CBSE Sample Papers for Class 10 Maths Standard Set 8 with Solutions

Students must start practicing the questions from CBSE Sample Papers for Class 10 Maths with Solutions Set 8 are designed as per the revised syllabus.

CBSE Sample Papers for Class 10 Maths Standard Set 8 with Solutions

Time Allowed: 3 Hours
Maximum Marks: 80

General Instructions:

This Question Paper has 5 Sections A-E.
Section A has 20 MCQs carrying 1 mark each
Section B has 5 questions carrying 02 marks each.
Section C has 6 questions carrying 03 marks each.
Section D has 4 questions carrying 05 marks each.
Section E has 3 case based integrated units of assessment (04 marks each) with sub-parts of the values of 1, 1 and 2 marks each respectively.
All Questions are compulsory. However, an internal choice in 2 Qs of 5 marks, 2 Qs of 3 marks and 2 Questions of 2 marks has been provided. An internal choice has been provided in the 2 marks questions of Section E
Draw neat figures wherever required. Take π =22/7 wherever required if not stated.

Section – A
(Section A consists of 20 questions of 1 mark each.)

Question 1.
A prime number greater than 91 but less than 100 is: (1)
(a) 94
(b) 97
(c) 96
(d) 95
Answer:
(b) 97

Explanation: prime number greater than 91 but less than 100 is 97.

Question 2.
Find a zero of the polynomial x³ – 8. (1)
(a) \( \sqrt{2} \)
(b) –\( \sqrt{2} \)
(c) 3
(d) 2
Answer:
(d) 2

Explanation: Here, x³ – 8 = 0 gives x³ = 8 i.e. x= \(\sqrt[3]{8}\) = 2

Question 3.
Write a quadratic polynomial whose sum of zeroes is \(\frac{-1}{4}\) and product of zeroes is \(\frac{1}{4}\) . (1)
(a) 4x² + x + 1
(b) x² + 4x – 1
(c) 2x² + 3x – 1
(d) x² – 2x + 1
Answer:
(a) 4x² + x + 1

Explanation:
Sum of zeroes = \(\frac{-1}{4}\)
Product of zeroes = \(\frac{1}{4}\)
∴ Quadratic Polynomial is
x² – (sum of zeroes) x + product of zeroes = 0
∴ x² – \(\frac{-1}{4}\) x + \(\frac{1}{4}\) = 0
⇒ 4x² + x + 1 = 0

Question 4.
Determine the roots of the equation S x² – 2x – 73 =0 (1)
(a) \(\sqrt{3},-\frac{1}{\sqrt{3}}\)
(b) 1, \(\frac{-1}{2}\)
(c) \(\sqrt{3}\), -1
(d) \(\frac{1}{\sqrt{3}}\), -1
Answer:
(a) \(\sqrt{3},-\frac{1}{\sqrt{3}}\)

Explanation:
The given equation has roots, whose sum is \(\frac{2}{\sqrt{3}}\) and product is -1. This is possible only with \(\sqrt{3}\) and \(-\frac{1}{\sqrt{3}}\).

Question 5.
Find the 15th term of the AP: x – 7, x – 2, x + 3….. (1)
(a) x + 67
(b) x + 4
(c) x + 36
(d) x + 63
Answer:
(d) x + 63

Explanation:
Here, a = x – 7, d = 5
So, 15th term = a + 14 d = (x – 7) + 14 (5)
= x + 63

Question 6.
Find The discriminant of the equation (x + 1)³ = 4 – 1 + x³ (1)
(a) 52
(b) 53
(c) 64
(d) 72
Answer:
(a) 52

Explanation:
Given, equation is :
(x + 1)³ = 4 – x + x³
⇒ x³ + 1 + 3x(x + 1) = 4 – x + x³
⇒ 1 + 3x² + 3x = 4 – x
⇒ 3x² + 4x – 3 = 0
On comparing this equation with ax² + bx + c = 0, we get a = 3, b = 4, c = – 3
∴ Discriminant, D = b² – 4ac
= 4² – 4 ×3 × (-3)
= 16 + 36
= 52

