Students must start practicing the questions from CBSE Sample Papers for Class 10 Maths with Solutions Set 2 are designed as per the revised syllabus.
CBSE Sample Papers for Class 10 Maths Standard Set 2 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.
The median class of the following distribution is: (1)
| Class | 40-45 | 45-50 | 50-55 | 55-60 | 60-65 | 65-70 | 70-75 |
| Frequency | 2 | 3 | 8 | 6 | 6 | 3 | 2 |
(a) 45 – 50
(b) 65 – 70
(c) 40 – 45
(d) 55-60
Answer:
(d) 55-60
Explanation:
| Class | Frequency | Cumulative Frequency |
| 40-45 | 2 | 2 |
| 45-50 | 3 | 8 |
| 50-55 | 8 | 13 |
| 55-60 | 6 | 19 |
| 60-65 | 6 | 25 |
| 65-70 | 3 | 28 |
| 70-75 | 2 | 30 |
So, N = 30 and \(\frac{N}{2}\) = 15
The cumulative frequency, just greater than 15, is 19 which belongs to class interval 55-60.
Hence, the median class is 55-60.
Question 2.
An integer is chosen at random between 1 and 100. The probabilitg that the chosen number is divisible bg 10 is: (1)
(a) \(\frac{9}{98}\)
(b) \(\frac{8}{95}\)
(c) \(\frac{7}{97}\)
(d) \(\frac{5}{97}\)
Answer:
(a) \(\frac{9}{98}\)
Explanation: Numbers divisible by 10 between 1 and 100 are:
10, 20, 30, 40 90
So, required probability = \(\frac{9}{98}\)
Question 3.
Two different dice are rolled together. The probability of getting a sum of 10 of the numbers on the two dice is: (1)
(a) \(\frac{2}{13}\)
(b) \(\frac{5}{14}\)
(c) \(\frac{1}{12}\)
(d) \(\frac{1}{13}\)
Answer:
(c) \(\frac{1}{12}\)
Explanation:
The pairs having sum of 10 are (4, 6), (5, 5) and (6, 4), out of (6 × 6), i.e. 36 pairs.
So, the required probability is \(\frac{3}{36}\) or \(\frac{1}{12}\)
Question 4.
The area of the largest triangle that can be inscribed in a semi-circle of radius r units is: (1)
(a) 2r2
(b) r2
(c) \(\frac{r^2}{2}\)
(d) 1
Answer:
(b) r2
Explanation:
Area of triangle = \(\frac{1}{2}\) × 2r × r = r2
Question 5.
The total surface area of a quadrant of a wooden sphere of radius 3.5 cm is: (1)
(a) 56 cm2
(b) 35 cm2
(c) 77 cm2
(d) 22
cm2
Answer:
(c) 77 cm2
Explanation: Radius of sphere, r = 3.5 cm
∴ T.S.A = \(\frac{4 \pi r^2}{4}+\frac{\pi r^2}{2}+\frac{\pi r^2}{2}\)
= πr2 + πr2
= 2πr2 = 2π(3.5)2
= 77 cm2
Question 6.
If sin θ – cos θ = 0, then the value of sin4 θ + cos4 θ is: (1)
(a) \(\frac{1}{2}\)
(b) \(\frac{2}{3}\)
(c) \(\frac{\sqrt{3}}{2}\)
(d) \(\frac{1}{\sqrt{2}}\)
Answer:
(a) \(\frac{1}{2}\)
Explanation:
Given, sin θ – cos θ
⇒ \(\frac{sin θ}{cos θ}\) = 1
or tan θ = 1
∴ θ = 45°
Now, sin4 θ + cos4 θ
= sin4 45° + cos4 45°
= \(\left(\frac{1}{\sqrt{2}}\right)^4+\left(\frac{1}{\sqrt{2}}\right)^4\)
= \(\frac{1}{4}+\frac{1}{4}=\frac{2}{4}=\frac{1}{2}\)
Question 7.
