Students must start practicing the questions from CBSE Sample Papers for Class 10 Maths Basic with Solutions Set 9 are designed as per the revised syllabus.
CBSE Sample Papers for Class 10 Maths Basic Set 9 with Solutions
Time: 3 Hours
Maximum Marks: 80
General Instructions:
Read the following instructions very carefully and strictly follow them:
- This question paper comprises four sections – A, B, C and D.
This question paper carries 40 questions. All questions are compulsory. - Section A – Question no. 1 to 20 comprises of 20 questions of one mark each.
- Section B – Question no. 21 to 26 comprises of 6 questions of two marks each.
- Sections C – Question no. 27 to 34 comprises of 8 questions of three marks each.
- Sections D -Question no. 35 to 40 comprises of 6 questions of four marks each.
- There is no overall choice in the question paper. However, an inrernal choice has been provided in 2 questions of one mark, 2 questions of two marks, 3 questions of three marks and 3 questions of four marks. You have to attempt only one of the choices in such questions.
- In addition to this, separate instructions are given with each section and question, wherever necessary.
- Use of calculators is not permitted.
Section – A (20 Marks)
Question number 1 to 10 are multiple choice questions of 1 mark each. Select the correct option.
Question 1.
HCF of two numbers is 27 and their LCM is 162. If one of the number is 54 then the other number is: 1
(a) 36
(b) 35
(c) 9
(d) 81
Answer:
(d) 81
Explanation: We know that for any two numbers,
HCF × LCM = Product of numbers
⇒ 27 × 162 = 54 × other number
⇒ Other number = \(\frac{27 \times 162}{54}\) i.e., 81
Question 2.
The cumulative frequency table is useful in determining: 1
(a) Mean
(b) Median
(c) Mode
(d) All of these
Answer:
(b) Median
Explanation: Since formula for median is:
Me = l + (\(\frac{\frac{N}{2}-c f}{f}\)) × h
∴ The cumulative frequency table is useful in determining median.
Question 3.
In Figure, O is the centre of circLe. PQ is a chord and PT is tangent at P which makes an angle of 50° with PQ. ∠POQ is:
(a) 130°
(b) 90°
(c) 100°
(d) 75°
Answer:
(c) 100°
Explanation: We know tangent is perpendicular to radius .at the point of contact.
So, ∠OPT = 90°
In ΔPOQ,
∠QPO = 90° – 50° = 40°
Also, ∠Q = ∠QPO = 40°
(as, OP = OQ = radii)
∴ ∠POQ = 180° – (40° + 40°)
= 100° [By angle sum property of triangle]
Question 4.
2√3 is: 1
(a) an integer
(b) a rational umber
(c) an irrational number
(d) a whole number
Answer:
(c) an irrational number
Explanation: Since √3 is an irrational number, 2√3 is also an irrational number.
Question 5.
Two coins are tossed simultaneously. The probability of getting at most one head is: 1
(a) \(\frac{1}{4}\)
(b) \(\frac{1}{2}\)
(c) \(\frac{2}{3}\)
(d) \(\frac{3}{4}\)
Answer:
(d) \(\frac{3}{4}\)
Explanation: Possible outcomes
= {HH, HT, TH, TT}
Favourable outcomes = {TT, HT, TH]
So, required probability = \(\frac{3}{4}\)
Question 6.
If one zero of the polynomial (3x2 + 8x + k) is the reciprocal of the other, then value of k is: 1
(a) 3
(b) -3
(c) \(\frac{1}{3}\)
(d) –\(\frac{1}{3}\)
Answer:
(a) 3
Explanation: Let one zero of given polynomial be α.
Then, other zero = \(\frac{1}{\alpha}\)
So, product of zeroes = α × \(\frac{1}{\alpha}\) = 1
Given, polynomial equation is 3x2 + 8x + k.
product of zeroes = \(\frac{k}{3}\)
⇒ \(\frac{k}{3}\) = 1
⇒ k = 3
Question 7.
