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

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

Time Allowed : 3 hours
Maximum Marks : 80

General Instructions:

1. This Question Paper has 5 Sections A, B, C, D, and E.
2. Section A has 20 Multiple Choice Questions (MCQs) carrying 1 mark each.
3. Section B has 5 Short Answer-I (SA-I) type questions carrying 2 marks each.
4. Section C has 6 Short Answer-Il (SA-II) type questions carrying 3 marks each.
5. Section D has 4 Long Answer (LA) type questions carrying 5 marks each.
6. Section E has 3 Case Based integrated units of assessment (4 marks each) with sub-parts of the values of 1,1 and 2 marks each respectively.
7. All Questions are compulsory. However, an internal choice in 2 Qs of 2 marks, 2 Qs of 3 marks and 2 Questions of 5 marks has been provided. An internal choice has been provided in the 2 marks questions of Section E.
8. Draw neat figures wherever required. Take n = \(\frac{22}{7}\) wherever required if not stated.

Section – A
(Section A consists of 20 Questions of 1 marks each.)

Question 1.
If two positive integers a and b are written as a = x³y² and b = xy³; x, y are prime numbers, then HCF (a, b) is : [1]
(A) xy
(B) xy²
(C) x³y³
(D) x²y²
Answer:
(B) xy²

Explanation :
Since a = x³y² = x × x × x × y × y and
b = xy³ =x × y × y × y
Thus, HCF of a and k = x × y × y = xy²

Question 2.
The pair of equations x = a and y = b graphically represents lines which are : [1]
(A) parallel
(B) intersecting at (b, a)
(C) coincident
(D) intersecting at (a, b)
Answer:
(D) intersecting at (a, b)

Explanation:
(x = a) is the equation of a straight line parallel to the y-axis at a distance ‘a’ from it. Again, y = b is the equation of a straight line parallel to the x-axis at a distance ‘b’ from it. So, the pair of equations x = a and y = b graphically represents lines which are intersecting at (a, b).

Question 3.
Which of the following is not a quadratic equation? [1]
(A) 2(x – 1)² = 4x² – 2x + 1
(B) 2x – x² = x² + 5
(C) (\(\sqrt{2}\)x + \(\sqrt{3}\))² + x² = 3x² – 5x
(D) (x2 + 2x)² = x⁴ + 3 + 4x³
Answer:
(C) (\(\sqrt{2}\)x + \(\sqrt{3}\))² + x² = 3x² – 5x

Explanation:
(\(\sqrt{2}\)x + \(\sqrt{3}\))² + x² = 3x² – 5x
(\(\sqrt{2}\)x)² + (\(\sqrt{3}\))² + 2 × \(\sqrt{2}\)x × \(\sqrt{3}\) + x² = 3x² – 5x
2x² + 3 + 2\(\sqrt{6}\)x + x² = 3x² – 5x
3x² + 2\(\sqrt{6}\)x + 3 = 3x² – 5x
x(5 + 2\(\sqrt{6}\)) + 3 = 0
It is not of the form of ax² + bx + c = 0.

Question 4.
The sum of first five positive integers divisible by 6 is : [1]
(A) 180
(B) 90
(C) 45
(D) 30
Answer:
(B) 90

Explanation:
Positive integers divisible by 6 are 6, 12,18,24,30
Since difference is same, its an AP
We need to find sum of first 5 integers.
We can use formula,
S_n = f [2a + (n-1) d]
Here, n = 5, d = 6, a = 6
S₅ = \(\frac{5}{2}\) [2 × 6 + (5 -1) × 6]
S₅ = \(\frac{5}{2}\) (12 + 24)
S₅ = \(\frac{1}{2}\) × 36 = 90

Question 5.
The distance between the points A (0, 6) and B (0, -2) is : [1]
(A) 6
(B) 8
(C) 4
(D) 2
Answer:
(B) 8

Explanation:
The distance between two points (x₁, y₁) and (x₂ y₂) is given as,
d = \(\sqrt{(x_2 – x_1)^2+(y_2-y_1)^2}\)
Where, x₁ = 0, y₁ = 6 and x₂ = 0, y₂ = -2
So, distance between A (0,6) and B (0, -2):
AB = \(\sqrt{(0-0)^2 + (-2 – 6)^2}\)
= \(\sqrt{0+(-8)^2}\)
= \(\sqrt{8^2}\)
= 8

