These Sample papers are part of CBSE Sample Papers for Class 10 Maths. Here we have given CBSE Sample Papers for Class 10 Maths Paper 2. According to new CBSE Exam Pattern, MCQ Questions for Class 10 Maths Carries 20 Marks.
CBSE Sample Papers for Class 10 Maths Paper 2
Board | CBSE |
Class | X |
Subject | Maths |
Sample Paper Set | Paper 2 |
Category | CBSE Sample Papers |
Students who are going to appear for CBSE Class 10 Examinations are advised to practice the CBSE sample papers given here which is designed as per the latest Syllabus and marking scheme as prescribed by the CBSE is given here. Paper 2 of Solved CBSE Sample Paper for Class 10 Maths is given below with free PDF download Solutions.
Time Allowed: 3 hours
Maximum Marks: 80
General Instructions:
- All questions are compulsory.
- The question paper consists of 30 questions divided into four sections A, B, C and D.
- Section A contains 6 questions of 1 mark each. Section B contains 6 questions of 2 marks each. Section C contains 10 questions of 3 marks each. Section D contains 8 questions of 4 marks each.
- There is no overall choice. However, an internal choice has been provided in four questions of 3 marks each and three questions of 4 marks each. You have to attempt only one of the alternatives in all such questions.
- Use of calculators is not permitted.
SECTION-A
Question 1.
State the Fundamental Theorem of Arithmetic.
Question 2.
What is the nature of roots of the quadratic equation 4×2 – 12x – 9 = 0 ?
Question 3.
Find the common difference of an AP in which a18 – a14 = 32.
Question 4.
If the points A (1, 2), B (0, 0) and C (a, b) are collinear, then what is the relation between a and b?
Question 5.
A and B are respectively the points on the sides PQ and PR of a APQR such that PQ = 12.5 cm, PA = 5 cm, BR = 6 cm and PB = 4 cm. Is AB || QR ? Give reason.
Question 6.
Express sin 67° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°.
SECTION-B
Question 7.
From your pocket money, you save ₹1 on day 1, ₹2 on day 2, ₹3 on day 3 and so on. How much money will you save in the month of March 2008?
Question 8.
Find the values of x for which the distance between the point P (2, -3) and Q (x, 5) is 10 units.
Question 9.
All cards of ace, jack and queen are removed from a deck of playing cards. One card is drawn at random from the remaining cards. Find the probability that the card drawn is
- a face card.
- not a face card.
Question 10.
Does the following pair of equations represent a pair of coincident lines? Justify your answer.
\(\frac { x }{ 2 } +y+\frac { 2 }{ 5 } =0,\quad 4x+8y+\frac { 5 }{ 16 } =0\)
Question 11.
Two dice are thrown at the same time and the product of numbers appearing on them is noted. Find the probability that the product is a prime number.
Question 12.
Can the number 4n, n being a natural number, end with the digit 0? Give reason.
SECTION—C
Question 13.
Find the zeros of the quadratic polynomial x2 + 5x+ 6 and verify the relationship between the zeros and the coefficients.
Question 14.
Prove that \(5+\sqrt { 2 } \) is an irrational number.
Question 15.
For what value of ‘k’ will the following pair of linear equations have infinitely many solutions?
Question 16.
Prove that
\(\sqrt { \frac { secA-1 }{ secA+1 } } +\sqrt { \frac { secA+1 }{ secA-1 } } =2cosecA\)
Question 17.
Prove that the points A(-3,0), B(l, -3) and C(4,1) are the vertices of an isosceles right triangle.
OR
For what value of ‘k’ the points A (1,5), B (k, 1) and C (4,11) are collinear?
Question 18.
The incircle of AABC touches the sides BC, CA and AB at D, E, and F respectively. If AB = AC, prove that BD = CD.
Question 19.
PQRS is a square land of side 28 m. Two semicircular grass covered portions are to be made on two of its opposite sides as shown in the figure. How much area will be left uncovered? \(\left( Take\Pi =\frac { 22 }{ 7 } \right) \)
Question 20.
Using converse of Basic Proportionality Theorem, prove that the line joining the mid-points of any two sides of a
triangle is parallel to the third side.
OR
In Fig. \(\frac { AO }{ OC } =\frac { BO }{ OD } =\frac { 1 }{ 2 } \) and AB = 5 cm. Find the value of DC.
Question 21.
Two spheres of same metal weigh 1 kg and 7 kg. The radius of the smaller sphere is 3 cm. The two spheres are melted to form a single big sphere. Find the diameter of the new sphere.
OR
How many silver coins, 1.75 cm in diameter and of thickness 2 mm, must be melted to form a cuboid of dimensions 5.5 cm x 10 cm x 3.5 cm?
Question 22.
