# CBSE Sample Papers for Class 10 Maths Paper 1

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.

Time Allowed: 3 hours
Max. 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 4 questions of 3 marks each and 3 questions of 4 marks each. You have to attempt only one of the alternatives in all such questions.
• Use of calculator is not permitted.

SECTION-A

Question 1.
After how many decimal places will the decimal expansion of the number 47/23 . 52
terminate ? [1]

Question 2.
If two zeroes of polynomial ƒ(x) = x3 – 4x² – 3x + 12 are √3 and – √3, then find third zero. [l]

Question 3.
Find the smallest positive value of k for which the equation x² + kx + 9 = 0 has real roots. [1]

Question 4.
The coordinates of the points P and Q are (4, – 3) and (- 1, 7), respectively. Find the abscissa of a point R on the line segment PQ such that $$\frac { PR }{ PQ }$$ = 3/5. [1]

Question 5.
In the adjoining figure, PA and PB are tangents from a point P to a circle with centre O. Show that quadrilateral that OAPB is cyclic. [1]

Question 6.
If for some angle θ, cot 2θ = 1/√3, then find the value of sin 30, where 2θ < 90°. [1]

SECTION-B

Question 7.
Is there any natural number n for which 6n ends with the digit 0 ? Give reasons in support of your answer. [2]

Question 8.
Which term of the A.P. : 5, 17, 29, 41, … will be 120 more than its 15th term ? [2]

Question 9.
The length of a line segment is 10 units. If one end is (2, – 3) and the abscissa of the other end is 10, then its ordinate is either 3 or – 9. Give justification for the two answers. [2]

Question 10.
Find the value of c for which the pair of equations cx – y = 2 and 6x – 2y = 4 will have infinitely many solutions. [2]

Question 11.
More than ogive and less than ogive for a given data with mean 23 meet at (20, 22). Find its mode. [2]

Question 12.
A letter of English alphabets is chosen at random. Find the probability that it is a letter of the word ‘MATHEMATICS’. [2]

SECTION-C

Question 13.
Find the zeroes of the polynomial p(x) = 4√3x² – 2√3x – 2√3 and verify the relationship between the zeroes and the coefficients. [3]
OR
On dividing the polynomial ƒ(x) = x3 – 5x² + 6x – 4 by a polynomial g(x), the quotient q(x) and remainder r(x) are x – 3 and – 3x + 5, respectively. Find the polynomial g(x).

Question 14.
If the sum of the first n terms of an A.P. is 4n – n² , what is the 10th term and the nth term ? [3]
OR
Flow many terms of the A.P. : 9, 17, 25, … must be taken to give a sum 636 ?

Question 15.
If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find the value of x and y. [3]

Question 16.
The sides AB, BC and median AD of a ∆ABC are respectively proportional to the sides PQ, QR and the median PM of ∆PQR. Show that ∆ABC ~ ∆PQR. [3]
OR

ABC is an isosceles triangle right angled at B. Two equilateral triangles ACD and ABE are constructed on the sides AC and AB, respectively. Find the ratio of the area of ∆ABE and ∆ACD.

Question 17.
A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm, respectively. Find the sides AB and AC. [3]
OR

Two tangents TP and TQ are drawn to a circle with centre O from an external point T. Prove that ∠PTQ = 2∠OPQ.

Question 18.
Evaluate:

Question 19.
In the figure, ABC is a triangle right angled at A. Semi-circles are drawn on AB, AC and BC as diameters. Find the area of the shaded region. [3]

Question 20.
A bag contains white, black and red balls only. A ball is drawn at random from the bag. The probability of getting a white ball is 3/10 and that of a black ball is 2/5. Find the probability of getting a red ball. If the bag contains 20 black balls, then find the total number of balls in the bag. [3]

Question 21.
Show that the square of an odd positive integer can be of the form 6q + 1 or 6q + 3 for some integer. [3]

Question 22.
A rectangular water tank of base 11 m × 6 m contains water upto a height of 5 m. If the water in the tank is transferred to a cylindrical tank of radius 3.5 m, find the height of the water level in the tank. [3]

SECTION-D

Question 23.
Construct an isosceles triangle whose base is 6 cm and altitude 5 cm and then construct another triangle whose sides are $$\frac { 7 }{ 5 }$$ of the corresponding sides of the isosceles triangle. [4]

Question 24.
Prove that :

Question 25.
If the price of a book is reduced by ₹5, a person can buy 5 more books for ? 300. Find the original list price of the book. [4]
OR
Solve for x :

Question 26.
State and prove the basic proportionality theorem. [4]
OR
Prove that the ratio of the areas of two similar triangles is equal to the ratio of the squares of the corresponding sides.