Question 6.
In the given figure, AT is a tangent to the circle with centre ‘O’ such that OT = 8 cm and ZOTA = 60°. Then AT is equal to: [1]
(A) 4 cm
(B) 2 cm
(C) 2\(\sqrt{3}\) cm
(D) 4\(\sqrt{3}\) cm

Answer:
(A) 4 cm

Explanation:
Join OA. OA is radius and AT is tangent at contact point A.
OAT = 90°,
Given that, OT = 8 cm
Now, \(\frac{A T}{8}=\frac{\text { base }}{\text { hypotenuse }}=\cos 60^{\circ}\)
⇒ AT = 8 × \(\frac{1}{2}\) = 4 cm.

Question 7.
The line segment joining the points P(-3,2) and Q(5,7) is divided by the Y-axis in the ratio [1]
(A) 3 : 1
(B) 3 : 4
(C) 3 : 2
(D) 3 : 5
Answer:
(D) 3 : 5

Explanation:
Let the point on Y-axis which divides the line PQ is M(0, y) and the ratio be
According to the section formula,
M(x,y) = \(\frac{1}{2}\)
M(0,y) = \(\frac{5k + (-3)}{k+1}, \frac{k(7)+ 1(2)}{k+1}\)
On comparing, we get
o = \(\frac{5k + (-3)}{k+1}\)
or 5k – 3 = 0
or k = \(\frac{3}{5}\)
Required ratio = 3:5

Question 8.
In the given figure, AB is a chord of the circle and AOC is its diameter, such that ∠ACB = 50°. If AT is the tangent to the circle at the point A, then ∠BAT is equal to : [1]
(A) 65°
(B) 60°
(C) 50°
(D) 40°

Answer:
(C) 50°

Explanation :
Since, the angle between chord and tangent is equal to the angle subtended by the same chord in alternate segment of circle.
⇒ ∠BAT = 50°

Question 9.
If 4 tan θ = 3, then \(\left(\frac{4 \sin \theta-\cos \theta}{4 \sin \theta+\cos \theta}\right)\) is equal to: [1]
(A) \(\frac{1}{3}\)
(B) \(\frac{1}{3}\)
(C) \(\frac{1}{2}\)
(D) \(\frac{3}{4}\)
Answer:
(C) \(\frac{1}{2}\)

Explanation:
Given, 4 tan θ = 3 that \(\frac{4 sin θ-cos θ}{4 sin θ+cos θ}\) is equal to:
tan θ = \(\frac{3}{4}\)

Question 10.
Given that sin θ = \(\frac{a}{b}\) then cos θ is equal to: [1]
(A) \(\frac{b}{\sqrt{b^2-a^2}}\)
(B) \(\frac{b}{a}\)
(C) \(\frac{\sqrt{b^2-a^2}}{b}\)
(D) \(\frac{a}{\sqrt{b^2-a^2}}\)
Answer:
(C) \(\frac{\sqrt{b^2-a^2}}{b}\)

Explanation:
Given, sin θ = \(\frac{a}{b}\)

Question 11.
The value of sin² 60° + 2 tan 45° – cos² 30° is: [1]
(A) 1
(B) 2
(C) 3
(D) 0
Answer:
(B) 2

Explanation:
sin² 60° + 2 tan 45° – cos² 30°
= \(\left(\frac{\sqrt{3}}{2}\right)^2+2(1)-\left(\frac{\sqrt{3}}{2}\right)^2\)
[For any two correct values]
= 2

Question 12.
The area of the square that can be inscribed in a circle of radius 8 cm is : [1]
(A) 256 cm²
(B) 128 cm²
(C) 64\(\sqrt{2}\) cm²
(D) 64cm²
Answer:
(B) 128 cm²

Explanation:
Given, radius of circle, r = OC = 8 cm
Diameter of the circle = AC = 2 × OC = 2 × 8= 16 cm
which is equal to the diagonal of a square.

Let side of square be ‘a’.
Using Pythagoras theorem,
AB² + BC² = AC²
a² + a² = 16²
a² =256
a² = 128 cm².
Hence, area of square is 128 cm².