Question 7.
The next term of the A.P.: 3, 3 + \(\sqrt{2}\), 3 + 2 \(\sqrt{2}\), 3 + 3\(\sqrt{2}\) ,…. is: (1)
(a) 0
(b) 3 + 4\(\sqrt{2}\)
(c) 3 – 4\(\sqrt{2}\)
(d) 1
Answer:
(b) 3 + 4\(\sqrt{2}\)

Explanation:
Here,
d= \(\sqrt{2}\)
So, next term is (3+ 3\(\sqrt{2}\)) + \(\sqrt{2}\) i.e. 3 + 4\(\sqrt{2}\).

Question 8.
The value of x and y : x + 2y = 9, 2x – y = 8 is: (1)
(a) 0, 0
(b) 3, 5
(c) 5, 2
(d) 0,1
Answer:
(c) 5, 2

Explanation:
Here, x + 2y = 9 …(i)
2x – y = 8 …(ii)
Multiply equation (ii) by 2 and add both the equations.
x + 2y = 9
4x – 2 y = 16
5x = 25
x = \(\frac{25}{5}\) = 5
Then,
y = \(\frac{9 – x}{2}\) = \(\frac{9 – 5}{2}\) = 2

Question 9.
The condition for the points (a, 0), (0, b) and (1,1) to be collinear, is:
(a) \(\frac{1}{a} – \frac{1}{b}\) = 1
(b) \(\frac{1}{ab}\) = 1
(c) 0
(d) \(\frac{1}{a} + \frac{1}{b}\) = 1
Answer:
(d) \(\frac{1}{a} + \frac{1}{b}\) = 1

Explanation:
As the given points are collinear, a(b – 1) + 0 (1 – 0) + 1 (0 – b) = 0
⇒ ab – a – b = 0
⇒ ab = a + b
⇒ \(\frac{1}{a} + \frac{1}{b}\) = 1

Question 10.
The length of the altitude of an isosceles triangle of sides 6 cm, 6 cm and 4 cm is:
(a) 4\(\sqrt{2}\) cm
(b) 5\(\sqrt{2}\) cm
(c) 6\(\sqrt{2}\) cm
(d) 2\(\sqrt{2}\) cm
Answer:
(a) 4\(\sqrt{2}\) cm

Explanation:
Since, AD is an altitude in isosceles ∆ABC.

Question 11.
Find the coordinates of a point on y-axis which is equidistant from the points (6,5)and (-4, 3).
(a) (4, 2)
(b) (3, 2)
(c) (0,9)
(d) (9,2)
Answer:
(c) (0,9)

Explanation:
Let, the coordinate on y axis be P(0, y)
The given points are A(6, 5) and B(-4, 3)
∴ PA = PB
⇒ PA² = PB²
⇒ A (0 – 6)² + (y – 5)² = (0 + 4)² + (y – 3)²
⇒ 36 + y² + 25 – 10y = 16 + y² + 9 – 6y
⇒ – 4y = – 36
⇒ y = 9
Then, coordinate on y-axis is (0, 9).

Question 12.
From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. The radius of the circle is: (1)
(a) 7 cm
(b) 4 cm
(c) 9 cm
(d) 10 cm
Answer:
(a) 7 cm

Explanation:
Here, OP is a tangent to a circle with radius ‘r’ of length 24 cm.