Question 27.
A building is made by keeping the lower window of a building at a height of 2 m above the ground and its upper window 6 m vertically above the lower window in view to have proper sunlight. At certain instant, the angles of elevation of a balloon from these windows are observed to be 60° and 30° respectively. Find the height of the balloon above the ground. [4]

Question 28.
A well of diameter 3 m and 14 m deep is dug. The earth, taken out of it, has been evenly spread all around it in the shape of a circular ring of width 4 m to form an embankment. Find the height of the embankment. [4]

Question 29.
Solve the equations 5x – y = 5 and 3x – y = 3 graphically. Find area of the triangle formed between the lines and Y-axis. [4]
OR
A two-digit number is obtained by either multiplying the sum of the digits by 8 and then subtracting 5 or by multiplying the difference of the digits by 16 and then adding 3. Find the number.

Question 30.
The following table shows the ages of the patients admitted in a hospital during a month :

 Age (in years) Number of patients 5-15 6 15-25 11 25-35 21 35-45 23 45-55 14 55-65 5

Find the mean and the mode of the data given above. [4]

SOLUTIONS
SECTION-A

Let

x will terminate after 3 decimal places.

Let α = √3, β = √3 be the given zeroes and γ be the third zero, then
α + β + γ = $$-\frac { -4 }{ 1 }$$
√3 – √3 + γ = 4
γ = 4
Hence, third zero is 4.

Given equation is,
Here A = 1, B = k, C = 9
∵ Roots are real
∵ D ≥ 0
⇒ B² – 4AC ≥ 0
⇒ k² -4 × 1 × 9 ≥ 0
⇒ k²- 36 ≥ 0
⇒ k² ≥ 36
⇒ k² ≥ 6²
⇒ k ≥ 6 or k ≤ -6
So,the least positive value for which the given equation has real roots is k = 6.

The coordinates of P and Q are (4,-3) and (-1,7) respectively.
Let coordinate of R be (x,y).

Now,

By section formula,

Thus, abscissa of point R is 1.

In a circle, radius is perpendicular to the tangent at the point of contact.
∵ ∠OAP = 90°
and ∠OBP = 90°
=> ∠OAP + ∠OBP = 90° + 90° = 180°
∵ Opposite angles are supplementary .

Here cot 2θ = $$\frac { 1 }{ \sqrt { 3 } }$$ = cot 60°
⇒  2θ = 60° ⇒ θ = 30°
∵ sin 3θ = sin 90° = 1.

SECTION-B

6n = (2 x 3)n = 2n x 3n
6n cannot end in zero for any natural number ‘n’.
To a number end in zero, ‘2′ and ‘5’ both must be factor of the number.
Factor ‘5’ is not present in 6n so it will not end in ‘0’ for any natural number ‘n’

Given A.P. is 5, 17, 29, 41..
Here,
a = 5, d = 17 – 5 = 12
Let nth term of the A.P. is 120 more than its 15th term.
Then, according to the question,
an = 120 + a15
a + (n-1)d = 120 + a + 14d
(n-1) × 12 = 120 + 14 × 12
(n-1) × 12 = 288
n-1 = 24
n = 25
∵ Required term is 25th term.

Let AB be the line segment of length 10 cm having coordinates A(2, – 3) and B(10, b).
Here

For x = 10, there will be two points which are at a distance of 10 units from A.

Given pair of equations are,
cx – y = 2
6x – 2y =4
On comparing the equations with a1x + b1y + c1 = 0’and a2x + b2y + c2, respectively, we get
a1= c : b1 = – 1, c1 = 2
a2 = 6 : b2 = – 2, c2 = 4
System has infinitely many solutions, if
$$\frac { { a }_{ 1 } }{ { a }_{ 2 } } =\frac { { b }_{ 1 } }{ { b }_{ 2 } } =\frac { { c }_{ 1 } }{ { c }_{ 2 } }$$
⇒ $$\frac { c }{ 6 } =\frac { 1 }{ 2 } =\frac { 2 }{ 4 }$$
⇒ c = 3.