The ratio of the height of a tower and the length of its shadow on the ground is \(\sqrt{3}\) : 1 What is the angle of elevation? (1)
(a) 30°
(b) 45°
(c) 60°
(d) 90°
Answer:
(c) 60°
Explanation: Let OA be the tower, OB be its shadow and 0 be the angle of elevation of the sun at that instant
Then, in triangle OAB, we have
tan θ = \(\frac{AO}{OB}\)
tan θ = \(\frac{\sqrt{3}}{1}\)
⇒ θ = 60°
Hence, angle of elevation of the sun is 60°.
Question 8.
The zeroes of the polynomial p(x) = x3 – 4x is: (1)
(a) 1, 4, 2
(b) 2, 0, 3
(c) 0, -2, 2
(d) 0, 2, 2
Answer:
(c) 0, -2, 2
Explanation:
p(x) = x3 – 4x = x(x2 – 4)
= x(x + 2)(x – 2)
Thus, the zeros of p(x) are 0,-2 and 2.
Question 9.
The first negative term of the AP: 20, 19\(\frac{1}{4}\), 18\(\frac{1}{2}\), 17\(\frac{3}{4}\), is: (1)
(a) 27
(b) 24
(c) 25
(d) 28
Answer:
(d) 28
Explanation:
Here, a = 20 and d = –\(\frac{3}{4}\)
If the nth term be the first negative term, then
a + (n – 1) d < 0
i.e, 20 + (n – 1) (-\(\frac{3}{4}\)) < 0 ⇒ n > \(\frac{83}{3}\) or 27\(\frac{2}{3}\)
which is true for n = 28.
Hence, 28th is the first negative term of the given A.P.
Question 10.
The value of k for which the equation x2 + 4x + k = 0 has real roots is: (1)
(a) k = 4
(b) k < 4 (c) k > 4
(d) k ≤ 4
Answer:
(b) k ≤ 4
Explanation: Equation will have real roots
when (4)2 – 4k ≥ 0, i.e. k ≤ 4
Question 11.
Write the solution of the following pair of equations: x – 3y = 2; 3x – y = 14 is: (1)
(a) x = 5, y = 1
(b) x = 3, y = 5
(c) x = 2, y = 1
(d) x = 3, y = 2
Answer:
(a) x = 5, y = 1
Explanation:
Given, equations are
x – 3y =2 …(i)
and 3x – y =14 …(ii)
Then, x = 2 + 3y [From equation (i)]
Put the value of x’ in equation (ii)
3(2 + 3y) – y = 14
⇒ 6 + 9y – y = 14
⇒ 6 + 8y = 14
⇒ 8y = 8
⇒ y = 1
Then, x = 2 + 3 × 1 = 5
Then values of x and y are 5 and 1 respectively.
Question 12.
Write a quadratic polynomial for which sum and product of the zeros are 3 and -10 respectively. (1)
(a) 2x2 + 3x+5
(b) x2 – 3x – 10
(c) x2 + 3x + 10
(d) x2 – 5x + 4 1
Answer:
(b) x2 – 3x – 10
Explanation:
Sum of zeroes = 3
i.e., a + b = 3
and product of zeroes = – 10
i.e., ab = – 10
Then, quadratic polynomial is
x2 – (a + b) x + ab
i.e., x2 – 3x – 10
Question 13.
The pairs of equations x + 2y + 5 = 0 and 5x + 10y + 25 = 0 have: (1)
(a) unique solution
(b) exactly two solutions
(c) Infinitely many solutions
(d) No solution
Answer:
(c) Infinitely many solutions
Explanation:
\(\frac{a_1}{a_2}=\frac{1}{5}\)
\(\frac{b_1}{b_2}=\frac{2}{10}=\frac{1}{5}\)
\(\frac{c_1}{c_2}=\frac{2}{25}=-\frac{1}{5}\)
This shows:
\(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\)
Therefore, the pair of equations has infinitely many solutions.
Question 14.
The perimeter of a circle having radius 7cm is equal to: (1)
(a) 30 cm
(b) 3.14 cm
(c) 31.4 cm
(d) 44 cm
Answer:
(d) 44 cm
Explanation:
The perimeter of the circle is equal to the circumference of the circle.
