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 1.
CBSE Sample Papers for Class 10 Maths Paper 1
Board | CBSE |
Class | X |
Subject | Maths |
Sample Paper Set | Paper 1 |
Time Allowed | 3 hours |
Maximum Marks | 80 |
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 1 of Solved CBSE Sample Paper for Class 10 Maths is given below with free PDF download solutions.
General Instructions
All questions are compulsory.
- This questions paper consists of 30 questions, distributed in four sections – A, B, C and D.
- Section A contains six questions of 1 mark each; Section B contains six questions of 2 marks each; Section C contains ten question of 3 marks each; and Section D contains eight question of 4 marks each.
- There is no overall choice. However, an internal choice has been provided in four questions of 3 marks each; and in three questions of 4 marks each. You have to attempt-only one of the two alternatives given in these questions.
- Use of calculators is not permitted.
SECTION – A
Question 1.
Given the HCF (306,657) = 9, find the LCM (306,657) (1)
Question 2.
Check whether -150 is a term of the AP: 11, 8, 5, 2 (1)
For more information about NCERT Solutions Class 10 Maths
Question 3.
The sum of two numbers is 35 and their difference is 13. Find the two numbers. (1)
Question 4.
In ∆ABC, DE || BC. If AD: DB = 3: 5 and AC = 4.8 cm, find AE. (1)
Question 5.
The length of a tangent from a point A at a distance 5 cm from the centre of a circle is 4 cm. Find the radius of the circle. (1)
Question 6.
What is the lower limit of the median class of the following frequency distribution?
Age (in years) | 0 – 10 | 10 – 20 | 20 – 30 | 30 – 40 | 40 – 50 | 50 – 60 |
Number of Patients | 16 | 13 | 6 | 11 | 27 | 18 |
SECTION – B
Question 7.
Show that any positive integer is of the form 3q or 3q + 1 or 3q + 2, for some integer q. (2)
Question 8.
Find the degree of the polynomial (x + 1) (x^{2} – x – x^{4}+ 1). (2)
Question 9.
Find the roots of the quadratic equation: 6x^{2} – x – 2 – 0. (2)
Question 10.
In Fig. 2, PQ || BC and AP: PB = 1: 2.
Question 11.
Power that \(\sqrt { \frac { 1+\sin { A } }{ 1-\sin { A } } }\) = sec A + tan A
Question 12.
In Fig. 3, find the perimeter of the shaded region where ADC, AEB and BFC are semi-circles on diameters AC, AB and BC respectively.
SECTION – C
Question 13.
Prove that \(\sqrt { 5 }\) is an irrational number. (3)
Question 14.
Find a quadratic polynpmial with \(\frac { 1 }{ 4 } \) and \(\frac { 1 }{ 4 } \) as the sum and product of its zeroes respectively. (3)
Question 15.
Show graphically that the following system of equations
2x + 4y = 10 and 3x + 6y = 12 has no solution. (3)
Question 16.
Determine if the points (1, 5), (2, 3) and (-2, -11) are collinear. (3)
Question 17.
Find the ratio in which the j’-axis divides the line segment joining the points (-4, -6) and (10, 12). Also, find the coordinates of the point of division. (3)
Question 18.
Prove that the tangent to a circle is perpendicular to the radius through the point of contact. (3)
OR
In Fig. 4, a quadrilateral ABCD is drawn to circumscribe a circle. Prove that AB + CD = AD + BC (3)
Question 19.
if sin(x – 20)° = cos (3x – 1O)°, then find the value of x (3)
Question 21.
A box contains 7 red, 8 green and 5 white balls. A ball is drawn at random from the box. Find the probability that the ball drawn is (a) white (b) neither red nor white. (3)
OR
A die has its six faces marked 0, 1, 1, 1, 6, 6. Two such dice are thrown together and the total score is recorded.
- How many different scores are possible?
- What is the probability of getting a total of 7?
Question 22.
A bag contains 5 red balls and some blue balls. If the probability of drawing a blue ball from the bag is thrice that of a red ball, find the number of blue balls in the bag. (3)
SECTION – D
Question 23.
Find a natural number whose square diminished by 84 is equal to thrice of 8 more than the given number. (4)
OR
Find whether the following equation has real roots. If real roots exist, find them.
x^{2} + 5 \(\sqrt { 5 } \) x – 70 = 0
Question 24.
The 8^{th} term of an AP is half its 2^{nd} term and the 11th term exceeds one-third of its 4^{th} term by 1. Find the 15^{th} term. (4)
OR
Find the sum of those integers from 1 to 500 which are multiples of 2 or 5.
Question 25.
In Fig. 5, BAD is a triangle-right angled at A and AC ⊥ BD. Show that
- AB^{2} = BC.BD
- AD^{2} = BD.CD
- AC^{2} = BC.DC
Question 26.
Draw a triangle ABC in which AB = 4 cm, BC = 6 cm and AC to ∆ABC with scale factor \(\frac { 3 }{ 2 } \) Are the two triangles congruent (4)
Question 27.