And OQ = 25 cm
Now, ∠QPO = 90°, as tangent at any point to a circle is ⊥r to the radius.
∴ In AOPQ by pythagoras theorem
OQ² = OP² + PQ²
25² = r² + 24²
⇒ r² = 625 – 576
= 49
⇒ r = 7 cm

Question 13.
If 3cos A = 1, then find the value of cosec A. (1)
(a) \(\frac{1}{\sqrt{2}}\)
(b) \(\frac{5}{\sqrt{2}}\)
(c) \(\frac{2 \sqrt{2}}{3}\)
(d) \(\frac{3}{2 \sqrt{2}}\)
Answer:
(d) \(\frac{3}{2 \sqrt{2}}\)

Explanation:
Here, 3 cos A = 1
cos A = \(\frac{1}{3}\)
Then cos² A = \(\frac{1}{9}\)
Then, sin² A = 1 – cos² A
= 1 – \(\frac{1}{9}\) = \(\frac{8}{9}\)
Sin A = \(\frac{2 \sqrt{2}}{3}\)
Then cosec A = \(\frac{1}{sin A}\) = \(\frac{3}{2 \sqrt{2}}\)

Question 14.
The perimeter of a quadrant of a circle of radius V is: (1)
(a) \(\frac{r^{2}}{2}\)
(b) π+ 4
(c) \(\frac{r}{2}\) (π + 4)
(d) \(\frac{r}{2}\)
Answer:
(c) \(\frac{r}{2}\) (π + 4)

Explanation:
Perimeter of quadrant,

Perimeter = OB + OA + \(\overparen{\mathrm{AB}}\)
= r + r + \(\frac{2πr}{4}\)
= 2r + \(\frac{π}{2}\) r
= \(\frac{r}{2}\)(π + 4)

Question 15.
The total surface area of a quadrant of a sphere of radius V is:
(a) πr²
(b) 2πr²
(c) 2πr
(d) \(\frac{πr^{2}}{2}\)
Answer:
(b) 2πr²

Explanation:
Total surface area of quadrant
= \(\frac{4 \pi r^2}{4}+\frac{\pi r^2}{2}+\frac{\pi r^2}{2}\)
= πr² + πr²
= 2πr²

Question 16.
The probability of drawing at random a green coloured ball from a bag containing 6 red and 5 black balls is: (1)
(a) 0
(b) \(\frac{1}{2}\)
(c) \(\frac{3}{5}\)
(d) 1
Answer:
(a) 0

Explanation:
Explanation: From the 11(6+5) balls in the bag, no ball is of green colour.
P(a green ball)=\(\frac{0}{11}\) i.e. 0

Question 17.
The median of the first 50 even natural numbers is: (1)
(a) 48
(b) 49
(c) 50
(d) 51
Answer:
(d) 51

Explanation:
First 50 even natural number are:
2, 4, 6, …………, 98,100.
As median is the middle-most value,
median = \(\frac{50 + 52}{2}\) = 51

Question 18.
The value of \(\frac{1+\tan ^2 \theta}{1+\cot ^2 \theta}\) is:
(a) tan θ
(b) cos² θ
(c) 1 + sin θ
(d) 0
Answer:
(a) tan θ

Explanation:
\(\frac{1+\tan ^2 \theta}{1+\cot ^2 \theta}\)

DIRECTION: In the question number 19 and 20, a statement of assertion (A) is followed by a statement of reason (R).
Choose the correct option as:
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A)
(b) Both assertion (A) and reason (R) are true and reason (R) is not the correct explanation of assertion (A)
(c) Assertion (A) is true but reason (R) is false.
(d) Assertion (A) is false but reason (R) is true.

Question 19.
Statement A (Assertion): There are infinite number of tines which passes through (1, 13).
Statement R (Reason): A Linear equation in two variabLes has infiniteLy many soLutions. (1)
Answer:
(b) Both assertion (A) and reason (R.) are true and reason (T) is not the correct explanation of assertion (A)

Explanation:
Through o point infinite tines can be drown. Through (1,13), infinite number of tines can be drawn.
Also o Line has infinite points on it, hence a linear equation representing a tine has infinite solutions.