The following data gives the information on the observed lifetimes (in hours) of 225 electrical components:
Lifetimes (in hours) | 0-20 | 20-40 | 40-60 | 60-80 | 80-100 | 100-120 |
Frequency | 10 | 35 | 52 | 61 | 38 | 29 |
Determine the modal lifetimes of the components.
SECTION-D
Question 23.
There are two poles, one each on either bank of a river, just opposite to each other. One pole is 60 m high. From the top of this pole, the angles of depression of the top and the foot of the other pole are 30° and 60° respectively. Find the width of the river and the height of the other pole.
Question 24.
Prove that the ratio of areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
Use the above theorem, in the following:
The areas of two similar triangles are 81 cm2 and 144 cm2. If the largest side of the smaller triangle is 27 cm, find the largest side of the larger triangle.
OR
Prove that in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
Use the above theorem, in the following:
If ABC is an equilateral triangle with AD ⊥ BC, then prove that AD2 = 3 DC2.
Question 25.
An iron pillar has lower part in the form of a right circular cylinder and the upper part in the form of a right circular cone. The radius of the base of each of the cone and cylinder is 8 cm. The cylindrical part is 240 cm high and the conical part is 36 cm high. Find the weight of the pillar if 1 cm3 of iron weighs 7.5 grams. \(\left( Take\Pi =\frac { 22 }{ 7 } \right) \)
OR
A container (open at the top) made up of a metal sheet is in the form of a frustum of a cone of height 16 cm with radii of its lower and upper ends as 8 cm and 20 cm respectively. Find
- the cost of milk when it is completely filled with milk at the rate of ₹15 per litre.
- the cost of metal sheet used, if it costs ₹5 per 100 cm2 . (Take π = 3.14).
Question 26.
The median of the following data is 20.75. Find the missing frequencies x and y, if the total frequency is 100.
Class Interval | 0-5 | 5-10 | 10-15 | 15-20 | 20-25 | 2530 | 30-35 | 35-40 |
Frequency | 7 | 10 | x | 13 | y | 10 | 14 | 9 |
Question 27.
If the price of petrol is increased by ₹2 per litre, a person will have to buy 1 litre less petrol for ₹1740. Find the original price of petrol at that time.
- Why do you think the price of petrol is increasing day-by-day?
- What should we do to save petrol?
Question 28.
Determine an AP whose 3rd term is 16 and when 5th term is subtracted from 7th term, we get 12.
OR
Find the sum of all three digit numbers which leave the remainder 3 when divided by 5.
Question 29.
Construct a triangle similar to given ∆ABC in which AB = 4 cm, BC = 6 cm and ∠ABC = 60°, such that each side of the new triangle is 3/4 of given ∆ABC.
Question 30.
Prove that: \(\frac { tan\theta }{ 1-cot\theta } +\frac { cot\theta }{ 1-tan\theta } =1+sec\theta cosec\theta =1+tan\theta +cot\theta \) .
Answers
SECTION—A
Answer 1.
Every composite number can be factorised as a product of prime numbers, and this factorization is unique, apart from the order in which the prime factors occur.
Answer 2.
The given quadratic equation is
4x2 – 12x – 9 = 0
Here, a = 4, b = – 12, c = – 9
D = b2 – 4 ac
= (-12)2-4x4x (-9) = 144 +144 = 288
Since D > 0, therefore, roots of the given equation are real and distinct.
Answer 3.
Given, a18 – a14 = 32
=> (a + 17d) – (a + 13d) = 32 => 17d – 13d = 32 or \(d=\frac { 32 }{ 4 } =8\)
Answer 4.
Points A, B and C are collinear
=> 1(0 – b) + 0 ( b – 2) + a (2 – 0) = 0
=> -b+2a=0 or 2a=b
Answer 5.
Yes, \(\frac { PA }{ AQ } =\frac { 5 }{ 12.5-5 } =\frac { 5 }{ 7.5 } =\frac { 2 }{ 3 } \)
\(\frac { PB }{ BR } =\frac { 4 }{ 6 } =\frac { 2 }{ 3 } \)
Since \(\frac { PA }{ AQ } =\frac { PB }{ BR } =\frac { 2 }{ 3 } \)
∴ AB||QP
Answer 6.
sin67°+cos75°
Answer 7.
The series is in the form 1,2, 3,4,…, 31.
∴ Money saved = 1 + 2 + 3 + ………………. + 31
Here, a = 1, d = 1, n = 31
Then, Sn =\(\frac { n }{ 2 } \){2a + (n-1)d}
=\(\frac { 31 }{ 2 } \){2×1+(31-1)x1} =\(\frac { 31 }{ 2 } \)[2+30]=\(\frac { 31 }{ 2 } \)x32=₹496
Money saved =₹496
Answer 8.
Here, P = (2, – 3), Q = (x, 5), PQ = 10 units
xi = 2, y1 = – 3, x2 = x, y2 = 5 B
y using distance formula,
Answer 9.