Determine the height of a mountain if the angle of elevation of its top at an unknown distance from the base is 30° and at a distance 10 km further off from the mountain, along the same line, the angle of elevation is 15°. [Use tan 15° = 0.27] (4)
OR
A person standing on the bank of a river observes that the angle of elevation of the top of a tree standing on the opposite bank is 60°. When he moves 40 metres away from the bank, he finds the angle of elevation is 30° Find the height of the tree and the width of the river.
Question 28.
In Fig. 6, arcs have been drawn of radius 21 cm each with vertices A, B, C and D of a quadrilateral ABCD as centres. Find the area of the shaded region.
Question 29.
The length of 40 leaves of a plant are measured correct to the nearest millimetre and the data
obtained is represented in the following table :
Find the median length of the leaves. (4)
Question 30.
A copper wire, 3 mm in diameter, is wounded about a cylinder whose length is 12 cm, and diameter 10 cm, so as to cover the curved surface of the cylinder. Find the length and mass of the wire, assuming the density of copper to be 8.88 g per cu. cm.
Solutions
SECTION – A
Solution 1.
We know that
HCF × LCM = Product of two numbers
⇒ LCM = \(\frac { 306\times 657 }{ 9 } =22338\)
Thus, LCM (306,657) = 22338
Solution 2.
Assume that -150 is the n^{th} term of the given AP. Then,
Solution 3.
Let the two numbers bep and q (p > q). Then, as per the question,
p + q = 35 and p – q = 13
Solving the two equations simultaneously, we have
2p = 48 and 2q = 22
p = 24 and q = 11
Thus, the two numbers are 24 and 11. (1)
Solution 4.
Since DE || BC, by B.P. Theorem,
Solution 5.
We know that
OP ⊥ AP
So, in ∆OPA,
OA^{2} = OP^{2} + AP^{2}
⇒ OP^{2} = OA^{2} – AP^{2} = 5^{2} – 4^{2} = 25 – 16 = 9
⇒ OP = \(\sqrt { 9 } \) =3cm.
Thus, the radius of the circle is 3 cm.
(1)
Solution 6.
Here, cumulative frequencies are :
16, 29, 35, 46, 73, 91 .
So, N = 91 and \(\frac { N }{ 2 } \) = 45.5
Thus, lower limit of the median class is 30. (1)
SECTION – B
Solution 7.
Let V be an integer and b = 3.
Applying Euclid’s division lemma with a and b = 3, we have
a = 3q + r, where 0 ≤ r < 3 and q is some integer
⇒ a = 3q or a = 3q + 1 or a = 3q + 2 for more integer q. (2)
Solution 8.
(x + 1 ) (x^{2} – X- X^{4} + 1)
= (x^{3} – x^{2} – x^{5} + x) + (x^{2} – x – x^{4} + 1)
= -x^{5} – x^{4} + x^{3} + 1
Thus, the degree of the given polynomial is 5. (2)
Solution 9.
We have
Solution 10.
As PQ || BC, we have
∠APQ = ∠ABC (Corr. angles)
∠PAQ = ∠BAC (Common)
∠AQP = ∠ACB (Alt. angles)
Solution 11.
Here
Hence, proved. (2)
Solution 12.
From the given figure,
Perimeter of the shaded region
SECTION – C
Solution 13.
Let us assume, to the contrary, that \(\sqrt { 5 }\) is a rational number
Then, there exists co-prime positive integers a and b such that :
It means 5 divides b^{2} and so 5 divides b.
So, 5 is a common factor of both a and b, which is a contradiction.
So, our assumption that \(\sqrt { 5 }\) is rational is wrong.
Hence, we conclude that \(\sqrt { 5 }\) is an irrational number. (3)
Solution 14.
Let α , β be the two zeroes of the quadratic polynomial.
Here, we have
Solution 15.
We have, 2x + 4y = 10
⇒ y = \(\frac { 5-x }{ 2 }\)
Thus, we have the following table of values :
Plot the points A (1, 2), B (3,1) and C (5, 0) on a graph paper. Join A, B and C and get a line, the graph of the equation 2x + 4y = 10 Further, we have, 3x + 6y = 12
⇒ y = \(\frac { 12-3x }{ 6 }\)
Thus, we have the following table :
Plot the point P(0, 2), Q(2, 1) and R(4, 0) on the same graph paper. Join P, Q and R and get a line, the graph of the equation 3x + 6y = 12
From the graph, we find that the two lines represented by 2x + 4y = 10 and 3x + 6y = 12 are parallel.
So, two lines do not have any common point.
Hence, the given system of equations has no solution. (3)
Solution 16.
Let the given three points be denoted by A, B and C respectively, i.e.
A (1, 5), B (2, 3), C (-2, -11)
The points A, B and C will be collinear if the sum of the lengths of the two line segments is equal to the length of the third line segment.
AB + BC = AC or AB + AC = BC or AC + BC = AB
Solution 17.
Let the ratio be k: 1.