Circumference = 2πr
= 2 × \(\frac{22}{7}\) × 7
= 44 cm
Question 15.
The chord of a circle of radius 10 cm subtends a right angle at its centre. The length of the chord is: (1)
(a) 10 cm
(b) 20 cm
(c) 10\(\sqrt{2}\)cm
(d) 10\(\sqrt{3}\) cm
answer:
(c) 10\(\sqrt{2}\)cm
Explanation:
Here, AB is a chord, subtending a right angle at the centre O.
Given, radius of the circle, r = 10 cm.
In AAOB, by Pythagoras theorem
AB2 = OA2 + OB2
= 102 + 102
= 100 + 100 = 200
AB = \(\sqrt{200}\) = 10\(\sqrt{2}\) cm
Question 16.
If 6 times the 6th terms of on A.P. is equal to 9 times the 9h term, then find its 15th terms: (1)
(a) 10
(b) 31
(c) 22
(d) 0
Answer:
(d) 0
Explanation:
Let, the first term of an A.P. be ‘a’ and its common difference be ‘d’.
Then, 6 (a6) = 9(a9) (given)
6(a + 5 d) = 9(a + 8d)
6 a + 30 d = 9a + 72 d
-3 a = 42 d
a = – 14d …(i)
15th term, = a + 14d
= – 14d + 14d [using (i)]
= 0
Hence, the 15th term of A.P. is 0.
Question 17.
The mid-point of the line segment joining the points (-2, 4) and (6,10) is: (1)
(a) (2, 7)
(b) (5, 2)
(c) (3, 5)
(d) (4, 5)
Answer:
(a) 2, 7
Explanation:
The mid-point of the line -2 + 6 4 + 10
segment is \(\left(\frac{-2+6}{2}, \frac{4+10}{2}\right)\) i.e., (2,7)
Question 18.
The value of ‘a’, if HCF (a, 18) = 2 and LCM (a, 18) = 36, is: (1)
(a) 2
(b) 5
(c) 7
(d) 4
Answer:
(d) 4
Explanation:
Given, HCF (a, 18) = 2 and
LCM (a, 18) = 36
HCF (a, b) × LCM (a, b) = a × b
2 × 36 = a × 18
a = 4
Hence, value of ‘a’ is 4.
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): The graph of the linear equations 5x + 3y = 12 and 7x – 5y = 4 gives a pair of intersecting lines.
Statement R (Reason): The graph of linear equations a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 gives a pair of intersecting lines if \(\frac{a_1}{a_2}\) ↑\(\frac{b_1}{b_2}\) (1)
Answer:
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A)
Explanation:
The graph of linear equations 5x + 3y = 12, and 7x – 5y = 4 gives a pair of intersecting lines.
5x + 3y = 12
7x – 5y = 4
\(\frac{5}{7}\) ≠ \(\frac{3}{-5}\)
Hence lines intersects each others.
The graph of linear equations a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 gives a pair of
Intersecting lines if \(\frac{a_1}{a_2}\) ↑\(\frac{b_1}{b_2}\)
Question 20.
Statement A (Assertion): If the circumference of a circle is 176 cm, then its radius is 22 cm.
Statement R (Reason): Circumference = 2π × radius. (1)
Answer:
(d) Assertion (A) is false but reason(R) is true.
Explanation:
Circumference (c) = 176 cm
∴ 2 nr= 176
∴ 2 × \(\frac{22}{7}\) × r = 176
∴ \(\frac{44}{7}\) × r = 176
∴ r = 176 × \(\frac{7}{44}\) = 28 cm
∴ The radius of the circle is 28 cm.
Section – B
(Section B consists of 5 questions of 2 marks each.)
Question 21.
If n is a positive odd integer, then show that n2 – 1 is divisible by 8.