Question 13 .
The surface areas of two spheres are in the ratio 16 : 9. The ratio of their volumes is : [1]
(A) 64 : 27
(B) 16 : 9
(C) 4 : 3
(D) 163 : 93
Answer:
(A) 64 : 27

Explanation:

Question 14.
\(\sqrt{2}\) – \(\sqrt{3}\) is : [1]
(A) Rational Number
(B) Integer
(C) Irrational Number
(D) None of these
Answer:
(C) Irrational Number

Explanation:
We know that \(\sqrt{2}\) and \(\sqrt{3}\) both are irrational numbers and difference of two irrational number is always irrational.

Question 15.
The area of a quadrant of a circle where the circumference of circle is 176 m, is [1]
(A) 2464 m²
(B) 1232 m²
(C) 616 m²
(D) 308 m²
Answer:
(C) 616 m²

Explanation:
It is given that circumference of the circle is 176 cm²
2πr = 176
2 × \(\frac{22}{7}\) × r = 176
r = \(r=\frac{176 \times 7}{2 \times 22}=28\) cm
Also, in a quadrant θ : 90°
Area of quadrant = \(\frac{θ \times πr^2}{360°}\)
= \(\frac{90^{\circ}}{360^{\circ}} \times \frac{22}{7} \times 28 \times 28\)
= 616 cm²

Question 16.
For the following distribution : [1]

Marks	Number of students
Below 10	3
Below 20	12
Below 30	27
Below 40	57
Below 50	75
Below 60	80

the modal class is:
(A) 10-20
(B)20-30
(C) 30-40
(D) 50-60
Answer:

Explanation:

Marks	Number of students	f1
0-10	3 – 0 = 3	3
10-20	12 – 3 = 9	9
20-30	27 – 12 = 15	15
30-40	57 – 27 = 30	30
40-50	75 – 57 = 18	18
50-60	80 – 75 = 75	5

Modal class has maximum frequency (30) in class 30 – 40.

Question 17.
The probability of getting an even number, when a die is thrown once, is : [1]
(A) \(\frac{1}{2}\)
(B) \(\frac{1}{3}\)
(C) \(\frac{5}{6}\)
(D) \(\frac{5}{9}\)
Answer:
(A) \(\frac{1}{2}\)

Explanation:
Even number = 2,4,6
Total outcomes = 1,2, 3,4, 5,6
Required Probability = \(\frac{\text { Favourable outcomes }}{\text { Total outcoems }}\)
= \(\frac{3}{6}\) = \(\frac{1}{2}\)

Question 18.
Two identical fair dice have numbers 1 to 6 written on their faces. Both are tossed simultaneously. What is the probability that the product of the numbers that turn up is 12? [1]
(A) \(\frac{1}{36}\)
(B) \(\frac{1}{9}\)
(C) \(\frac{1}{6}\)
(D) \(\frac{1}{3}\)
Answer:
(B) \(\frac{1}{9}\)

Explanation:
When dice is rolled twice, then out of events in which the product of the numbers that turn up 12.
= {(2,6), (6,2), (3,4), (4,3)}
No. of events = 4
Total no. of events = 36
Probability (product of 12) = \(\frac{4}{36}\) = \(\frac{1}{9}\)

Directions for question 19 and 20 : In the following questions, A statement of Assertion (A) is followed by a statement of Reason (R). Choose the correct option.
(A) Both (A) and (R) are true and (R) is the correct explanation of (A).
(B) Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
(C) (A) is true but (R) is false.
(D) (A) is false and (R) is true.

Question 19.
Assertion (A): If HCF (336, 54) = 9, then LCM (336,54) = 2016.
Reason (R): The sum of exponents of prime factors in the prime factorisation of 196 is 4. [1]
Answer:
(B) Both (A) and (R) are true but (R) is NOT the correct explanation of (A).

Explanation:
Let us consider the assertion,
∵ HCF × LCM = Product of numbers
∴ 9 × LCM = 336 × 54
⇒ LCM = \(\frac{336 × 54}{9}\)
= 2016
Thus, the assertion is correct.
Now, let us consider the reason:
Prime factors of 196 = 2² × 7²
∴ The sum of exponents of prime factors =2+2=4
So, the reason is correct.
Thus, assertion is correct and reason is also correct.

Question 20.
Assertion (A): If the distance between the point (4, p) and (1,0) is 5, then the value of p is 4.
Reason (R): The point which divides the line segment joining the points (7, – 6) and (3, 4) in ratio 1 : 2 internally lies in the fourth quadrant. [1]
Answer:
(D) (A) is false and (R) is true.

