Students must start practicing the questions from CBSE Sample Papers for Class 10 Maths with Solutions Set 9 are designed as per the revised syllabus.
CBSE Sample Papers for Class 10 Maths Standard Set 9 with Solutions
Time: 3 hrs
Max. Marks: 80
Instructions
- This question paper has 5 Sections A-E.
- Section A has 20 MCQs carrying 1 mark each.
- Section B has 5 questions carrying 2 marks each.
- Section C has 6 questions carrying 3 marks each.
- Section D has 4 questions carrying 5 marks each.
- Section E has 3 Case Based integrated units of assessment (4 marks each).
- All questions are compulsory. However, an internal choice in 2 questions of 2 marks, 2 questions 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.
- 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.
In a ∆ABC, D and E are point on the sides AB and AC, respectively such that DE || BC. If \(\frac{A D}{D B}\) = \(\frac{2}{3}\) and AC = 18 cm, then the value of AE is [1]
(a) 5 cm
(b) 6.2 cm
(c) 8 cm
(d) 7.2 cm
Answer:
(d) 7.2 cm
In ∆ABC \(\frac{A D}{D B}\) = \(\frac{A E}{E C}\)
[by basic proportionality theorem]
Question 2.
The value of sec 45° + cosec 45° is [1]
(a) 2\(\sqrt{2}\)
(b) \(\frac{\sqrt{3}+1}{2}\)
(c) 1
(d) \(\sqrt{2}\)
Answer:
(a) 2\(\sqrt{2}\)
sec 45° + cosec 45° = \(\sqrt{2}\) + \(\sqrt{2}\) = 2\(\sqrt{2}\)
Question 3.
In the following figure, then the value of ∠F is [1]
(a) 0°
(b) 30°
(c) 45°
(d) 60°
Answer:
(d) 60°
In ∆ABC and ∆DEF,
\(\frac{A B}{D F}\) = \(\frac{B C}{F E}\) = \(\frac{C A}{E D}\) = \(\frac{1}{2}\)
∴ By SSS criterion of similarity, we have
∆ABC – ∆DFE
⇒ ∠A = ∠D = ∠B = ∠F and ∠C = ∠E
∴ ∠F = 60°
Question 4.
Factorise the number 98 through factor tree [1]
(a) 2
(b) 2 × 73
(c) 72
(d) 2 × 72
Answer:
(d) 2 × 72
Factor tree is as follows
∴ 98 = 2 × 7 × 7 = 2 × 72
Question 5.
A tower stands near an airport. The angle of elevation θ of the tower from a point on the ground is such that its tangent is \(\frac{5}{12}\).
The height of the tower, if the distance of the observer from the tower is 120 m is [1]
(a) 60 m
(b) 50 m
(c) 40 m
(d) None of these
Answer:
(b) 50 m
Let BC = h m be the height of the tower and let A be the point on the ground such that AB = 120 m and ∠BAC = θ.
∴ h = \(\frac{5 \times 120}{12}\) = 5 × 10 = 50 m
Question 6.
In the figure, M is the mid-point of line LN. Then, x + y is equal to [1]
(a) 7.5
(b) 3.5
(c) 4.5
(d) 5.5
Answer:
(a) 7.5
From the figure, mid-point of LN is given by
(x, y) = \(\left(\frac{7+1}{2}, \frac{3+4}{2}\right)\) = (4, 3.5)
Now, value of x + y = 4 + 3.5 = 7.5
Question 7.
If four vertices of a parallelogram taken in order are (- 3, -1), (a, b), (3,3) and (4, 3), then a : b is equal to [1]
(a) 1 : 4
(b) 4 : 1
(c) 1 : 2
(d) 2 : 1
Answer:
(b) 4 : 1
Let points be A(-3, -1), B(a, b), C(3, 3) and D(4, 3).
Coordinates of the mid-point of AC = Coordinates of the mid-point of BD
[∵ in parallelogram, diagonals bisect each other]
Question 8.
The area of a circle whose circumference is 52.8 cm is [1]
(a) 221.76 cm2
(b) 224.75 cm2
(c) 220.76 cm2
(d) None of these
Answer:
(a) 221.76 cm2
Given, 2πr = 52.8
⇒ 2 × \(\frac{22}{7}\) × r = 52.8
⇒ r = 52.8 × \(\frac{7}{44}\)
⇒ r = 8.4 cm
Area of circle = πr2
= (\(\frac{22}{7}\) × 8.4 × 8.4)cm2
= 221.76 cm2
Question 9.