OR
Check whether 15n can end with digit zero for any natural number n. (2)
Answer:
We know that every positive odd integer is of the form 2q + 1, where q is a whole number. Put n = 2q + 1 and get
n2 – 1 = (2g + 1)2 -1
= (2q + 1 – 1) (2q + 1 + 1)
= 2q(2q + 2)
= 4 q(q + 1)
Clearly, for q = 0, 1, 2 4q{q + 1) is divisible by 8.
OR
No; because 15n = (3 × 5)n = 3n × 5n.
So, the only primes in the factorisation of 15n are 3 and 5, and not 2 and 5.
Hence, 15n cannot end with the digit 0.
Question 22.
If P(5, 7), Q(x, – 2) and R(- 3, y) are collinear points such that PR = 2PQ, calculate the values of x and y.
OR
Show that the roots of the quadratic equation: (b – c)x2 + (c – a) x + (a – b) = 0 are equal if c + a = 2b. (2)
Answer:
Since P, Q, R are collinear and PQ = \(\frac{1}{2}\) PR,
So, Q is the mid-point of PR.
(x, -2) =\(\left(\frac{5-3}{2}, \frac{7+y}{2}\right)\)
⇒ (x, -2) = (1, \(\frac{7+y}{2}\))
⇒ x = 1 and \(\frac{7+y}{2}\) = -2
⇒ x = 1 and y = -11
OR
The roots of the given equation will be equal, if
(c – a)2 = 4(b – c) (a – b)
For c + a = 2b, we have:
c-a = 2b-a-a ⇒ c-a=2(b-a)
⇒ (c – a)2 = 4(a – b)2 …(i)
Also, c + a = 2b
gives b – c = a – b
So, 4(b – c)(a – b) = 4(a – b)2 …(ii)
From (i) and (ii), we have
(c – a)2 = 4(b – c)(a – b) = 4(a – b)2
Question 23.
Determine the value of “k” so that the ratio of the zeros in the quadratic polynomial 3x2 – kx + 14 is 7:6. (2)
Answer:
Let the zeros are 7p and 6p.
3x2 – k + 14
7p + 6p = \(\frac{-(-k)}{3}\) = \(\frac{k}{3}\)
and 7p × 6p = \(\frac{14}{3}\)
⇒ 42 p2=\(\frac{14}{3}\)
p = 3
⇒ 39 p = k
∴ k = 39 × 3
∴ k = 117
Question 24.
The sum of circumferences of two circles is 132 cm. If the radius of one circle is 14 cm, find the radius of the other circle. (2)
Answer:
Let r = 14 cm and R cm be the radii of two circles.
Then,
2π(14) + 2πR = 132
⇒ 2πR = 132 – 88
R = \(\frac{44 × 7}{2 × 22}\) = 7
Thus, the radius of the other circle is 7 cm.
Question 25.
In a car park, there are 125 cars, 3p motorbikes, 2q lorries and 20 buses. One of the vehicles leaves the car park at random. Given that the probability that the 3 vehicle is a motorbike is \(\frac{3}{40}\) and probability that the vehicle is a bus is \(\frac{1}{10}\), form a pair of linear equations in p and q. (2)
Answer:
Section – C
(Section C consists of 6 questions of 3 marks each.)
Question 26.
Find the HCF and the LCM of 72 and 120, using prime factorisation method. (3)
Answer:
72 = 2 × 2 × 2 × 3 × 3. or 23 × 32
120 = 2 × 2 × 2 × 3 × 5, or
23 × 31 × 51
So, HCF(72,120) = 23 × 31, i.e. 24.
Question 27.
The diagram given below shows a sequence of square wire frames. The lengths of a side of these frames are ‘x* cm, (x + 3) cm, (x + 6)
cm, respectively. The sum of the areas of the first three squares is 525 cm².
(A) Express the length of a side of the nth frame in terms of x and n.
(B) Find the value of x.
(C) A piece of wire is 99 cm long. It is cut and bent into a frame in the sequence. Find the length of a side of the largest frame than can be formed.