Question 20.
Statement A (Assertion): Two simiLar triangLes are always congruent.
Statement R (Reason): Two similar triangles are said to be congruent if their areas are equaL (1)
Answer:
(d) Assertion (A) is false but reason(R) is true.

Explanation: Two similar triangles are not congruent generally.
If the area of two similar triangles are equal then the triangles ore congruent.

SECTION – B
(Section B consists of 5 questions of 2 marks each.)

Question 21.
Write the prime factorisation of 8190.
OR
Find the HCFof2205, 5145 and 4410. (2)
Answer:
The prime factorisation of 8190 is:
8190 = 2 ×3 × 3 × 5 × 7 × 13.

OR
HCF of 2205, 5145 and 4410

2205 = 3 × 3 × 5 × 7 × 7
5145 = 3 × 5 × 7 × 7 × 7
4410 = 2 × 3 × 3 × 5 × 7 × 7
HCF = 3 × 5 × 7 × 7 = 735

Question 22.
In an A.P., a = 5, d = 2 and n = 50, find an. (2)
Answer:
In given A.P.
Where a is 1st term, d is common difference and n is number of terms.
a_n = a + (n – 1 )d
= 5 + (50 – 1) × 2
= 5 + 49 × 2
= 5 + 98
= 103

Question 23.
If Q (0, 1) is equidistant from P(5, -3) and R (x, 6), find the values of ‘x’. Also, find the distances of QR and PR.
OR
Find the ratio in which P(4, p) divides the line segment joining the points A(2, 3) and B(6, 3). Hence find the value of p. (2)
Answer:
Since, Q(0, 1) is equidistant from P(5, -3) and R(x, 6),
PQ = QR
⇒ PQ² = QR²
i.e (5 – 0)² + (-3 – 1)² = (x – 0)² + (6 – l)²
i.e 25 + 16 = x² + 25
i.e x² = 16
or x = ±4

Thus, R(4, 6) or R (-4, 6)
For R (4, 6),
QR = \(\sqrt{(4-0)^2+(6-1)^2}\)
= \(\sqrt{16+25}\) = \(\sqrt{41}\)
and PR = \(\sqrt{(4-5)^2+(6+3)^2}\)
= \(\sqrt{1+81}\) = \(\sqrt{82}\)

For R(-4,6),
QR = \(\sqrt{(-4-0)^2+(6-1)^2}\)
= \(\sqrt{16+25}\) = \(\sqrt{41}\)
and PR = \(\sqrt{(-4-5)^2+(6+3)^2}\)
= \(\sqrt{81+81}\) = \(\sqrt{162}\)
= \(9 \sqrt{2}\)
OR
Given, point P(4, p) and line segment A(2,3) and B(6, 3)
Let the point P divide given line segment AB in the ratio of k: 1.
Then, P(4,p) = [\left(\frac{k \times 6+2}{k+1}\right) \cdot\left(\frac{k \times 3+3}{k+1}\right)]
∴ On comparing, we get
\(\frac{6k + 2}{k+1}\) = 4
⇒ 6k + 2 = 4k + 4
⇒ k = 1
Then point P divides line AB in the ratio of 1:
Now, p = \(\frac{1×3 + 3}{1+1}\) = \(\frac{6}{2}\)
⇒ p = 3
Hence, value of p is 3.

Question 24.
A tree casts a shadow of 4\(\sqrt{3}\) m on ground. If the sun’s elevation is 60°. then find the height of the tree. (2)
Answer;

Let PQ be the tree and RQ be its shadow.
∴ In ∆PQR
tan 60° = \(\frac{PQ}{QR}\)
\(\sqrt{3}\) = \(4 \sqrt{3}\)
PQ = 12 m.
Hence, the height of the tree is 12 m.