Total number of cards in a deck of playing cards = 52
Cards removed = 4 + 4 + 4 = 12
Remaining cards = 52 – 12 = 40
(a) Number of possible outcomes = 40
Number of favourable outcomes = 4 which are all kings.
P (getting a face card) \(=\frac { 4 }{ 40 } =\frac { 1 }{ 10 } \)
(b) Number of favourable outcomes = 40 – 4 = 36
P (not getting a face card) \(=\frac { 36 }{ 40 } =\frac { 9 }{ 10 } \)
Answer 10.
∴The given system does not represent a pair of coincident lines.
Answer 11.
Product of the number on the dice is prime number, i.e., 2, 3,5.
The possible ways are, (1,2), (2,1), (1, 3), (3,1), (5,1), (1, 5)
So, number of possible ways = 6
∴ Required probability \(=\frac { 6 }{ 36 } =\frac { 1 }{ 6 } \)
Answer 12.
If 4n ends with 0, then it must have 5 as a factor. But, (4)n = (22)n = 22n , i.e., the only prime factor of 4n is 2. Also, we know from the fundamental theorem of arithmetic that the prime factorisation of each number is unique.
.∴ 4n can never end with 0.
Answer 13.
The zeros of f(x) are given by
Answer 14.
Let us assume that \(5+\sqrt { 2 } \) is a rational number. Then there exist co-prime positive integers a and b such that \(5+\sqrt { 2 } =\frac { a }{ b } \)
Answer 15.
Answer 16.
Answer 17.
Here, A = (- 3, 0), B = (1, – 3), C = (4,1)
Now, by using distance formula
Answer 18.
Since tangents from an exterior point to a circle are equal in length.
Answer 19.
In square PQRS, side a – 28 m.
In two semi-circles, PQ and RS are diameters. So, PQ = RS = 28 m
Answer 20.
Given: ∆ABC in which D and E are the mid-points of sides AB and AC respectively.
To prove: DE||BC
Proof: Since, D and E are the mid-points of AB and AC respectively
Answer 21.
Answer 22.
Here, the maximum class frequency is 61 and the class corresponding to this frequency is 60-80. So, the modal class is 60-80.
Hence, modal lifetime of the components is 65.625 hours.
Answer 23.
Let AB be the pole of height 60 m. Let the height of the other pole CD be h m and BC be the width of the river. If BC = xm, then DE=BC=xm.
Hence, width of the river = 34.6 m and height of the other pole = 40 m.
Answer 24.
Given: Two triangles ABC and PQR such that ∆ABC ~ ∆PQR
Answer 25.
Let r1 cm and r2 cm denote the radii of the base of the cylinder and cone respectively. Then,
r1=r2=8cm
Let h1 and h2 cm be the heights of the cylinder and the cone respectively. Then,
h1 = 240 cm and h2 = 36 cm.
Answer 26.
Class interval |
Frequency (l) |
Cumulative frequency (cf) |
0-5 |
7 |
7 |
5-10 |
10 |
17 |
10-15 |
x |
17+x |
15-20 |
13 |
30+x |
20-25 |
y |
30+x+y |
25-30 |
10 |
40+x+y |
30-35 |
14 |
54+x+y |
35-40 |
9 |
63+x+y |
Total |
100 |
n=100 (Given)
So, 63+x+y=100 ….(i)
i.e., x+y=37
The median is 20.75 which lies in the class interval 20-25. So, median class is 20 – 25.
Answer 27.
Let the original price of the petrol be ₹x per litre.
Then, quantity of petrol that can be purchased = \(\frac { 1740 }{ x } \) litre
According to question
∴ Original cost of petrol was ₹58 per litre.
Value:
- Petrol is a natural resource which is depleting day-by-day. So, due to more demand and less supply, its price is increasing.
- We should use more of public transport and substitute petrol with CNG or other renewable resources.
Answer 28.
Let a be the first term and d be the common difference of the AP.
Answer 29.
Steps of construction:
Step I: Draw a line segment BC = 6 cm.
Step II: Draw ∠YBC = 60° at B.
Step III: Mark an arc of BA = 4 cm.
Step IV: Join A to C to get the triangle.
Step V: Draw any ray BX making an acute angle with BC on the side opposite to the vertex A.
Step VI: Locate 4 (the greater of 3 and 4 in 3/4) points B1, B2, B3 and B4 on BX so that BB1 = B1B2 = B2B3 = B3B4.
StepVII: Join B4 to C and draw a line through B3 (the 3rd point 3 being smaller of 3 and 4 in 3/4) parallel to B4C to intersect BC at C’.
StepVIII: Draw a line through C’ parallel to the line CA to intersect BA at A’. Then, AA’ BC’ is the required triangle.
Answer 30.
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