The distance of the (-12, 5) from the origin is: 1
(a) 12
(b) 5
(c) 13
(d) 169
Answer:
(c) 13
Explanation: Distance of (-12, 5) from (0, 0)
= \(\sqrt{(0+12)^2+(0-5)^2}\)
= \(\sqrt{144+25}\)
= \(\sqrt{169}\). i.e. 13 units.
Question 8.
If the centre of a circle is (3, 5) and end points of a diameter are (4, 7) and (2, y), then the value of y is:
(a) 3
(b) -3
(c) 7
(d) 4
Answer:
(a) 3
Explanation: Since the centre of a circle is the mid-point for its every diameter, i.e., O is the mid-point of AB.
So, (3, 5) = (\(\frac{4+2}{2}, \frac{7+y}{2}\))
⇒ \(\frac{7+y}{2}\) = 5
⇒ y = 3
Question number 11 to 15, fill in the blanks.
Question 9.
The area of triangle formed with the origin and the points (4, 0) and (0, 6) is …………… .
OR
The co-ordinate of the point dividing the line segment joining the points A(1, 3) and B(4, 6) in the ratio 2 : 1 is ……………… . 1
Answer:
From graph,
OA = 4 units
OB = 6 units
∴ Area of ΔOAB = \(\frac{1}{2}\) × OA × OB
= \(\frac{1}{2}\) × 4 × 6sq. units
= 12 sq. units
OR
Let P divides AB in the ratio 2 : 1. So,
P(\(\frac{2 \times 4+1 \times 1}{2+1}, \frac{2 \times 6+1 \times 3}{2+1}\))
i.e., P(3, 5)
Question 10.
Value of the roots of the quadratic equation, x2 – x – 6 = 0 are ……………. . 1
Answer:
x2 – x – 6 = 0
⇒ x2 – 3x + 2x – 6 = 0
⇒ x(x – 3) + 2(x – 3) = 0
⇒ (x – 3)(x – 2) = 0
⇒ x – 3 = 0, x + 2 = 0
Hence, the two roots are 3 and -2.
Question 13.
If sin θ = \(\frac{5}{13}\) then the value of tan θ is …………… . 1
Answer:
Given, sin θ = \(\frac{5}{13}\)
⇒ cos θ = \(\sqrt{1-\sin ^2 \theta}=\sqrt{1-\frac{25}{169}}\)
= \(\sqrt{\frac{144}{169}}=\frac{12}{13}\)
∴ tan θ = \(\frac{\sin \theta}{\cos \theta}=\frac{\frac{5}{13}}{\frac{12}{13}}=\frac{5}{12}\)
Question 11.
The value of (tan2 60° + sin2 45°) is ……………. . 1
Answer:
tan2 60° + sin2 45° = (√3)2 + (\(\frac{1}{\sqrt{2}}\))2
= 3 + \(\frac{1}{2}\)
= \(\frac{7}{2}\)
Question 12.
A card is drawn at random from a well shuffled deck of 52 playing cards. What is the probability of getting a black king? 1
Answer:
In the pack of 52 cards, there are two black kings.
So, P (a black king) = \(\frac{2}{52}\), i.e \(\frac{1}{26}\)
Question 13.
A ladder 25 m long just reaches the top of a building 24 m high from the ground. What is the distance of the foot of ladder from the base of the building? 1
Answer:
Let AC be the ladder and BC be the building.
So, ΔABC is right-angled triangle where in∠B = 90°.
So, by Pythagoras theorem,
AB2 = AC2 – BC2
= 252 – 242
= 625 – 576
= 49
∴ AB = 7cm,
Thus, the distance of the foot of the ladder from the base of the building is 7 metres.
Question 14.
If 3k – 2,4k – 6 and k + 2 are three consecutive terms of A.P., then find the value of k. 1
Answer:
Here, 3k – 2, 4k – 6 and k + 2 are three consecutive terms of A.P. So,
(3k – 2) + (k + 2) = 2 (4k – 6)
⇒ 4k = 8k – 12
⇒ 4k = 12
i.e., k = 3
Section – B (12 Marks)
Question numbers 21 to 26 carry 2 marks each.
Question 15.
In a lottery, there are 10 prizes and 25 blanks. What is the probability of getting a prize? 2
Answer:
Total outcomes = 10 + 25 = 35
Favourable outcomes = 10
P (a prize) = \(\frac{10}{35}\) = \(\frac{2}{7}\)
Question 16.