Question 25.
Find the mode of the following frequency distribution: (2)

Class	15-20	20-25	25-30	30-35	35-40	40-45
Frequency	3	8	9	10	3	2

Answer:
In the given frequency distribution, modal class
Is 30-35, with maximum frequency 10.
Here tower Limit of modal class. I = 30
frequency of class preeceding modal class, f_o = 9
frequency of modal doss, f₁ = 10
frequency of cLass succeeding modal class, f₂ = 3
\(M_0=l+\frac{f_1-f_0}{2 f_1-f_0-f_2} \times h\)
\(=30+\frac{10-9}{20-9-3} \times 5\)
= 30 + \(\frac{5}{8}\) = 30 + 0.625
= 30.625

SECTION – C
(Section C consists of 6 questions of 3 marks each.)

Question 26.
Prove that an irrational number, using the fact that 2\(\sqrt{3}\) – 4 is an irrational number. (3)
Answer:
Let us assume on the contrary, that 2\(\sqrt{3}\) – 4 be a rational number.
Then,
2\(\sqrt{3}\) – 4 = \(\frac{p}{q}\). where p and q are co-primes and q ≠ 0.
⇒ \(\sqrt{3}=\frac{1}{2}\left(\frac{p}{q}+4\right)\)
Since p and q are integers, \(\frac{1}{2}\left(\frac{p}{q}+4\right)\) is rational and s0 \(\sqrt{3}\) is rationaL But this contradicts the fact that \(\sqrt{3}\) is irrational.
Hence, 2\(\sqrt{3}\) – 4 is an irrational number.

Question 27.
The sum of two numbers, as well as, the difference of their squares is 9. Find the numbers.
OR
Find the values of k for which the following equations have an infinite number of solutions: 2x + 3y = 7; (k – 1)x + (k + 2) y = 3k (3)

Answer:
Let the two numbers be x and y. (x > y).
Then, x + y = 9 and x² – y² = 9
⇒ x + y = 9 and (x – y) (x + y) = 9
⇒ x + y = 9 and x – y = 1
Solving the two equations, we get x = 5 and y = 4
Thus, the two numbers are 5 and 4.

Given, equations are
2x + 3y = 7
(k – 1)x + (k + 2)y = 3k
Here, a₁ = 2, b₁ = 3, c₁ = 7
a₂ = k – 1, b₂ = k + 2 c₂ = 3k
Since, given system of equation has infinite solution.
∴ \(\begin{gathered}
\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2} \\
\frac{2}{k-1}=\frac{3}{k+2}=\frac{7}{3 k}
\end{gathered}\)
On comparing the two ratios
⇒ 2(k+2) = 3(k-1)
⇒ 2k + 4 = 3k – 3
⇒ -k = -7
⇒ k = 7
and, \(\frac{3}{k+2}=\frac{7}{3 k}\)
⇒ 9k = 7k + 14
⇒ 2k = 14
⇒ k = 7
Hence, the value of k is 7.

Question 28.
Show that ∆ABC with vertices A(-2, 0), B(2, 0) and C(0, 2) is similar to A PQR with vertices P(-4, 0), Q(4, 0) and R(0,4). (3)
Answer:
Here, vertices of ∆ABC are A(-2, 0), B(2, 0) and C(0, 2)
Another ∆PQR, with vertices P(-4, 0), Q(4, 0) and R(0, 4).
In ∆ABC,

Then, lenth of AB = \(\sqrt{(2+2)^2+(0-0)^2}\)
= 4 units

lenth of BC = \(\sqrt{(0-2)^2+(2-0)^2}\)
= \(\sqrt{4+4}\)
= \(\sqrt{8}\)
= \(2 \sqrt{2}\)

lenth of AC = \(\sqrt{(-2-0)^2+(0-2)^2}\)
= \(\sqrt{4+4}\)
= \(2 \sqrt{2}\)