Explanation :
In case of assertion:
Distance between two points (x₁, y₁) and (x₂, y₂) is given as,
d = \(\sqrt{(x_2+x_1)^2 + (y_2+y_1)^2}\)
where,
(x₁, y₁) = (4, p)
(x₂, y₂) = (1,0)
And, d = 5
Put the values, we have
5² = (1 – 4)² + (0 – p)²
25 = (-3)² + (-p)
25 – 9 = p²
16 = p²
p = ± 4
∴ Assertion is incorrect.
In case of reason:
Let (x, y) be the point
Then, x = \(\frac{m x_2+n x_1}{m+n}\) and y = \(\frac{m y_2+n y_1}{m+n}\)
Here, x₁ = 7, y₁ = -6, x₂ = 3, y₂ = 4, m = 1 and n = 2
x = \(\frac{1(3)+2(7)}{1+2}\)
x = \(\frac{3+14}{3}\) = \(\frac{17}{3}\)
And y = \(\frac{1(3)+2(7)}{1+2}\)
y = \(\frac{4-12}{3}\) = \(\frac{-8}{3}\)
So, the required point (x, y) = (\(\frac{17}{3}\), \(\frac{-8}{3}\)) lies in IV^th quadrant.
Reason is correct.
Hence, assertion is incorrect but reason is correct.

Section – B
Section B consists of 5 Questions of 2 marks each.

Question 21.
If x = (-\(\frac{1}{2}\)) is a solution of the quadratic equation 3x² + 2kx – 3 = 0, find the value of k.
OR
Find the quadratic polynomial whose sum and product of the zeroes are \(\frac{21}{8}\) and \(\frac{5}{16}\) respectively. [2]
Answer:
Putting x = \(\frac{1}{2}\) in 3x² + 2kx – 3 = 0
3(-\(\frac{1}{2}\)) + 2k(-\(\frac{1}{2}\)) – 3 = 0
\(\frac{3}{4}\) – k – 3 = 0
k = \(\frac{3}{4}\) – 3
k = \(\frac{3-12}{4}\)
Hence, k = \(\frac{-9}{4}\)
OR
Given,
Sum of zeroes = \(\frac{21}{8}\)
and Product of zeroes = \(\frac{5}{16}\)
So, quadratic polynomial = x² – (Sum of zeroes) x + Product of zeroes
= \(x^2-\left(\frac{21}{8}\right) x+\frac{5}{16}\)
= \(\frac{1}{16}\left(16 x^2-42 x+5\right)\)

Question 22.
PQ is a tangent drawn from an external point P to a circle with centre O and QOR is the diameter of the circle. If ∠POR = 120°, What is the measure of ∠OPQ ? [2]

Answer:
In AOQP
∠POR = ∠OQP + ∠OPQ
(Exterior angle)
∴ ∠OPQ = ∠POR – ∠OQP
= 120° – 90°
= 30°

Question 23.
In Figure, AP and BP are tangents to a circle with centre O, such that AP = 5 cm and ZAPB = 60°. Find the length of chord AB. [2]

Answer:
PA = PB
or, ∠PAB = ∠PBA = 60°
∴ ∆PAB is an equilateral triangle.
Hence, AB = PA = 5 cm

Question 24.
In figure, a tower AB is 20 m high and BC, its shadow on the ground, is 20\(\sqrt{3}\) m long. Find the Sun’s altitude. [2]

Answer:
Let the ∠ACB be θ,
∠B = 90°
AB BC 20
tan θ = \(\frac{AB}{BC}\)
tan θ = \(\frac{20}{20 \sqrt{3}}=\frac{1}{\sqrt{3}}\) = tan 30°
⇒ θ = 30°
Thus, the Sun’s altitude is 30°.

Question 25.
A solid sphere of radius 3 cm is melted and then cast into small balls each of radius 0.3 cm. Find the number of balls thus obtained. [2]
OR
If the circumferences of two concentric circles forming a ring are 88 cm and 66 cm respectively. Find the width of the ring.
Answer:
Let the number of balls be n.
Volume of n small balls = Volume of sphere
\(n \times \frac{4}{3} \pi \times(0.3)^3=\frac{4}{3} \times \pi \times(3)^3\)
⇒ \(n=\frac{3 \times 3 \times 3}{0.3 \times 0.3 \times 0.3}=1000\)
OR
Circumference of the outer circle,
2πr₁ = 88 cm
r₁ = \(\frac{81 \times 7}{2 \times 22}=14\)
Circumference of the inner circle, 2πr₂ = 66 cm
r₂ = \(\frac{66 \times 7}{2 \times 22}=\frac{21}{2}\)
= 10.5 cm
Width of the ring = r₁ – r₂
= 14 – 10.5 cm
= 3.5 cm

Section – C
Section C consists of 6 Questions of 3 marks each.