In a family of three children, find the probability of having at least two boys.
OR
Two dice are tossed simultaneously. Find the probability of getting:
(A) an even number on both dice.
(B) the sum of two numbers more than 9. 2
Answer:
Let B denote a boy and G denote a girl.
Then, possible outcomes
= {BBB, BBG, BGB, GBB, GGB, GBG, BGG, GGG}
Favourable outcomes
= {BBG, BGB, GBB, BBB}
∴ P (at least 2 boys) = \(\frac{4}{8}\) i.e., \(\frac{1}{2}\)
OR
Possible outcomes = 36
(A) Favourable outcomes
= 1(2, 2), (2, 4), (2, 6), (4, 2), (4, 4), (4, 6), (6, 2) (6, 4) (6, 6)}
So, required probability
= \(\frac{9}{36}\) i.e., \(\frac{1}{4}\)
(B) Favourable outcomes
= {(4,6), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6)}
So, required probability
= \(\frac{6}{36}\) i.e., \(\frac{1}{6}\)
Let C1 and C2 be two concentric circle with centre O.
Question 17.
Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of larger circle which touches the smaller circle. 2
Answer:
Let AB be the chord of larger circle, with radius 5 cm, which touches the smaller circle, with radius 3 cm, at point P.
So, AB is tangent to smaller circle at point P.
Therefore, OP is perpendicular to AB.
In ΔAPO, OP = 3 cm and OA = 5 cm
So, AP = \(\sqrt{O A^2-O P^2}\)
= \(\sqrt{5^2-3^2}\)
= \(\sqrt{25-9}\)
= \(\sqrt{16}\), i.e., 4 cm
Further, AB = 2 × AP = (2 × 4cm) = 8cm.
Thus, the length of the chord of Larger circle which touches the smaller circle is 8 cm.
Question 18.
Prove that: \(\frac{1}{1+\sin \theta}+\frac{1}{1-\sin \theta}\) = 2sec2 θ
OR
Prove that: \(\frac{1-\tan ^2 \theta}{1+\tan ^2 \theta}\) = cos2 θ – sin2 θ 2
Answer:
Question 19.
The wheel of a motorcycle is of radius 35 cm. How many revolutions are required to travel a distance of 11 m? 2
Answer:
Distance covered by wheel in one revolution = Circumference of wheel
= 2π (35) cm = 70π cm
So, number of revolutions required to cover a distance of 11m
= \(\frac{11 \times 100}{70 \pi}=\frac{11 \times 100}{7 \times \frac{22}{7}}\)
= 5
Thus, in 5 revolutions, a distance of 11 m will be travelled by the motorcycle.
Section – C (24 Marks)
Question numbers 27 to 34 carry 3 marks each.
Question 20.
If α and β are the zeroes of the polynomial f(x) = 5x2 – 7x + 1, then find the value of
(\(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}\))
Answer:
Given equation is f(x) = 5x2 – 7x + 1
Question 21.
The minute hand of a clock is 21 cm long. Calculate the area swept by it and the distance travelled by its tip in 20 minutes. 3
Answer:
Length of minute hand of clock = 21 cm
Angle covered by minute hand in 60 min = 360°
So, angle covered by minute hand in 20 min
= 360° × \(\frac{20^{\circ}}{60^{\circ}}\) = 120°
∵ length of arc of centre angle θ = \(\frac{\theta}{360^{\circ}}\) × 2πr
Length of arc = \(\frac{120^{\circ}}{360^{\circ}}\) × 2π × 21
= 3 × 2 × \(\frac{22}{7}\) × 21
= 44 cm
The distance travelled minute hand
= 44 cm
∴ Area of sector with centre angle θ
= \(\frac{\theta}{360^{\circ}}\) × πr2
∴ Required area = \(\frac{120}{360}\) × π × 21 × 21
= 462 cm2
Question 22.
If x = 3 sin θ + 4 cos θ and y = 3 cos θ – 4 sin θ then prove that x2 + y2 = 25.