If Σfi is the sum of frequency and it is equal – to the sum of first 10 natural numbers and Σfixi = 120, then the mean is [1]
(a) 2
(b) 0
(c) 2.18
(d) 1.18
Answer:
(c) 2.18
Given, Σfi = 1 + 2 + ….. + 10
= \(\frac{10}{2}\)[2 × 1 + (10 – 1) × 1]
[Sn = \(\frac{n}{2}\)[2a + (n – 1)d]]
= 5[2 + 9] = 55
∴ Mean, \(\frac{\Sigma f_i x_i}{\Sigma f_i}\) = \(\frac{120}{55}\) = 2.18
Question 10.
If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of 80°, then the value of ∠POA is [1]
(a) 60°
(b) 90°
(c) 0°
(d) 50°
Answer:
(d) 50°
We have, OA ⊥ AP
[tangent is perpendicular to the radius at the point of contact]
⇒ ∠OAP = 90°
[∵ the centre of a circle lies on the bisector of the angle between the two tangents]
= \(\frac{1}{2}\) × 80° = 40°
In ∆APO, ∠OAP + ∠APO + ∠POA = 180°
⇒ 90° + 40° + ∠POA = 180°
⇒ ∠POA = 50°
Question 11.
A cylindrical vessel 32 cm high and 18 cm as the radius of the base, is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 24 cm, then the radius of its base is [1]
(a) 30 cm
(b) 36 cm
(c) 34 cm
(d) 35 cm
Answer:
(b) 36 cm
Given cylindrical vessel having radius, r = 18 cm and height h = 32 cm
Also given, height of conical heap, H = 24 cm
Now, volume of cylindrical vessel = πr2h
= π(18)2 × 32
Now, volume of conical heap = \(\frac{1}{3}\)πr2H
= \(\frac{1}{3}\) × π × R2 × (24)
But according to the given condition,
Volume of cylindrical vessel = Volume of conical heap
∴ π(18)2 × 32 = \(\frac{1}{3}\)πr2 × 24
⇒ R2 = \(\frac{(18)^2 \times 3 \times 32}{24}\)
= (18)2 × 4
∴ R2 = (36)2
⇒ R = 36 cm
Question 12.
The median of a given data is 20. If each item is increased by 2, then the new median will be [1]
(a) 40
(b) 10
(c) 22
(d) None of these
Answer:
(c) 22
Since, median is the middle most item and each item is increased by 2, then the new median will be 20 + 2 = 22.
Question 13.
For some integer P, every even integer is of the form [1]
(a) P
(b) P + 1
(c) 2P
(d) 2P + 1
Answer:
(c) 2P
We know that, even integer are 2, 4, 6,……
So, it can be written in the form of 2P.
Where, P = integer = Z
[since, integer is represented by Z]
or P = ……, -1, 0, 1, 2, 3, ………….
∴ 2P = …….., -2, 0, 2, 4, 6 ……….
Question 14.
If the first three terms of an AP are x – 1, x + 1, 2x + 3, then the value of x is [1]
(a) 1
(b) 2
(c) -1
(d) 0
Answer:
(d) 0
We have, x + 1 – (x – 1) = (2x + 3) – (x + 1)
[∵ common difference remains same]
⇒ 2 = x + 2 ⇒ x = 0
Question 15.
If tan (θ1 + θ2) = \(\sqrt{3}\) and sec (θ1 – θ2) = \(\frac{2}{\sqrt{3}}\), then the value of sin 2θ1 + tan 3θ2 is [1]
(a) 0
(b) 1
(c) 2
(d) 3
Answer:
(c) 2
Given, tan(θ1 + θ2) = \(\sqrt{3}\)
⇒ θ1 + θ2 = 60° …… (i)
Also, sec(θ1 – θ2) = \(\frac{2}{\sqrt{3}}\) ⇒ θ1 – θ2 = 30° …….(ii)
On adding Eqs. (i) and (ii), we get
2θ1 = 90° ⇒ θ1 = 45° and θ2 = 15°
∴ sin 2 θ1 + tan 3θ2 = sin 90° + tan 45° = 1 + 1 = 2
Question 16.
The ratio, in which point P(1, 2) divides the join of A(-2, 1) and B(7, 4), is [1]
(a) 1 : 2
(b) 2 : 1
(c) 2 : 2
(d) None of these
Answer:
(a) 1 : 2
Let the required ratio is k : 1.
Then, \(\frac{7 k-2}{k+1}\) =1 [by section formula]
⇒ 7k – 2 = k + 1 ⇒ 6k = 3 ∴ k = \(\frac{1}{2}\)
∴ Required ratio = 1 : 2
Question 17.
If \(\frac{1}{2}\) is a root of the equation x2 + kx – \(\frac{5}{4}\) = 0, then the value of k is [1]
(a) 2
(b) -2
(c) \(\frac{1}{4}\)
(d) \(\frac{1}{2}\)
Answer:
(a) 2
Since, \(\frac{1}{2}\) is a root of the quadratic equation x2 + kx – \(\frac{5}{4}\) = 0.
