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. According to new CBSE Exam Pattern, MCQ Questions for Class 10 Maths Carries 20 Marks.
CBSE Sample Papers for Class 10 Maths Paper 1
|Sample Paper Set||Paper 1|
|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.
Time Allowed: 3 hours
Max. Marks: 80
- 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.
Write the condition to be satisfied by q so that a rational number has a terminating decimal expansion.
For what value of k the quadratic equation x2 – kx + 4 = 0 has equal roots?
Given that tan . What is the value of ?
Which term of the sequence 114,109,104 … is the first negative term?
In the given figure, DE is parallel to BC and AD = 1 cm, BD = 2 cm. What is the ratio of the area of ∆ ABC to the area of ∆ ADE?
AOBC is a rectangle whose three vertices are A (0, 3), O (0, 0) and B (5, 0). Find the length of its diagonal.
Without drawing the graph, state whether the following pair of linear equations will represent intersecting lines, coincident lines or parallel lines:
6x – 3y + 10 = 0
2x – y + 9 = 0
Justify your answer.
Find a point on the y-axis which is equidistant from the points A (6,5) and B (- 4, 3).
One card is drawn from a well shuffled deck of 52 playing cards. Find the probability of getting
- a non-face card.
- a black king or a red queen.
Write whether the square of any positive integer can be of the form 3m + 2, where m is a natural number. Justify your answer.
The 4th term of an AP is zero. Prove that the 25th term of the AP is three times its 11th term.
A card is drawn at random from a pack of 52 playing cards. Find the probability that the card drawn is neither an ace nor a king.
Using Euclid’s division algorithm, find the HCF of 56,96 and 404.
Prove that is an irrational number.
If two zeros of the polynomial x4 + 3x3 – 20x2 – 6x + 36 are and , find the other zeros of the polynomial.
Draw the graph of the following pair of linear equations
x + 3y = 6
2x-3y = 12
Hence find the area of the region bounded by x = 0, y = 0 and 2x-3y = 12
Observe the graph given below and state whether triangle ABC is scalene, isosceles or equilateral. Justify your answer. Also find its area.
Find the area of the quadrilateral whose vertices taken in order are
A(- 5, – 3), B (- 4, – 6), C (2, -1) and D (1, 2).
Prove that the intercept of a tangent between two parallel tangents to a circle subtends a right angle at the centre of the circle.
A square field and an equilateral triangular park have equal perimeters. If the cost of ploughing the field at rate of ₹5/m2 is ₹720, find the cost of maintaining the park at the rate of ₹10/m2.
An iron solid sphere of radius 3 cm is melted and recast into small spherical balls of radius 1 cm each. Assuming that there is no wastage in the process, find the number of small spherical balls made from the given sphere.
A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.
The inner circumference of a circular track [Fig.] is 220 m. The track is 7 m wide everywhere. Calculate the cost of putting up a fence along the outer circle at the rate of ₹2 per metre.
Find the mean of the following distribution:
Some students arranged a picnic. The budget for food was ₹240. Because four students of the group failed to go, the cost of food to each student got increased by ₹ 5. How many students went for the picnic?
A plane left 30 minutes late than its scheduled time and in order to reach the destination 1500 km away in time, it had to increase the speed by 250 km/h from the usual speed. Find its usual speed.
From the top of a building 100 m high, the angles of depression of the top and bottom of a tower are observed to be 45° and 60° respectively. Find the height of the tower. Also find the distance between the foot of the building and bottom of the tower.
The angle of elevation of the top of a tower at a point on the level ground is 30°. After walking a distance of 100 m towards the foot of the tower along the horizontal line through the foot of the tower on the same level ground, the angle of elevation of the top of the tower is 60°. Find the height of the tower.
Prove that in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
Using the above, solve the following:
A ladder reaches a window which is 12 m above the ground on one side of the street. Keeping its foot at the same point, the ladder is turned to the other side of the street to reach a window 9 m high. Find the width of the street if the length of the ladder is 15 m.
The interior of building is in the form of a right circular cylinder of radius 7m and height 6 m, surmounted by a right circular cone of same radius and of vertical angle 60°. Find the cost of painting the building from inside at the rate of ₹30/m2.
The following table shows the marks obtained by 100 students of class X in a school during a particular academic session. Find the mode of this distribution.
|Marks||No. of Students|
|Less than 10||7|
|Less than 20||21|
|Less than 30||34|
|Less than 40||46|
|Less than 50||66|
|Less than 60||77|
|Less than 70||92|
|Less than 80||100|
Do you think marks obtained is the appropriate way to evaluate a student?
Divide 56 into four parts which are in AP such that the ratio of product of extremes to the product of means is 5 : 6.
If Sn denotes the sum of the first n terms of an AP, prove that S30= 3 (S20 – S10).
Construct a ∆ ABC in which CA = 6 cm, AB = 5 cm and ∠ BAC = 45°, then construct a triangle similar to the given triangle whose sides are of the corresponding sides of the ∆ ABC.
If x sin3 θ + y cos3 θ = sin θ cos θ and x sin θ = y cos θ, prove that x2 + y2 = 1.
The prime factorisation of q is of the form 2n5m, where n, m are non-negative integers.
We have x2-kx+4=0
Here, a=1, b=-k, c=4
For equal roots, D=0
⇒ b2-4ac=0 ⇒ (-k)2-4x1x4=0
⇒ k2-16=0 ⇒ k2=16
The given sequence is an AP in which 1st term is 114 and common difference is – 5. Let the nth term of the given AP be the first negative term i.e., an < 0.
Thus, 24th term of the given sequence is the first negative term.