OR
If sin θ + sin2 θ = 1; then prove that cos2 θ + cos4 θ = 1. 3
Answer:
Given,
x = 3 sin θ + 4 cos θ
y = 3 cos θ – 4 sin θ
Here
x2 + y2 = (3 sin θ + 4 cos θ)2 + (3 cos θ – 4 sin θ)2
= 9 sin2 θ + 16 cos2 θ + 24 sin θ cos θ + 9 cos2 θ + 16 sin2 θ – 24 sin θ cos θ
= 9 (sin2 θ + cos2 θ) + 16 (sin2 θ + cos2 θ)
= 9 + 16 (as, sin2 θ + cos2 θ = 1)
= 25
OR
Given, sin θ + sin2 θ = 1,
⇒ sin θ = 1 – sin2 θ = cos2 θ
i.e., cos2 θ = sin θ
Now, cos2 θ + cos4 θ = 1 – sin2 θ + (cos2 θ)2
= 1 – sin2 θ + (sin θ)2
= 1 – sin2 θ + sin2 θ
= 1
Question 23.
Prove that √3 is an irrational number. 3
Answer:
Let us assume, to the contrary, that √3 is a rational number.
Then, √3 = \(\frac{p}{q}\), where p and q are integers
having no common factor, and q ≠ 0.
On squaring both sides, we get
3 = \(\frac{p^2}{q^2}\)
⇒ 3q2 = p2
⇒ 3 divides p2
⇒ 3 divides p
(If a is a prime number and if a divides p2, then a divides p2, where p is an positive integer.)
Let p = 3n, where n is an integer.
Also 3q2 = p2
⇒ 3q2 = (3n)2 = 9n2
⇒ q2 = 3n2
Arguing as above, we get 3 divides q.
Thus, both p and q have a common factor 3 which is a contradiction.
Hence, √3 is an irrational number.
Question 24.
In a classroom, 4 friends are seated at the points A, B, C and D as shown in Figure, Champa and Chameli walk into the class and after observing for a few minutes Champa asks Chameli, “Don’t you think ABCD is a square?” Chameli disagrees. Using distance formula, find which of them is correct.
Answer:
The co-ordinates of A, B, C and D are A (3, 4), B (6, 7), C (9,4) and D (6,1) Now, using distance formula:
AB = \(\sqrt{(6-3)^2+(7-4)^2}\)
Thus, AC = BD.
Hence, ABCD is a square.
Question 25.
Solve graphically:
2x – 3y + 13 = 0; 3x – 2y + 12 = 0 3
Answer:
Given equations are: 2x – 3y + 13 = 0 and 3x- 2y + 12 = 0
Table of values for 2x – 3y + 13 = 0
x | -6.5 | -5 | -2 |
y | 0 | 1 | 3 |
Table of values for 3x – 2y + 12 = 0
x | -4 | 0 |
y | 0 | 6 |
The two equations of lines intersect at point (-2, 3).
Hence, the solution of the two equations is x = -2 y = 3
Section – D (24 Marks)
Question numbers 35 to 40 carry 4 marks each.
Question 26.
The product of two consecutive positive integers is 306. Find the integers. 4
Answer:
Let two consecutive positive integers be x and (x + 1).
ATQ x(x + 1) = 306
x2 + x – 306 = 0
x2 + 18x – 17x – 306 = 0
x(x + 18) – 17(x + 18) = 0
(x + 18) (x – 17) = 0
x + 18 = 0
x = -18
x – 17 = 0
x = 17
Since, x is a positive integer.
∴ x ≠ -18
⇒ x = 17
Hence, integers are 17 and 18.
Question 27.
The 17th term of an A.P. is 5 more than twice its 8th term. If 11th term of A.P. is 43; then find its nth term.
OR
How many terms of A.P. 3, 5, 7, 9,… must be taken to get the sum 120? 4
Answer:
Let ‘a’ be the first term and ‘d the common difference of the AP. Then,
17th term = a + 16d
8th term =a + 7d
11th term = a + 10 d
and nth term = a + (n – 1 )d.