Then, (\(\frac{1}{2}\))2 + k(\(\frac{1}{2}\)) – \(\frac{5}{4}\) = 0
⇒ \(\frac{1}{4}\) + \(\frac{k}{2}\) – \(\frac{5}{4}\) = 0
⇒ \(\frac{1+2 k-5}{4}\) = 0
⇒ 2k – 4 = 0 ⇒ 2k = 4 ⇒ k = 2
Question 18.
If the perimeter of a circle is equal to that of a square, then the ratio of their areas is [1]
(a) 22 : 7
(b) 14 : 11
(c) 7 : 22
(d) 11 : 14
Answer:
(b) 14 : 11
Let r be the radius of a circle and a be the side of a square.
Given, Perimeter of a circle = Perimeter of a square
∴ 2πr = 4a ⇒ a = \(\frac{\pi r}{2}\) ….. (i)
Area of circle = πr2
Area of square = a2
Directions In the question number 19 and 20, a statement of Assertion (A) is followed by a statement of Reason (R). Choose the correct option.
Question 19.
Assertion (A) HCF of (11, 17) is 1.
Reason (R) If p and q are prime, then HCF (p, q) = 1
(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 but 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. [1]
Answer:
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
As both the given numbers (11, 17) are prime numbers
∴ HCF of (11, 17) = 1
Hence, both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
Question 20.
Assertion (A) 4x + 3y = 18 is a line which is parallel to X-axis.
Reason (R) The graph of linear equation ax = b, where a ≠ 0 is parallel to X-axis. [1]
(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 but 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.
Answer:
(d) Assertion (A) is false but Reason (R) is true.
The graph of the equation 4x + 3y = 18 is shown below
Hence, Assertion (A) is false.
Given, ax = b ⇒ x = \(\frac{b}{a}\)
We know that, graph of x = \(\frac{b}{a}\) is a straight line parallel to Y-axis.
So, the given Reason is true.
Section B
(Section B consists of 5 questions of 2 marks each.)
Question 21.
In ∆ABC, right angled at B, if tan A = \(\frac{1}{\sqrt{3}}\), then find the value of
(i) sin A cos C + cos A sin C.
(ii) cos A cos C – sin A sin C. [2]
Answer:
Given, tan A = \(\frac{1}{\sqrt{3}}\)
⇒ tan A = tan 30° [∵ tan 30° = \(\frac{1}{\sqrt{3}}\)]
⇒ ∠A = 30° ……….. (i)
On applying angle sum property in right angled ∆ABC, we get
∠A + ∠B + ∠C = 180°
⇒ 30° + 90° + ∠C = 180°
[∵ from Eq. (i) and given, ∠B = 90° ]
⇒ ∠C = 180° – 120°
⇒ ∠C = 60° ……… (ii)
(i) sinAcosC + cosAsinC = sin30°cos60° + cos 30° sin 60°
[from Eqs. (i) and (ii)]
= \(\frac{1}{2}\) × \(\frac{1}{2}\) + \(\frac{\sqrt{3}}{2}\) + \(\frac{\sqrt{3}}{2}\)
= \(\frac{1}{4}\) + \(\frac{3}{4}\) = \(\frac{4}{4}\) = 1
(ii) cosAcosC – sinAsinC
= cos 30°cos60° – sin30°sin60°
[from Eqs. (i) and (ii)]
= \(\frac{\sqrt{3}}{2}\) × \(\frac{1}{2}\) – \(\frac{1}{2}\) × \(\frac{\sqrt{3}}{2}\) = \(\frac{\sqrt{3}}{4}\) – \(\frac{\sqrt{3}}{4}\) = 0
Question 22.
Shivangi started work in 1991 at an annual salary of ₹ 5000 and received an increment of ₹ 200 each year. In which year did his income reach ₹ 7000? [2]
Answer:
The annual salary received by Shivangi in the years 1991, 1992, 1993 etc., is ₹ 5000, ₹ 5200, ₹ 5400
₹ 7000
Hence, the list of numbers 5000, 5200, 5400, ……… 7000 form an AP
∵ a2 – a1 = a3 – a2 = 200
⇒ Common difference, d = 200
Let nth term of an AP
an = 7000
⇒ 7000 = a + (n – 1) [∵ an = a + (n – 1)d]
⇒ 7000 = 5000 + (n – 1)(200)
⇒ 200 (n – 1) = 7000 – 5000 = 2000
⇒ n – 1 = \(\frac{2000}{200}\) = 10
⇒ n = 10 + 1 = 11
Thus, 11th yr of his service or in 2001 Shivangi received an annual salary ₹ 7000.
Or
If five times the fifth term of an AP is equal to 8 times its eight term, show that its 13th term is zero.