In ∆PQR,

lenth of PQ = \(\sqrt{(4+4)^2+(0-0)^2}\)
= \(\sqrt{8^2}\) = 8 units

lenth of QR = \(\sqrt{(4-o)^2+(o-4)^2}\)
= \(\sqrt{4^2+4^2}\)
= \(\sqrt{32}\)
= \(4 \sqrt{2}\)

lenth of AC = \(\sqrt{(-4)^2+(4)^2}\)
= \(4 \sqrt{2}\) units

∴ \(\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{4}{8}=\frac{1}{2}\)
\(\frac{\mathrm{BC}}{\mathrm{QR}}=\frac{2 \sqrt{2}}{4 \sqrt{2}}=\frac{1}{2}\)
and \(\frac{\mathrm{AC}}{\mathrm{PR}}=\frac{2 \sqrt{2}}{4 \sqrt{2}}=\frac{1}{2}\)
Now, \(\frac{A B}{P Q}=\frac{B C}{Q R}=\frac{A C}{P R}=\frac{1}{2}\)
Hence, ∆ABC ~ ∆PQR (by SSS – Similarity)

Question 29.
What is the ratio between the areas of the circle and the square when a square is inscribed in a circle? (3)
Answer:
Let,
radius of the circle be r units.
Then, diagonaL of the square = 2r
⇒ Side of the square = \(\frac{2 r}{\sqrt{2}}=\sqrt{2} r=\frac{\pi}{2}\)
∴ \(\frac{\text { Area of the circle }}{\text { Area of the square }}=\frac{\pi r^2}{(\sqrt{2} r)^2}=\frac{\pi r^2}{2 r^2}\)
= π : 2

Question 30.
If sum and product of quadratic polynomial are 2 and -8 respectively, then find a quadratic polynomial and zeroes of the polynomial so obtained.
OR
The area of a sector of a circle of radius 36 cm is 54 n. sq cm. Find the length of the corresponding arc of the sector. (3)
Answer:
Let α and β be the zeroes of the polynomial.
Given: Sum of zeroes, α + β = 2
Product of zeroes, α × β = -8
Equation of polyomial:
x² – (sum of zeroes)x + product of zeroes
x² – (2)x + (- 8)
⇒ x² – 2x – 8
∠eores of polynomial
Let P(x) = x² – 2x – 8
= x² – 4x + 2x – 8
= x(x – 4) + 2(x – 4)
= (x – 4)(x + 2)
Hence, the zeroes of the polynomial are 4 and -2.

Let 9 be the angle of the sector. Then,

Question 31.
Find the mean of the following frequency distribution: (3)

Marks	Below 10	Below 20	Below 30	Below 40	Below 60	Below 70	Below 80	Below 90	Below 100
Number of students	12	22	35	50	86	97	104	109	115

Answer:

SECTION – D
(Section D consists of 4 questions of 5 marks each.)

Question 32.
if tan q + sin q = m and tan q – sin q = n, show that: m² – n² = \(4 \sqrt{mn}\)
OR
The angle of elevation of the top of a building from the foot of a tower is 30°, and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building. (5)
Answer:
We have,

Question 33.
A circle is inscribed in a DABC having sides 8 cm, 10 cm and 12 cm as shown in the figure. Find AD, BE and CF.

OR
The 6th term of an AP is five times the 1st term and the 11th term exceeds twice the 5th term by 3. Find the 8th term of the AP. (5)
Answer:
We know that the tangents drawn from an external point to a circle are equal. Therefore,

gives y = 5
Hence, AD = x = 7 cm, BE = y = 5 cm and CF = z = 3 cm.

Let a and d be the first term and the common difference of the AP respectively. Then,
a₆ = a + 5d = 5a …(i)
and a₁₁ = 2 a₅ + 3
⇒ a + 10d = 2 (a + 4d) + 3 …..(ii)
Simplifying (i) and (ii), we get
4a = 5d and a – 2d + 3 = 0
Solving these equations, we get d = 4 and a = 5
Thus, a₈ = a + 7d = 5 + 7(4) = 33

Question 34.
On the ground, a tree with a length of 6 m creates a shadow that is 4 m long, while another tree creates a shadow that is 28 m long. Find the tree’s height.