OR
If the roots of the equation x2 +2cx + ab = 0 are real and unequal, prove that the equation x2 – 2(a + b)x + a2 + b2 + 2c2 = 0 has no real roots. (3)
Answer:
The length of the sides of the sequence of squares are: x, x + 3, x + 6
It is an A.P. with a = x, d = 3
(A) So, side of the nth square = a + (n – 1 )d
= x + (n – 1)(3)
= x + 3n – 3
(B) It is given that
x2 + (x + 3)2 + (x + 6)2 = 525
⇒ 3x2 + 18x + 45 = 525
⇒ x2 + 6x- 160 = 0
⇒ x2 + 16x – 10x – 160 = 0
⇒ x (x + 16) – 10 (x + 16) = 0
⇒ (x + 16)(x – 10) = 0
⇒ x = 10 (∴ x ≠ -16)
(C) Side of the square formed with a wire of length 99 cm at the most can be \(\frac{99}{4}\) cm.
Thus, the side of the square with largest frame = 24 cm.
OR
The two equations are
x2 + 2cx + ab = 0 …(i)
and x2 – 2(a + b) x + a2 + b2 + 2c2 = 0 …(ii)
Let D1 and D2 be the discriminants of equations (i) and (ii), respectively.
Then,
D1 = (2c)2 – 4 × 1 × ab = 4 (c2 – ab)
D2 = (-2 (a + b))2 – 4 × 1 × (a2 + b2 + 2c2)
= 4 (a + b)2 – 4 (a22 + b2 + 2c2)
= 4 (a2 + b2 + 2ab) – 4a2 – 4b2 – 8c2
= 8ab – 8c2
= – 8 (c2 – ab)
Since, the roots of equation (i) are real and unequal Therefore,
D1 > 0
⇒ 4 (c2 – ab) > 0
⇒ c2 – ab > 0
⇒ – 8 (c2 – ab) < 0
⇒ D2 < 0
∴ Roots of equation (ii) are not real
Question 28.
Let A(4,2), B(6,5) and C(1, 4) be the vertices of ∆ABC. The median AD from A meets BG in D. Find the coordinates of the point P on AD such that, AP: PD = 2 : 1. (3)
Answer:
Question 29.
If tan θ = \(\frac{12}{13}\), evaluate \(\frac{2 \sin \theta \cos \theta}{\cos ^2 \theta-\sin ^2 \theta}\)
Answer:
Question 30.
There are 4 green marbles, 8 white marbles, and 5 red marbles in a box. Randomly, one marble is taken out of the box. What is the probability that the marble will be one of the following:
(A) Red
(B) White
(C) Not green
OR
A sphere of diameter 6 cm is dropped in a right circular cylindrical vessel partly filled with water. The diameter of the cylindrical vessel is 12 cm. If the sphere is completely submerged in water, by how much will the level of water rise in the cylindrical vessel? (3)
Answer:
Total number of marbles in the box = 5 + 8 + 4 = 17
Total number of elementary events = 17
(A) There are 5 red marbles in the box. Favourable number of elementary events = 5
P(getting a red marble) = \(\frac{5}{17}\)
(B) There are 8 white marbles in the box. Favourable number of elementary events = 8
P(getting a white marble) = \(\frac{8}{17}\)
(C) There are 5 + 8 = 13 marbles in the box, which are not green.
Favourable number of elementary events = 13
P(not getting a green marble) = \(\frac{13}{17}\)
OR
We have, Radius of the sphere = 3 cm
Volume of the sphere = \(\frac{4}{3}\)π(3)3 cm3
Radius of the cylindrical vessel = 6 cm
Suppose water level rises by ‘h’ cm in the cylindrical vessel. Then,
Volume of the cylinder of height ‘h’ cm and radius 6 cm
= π(6)2 h cm3, i.e. 36π h cm3
Clearly, volume of water displaced by the sphere is equal to the volume of the sphere.
⇒ 36π h = 36π
⇒ h = 1 cm
Hence, water level rises by 1 cm.
Question 31.
If sec θ + tan θ = m, prove that \(\frac{m^2 – 1}{m^2 + 1}\) =sin θ.
Answer:
Proof:
Section – D
(Section D consists of 4 questions of 5 marks each.)
Question 32.