According to the question,
a + 16d = 2 (a + 7d) + 5
and a + 10d = 43
⇒ a – 2d + 5 = 0 and a + 10d = 43
Solving the two equations, we get
d = 4 and a = 3
Thus, nth term is 3 + (n – 1) (4), i.e., 4n – 1
OR
Given A.P., 3, 5, 7, 9
Here, a = 3, d = 2
Let the A.P. contains n terms. Then,
Sn = \(\frac{n}{2}\)[2a + (n – 1)d]
120 = \(\frac{n}{2}\)[2 × 3 + (n – 1)2]
⇒ n[2n + 4] = 240
⇒ n[n + 2] = 120
⇒ n2 + 2n- 120 = 0
⇒ n2 + 12n – 10n – 120 = 0
⇒ n(n + 12) – 10 (n + 12) = 0
⇒ (n + 12) (n – 10) = 0
⇒ n – 10 = 0 (∵ n + 12 ≠ 0)
⇒ n = 10
Thus, first 10 terms of the A.P. give the sum 120.
Question 28.
A person standing on the bank of a river observes that the angle of elevation of the top of a tree standing on opposite bank is 60°. When he moves 30 m away from the bank, he finds the angle of elevation to be 30°. Find the height of the tree and width of the river. [Take √3 = 1.732] 4
Answer:
Let XY denote the tree, BX the river and A the point, 30 m away from the bank B of the river BX.
Let ‘h’m be the height of the tree, and ld be the width of the river.
Eliminating ‘h’ from eqs. (i) and (ii), we have
\(\frac{\sqrt{3} d}{d+30}=\frac{1}{\sqrt{3}}\)
⇒ 3d = d + 30
⇒ d = 15 m
Substituting this value in eq. (ii), we have
h = 15√3 = 15 × 1.732
= 25.98 m.
Hence, height of the tree is 25.98 m; and the width of the river is 15 m.
Question 29.
Prove that the length of tangents drawn from an external point to a cricle are equal. 4
Answer:
Given a circle with centre O, a point P lying outside the circle and two tangents PQ and PR on the circle from point P.
To prove: PQ = PR.
Construction : join OP, OQ and OR.
Proof: Since, the tangent at any point of circle is perpendicular to the radius through the point of contact.
∴ ∠OQP = ∠ORP = 90°
Now, in right triangles OQP and ORP,
OQ = OR (radii of the same circle)
OP = OP (common)
∠OQP = ∠ORP (each 90°)
Therefore, ΔOQP ~ ΔORP (RHS similarity criterion)
This gives PQ = PR. (C.P.S.T.)
Question 30.
From a solid cylinder whose height is 15 cm and the diameter is 16 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of remaining solid. (Give your answer in terms of π)
Answer:
From figure,
Total surface area of remaining solid
= curved suface area of cylinder + curved suface area if conical cavity + Surface area of top circular base
= 2πrh + πrl + πr2
∵ Slant height of conical cavity,
l = \(\sqrt{h^2+r^2}\)
= \(\sqrt{(15)^2+(8)^2}\)
= \(\sqrt{225+64}\)
= \(\sqrt{289}[latex] = 17 cm.
∵ Required suface area = πr (2h + l + r)
= π × 8(2 × 15 + 17+ 8) cm2
= 8π (30 + 25) cm2
= 440π cm2
Question 31.
The mode of the following frequency distribution is 36. Find the missing frequency (f).
Class | Frequency |
0 – 10 | 8 |
10 – 20 | 10 |
20 – 30 | f |
30 – 40 | 16 |
40 – 50 | 12 |
50 – 60 | 6 |
60 – 70 | 7 |
Answer:
Since the mode is 36. the modal doss is 3D – 4D. For this class,
l = 30,f1 = 16, f0 = f, f2 = 12 and h = 10
Using the formula.
mode = l + [latex]\frac{f_1-f_0}{2 f_1-f_0-f_2}\) × h
⇒ 36 = 30 + \(\frac{16-f}{2(16)-f-12}\) × 10
⇒ \(\frac{16-f}{20-f}=\frac{6}{10}=\frac{3}{5}\)
⇒ 80 – 5f = 60 – 3f
⇒ f = 10
Thus, the missing frequency (f) is 10.