Answer:
Let first term and common difference of an AP be a and d respectively.
According to the given condition,
5a5 = 8a8
⇒ 5[a + (5 – 1)d] = 8 [a + (8 – 1)d]
[∵ an = a + (n – 1)d]
⇒ 5[a + 4d] = 8[a + 7d]
⇒ 5a + 20d = 8a + 56d ⇒ 3a + 36 d = 0
⇒ 3(a + 12d) = 0 ⇒ [a + (13 – 1)d] = 0
⇒ a13 = 0 Hence proved.
Question 23.
The radii of two circles are 17 cm and 8 cm, respectively. Find the radius of the circle, which has circumference equal to the sum of the circumferences of the two circles. [2]
Answer:
Let the radii of two circles be r1 = 17 cm and r2 = 8 cm.
Sum of circumferences of two circles
= 2πr1 + 2πr2 = 2π(r1 + r2)
= 2π(17 + 8)
= 2π × 25 = 50π
Let R be the radius of the circle whose circumference is equal to the sum of the circumferences of two circles.
∴ According to the question,
2πR = 50π
⇒ R = 25 cm
Question 24.
If two tangents are inclined at 60° are drawn to a circle of radius 3 cm, then find length of each tangent. [2]
Answer:
Let PA and PB are two tangents to the circle with centre O.
We know that, radius is perpendicular to the tangent at the point of contact.
[ lengths of two tangents from an external point to a circle are equal]
So, length of each tangent is 3\(\sqrt{3}\) cm.
Or
If PQ is a tangent to a circle with centre O at point P. If ∆OPQ is an isosceles triangle, then find ∠OQP. [2]
Answer:
Given, PQ is a tangent to a circle with centre 0 and ∆OPQ is an isosceles triangle.
Here, PO = PQ ⇒ ∠O = ∠Q
In ∆OPQ,
∠P + ∠Q + ∠O = 180°
[∵ sum of all angles of a triangle is 180°]
2∠Q + ∠P = 180°
2∠Q + 90° = 180°
[∠P = 90°, ∵ OP ⊥ PQ]
2∠Q = 90°
∠Q = 45°
Question 25.
PQ is a chord of length 8 cm of a circle of radius 5 cm. The tangents at P and Q intersect at a point T (see figure). Find the length TP. [2]
Answer:
Join OT. Let it intersect PQ at the point R. Then, ∆TPQ is isosceles and TO is the angle bisector of ∠PTQ. So, OT perpendicular PQ and therefore, OT bisects PQ which gives PR = RQ = 4 cm.
Also, OR = \(\sqrt{O P^2-P R^2}\) = \(\sqrt{5^2-4^2}\) cm = 3 cm
Now, ∠TPR + ∠RPO = 90° = ∠TPR + ∠PTR So, ∠RPO = ∠PTR
Therefore, right angled ∆TRP is similar to the right angled ∆PRO by AA similarity. This gives \(\frac{T P}{P O}\) = \(\frac{R P}{R O}\)
i.e. \(\frac{T P}{5}\) = \(\frac{4}{3}\) or TP = \(\frac{20}{3}\) cm = 666 cm
Section C
(Section C consists of 6 questions of 3 marks each.)
Question 26.
Find the ratio in which the line 2x + y = 4 divides the join of A(2, -2) and B(3, 7). Also, find the coordinates of the point of intersection. [3]
Answer:
Let the line 2x + y – 4 = 0 divides the line segment joining the points A (2, -2) and 6 (3,7) in the ratio k : 1 at the point P.
But P lies on 2x + y – 4 = 0.
So, coordinates of P satisfy the equation 2x + y – 4 = 0
∴ 2\(\left(\frac{3 k+2}{k+1}\right)\) + \(\frac{7 k-2}{k+1}\) – 4 = 0
⇒ 6k + 4 + 7k – 2 – 4k – 4 = 0
⇒ 9k – 2 = 0
⇒ 9k = 2 ⇒ k = \(\frac{2}{9}\)
So, the point P divides the line segment in the ratio 2 : 9.
Question 27.
Prove that
cot A\(\sqrt{\frac{1+\sin A}{1-\sin A}}\) = \(\frac{1+\sin A}{\sin A}\) [3]
Answer:
Question 28.