From the first floor of Qutab Minor, which 25 m from the level ground, a man observes the top of a building at an angle of elevation of 30° and the angle of depression of the base of the building to be 60°. Calculate the height, of the building. (5)
Answer:
Let the first floor of Qutab Minor AB be at the point x.
Question 33.
In the figure, ABC and DBC are two triangles is at a height of on the same base BC. If AD intersects BC at Q,
show that \(\frac{\operatorname{ar}(\triangle \mathrm{ABC})}{\operatorname{ar}(\triangle D B C)}=\frac{\mathrm{AD}}{\mathrm{DO}}\)
OR
Prove that the line segments joining the mid – points of the sides of a triangle form four triangles, each of which is similar to the original triangle. (5)
Answer:
Draw AM ⊥ BC and DN ⊥ BC
In As AOM and DON, we have:
∠M = ∠N [Each = 90°]
∠AOM = ∠DON
[Vertically opposite angles]
So, by AA similarity criterion,
∆AOM ~ ADON .
\(\frac{\mathrm{AM}}{\mathrm{DN}}=\frac{\mathrm{AO}}{\mathrm{DO}}\) …(i)
OR
Consider a triangle ABC with D, E, F as the mid-points of sides BC, CA, AB respectively.
Since, F and E are mid-points of AB and AC respectively,
So, by mid-point theorem,
FE || BC ⇒ ∠AFE = ∠B
Thus, in ∆AFE and ∆ABC, we have:
∠AFE = ∠B and ∠A is common in both.
So, ∆AFE ~ ∆ABC
Similarly, AFBD ~ ∆ABC and AEDC ~ ∆ABC
Now, we shall show that ADEF ~ ∆ABC
Clearly, ED || AF and DF || EA
∴ AFDE is a parallelogram.
⇒ ∠EDF = ∠A
Similarly, BDEF is a parallelogram.
∴ ∠DEF = ∠B
Thus, in As DEF and ABC, we have:
∠EDF = ∠A, ∠DEF = ∠B
So, by AA similarity criterion, ADEF ~ ∆ABC
Thus, each one of the triangles AFE, FBD, EDC and DEF is similar to triangle ABC.
Question 34.
A piece of cloth costs f 35. If the piece were 4 m longer and each metre costs ? 1 less, the cost would remain unchanged. How long is the piece? (5)
Answer:
Let ‘l’ metres be the length of the piece, with a total cost of ₹ 35.
So, cost of 1 metre long piece = ₹ (\(\frac{35}{l}\))
Also, “l + 4” metres long piece costs ₹ 35.
So, cost of 1 metre long piece = ₹ \(\frac{35}{l+4}\)
As per the question;
\(\frac{35}{l+4}\) + 1 = \(\frac{35}{l}\)
\(\frac{35+l+4}{l+4}\) = \(\frac{35}{l}\)
⇒ 39l + l2 = 35l + 140
⇒ l2+4l-140 = 0
⇒ l2 + 14l – 10l – 140 = 0
⇒ l(l + 14) – 10(l+ 14) = 0
⇒ (l + 14) (l – 10) = 0
⇒ l – 10 = 0
[(l+14)≠0]
Thus, 10 metres is the length of the piece.
Question 35.
PQ is a 16 cm long chord of a circle with a 10 cm radius in given figure. At a point T, the tangents at P and Q come together. Determine TP’s length.