Answer:
Given,
Length of the vertical pole = 6 m Shadow of the pole = 4 m Let the height of the tower be h m.
Length of the shadow of the tower = 28 m
In ∆ABC and ∆DFE,
∠C = ∠E (angle of elevation)
∠B = ∠F = 90°
By AA similarity criterion,
∆ABC ~ ∆DFE
We know that the corresponding sides of two similar triangles are proportional.
\(\frac{AB}{DF}\) = \(\frac{BC}{EF}\)
\(\frac{6}{h}\) = \(\frac{4}{28}\)
h = \(\frac{6 \times 28}{4}\)
h = 6 × 7
h = 42
Hence, the height of the tower = 42 m.

Question 35.
Two dice have the following numbers: 1, 2, 3, 4, 5, 6 and 1,1, 2, 2, 3, 3. The total of the numbers on them is calculated once they are thrown. Calculate the probability of getting each sum from 2 to 6 individually. (5)
Answer:
Number of total outcome = n(S) = 36
(i) Let E₁ be the event ‘getting sum 2’
Favourable outcomes for the event E₁ = {(1,1),(1,1)}
n(E₁) = 2
\(P\left(E_1\right)=\frac{n\left(E_1\right)}{n(S)}=\frac{2}{36}=\frac{1}{18}\)

(ii) Let E₂ be the event ‘getting sum 3’
Favourable outcomes for the event E₂ = {(1,2),(1,2),(2,1),(2,1)}
n(E₂) = 4
\(P\left(E_2\right)=\frac{n\left(E_2\right)}{n(S)}=\frac{4}{36}=\frac{1}{9}\)

(iii) Let E₃ be the event ‘getting sum 4’
Favourable outcomes for the event E₃ = {(2,2),(2,2),(3,1),(3,1)(1,3),(1,3)}
n(E₃) = 6
\(P\left(E_3\right)=\frac{n\left(E_3\right)}{n(S)}=\frac{6}{36}=\frac{1}{6}\)

(iv) Let E₄ be the event ‘getting sum 5’
Favourable outcomes for the event E₄ = {(2,3),(2,3),(4,1),(4,1)(3,2),(3,2)}
n(E₄) = 6
\(P\left(E_4\right)=\frac{n\left(E_4\right)}{n(S)}=\frac{6}{36}=\frac{1}{6}\)

(v) Let E₅ be the event ‘getting sum 6’
Favourable outcomes for the event E₅ = {(3,3),(3,3),(4,2),(4,2)(5,1),(5,1)}
n(E₅) = 6
\(P\left(E_5\right)=\frac{n\left(E_5\right)}{n(S)}=\frac{6}{36}=\frac{1}{6}\)

SECTION – E
(Case Study Based Questions)
(Section E consists of 3 questions. All are compulsory.)

Question 36.
On the roadway, Points A and B, which stand in for Chandigarh and Kurukshetra, respectively, are located nearly 90 kilometres apart. At the same time, a car departs from Kurukshetra and one from Chandigarh. These cars will collide in 9 hours if they are travelling in the same direction, and in 9/7 hours if they are travelling in the other direction. Let X and Y be two cars that are travelling at x and y kilometres per hour from places A and B, respectively.

On the basis of the above information, answer the following questions:
(A) When both cars move in the same direction, then find the situation can be represented algebraically.
OR
When both cars move in the opposite direction, then find the situation can be represented algebraically. (2)
(B) Find the speed of car x. (1)
(C) Find the speed of car y. (1)
Answer:
(A) Suppose two cars meet at point Q.
Then, Distance travelled by car X = AQ,
Distance travelled by car Y = BQ.
It is given that two cars meet in 9 hours.
Distance travelled by car X in 9 hours
= 9x km = AQ = 9x
Distance travelled by car Y in 9 hours
= 9y km = BQ = 9y