Solve the following equation for x,
9(x2 + \(\frac{1}{x^2}\)) – 9(x + \(\frac{1}{x}\)) – 52 = 0. [3]
Answer:
Given, 9\(\left(x^2+\frac{1}{x^2}\right)\) – \(\left(x+\frac{1}{x}\right)\) – 52 = 0
∵ x2 + \(\frac{1}{x^2}\) + 2 = \(\left(x+\frac{1}{x}\right)^2\)
x2 + \(\frac{1}{x^2}\) = \(\left(x+\frac{1}{x}\right)^2\) – 2
The given equation becomes
9(y2 – 2) – 9y – 52 = 0
⇒ 9y2 – 18 – 9y – 52 = 0
⇒ 9y2 – 9y – 70 = 0
⇒ 9y2 – 30y + 21y – 70 = 0 [by factorisation]
⇒ 3y(3y -10) + 7(3y – 10) = 0
⇒ (3y – 10)(3y + 7) = 0
⇒ 3y – 10 = 0 or 3y + 7 = 0
⇒ 3y = 10 or 3y = -7
∴ y = \(\frac{10}{3}\) or y = –\(\frac{7}{3}\)
When y = \(\frac{10}{3}\), then x + \(\frac{1}{x}\) = \(\frac{10}{3}\)
⇒ 3x2 + 3 = 10x
⇒ 3x2 – 9x – x + 3 = 0
⇒ 3x(x – 3) -1 (x – 3) = 0
⇒ (x – 3)(3x – 1) = 0
⇒ x – 3 = 0 or 3x – 1 = 0
⇒ x = 3 or 3x = 1
∴ x = 3 or x = \(\frac{1}{3}\)
When, y = –\(\frac{7}{3}\), then x + \(\frac{1}{x}\) = –\(\frac{7}{3}\)
⇒ 3x2 + 3 = -7x
⇒ 3x2 + 7x + 3 = 0
On comparing Eq. (i) with ax2 + bx + c = 0, we get
a = 3, b = 7 and c = 3
∴ D = b2 – 4ac = (7)2 – 4 × 3 × 3
= 49 – 36 = 13 > 0
Real roots exist, which are given by using quadratic formula,
x = \(\frac{-b \pm \sqrt{L}}{2 a}\) = \(\frac{-7 \pm \sqrt{13}}{2 \times 3}\) = \(\frac{-7 \pm \sqrt{13}}{6}\)
Hence, x = 3, \(\frac{1}{3}\) and \(\frac{-7 \pm \sqrt{13}}{6}\)
Or
Using quadratic formula, solve for x
9x2 – 3(a + b)x + ab = 0 [3]
Answer:
Given equation is 9x2 – 3(a + b)x + ab = 0.
On comparing with Ax2 + Bx + C = 0, we get A = 9, B = -3(a + b)and C = ab
Discriminant, D = B2 – 4AC
= 9(a + b)2 – 4(9) (ab)
= 9{(a + b)2 – 4ab}
= 9 (a – b)2
[∵ (a + b)2 – 4ab = (a – b)2]
Therefore, the two real roots of the equation are given by
Hence, the two roots are \(\frac{a}{3}\) and \(\frac{b}{3}\)
Question 29.
AD is the median of ∆ABC and bisectors of ∠ADB and ∠ADC are DE and DF, which meet AB at E and AC at F. Prove that EF || BC. [3]
Answer:
Given, AD is the median of ∆ABC and DE is the bisector of ∠ADB and DF is the bisector of ∠ADC.
To prove EF || BC
Proof In ∆ADB, DE is the bisector of ∠ADB.
[by converse of Thales theorem]
Hence proved.
Or
A girl of height 120 cm is walking away from the base of a lamp-post at a speed of 1.2 m/sec. If the lamp is 4.8 m above the ground, then find the length of her shadow after 6 sec. [3]
Answer:
Let AB be the lamp-post, CD be the girl and D be the position of girl after 6 s.
Again, let DE = x m be the length of shadow of the girl.
Given, CD = 120 cm = 1.2 m, AB = 4.8 m and speed of the girl = 1.2 m/sec
Distance of the girl from lamp-post after 6 sec.
BD = 1.2 × 6 = 7.2 m [∵ distance = speed × time]
In ∆ABE and ∆CDF,
∠B = ∠D [each 90°]
∠E = ∠E [common angle]
∴ ∆ABE – ∆CDF [by AA similarity criterion]
⇒ \(\frac{B E}{D E}\) = \(\frac{A B}{C D}\)
On substituting all the values in Eq. (i), we get
\(\frac{7.2+x}{x}\) = \(\frac{4.8}{1.2}\) [∵ BE = BD + DE = 72 + x]
⇒ \(\frac{7.2+x}{x}\) = 4 ⇒ 7.2 + x = 4x
⇒ 3x = 7.2
⇒ x = \(\frac{7.2}{3}\) = 2.4 m
Hence, the length of her shadow after 6 s is 2.4 m.
Question 30.
Three sets of English, Hindi and Mathematics book have to be stacked in such a way that all the books are stored topicwise and the height of each stack is the same. [3]
The number of English book is 96, the number of Hindi books is 240 and the number of Mathematics books is 336. Assuming that the books are of the same thickness, determine the number of stacks of English, Hindi and Mathematics books.