OR
Find the mean and mode of the following frequency distribution:
| Marks | 0-9 | 10-19 | 20 -29 | 30 -39 | 40 -49 | 50-59 |
| Number of students | 4 | 6 | 12 | 6 | 7 | 5 |
Answer:
Given PQ = 16 cm
PO= 10 cm
To Find: TP
PR = RQ = \(\frac{16}{2}\) = 8 cm
[Perpendicular from the centre bisects the chord]
In ∆OPR
OR= \(\sqrt{OP^2-PR^2}\)
= \(\sqrt{10^2-8^2}\)
= \(\sqrt{100-64}\)
= \(\sqrt{36}\) = 6 cm
Let ∠POR be θ
In ∆POR, tan θ = \(\frac{PR}{RO}\) = \(\frac{8}{6}\)
tan θ = \(\frac{4}{3}\)
We know, OP ⊥ TP (Point of contact of a tangent is perpendicular to the line from the centre)
In ∆OTP, tan θ = \(\frac{OP}{TP}\)
\(\frac{4}{3}\) = \(\frac{10}{TP}\)
TP = \(\frac{10 × 3}{4}\) = \(\frac{15}{2}\) = 7.5 cm
OR
The given frequency distribution in “exclusive” form is:
| Marks | Frequency |
| 0.5-9.5 | 4 |
| 9.5-19.5 | 6 |
| 19.5-29.5 | 12 |
| 29.5-39.5 | 6 |
| 39.5-49.5 | 7 |
| 49.5-59.5 | 5 |
Calculation of Mode:
Here, class with highest frequency is 19.5 – 29.5.
So, modal class = 19.5 – 29.5
Here modal class is 19.5 – 29.5
So, l = 19.5, f1 = 12, cf = 6, f0 = 6, f2 = 6 and h = 10
Calculation of Mean:
Let the assumed mean(A) be 34.5.
| Class interval | Mid-value (xi) | fi | di = fi– A A = 34.5 | fidi |
| 0.5-9.5 | 4.5 | 4 | -30 | -120 |
| 9.5-19.5 | 14.5 | 6 | -20 | -120 |
| 19.5-29.5 | 24.5 | 12 | -10 | -120 |
| 29.5-39.5 | 34.5 | 6 | 0 | 0 |
| 39.5-49.5 | 44.5 | 7 | 10 | 70 |
| 49.5-59.5 | 54.5 | 5 | 20 | 100 |
| fi = 40 | fidi = -190 |
Then, Mean = A + \(\frac{\Sigma f_i d_i}{\Sigma f_i}\)
Thus, Mean = 34.5 + \(\frac{(-190)}{40}\)
= 34.5 + \(\frac{(-19)}{4}\)
= 34.5 – 4.75
= 29.75
Section – E
(Case Study Based Questions)
(Section E consists of 3 questions. Ail are compulsory.)
Question 36.
Formula one Portugese Grand Prix technical team at the Algarve International Circuit are analysing last year data of drivers’ performance to provide valuable inferences to commentators on how the drivers can improve this year.
The length of time taken by 80 drivers to complete a journey is given in the table below:
| Times (in minutes) | 70-80 | 80-90 | 90-100 | 100-110 | 110-120 | 120-130 |
| Number of drivers | 4 | 10 | 14 | 20 | 24 | 8 |
On the basis of the above information, answer the following questions:
(A) In which interval does the median of the distribution lie?
OR
Find the estimate of the mean time (in minutes) taken to complete the journey. (2)
(B) One driver is chosen at random. Find the probability that he took 90 minutes or less for the journey. (1)
(C) Two drivers are chosen at random. Find the probability that one took 80 minutes or less and other took more than 120 minutes for the journey. (1)
Answer:
(A)
| Time | f | c.f. |
| 70-80 | 4 | 4 |
| 80-90 | 10 | 14 |
| 90-100 | 14 | 28 |
| 100-110 | 20 | 48 |
| 110-120 | 24 | 72 |
| 120-130 | 8 | 80 |
| Total | 80 |
Then, \(\frac{N}{2}\) = 40
Cumulative frequency just greater than 40 is 48, which lies in the class interval 100¬110.
So, median class is 100-110.
OR
| Time | f | xi | fixi |
| 70-80 | 4 | 75 | 300 |
| 80-90 | 10 | 85 | 850 |
| 90-100 | 14 | 95 | 1330 |
| 100-110 | 20 | 105 | 2100 |
| 110-120 | 24 | 115 | 2760 |
| 120-130 | 8 | 125 | 1000 |
| Total | 80 | 8340 |
Then, Mean = A + \(\frac{\Sigma f_i d_i}{\Sigma f_i}\)
Mean = 34.5 + \(\frac{8340}{80}\)
= 104.25
(B) P(takes less than 90 minutes),
= \(\frac{4+10}{80}\)
= \(\frac{14}{80}\)
= \(\frac{7}{40}\)
(C) Required probability,
= \(\frac{4}{80}\) × \(\frac{8}{80}\)
= \(\frac{1}{20}\) × \(\frac{1}{10}\)
= \(\frac{1}{200}\)
Question 37.