Question 26.
Prove that \(\sqrt{2}\) is an irrational number. [3]
Answer:
Let √2 be a rational number.
∴ √2 = p/q,
where ‘p’ and ‘q’ are integers, q ≠ 0
On squaring both sides, we get
or 2 = \(\frac{p^{2}}{q^{2}}\)
or p² = 2q²
p² is divisible by 2.
∴ p is divisible by 2.
Let p = 2r for some positive integer r
or p² = 4r²
2q² = 4r²
or q² = 2r²
or q² is divisible by 2.
∴ q is divisible by 2.
From (i) and (ii), p and q are divisible by 2, which contradicts the fact that p and q are co-primes.
Hence, our assumption is false.
∴ \(\sqrt{2}\) is an irrational number.

Question 27.
If the term of an A.P. is \(\frac{1}{q}\) and qtb term is \(\frac{1}{p}\). Prove that the sum of first pq term of the A.P. is \(\frac{pq+1}{2}\) [3].
OR
If the sum of first 14 terms of an A.P is 1050 and its first term is 10, find the 20th term.
Answer:
Let first term and common difference of given A.E be a and d respectively.

Hence Proved
OR
Let common difference be d.
⇒ \(\frac{14}{2}\)[2(10) + (14 – 1)d] =1050
⇒ d = 10
a₂₀ = a + 19 d
= 10 + 19(10) = 200

Detailed Solution:
Given,
In A.E, S₁₄ = 1050
a= 10
a₂₀ = ?
S_n = \(\frac{n}{2}\)[2a + (n-1)d]
⇒ S14 = \(\frac{14}{2}\)[2 × 10 + (14 – 1)d]
⇒ 1050 = 7(20 + 13d)
⇒ 150 = 20 + 13d
⇒ 130 = 13d ⇒ d = 10
a_n = a + (n – 1 )d
∴ a₂₀ = 10 + (20 – 1) × 10
= 10 + 19 × 10
= 200

Question 28.
Find the sum of first 16 terms of the A.P 10,6,2, …… .
OR
For what value of p will the following system of equations have no solution ?
(2p – 1)x + (p – 1)y = 2p + 1; y + 3x – 1 = 0
Answer:
Here, a= 10,
d = 6 – 10 = -4
and n = 16
S_n = \(\frac{n}{2}\)[2a + (n-1)d]
S₁₆= \(\frac{16}{2}\)[2 × 10 + (16 – 1)(-4)]
= 8[20 + 15 × (-4)]
= 8(20 – 60)
= 8 × (-40)
= -320

OR

For (2p – 1)x + (p – 1)y – (2p + 1) = 0
a₁ = 2p -1, b₁ = p -1
and c₁ = – (2p + 1) Yz
and for 3x + y -1 = 0
a₂ = 3, b₂ = 1 and c₂ = -1
The condition for no solution is

Hence, system has no solution when p = 2.

Question 29.
In the given figure, ABC is a right angled triangle at DB = 90°. D is the mid-point of BC.

Show that AC² = AD² + 3CD².
Answer:
Given,
BD = CD = \(\frac{BC}{2}\)
or, BC = 2 BD
Using Pythagoras theorem in the right AABC, we have
AC² = AB² + BC²
= AB² + 4BD²
= (AB² + BD²) + 3 BD²
AC² = AD² + 3 CD².
[∴ BD = CD]

Question 30.
The angle of elevation of the top of a building from the foot of the 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 60 m high, find the height of the building. 3
Answer:

Question 31.
A bag contains cards numbered 1 to 49. Find the probability that the number on the drawn card is:
(i) an odd number
(ii) a multiple of 5
(iii) even prime
Answer:
Total cards = 49
Odd numbers from 1 to 49 are 1,3,5,7,9,11,13,15, 17,19,21, 23,25,27,29, 31, 33,35, 37, 39,41,43, 45, 47 and 49.
(i) No. of favourable outcomes = 25
P(odd number) = \(\frac{25}{49}\).
(ii) Multiple of 5 are 5,10,15,20,25,30,35,40 and 45.
No. of favourable outcomes = 9
P(multiple of 5) = \(\frac{9}{49}\).
(iii) Only even prime number is 2.
P(even prime) = \(\frac{1}{49}\).