In ∆ ABC and ∆ ADE
Length of diagonal
The given system of equations is
6x – 3y + 10 = 0 and 2x – y + 9 = 0
The given system of equations is of the form
a1X +b1y + c1 = 0
a2x +b2y+ c2 = 0
So, the given system of equations will represent parallel lines.
Here we have A = (6,5), B = (-4,3)
Let P (0, y) be the point on the y-axis which is equidistant from the points A (6,5) and B (-4, 3).
By using distance formula
Hence, the required point is (0, 9).
Number of possible outcomes = 52
Total face cards – 12
Number of favourable outcomes = 52 -12 = 40
Number of possible outcomes = 52
Number of favourable outcomes = 2+2= 4
No, because any positive in tegercanbewrittenas3q, 3q+1,3q+2, therefore, square will be 9q2 = 3m, 9q2 + 6q + 1 = 3 (3q2 + 2q) + 1 = 3m + 1,
9q2 + 12q + 4 = 3 (3(q2 + 4q + 1) + 1 = 3m + 1.
⇒a+3d=0 ⇒ a=-3d
∴ a25=3a1 Hence Proved
Let E be the event card drawn is neither an ace nor a king.
Then, the number of outcomes favourable to the event E = 44 (4 kings and 4 aces are not there)
Given integers are 56,96 and 404.
First we find the HCF of 56 and 96.
Applying Euclid’s division algorithm,
we get 96 = 56 x 1 + 40
Since the remainder 40 ≠ 0, so we apply the division lemma to 56 and 40.
56 = 40 x 1 + 16
Since the remainder 16 ≠ 0, so we apply the division lemma to 40 and 16.
40 = 16 x 2 + 8
Since 8 ≠ 0, so we apply the division lemma to 16 and 8.
16 = 8×2 + 0
Clearly, HCF of 56 and 96 is 8.
Let us find the HCF of 8 and the third number 404 by Euclid’s algorithm.
Applying Euclid’s division, we get
404 = 50 x 8 + 4
Since the remainder is 4 ≠ 0. So we apply the division lemma to 8 and 4.
8 = 4 x 2 + 0
We observe that the remainder at this stage is zero. Therefore, the divisor of this stage i.e., 4 is the HCF of 56,96 and 404.
Let us assume on the contrary that is a rational number. Then, there exist co-prime positive integers a and b such that
This contradicts the fact that 75 is an irrational number. So, our assumption is wrong. Hence 3-75 is an irrational number.
We know that if x = a is a zero of a polynomial, then x-α is a factor of/(x).
Since and are zeros of/(x). Therefore, is a factor of/(x).
Now, we divide/(x) = x4 + 3x3 – 20x2 – 6x + 36 by g(x) = x2 – 2 to find the other zeros of f(x).
By division algorithm, we have
Hence, the other zeros of the polynomial are 3 and – 6.
From equation (i), we have
x = 6 – 3y
From equation (ii), we have
Plotting the points (3,1), (0, 2), (6, 0), (9, 2) and (0, – 4) on the graph paper on a suitable scale and drawing a line joining them, we obtain the graph of the lines represented by the equations x + 3y = 6 and 2x – 3y = 12 as shown in figure.
It is evident from the graph that the two lines intersect at point (6, 0).
Area of the region bounded by x – 0, y = 0 and 2x-3y = 12.
= Area of ∆ OAB
= x OA x OB = x 6 x 4 = 12 sq. units.
In the given graph the triangle A ABC is scalene. Justification: The coordinates of A, B and C are (- 3, – 4), (3,0) and (- 5,0) respectively.
Given: AB and CD are two tangents to a circle and AB || CD. Tangent ST subtends an angle SOT at the centre.
Let the side of the square field = a m
and the side of the equilateral triangular park = b m
According to question, perimeter of square = perimeter of equilateral triangle
Let AB be a vertical pole of length 6 m and BC be its shadow and DE be tower and EF be its shadow. Join AC and DF.
Let the inner and outer radii of the circular track be r m and R m respectively. Then, Inner circumference = 2πr = 220 m
Calculation of arithmetic mean
|Total||Σfi = 40||Σfixi = 360|
Let the number of students who went for picnic = x
The cost of food to each student =
When four students failed to go, then number of students = x – 4
New cost of food to each student =
According to question, we have
Let, AB be the building and CD be the tower.
Let, distance between the foot of the building and the bottom of the tower BC=xm.
∴ DE=x m.
Let, height of the tower DC=hm, then BE=hm. So,
Given: A right triangle ABC right-angled at B.
To Prove: AC2 = AB2 + BC2
Construction: Draw BD ⊥ AC
Proof: In ∆ADB and ∆ ABC
r1 = radius of the base of the cylinder = 7 m
r2 = radius of the base of the cone = 7 m
h1 = height of the cylinder = 6m
h2 = height of the cone.
In ∆ VAB,
Hence, the mode of distribution is 44.7.
Value: No, in my opinion marks obtained should not be the only criteria to evaluate a student.
Let the four parts be a – 3d, a – d, a + d, a + 3d.
Steps of construction: We follow the following steps:
Step I: Draw AB = 5 cm.
Step II: At A, draw ∠ BAQ = 45°
Step III: Cut AC = 6 cm from AQ.
Step IV: Join CB to get ∆ ABC.
Step V: Draw any acute angle ∠BAP at A on opposite side of vertex C of ∆ ABC.
Step VI: Along AP, locate 6 points A1, A2, A3, A4, A5 and A6 such that AA1 = A1A2 = … = A5A6.
Step VII: Join A5 to B.
Step VIII: At point A6 draw a line parallel to A5B, which meets AB produced at S’.
Step IX: From B’ draw a line B’C’ parallel to BC.
Step X: AB’C’is the required triangle whose sides are of the corresponding sides of ∆ ABC. 5
Justification: Since BC || B’C’
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