Then, by section formula, the coordinates of its point P which divides the line segment in the ratio k: 1 are
Solution 18.
Given : A circle C (o, r) and a tangent AB at a point P.
To Prove : OP ⊥ AB
Construction : Take any point Q, other than P, on the tangent AB. Join OQ. Let OQ meets the circle at R
Proof : To prove that OP ⊥ AB„ it is sufficient to prove that OP is shorter than any other segment joining O to any point of AB.
Thus, OP is shorter than any other segment joining O to any point on AB.
Hence, OP ⊥ AB (3)
OR
Since lengths of two tangents drawn from an external point to a circle are equal.
AP = AS, BP = BQ, DR = DS and CR = CQ
Adding all these, we have
(AP + BP) + (CR + RD) = (BQ + CQ) + (DS + AS)
⇒ AB + CD = BC + DA. (3)
Solution 19.
sin(x – 20)° = cos (3x – 10)° gives
cos [90° – (x – 20)°] = cos (3.x – 10)°
⇒ 90° – (x – 20)° = (3x – 10)°
⇒ 90° – x° + 20° = 3x° – 10°
⇒ 4x° = 120°
⇒ x° = 30°
Thus, x = 30 (3)
OR
Solution 20.
Solution 21.
Here, n(S) = 7 + 8 + 5 = 20.
(a) P (a white ball) = \(\frac { 50 }{ 20 }\) = \(\frac { 1 }{ 4 }\)
(b) P (neither red nor white) = P (a green ball) = \(\frac { 8 }{ 20 }\) i.e. \(\frac { 2 }{ 5 }\)
OR
Total number of possible outcomes = 36
i. Six different scores possible are 0, 1, 2, 6, 7 and 12
ii. Sample space = {(1, 6), (1, 6), (1, 6), (1, 6), (1, 6), (1, 6), (6, 1), (6, 1), (6, 1), (6, 1), (6, 1), (6, 1),}
∴ Number of favourable outcomes = 12
So, P(a total of 7) = \(\frac { 12 }{ 36 }\) = \(\frac { 1 }{ 3 }\) (3)
Solution 22.
Let there be V blue balls in the bag.
∴ Total number of balls in the bag = 5 + x
Now
SECTION – D
Solution 23.
Let the given natural number be x
According to question,
x^{2} – 84 = 3 (x + 8)
⇒ x^{2} – 3x – 108 = 0
⇒ x^{2} – 12x + 9x- 108 = 0
⇒ x (x – 12) + 9 (x – 12) = 0
⇒ (x + 9)(x – 12) = 0
Either x – 12 = 0 or x + 9 = 0
⇒ x = 12 or x = – 9
Since x is a natural number, x ≠ – 9
x = 12
Hence, the required number is 12. (4)
OR
Solution 24.
Let ‘a’ be the first terni and V’ be the common difference of AP. Then, according to the question,
Solution 25.
Solution 26.
Solution 27.
Let AB be the mountain of height ‘b’ km.
Let C be a point at a distance of x km, from the base of the mountain such that the angle of elevation of the top at C is 30°. Let D be a point at a distance of 10 km from C such that angle of elevation at D is of 15°.
In ∆CAB, we have
Hence, the mountain is 5 km high.
OR
Let AB be the tree of height ‘ti metres standing on the bank of a river. Let C be the position of man standing on the opposite bank of river such that BC = x m.
Let D be the new position of the man. It is given that CD = 40 m and the angles of elevation of the top of the tree at C and D are 60° and 30° respectively,
∠ACB = 60° and ∠ADB = 30°
From (i), we have x = 20 m
Thus, the height of the tree is 20\(\sqrt { 3 }\) m and width of the river 20 m. (4)
Solution 28.
It is given that radius of each sector (r) = 21 cm
Thus, the required area is 1386 sq cm. (4)
Solution 29.
First we have to convert the given class intervals into continuous classes as follows :
Class Interval | C.I | Number of leaves fi | C.f. |
118 – 126 | 117 – 126.5 | 3 | 3 |
127 – 135 | 126.5 – 135.5 | 5 | 8 |
136 – 144 | 135.5 – 144.5 | 9 | 17 |
145 – 153 | 144.5 – 153.5 | 12 | 29 |
154 – 162 | 153.5 – 162.5 | 5 | 34 |
163 – 171 | 162.5 – 171.5 | 4 | 38 |
172 – 180 | 171.5 – 180.5 | 2 | 40 |
Here, N = 40. So, \(\frac { N }{ 2 } \) = 20
Median class is 144.5 – 153.5
[Median class = The class interval corresponding to c.f ≥ \(\frac { N }{ 2 } \)].
Width of the median class (b) = 153.5 – 144.5 = 9
Lower limit of the median class (l) = 144.5
Frequency of the median class (f) = 12
Cumulative frequency of the pre-median class (c) = 17
Using Median Formula, we have
Hence, the median length of the leaves is 146.75 mm. (4)
Solution 30.
Radius of the cylinder (r) = \(\frac { 10 }{ 2 }\) = 5 cm
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