Section – D
Section D consists of 4 Questions of 5 marks each.

Question 32.
Solve graphically the pair of linear equations :
3x – 4y + 3 = 0 and 3x + 4y – 21 = 0.
Find the co-ordinates of the vertices of the triangular region formed by these lines and x-axis. Also, calculate the area of this triangle.
OR
Solve for: \(\left(\frac{2 x}{x-5}\right)^2+5\left(\frac{2 x}{x-5}\right)-24=0, x \neq 5\).
Answer:
Given, 3x-4y + 3 = 0
⇒ y = \(\frac{3x + 3}{4}\)

x	3	7	-1
y	3	6	0

and 3x + 4y – 21 = 0
⇒ y = \(\frac{21-3x}{4}\)

x	3	7	11
y	3	0	-3

(i) These lines intersect each other at point (3,3).
Hence, x = 3 and y = 3.
(ii) The vertices of triangular region are P (3, 3), Q (-1,0) and R (7,0).
(iii) Area of A PQR = \(\frac{1}{2}\) × 8 × 3 = 12 Sq units
Hence, area of obtained PQR is 12 sq. units.
OR
Given,
\(\left(\frac{2 x}{x-5}\right)^2+5\left(\frac{2 x}{x-5}\right)-24=0\)
Let \(\frac{2x}{(x-5)}\) = y
∴ y² + 5y – 24 = 0
⇒ (y + 8)(y-3) = 0
y = 3 or – 8
Putting y = 3, we get
\(\frac{2x}{(x-5)}\) = 3
2x = 3x – 15
x = 15
Again, for y = – 8, \(\frac{2x}{(x-5)}\) = – 8
2x = -8x + 40
10x = 40
x = 4
Hence, x = 15 or 4

Question 33.
In the given figure. APQR is an equilateral triangle of side 8 cm and D, E, F are centres of drcular arcs, each of radius 4 cm. Find the area of shaded region. (Use π = 3.14 and \(\sqrt{3}\) = 1.732) [5]

Answer:
Area of shaded region = Area of APQR – 3 (area of sector)

Question 34.
AB is a chord of drcle with centre O. At B, a tangent PB is drawn such that its length is 24 cm. The distance of P from the centre is 26 cm. If the chord AB is 16 cm, find its distance from the centre.

Given, AB is a chord of circle with centre O and tangent PB = 24 cm, OP = 26 cm.
Construction : Join O to B and draw OC ⊥ AB.
By Pythagoras theorem,

OB = \(\sqrt{(26)^2-(24)^2}\)
= \(\sqrt{676 – 576}\)
= \(\sqrt{100}\)
= 10 cm
Now, in ∆OBC, BC = \(\frac{1}{2}\) AB = \(\frac{16}{2}\) = 8 cm.
(Perpendicular drawn from the centre to a chord bisects it.)
OB = 10 cm
OC² = OB² – BC²
= 10² – 8²
OC² = 36
OC = 6 cm
Distance of the chord from the centre = 6 cm

Question 35.
Find the mean of the following distribution:

Class	3-5	5-7	7-9	9-11	11-13
Frequency	5	10	10	7	8

OR

Weekly income (in ₹)	Number of families
0-1000	250
1000-2000	190
2000-3000	100
3000-4000	40
4000-5000	15
5000-6000	5
Total	600

Answer:

Class	Frequency (f)	Mid-Value (x)	f×x
3-5	5	4	20
5-7	10	6	60
7-9	10	8	80
9-11	7	10	70
11-13	8	12	96
	Σf=40		Σfx = 326

∴ mean = \(\frac{Σfx}{Σf}\)
= \(\frac{326}{40}\) = 8.15
OR
For calculating the median of grouped data, we first form c.f. table.