Answer:
In order to arrange the books as required, we have to find the largest numbers that divides 96,240 and 336 exactly. Therefore, such a number is their HCF.
Now, 96 = 25 × 3, 240 = 24 × 3 × 5
and 336 = 24 × 3 × 7
∴ HCF of (96, 240, 336) = 24 × 3 = 48
So, there must be 48 books in each stack. Number of stacks of English books = \(\frac{96}{48}\) = 2
Number of stacks of Hindi books = \(\frac{240}{48}\) = 5
Number of stacks of Mathematics books = \(\frac{336}{48}\) = 7
Question 31.
If α and β are zeroes of x2 – p(x + 1) – c, then show that (β + 1)(β + 1) = 1 – c and also show that \(\frac{\alpha^2+2 \alpha+1}{\alpha^2+2 \alpha+c}\) + \(\frac{\beta^2+2 \beta+1}{\beta^2+2 \beta+c}\) = 1. [3]
Answer:
Section D
(Section D consists of 4 questions of 5 marks each.)
Question 32.
Find the mean of the following frequency distribution using assumed mean method. [5]
Class | 2-8 | 8-14 | 14-20 | 20-26 | 26-32 |
Frequency | 6 | 3 | 12 | 11 | 8 |
Answer:
Let us make the following table
Here, assumed mean, A = 17,
Σfi = 40 and Σfidi = 72
∴ Mean \((\bar{x})\) = A + \(\frac{\Sigma f_j d_i}{\Sigma f_i}\) = 17 + \(\frac{72}{40}\)
= 17 + 1.8 = 18.8
Or
The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median of the data. [5]
Monthly consumption (in units) |
Number of consumers |
65-85 | 4 |
85-100 | 5 |
105-125 | 13 |
125-145 | 20 |
145-165 | 14 |
165-185 | 8 |
185-205 | 4 |
Answer:
Table for cumulative frequency is given below
Here, number of observations,
N = 68
Now, \(\frac{N}{2}\) = \(\frac{68}{2}\) = 34
Since, cumulative frequency just greater than 34 is 42 and the corresponding class interval is 125 – 145.
Therefore, 125-145 is the median class.
Here, lower limit of median class, l = 125,
frequency of median class, f = 20,
cumulative frequency of class preceding the median class, cf = 22
and class size, h = 145 – 125 = 20
∴ Median = l + \(\left(\frac{\frac{N}{2}-c f}{f}\right)\) × h
= 125 + \(\left(\frac{34-22}{20}\right)\) × 20
= 125 + \(\frac{12}{20}\) × 20 = 125 + 12 = 137
Hence, the median of given data is 137.
Question 33.
A rocket is in the form of a right circular cylinder closed at the lower end and surmounted by a cone with the same radius as that of the cylinder. The diameter and height of the cylinder are 6 cm and 12 cm, respectively. If the slant height of the conical portion is 5 cm, then find the total surface area and volume of the rocket, [take π = 3.14] [5]
Answer:
Since, rocket is the combination of a right circular cylinder and a cone.
Given, diameter of the cylinder = 6 cm
∴ Radius of the cylinder = \(\frac{6}{2}\) = 3 cm
and height of the cylinder = 12 cm
∴ Volume of the cylinder = πr2h
= 3.14 × (3)2 × 12
= 339.12 cm3
Curved surface area of cylinder = 2πh
= 2 × 3.14 × 3 × 12
= 226.08 cm2
and area of base of cylinder = πr2
= 3.14 × 3 × 3
= 28.26 cm2
Now, in right angled ∆AOC,
h = \(\sqrt{5^2-3^2}\) = \(\sqrt{25-9}\) = \(\sqrt{16}\) = 4
∴ Height of the cone, h = 4 cm
and radius of the cone, r = 3 cm
Now, volume of the cone = \(\frac{1}{3}\)πr2h
= \(\frac{1}{3}\) × 3.14 × (3)2 × 4
= 3.14 × 3 × 4
= 37.68 cm3
and curved surface area = πrl = 3.14 × 3 × 5
= 47.1 cm2
Hence, total volume of the rocket
= Volume of the cylinder + Volume of the cone
= 339.12+ 37.68 = 376.8 cm3
and total surface area of the rocket = Curved surface area of cone + Curved surface area of cylinder + Area of base of cylinder
= 47.1 + 226.08 + 28.26 = 301.44 cm2
Question 34.
Draw the graphs of the equations x – y + 1 = 0 and 3x + 2y – 12 = 0. Determine the coordinates of the vertices of the triangle formed by these lines and the X-axis and shade the triangular region. [5]
Answer:
Given, pair of linear equations is x – y + 1 = 0 and 3x + 2y – 12 = 0.
