.36. An Art and craft teacher prepared a fan which looked like the sector of a circle

using ice-cream sticks and black sheet as shown in the given figure:

Given:

Outer radius of the fan, R = 10 cm

Inner radius of the black sheet, r = 4 cm

Central angle, θ = 60°

(i) The angle subtended at the centre of the sector is 60⁰. Find the area of one face of the fan.

The area of a sector of a circle is given by:

Area = (θ/360°) × πR²

Substituting the given values,

Area = (60°/360°) × π × 10²

Area = (1/6) × 100π

Area of one face of the fan = 50π/3 cm²

(ii) With the given sector angle of 60⁰ at the centre, find area of black sheet used in the fan.

The black sheet forms a sector of a ring. Its area is equal to the difference between the areas of the two sectors.

Area of black sheet = (θ/360°) × π(R² − r²)

= (60°/360°) × π(10² − 4²)

= (1/6) × π(100 − 16)

= 84π/6

Area of black sheet = 14π cm²

(iii)(A) The boundary of the black sheet region on one face is to be covered by a designer tape. What length of tape is required?

The boundary consists of:

• The outer arc of radius 10 cm
• The inner arc of radius 4 cm
• Two straight sides, each of length (10 − 4) cm

Length of outer arc = (60°/360°) × 2π × 10

= 10π/3 cm

Length of inner arc = (60°/360°) × 2π × 4

= 4π/3 cm

Length of each straight side = 10 − 4 = 6 cm

Total length of tape = 10π/3 + 4π/3 + 6 + 6

= 14π/3 + 12

Taking π = 22/7,

= 44/3 + 12

= 80/3 cm

Required length of designer tape = 26⅔ cm (approximately)

OR

(iii) (B) If 29.6 cm of designer tape is required to cover the boundary of the black sheet region on one face of the fan, then evaluate the central angle of the sector.

Given: Total boundary length = 29.6 cm

Let the central angle be θ°.

The total boundary length is:

29.6 = (θ/360°) × 2π(10 + 4) + 2(10 − 4)

29.6 = (θ/360°) × 28π + 12

17.6 = 28πθ/360

θ = (17.6 × 360)/(28π)

Taking π = 22/7,

θ = (17.6 × 360)/88

θ = 72°

Q.36. For Visually Impaired Candidates

A motion sensor (or motion detector) is an electronic device that is designed to detect and measure movement. The motion detector can detect movement over a sector of angle θ = 70° to a distance of 24m.

Given: Central angle θ = 70° and radius r = 24 m.

(i) Derive a relation between length of arc(l) and the area of the sector(A) enclosed by it?

Length of arc, l = (θ/360°) × 2πr

Area of sector, A = (θ/360°) × πr²

Dividing the area by the arc length,

A/l = [(θ/360°) × πr²] / [(θ/360°) × 2πr]

A/l = r/2

Therefore, A = ½rl

(ii) How much area is monitored by the motion detector?

Area of sector = (θ/360°) × πr²

= (70°/360°) × π × 24²

= (7/36) × π × 576

Area monitored = 112π m²

(iii)(A) To increase the area to be monitored by 50% with same range of distance covered, what should be the angle?

For the same radius, the area of a sector is directly proportional to its central angle.

Let the new angle be θ.

New area = 150% of the original area

θ/70° = 150/100

θ = 70° × 3/2

New angle = 105°

OR

(iii)(B) For θ = 91°, what range of distance is required for the detector to monitor 30% more area?

The original area is 112π m².

Required area = 130% of 112π

= 1.3 × 112π

= 145.6π m²

Let the required radius be R metres.

Area of sector = (91°/360°) × πR²

Therefore,

(91/360) × πR² = 145.6π

R² = (145.6 × 360)/91

R² = 576

Required range = 24 m

Q.37. Delhi's pollution is a severe, often seasonal issue caused by a combination of factors like high vehicular and industrial emissions, dust from construction, and crop burning in surrounding states. In a school, students of classes I to XII thought of planting trees in and around the school to reduce air pollution.
It was decided that the number of trees that each section of each class will plant be three more than the double of the class in which they are studying, e.g., a section of class IV will plant 11 trees. If in the given school, there are two sections of each class, then answer the following questions:

The number of trees planted by each section of a class is three more than twice the class number.

Number of trees planted by each section of Class n = 2n + 3

There are two sections in each class, from Class I to Class XII.

(i) If number of trees planted by students class-wise follow arithmetic progression, then find the common difference.

For Class I, number of trees = 2(1) + 3 = 5

For Class II, number of trees = 2(2) + 3 = 7

For Class III, number of trees = 2(3) + 3 = 9

Thus, the arithmetic progression is:

5, 7, 9, 11, ...

Common difference, d = 7 − 5 = 2

Common difference = 2

(ii) What is the number of trees planted by students of class VI?

Number of trees planted by each section of Class VI:

= 2(6) + 3

= 15

Since there are two sections,

Total trees planted by Class VI = 2 × 15

Total = 30 trees

(iii) (A) What is the total number of trees planted by the students of given school?

The number of trees planted by each section forms an AP with:

First term, a = 5

Common difference, d = 2

Number of terms, n = 12

Last term, a₁₂ = a + (n − 1)d

= 5 + (12 − 1) × 2

= 5 + 22 = 27

Sum of n terms of an AP:

S_n = n/2 × (a + l)

Sum for one section of each class:

S₁₂ = 12/2 × (5 + 27)

= 6 × 32

= 192

There are two sections in each class.

Total number of trees = 2 × 192

Total = 384 trees

OR

(iii)(B) Identify the class that planted 34 trees.

Let the class number be n.

Each section plants (2n + 3) trees. Since there are two sections,

2(2n + 3) = 34

4n + 6 = 34

4n = 28

n = 7

Therefore, Class VII planted 34 trees in total.

Q.38. A labourer prints 400 T-shirts in a day. The supervisor checked the T-shirts and found that 312 prints were good, 54 prints were with minor defects and rest of the prints were of major defects.
Harish, a customer will buy a T-shirt only if it is good but a trader will buy, if it has no major defect. If a random T-shirt is picked, then

Given:

Total number of T-shirts = 400

Good prints = 312

Minor defects = 54

Number of T-shirts with major defects:

= 400 − (312 + 54)

= 400 − 366

= 34

We know that:

Probability of an event = Number of favourable outcomes / Total number of outcomes

(i) Find the probability that Harish will buy a T-shirt.

Harish will buy a T-shirt only if it is good.

Favourable outcomes = 312

P(Harish buys a T-shirt) = 312/400

= 39/50

Required probability = 39/50

(ii) Find the probability that a trader will buy the T-shirt.

A trader will buy a T-shirt if it has no major defect.

Number of T-shirts with no major defect:

= 400 − 34

= 366

P(Trader buys a T-shirt) = 366/400

= 183/200

Required probability = 183/200

(iii)(A) Find the probability that the T-shirt is not good.

Number of T-shirts that are not good:

= 400 − 312

= 88

P(T-shirt is not good) = 88/400

= 11/50

Required probability = 11/50

OR

(iii)(B) Find the probability that neither Harish nor the trader will buy the T-shirt.

Harish will not buy a T-shirt that is not good. The trader will not buy a T-shirt with a major defect.

Therefore, neither of them will buy a T-shirt only when it has a major defect.

Number of T-shirts with major defects = 34

P(Neither buys the T-shirt) = 34/400

= 17/200

Required probability = 17/200