Weekly income (in ₹)	Number of families (f<sub>i</sub>)	c.f.
0-1000	250	250
1000-2000	190	440
2000-3000	100	540
3000-4000	40	580
4000-5000	15	595
5000-6000	5	600

Number of observations = 600 (even)
The median of 600 observations = mean of (300th and 301th observation)
= Median lies in range (1000-2000)
Therefore, median dass = 1000-2000
\(\text { Median }=l+\left[\frac{\frac{n}{2}-c.f}{f}\right] h\)
where,
l = Lower limit of median dass = 1000
n = Total number of observations = 600
c.f. = c.f. preceding the median dass = 250
h = Class size = 1000-2000 = 1000
f = Frequency of median class = 190

Median = \(\text { Median }=1000+\left[\frac{\frac{600}{2}-250}{190}\right] 1000\)
=1000(1 + \(\frac{50}{150}\))
= 1000 × 1.26315
= 1263.15
Hence, the median income of family is 1263.15 per week.

Section – E
Case study based questions are compulsory.

Question 36.
A small scale industry produces a certain boxes of candles in a day. Number of boxes prepared by each worked on a particular day was 2 more than thrice the number of workers working in the industry. The number of boxes produced in a particular day was 85.

I. If the number of workers working in the industry is x. What was the number of boxes of candles prepared by each worker on that particular day?
II. Write the quadratic equation for the given case.
III. Find the number of workers working in the industry.
OR
How many boxes will be prepared when number of workers are increased by 2.
Answer:
I. Let the workers are x,
Number of boxes prepared by the workers = 2 + 3x

II. x(2 + 3x) = 85
⇒ 2x + 3x² = 85
⇒ 3x² + 2x – 85 = 0

III. x(2 + 3x) = 85
2x + 3x² = 85
3x² + 2x – 85 = 0
3x² + 17x – 15x – 85 = 0
x(3x + 17) – 5 (3x + 17) = 0
(3x + 17) (x – 5) = 0
x = \(\frac{17}{3}\), x = 5
Number of workers cannot be negative; therefore x = 5
OR
Number of workers = x Number of workers when increased by
2 = x + 2
∴ Number of boxes prepared by the workers
(x + 2) = (x + 2) [2 + 3 (x + 2)]
= 7 [2+ 3 × 7] [∵ x = 5]
= 7 × 23
= 161

Question 37.
Authority wants to construct a slide in a city park for children. The slide was to be constructed for children below the age of 12 years. Authority prefers the top of the slide at a height of 4 m above the ground and inclined at an angle of 30° to the ground.

I. Find the distance of AB.
II. Find the value of sin² 30° + cos² 60°.
III. If cos A = -v/3 then find the value of 12 cot² A – 2.
OR
In the given figure, if AB + BC = 25 cm and AC = 5 cm, then the value of BC.
Answer:

OR
We have,
∠C = 90°
AB + BC = 25 cm and AC = 5 cm
Let BC be x cm, then AB = (25 – x) cm
By using Pythagoras theorem,
AB² = BC² + AC²
⇒ (25-x)² = x² + (5)²
⇒ 625 – 50x + x² = x² + 25
⇒ 50x = 600
⇒ x = \(\frac{600}{50}\) = 12
Hence, BC = 12 cm

Question 38.
In order to conduct Sports Day activities in a school, lines have been drawn with chalk powder at a distance of 1 m each, in a rectangular shaped ground ABCD, 100 flowerpots have been placed at a distance of 1 m from each other along AD, as shown given figure below. Niharika runs l/4th the distance AD on the 2nd line and posts a green (G) flag. Preet runs l/5th distance AD on the eighth line and posts a red (R) flag.

I. Find the position of green flag.
II. Find the position of red flag.
III. What is the distance between both the flags?
OR
If Rashmi has to post a blue flag exactly halfway between the line segment joining the two flags, where should she post her flag?
Answer:
I. From the given figure,
absdssa = 2
and ordinate \(\frac{1}{4}\)th of 100
= \(\frac{1}{4}\) × 100
= 25
So, required point = (2,25)

II. From the given figure,
absdssa = 8
and ordinate \(\frac{1}{5}\)th of 100
= \(\frac{1}{5}\) × 100
= 20
So, required point = (8,20).

III. Position of Green flag = (2,25)
Position of Red flag = (8,20)
Distance between both the flags
= \(\sqrt{(8 – 2)^2+(20 – 25)^2}\)
= \(\sqrt{(6)^2+(-5)^2}\)
= \(\sqrt{36+25}\)
= \(\sqrt{61}\) units
OR
Position of blue flag
Mid-point of line segment joining the green and red flags
= \(\frac{2+8}{2},\frac{25+20}{2},\)
= (5, 22.5)