Table for x – y + 1 = 0 or y = x + 1 is
x | 0 | 4 | -1 |
y = x + 1 | 1 | 5 | 0 |
Points | A(0, 1) | B(4, 5) | C(-1, 0) |
Table for 3x + 2y – 12 = 0 or y = \(\frac{12-3 x}{2}\) is
x | 0 | 2 | 4 |
y = \(\frac{12-3 x}{2}\) | 6 | 3 | 0 |
Points | D(0, 6) | E(2, 3) | F(4, 0) |
Now, plot the points A (0, 1), 6(4, 5), C (-1,0) and join them to get a line CB.
Similarly, plot the points D(0, 6), E(2, 3), F (4, 0) and join them to get a line DF.
Clearly, the two lines intersect each other at the point E(2, 3). Hence, x = 2 and y = 3 is the solution of the given pair of equations. The line DE cuts X-axis at the point F (4, 0) and the line AB cuts X-axis at the point C(-1, 0).
Hence, the coordinates of the vertices of the ∆EFC, so formed are E(2, 3), F(4, 0) and C(-1, 0).
Or
The area of a rectangle gets reduced by 9 sq units, if its length is reduced by 5 units and breadth is increased by 3 units. If we increase the length by 3 units and the breadth by 2 units, then the area increased by 67 sq units. Find the dimensions of the rectangle.
Answer:
Let x units be the length and y units be the breadth of a rectangle, then area of rectangle = xy sq units.
In first case Area is reduced by 9 sq units, when length = (x – 5) units and breadth = (y + 3) units Then, area of rectangle = (x – 5)(y + 3) sq units
According to the question,
xy – (x – 5) × (y + 3) = 9
⇒ xy – (xy + 3x – 5y – 15) = 9
⇒ 3x + 5y + 15 = 9
⇒ 3x – 5y = 6 ……..(i)
In second case Area is increased by 67 sq units, when length = (x + 3) units and breadth = (y + 2) units According to the question,
(x + 3) × (y + 2) – xy = 67
⇒ xy + 2x + 3y + 6 – xy = 67
⇒ 2x + 3y = 61 ………(ii)
On multiplying Eq. (i) by 3 and Eq. (ii) by 5, we get 9x – 15y = 18 ………(iii)
and 10x + 15y = 305 ……..(iv)
On adding Eqs. (iii) and (iv), we get 19x = 323
⇒ x = 17
On substituting x = 17 in Eq. (ii), we get
34+ 3y = 61
⇒ 3y = 27
⇒ y = 9
Hence, the length of the rectangle is 17 units and breadth of the rectangle is 9 units.
Question 35.
D, E and F are respectively, the mid-points of sides AB, BC and CA of ∆ABC. Find the ratio of the areas of ∆DEF and ∆ABC. [5]
Answer:
We have, a ∆ABC in which D, E and F are mid points of AB, BC and CA respectively. D, E and F are joined to form ∆DEF.
Now, D is mid-point of AB
⇒ \(\frac{A D}{A B}\) = \(\frac{1}{2}\) …….. (i)
Similarly \(\frac{B E}{B C}\) = \(\frac{1}{2}\) ….. (ii)
From Eqs. (i) and (ii), we have
\(\frac{A D}{A B}\) = \(\frac{B E}{B C}\)
∴ Using the converse of the basic proportionally theorem, we have DE || AC
⇒ ∠BDE = ∠BAC ……. (iii) [corresponding angles]
Also ∠BED = ∠BCA … (iv) [corresponding angles]
From Eqs. (iii) and (iv), we have
∆ABC ~ ∆DEF [using AA similarity]
Section E
(Case Study Based Questions)
(Section E consists of 3 questions. All are compulsory.)
Question 36.
The below picture are few natural examples of parabolic shape which is represented by a quadratic polynomial. A parabolic arch is an arch in the shape of a parabola.
In structures, their curve represents an efficient method of load, and so can be found in bridges and in architecture in a variety of forms.
(i) In the standard form of quadratic polynomial, ax2 + bx + c, a, b and c are
(ii) If the roots of the quadratic polynomial are equal, where the discriminant D will be
Answer:
(i) In the standard form of quadratic polynomial ax2 + bx + c; ‘a’ is a non-zero real number, and b and c are any real number.
(ii) In a quadratic polynomial, if roots are equal, then discriminant, D = 0
Or
If α and β are the zeroes of the quadratic polynomial 2x2 – x + 8k , then find the value of k.
(iii) If the sum of the roots is -P and product of the roots is –\(\frac{1}{P}\), then find the quadratic polynomial.
Answer:
Given, α and \(\frac{1}{\alpha}\) are the zeroes of quadratic a polynomial 2x2 – x + 8k.