Clearly, AQ – BQ = AB
= 9x – 9y = 90
= x – y = 10

Suppose two cars meet at point P.
Then Distance travelled by car X = AP
and Distance travelled by car Y = BP.
In this case, two cars meet in 9/7 hours.
Distance travelled by car X in \(\frac{9}{7}\) hours
= \(\frac{9}{7}\) X km
⇒ AP = \(\frac{9}{7}\) X

Distance travelled by car Y in \(\frac{9}{7}\) hours
= \(\frac{9}{7}\) y km

clearly AP + BP = AB
⇒ \(\frac{9}{7}\) x + \(\frac{9}{7}\) y = 90
⇒ \(\frac{9}{7}\) (x + y) = 90
⇒ x + y = 70

(B) We have, x – y = 10
⇒ x + y = 70
Adding equations (i) and (ii), we get
2x = 80
⇒ x = 40
Hence, speed of car X is 40 km/hr.

Question 37.
Eshan purchased a new building for her business. Being in the prime location, she decided to make some more money by putting up an advertisement sign for a rental ad income on the roof of the building.

From a point P on the ground level, the angle of elevation of the roof of the building is 30° and the angle of elevation of the top of the sign board is 45°. The point P is at a distance of 24 m from the base of the building.
On the basis of the above information, answer the following questions:
(A) Find the height of the building (without the sign board).
OR
The height of the building (with the sign board) (2)
(B) Find the height of the sign board. (1)
(C) Find the distance of the point P from the top of the sign board. (1)
Answer:
(A) Without the sign board, the height of the shop is AB.
In ∆PAB,

OR
Considering, the diagram in the above question, AC as the new height of the shop including the sign-baard.
In ∆APC,
tan 45° = \(\frac{AC}{AP}\)
1 = \(\frac{AC}{24}\)
⇒ AC = 24 m

(B) From Q (i) and Q (ii).
Length of sign board, BC = AC – AB
= 24 – 14
= 10 m

Question 38.
In a toys manufacturing company, wooden parts are assembled and painted to prepare a toy. One specific toy is in the shape of a cone mounted on a cylinder.
For the wood processing activity center, the wood is taken out of storage to be sawed, after which it undergoes rough polishing, then is cut, drilled and has holes punched in it. It is then fine polished using sandpaper.

For the retail packaging and delivery activity center, the polished wood sub-parts are assembled together, then decorated using paint.
The total height of the toy is 26 cm and the height of its conical part is 6 cm. The diameters of the base of the conical part is 5 cm and that of the cylindrical part is 4 cm.
On the basis of the above information, answer the following questions:
(A) If its cylindrical part is to be painted yellow, find the surface area need to be painted. (1)
(B) if its conical part is to be painted green, find the surface area need to be painted.
OR
Rnd the volume of the wood used in making this toy. (2)
(C) If the cost of painting the toy is 3 paise per sq cm, then find the cost of painting the toy. (Use it = 3.14) (1)
Answer:
(A) C.S.A. of cylinder = 2nrH + πr2 = nr(2H + r)
= 2π(2 × 20 + 2)
= 84TI

(B) C.S.A. of cone = πrl + π(R² – r²)
= \(\begin{aligned}
&\pi\left[r \sqrt{r^2+h^2}+\left(R^2-r^2\right)\right] \\
&\pi\left[2.5 \sqrt{2.5^2+6^2}+\left(2.5^2-2^2\right)\right]
\end{aligned}\)
= π [2.5 × 6.5 + 0.5 × 4.5]
= π [16.25 + 2.25]
= 18.571 sq units

Volume of toy = Volume of cone + Volume of cylinder
= \(\frac{1}{3}\) πr²h + πR²H
= π[\(\frac{1}{3}\) × 2.5 × 2.5 × 6 + 2 × 2 × 20]
= π [12.5 + 80]
= 92.5TC cm³