Google maps cartography team is working on improving the scalability quality of maps when you use the app on your phones to zoom in using 4 fingers. They are using a proprietary tool called ‘MapMaker’ to figure out scalability factors. A mathematical model is created for a type of object (below cross¬section) to test its scalability on maps app.
In the diagram, AC = 8 cm, CE = 4 cm and the area of the triangle BECis 4.2 sq cm.
Another enlargement with centre E, maps AEBC onto AEFA. BC = 3.6 cm.
On the basis of the above information, answer the following questions:
(A) An enlargement, with centre A, maps ∆ABC onto ∆ADE, then find the scale factor of the enlargement. (1)
(B) Find the length of AF. (1)
(C) Calculate the area of ∆ABC.
OR
Find the area of ∆EFA. (2)
Answer:
(A) Scale factor = \(\frac{AC}{AE}\)
= \(\frac{AC}{AC+CE}\) = \(\frac{8}{8+4}\)
= \(\frac{8}{12}\) = \(\frac{2}{3}\)
(B) Since, ∆EBC ~ ∆EAF
\(\frac{EC}{EA}\) = \(\frac{BC}{AF}\)
\(\frac{4}{12}\) = \(\frac{3.6}{AF}\)
⇒ AF = 3.6 × 3
= 10.8 cm
(C) In ∆ABC, using Pythagoras theorem,
we have
AB2 = BC2 + AB2
AB2 = 64 – 12.96
= 51.04
⇒ AB = 7.15
∴ Area of ∆ABC = \(\frac{1}{2}\) × BC × AB
= \(\frac{1}{2}\) × 3.6 × 17.5
= 12.87
OR
Since, with centre E enlargement as done to ∆EBC and ∆EFA.
Question 38.
Amrita makes biscuits. The amount of mixture required to make one biscuit is 18 cu cm. Before it is cooked, the mixture is rolled into a sphere. After the biscuit is cooked, the biscuit becomes a cylinder of radius 3 cm and height 0.7 cm ( The increase in volume is due to air being trapped in the biscuit)
The cross – section of the box is a regular hexagon, containing 7 biscuits , arranged in Diagram I.
Three of the biscuits are shown in Diagram II. O is the centre of the hexagon and of the middle biscuits. B is the point where two biscuits touch. A and C are the centres of these biscuits. E is the mid – point of the side DF of the box.
On the basis of the above information, answer the following questions:
(A) Find the length of OB. (1)
(B) Find the length of OE. (1)
(C) Find the volume of the biscuits after it is cooked and also find the air trapped, while cooling the biscuit.
OR
Using the concept of similarity of triangles, find the length of a side of the box. (2)
Answer:
(A) Length of OB = \(\sqrt{OA^2 – AB^2}\)
(using Pythagoras theorem in ΔOBA)
= \(\sqrt{36 – 9}\)
= \(\sqrt{27}\)
= 5.19 5.2 cm
(B) Length of BE = 3 cm
Length of OE = OB + BE = 5.2 + 3
= 8.2 cm
(C) Volume of a biscuit, after cooking = Volume of cylinder = πr2h
= \(\frac{22}{7}\) × 3 × 3 × 0.7 7
= 22 × 0.9
= 19.8 cu cm
Air trapped = Volume of a biscuit after cooking – Volume of a biscuit before cooking
= 19.8 – 18
= 1.8 cu. cm
OR
Since, ∆OAB = ∆ODE
\(\frac{OB}{OE}\) = \(\frac{AB}{DE}\)
⇒ \(\frac{5.2}{8.2}\) = \(\frac{3}{DE}\)
∴ DE = 4.73
∴ DF = 2 × DE = 9.4 cm