Now, product of zeroes,
α × \(\frac{1}{\alpha}\) = \(\frac{\text { Constant term }}{\text { Coefficient of } x^2}\)
⇒ 1 = \(\frac{8 K}{2}\) ⇒ k = \(\frac{2}{8}\) = \(\frac{1}{4}\)
(iii) Given, sum of roots = – p
and product of roots = –\(\frac{1}{p}\)
Quadratic polynomial
= k[x2 – (Sum of roots) x + Product of roots]
= k[x2 – (-p)x + \(\left(-\frac{1}{p}\right)\)]
= k[x2 + px – \(\frac{1}{p}\)]
Question 37.
A girl 1.5 m tall spots a parrot sitting on the top of a building of height 58 m from the ground. The angle of elevation of the parrot from the eyes of girl at any instant is 60°.
The parrot flies away horizontally in such a way that it remained at a constant height from the ground. After 8 sec, the angle of elevation of the parrot from the same point is 30°.
Based on the above information, answer the following questions, (take \(\sqrt{3}\) = 1.73)
(i) Find the distance between the girl and the building. (1)
Answer:
Distance between girl and building = AB
Now, in ∆ABC,
tan 60° = \(\frac{B C}{A B}\)
⇒ \(\sqrt{3}\)AB = CH – BH = 58 – 1.5
⇒ AB = \(\frac{56.5}{\sqrt{3}}\)m
(ii) Find the distance of first position of the parrot from the eyes of the girl. (2)
Answer:
Distance of first position of parrot from the eyes of girl = AC
In ∆ABC, sin 60° = \(\frac{B C}{A C}\)
⇒ AC = \(\frac{C H-B H}{\sin 60^{\circ}}\) = \(\frac{58-1.5}{\sqrt{3} / 2}\) = \(\frac{113}{\sqrt{3}}\)m
Or
How much distance parrot covers? (2)
(iii) Find the speed of the parrot in 8 sec. (1)
Answer:
In ∆AED, tan 30° = \(\frac{D E}{A D}\)
⇒ AD = \(\sqrt{3}\)BC = 56.5\(\sqrt{3}\) m
[∵ DE = BC = 58 – 1.5 = 56.5]
Now, distance between two positions of parrot = EC
= BD = AD – AB
= 56.5\(\sqrt{3}\) – \(\frac{56.5}{\sqrt{3}}\)
= \(\frac{56.5(3-1)}{1.73}\)
= \(\frac{113}{1.73}\) = 65.24 m
(iii) Speed of parrot = \(=\frac{\text { Distance covered }}{\text { Time taken }}\)
= \(\frac{B D}{8}\) = \(\frac{65.24}{8}\)
= 8.15 m/sec
Question 38.
During the Lock-Down Period
During the lock-down period, people were very puzzled and they decided to play some game. Firstly, they collected the 17 cards and wrote the numbers 1 to 17 and put them in a box.
They betted for the chances of drawing the number either the prime, odd, or even number etc.
On the basis of above information, answer the following questions.
(i) Find the probability that the number on the card is an odd number. (1)
Answer:
There are 17 cards numbered 1, 2, 3, 4, 5,…….,17 in a box.
Out of 17 cards, one card can be drawn in 17 ways.
∴ Total number of outcomes = 17
There are 9 odd numbered cards, namely 1, 3, 5, 7, 9, 11, 13, 15 and 17.
∴ Number of favourable outcomes = 9
Hence, P (getting an odd number)
\(=\frac{\text { Number of favourable outcomes }}{\text { Total number of outcomes }}\)
= \(\frac{9}{17}\)
(ii) Find the probability that the number on the card is divisible by 2 and 3 both. (2)
Answer:
If a number is divisible by 2 and 3 both, then it must be a multiple of 6. In cards bearing number 1, 2, 3, 4, …, 17, there are only 2 cards which bear a number divisible by 2 and 3 both, i.e. by 6. These cards bearing numbers 6 and 12.
∴ Number of favourable outcomes = 2
Hence, P (getting a card bearing number divisible by 2 and 3 both)
\(=\frac{\text { Number of favourable outcomes }}{\text { Total number of outcomes }}\) = \(\frac{2}{17}\)
Or
Find the probability that the number on the card is a multiple of 3 or 5. (2)
(iii) What is called an event having only one outcome of the random experiment? (1)
Answer:
There are 7 numbered cards, which are multiple of 3 or 5 namely, 3, 6, 9, 12, 15, 5 and 10.
∴ Number of favourable outcomes = 7
Hence, P (getting a card bearing a number multiple of 3 or 5)
\(=\frac{\text { Number of favourable outcomes }}{\text { Total number of outcomes }}\) = \(\frac{7}{17}\)
(iii) An event having only one outcome of the random experiment is called elementary event.