Given : Mean, $$\bar { x }$$ = 23, Median, m0 = 20 (x-coordinate of point of intersection)
We know,
Mode, M0 = 3m0 – 2$$\bar { x }$$
=3 × 20 – 2 × 23 = 60 – 46 = 14

Favourable cases : M, A, T, H, E, I, C, S
=> No. of favourable cases = 8
=> n(E) = 8
Total cases = 26
=> n(S) = 26
Probability = $$\frac { n(E) }{ n(S) } =\frac { 8 }{ 26 } =\frac { 4 }{ 13 }$$

SECTION-C

Given polynomial is,
p(x) = 4√3x² – 2√3x – 2√3 = 0
=> 2√3 (2x² – x -1) =0
=> 2√3 (2x²-2x + x-1) =0
2√3 [2x(x – 1) + 1(x – 1)] =0
2√3 (2x + 1)(x – 1) =0
⇒  2x + 1 = 0 ⇒ x = $$-\frac { 1 }{ 2 }$$
and x – 1 = 0 ⇒ x = 1
∵ x = $$-\frac { 1 }{ 2 }$$, 1
Now
Sum of zeroes = $$-\frac { 1 }{ 2 } +1=\frac { 1 }{ 2 } =\frac { -Coefficient\quad of\quad x }{ { Coefficient\quad of\quad x }^{ 2 } }$$
Product of zeroes = $$-\frac { 1 }{ 2 } \times 1=\frac { 1 }{ 2 } =-\frac { 1 }{ 2 } =\frac { Constant\quad term }{ { Coefficient\quad of\quad x }^{ 2 } }$$
Hence, relationship between zeroes and coefficient is verified.
OR
We have,
Dividend : ƒ(x) = x3 – 5x² + 6x – 4
Divisor : g(x)
Quotient : q(x) = x – 3
Remainder : r(x) = – 3x + 5
We know,
Dividend = Divisor x Quotient + Remainde
OR
Divisor = $$\frac { Dividend-Remainder }{ Quotient }$$

15. Let the vertices of parallelogram be A( 1, 2), B(4, y), C(x, 6) and D(3, 5). We know, diagonals of a parallelogram bisect each other.
Mid-point of AC = Mid-point of BD
By Mid-point formula,

Given :
BD = 8 cm
DC = 6 cm
Radius of circle, r = 4 cm
AF = x
Let
We know, tangents to a circle from a same external point are equal
∵ BD = BF = 8 cm
AF = AE = x

DC = CE = 6 cm
AB = (x+8) cm
BC = (8+6) cm
AC = (6+x) cm

Now in ∆ABC
Semi-perimeter, s = $$\frac { AB+BC+CA }{ 2 }$$
$$\frac { X+8+14+x+6 }{ 2 } =\frac { 2x+28 }{ 2 } =x+14$$
We know,

OR
Given :
Circle with centre 0 with exterior point T having TP and TQ as tangents.
To prove : ∠PTQ = 2∠OPQ
Construction : Join PQ.
Proof : TP = TQ

[Tangents to a circle from an external point are equal]
⇒ ∆TPQ is isosceles.
⇒ ∠1 = ∠2
[Angles opposite to eqaul sides of a triangle are equal in length]
Let
∠PTQ = θ
In ∆PTQ,
∠1 + ∠2 + θ = 180°
∠1 + ∠1 = 180° – θ
2∠1 = 180° – θ
∠1 = $$\frac { { 180 }^{ 0 }-\theta }{ 2 } =({ 90 }^{ 0 }-\frac { \theta }{ 2 } )$$
∠TPO = 90°
[Radius is perpendicular to the tangent at the point of contact]
∵ ∠OPQ = 90° – ∠1 = 90° – $$-({ 90 }^{ 0 }-\frac { \theta }{ 2 } )=\frac { \theta }{ 2 }$$

From equation (i) and (ii),
∠OPQ = $$\frac { 1 }{ 2 }$$ ∠PTQ
OR
∠PTQ = 2∠OPQ

We have,

We have, AB = 3 cm, AC = 4 cm
in right ∆BAC, by Pythagoras theorem
BC² = AB² + BC²
= (3²) + (4²) = 9+16=25
BC = √25 = 5 cm

Given :
Rectangular tank :
Length = 11 m
Level of water, H = 5 m
Cylindrical tank :
Let the level of water in cylinderical tank be h.
Then,
Vol. of water in cylindrical tank = Vol. of water in rectangular tank
πr²h = L × B × H
$$\frac { 22 }{ 7 }$$ × 3.5 × h = 1 × 6 × 5
22 × 0.5 × 3.5 × h = 11 × 6 × 5
h = $$\frac { 11\times 6\times 5 }{ 22\times 0.5\times 3.5 }$$
h = 8.57 m.

SECTION-D

Steps of construction :

• Draw a line segment AB = 6 cm.
• Draw the perpendicular bisector XY of AB intersecting AB at D.
• With D as centre and radius 5 cm, draw an arc cutting XY at C.
• Join AC and BC. Thus, ∆ABC is obtained.
• Draw ray AX such that ∠BAX is acute.
• Mark A1 A2,…A7 on AX such that AA1 = A1A2 =…….= A6A7.
• Join A5 to B.
• From A7, draw a line parallel to A5B to meet AB produced at B’.
• From B’, draw a line parallel to BC to meet AC produced at C’.

Thus, ∆AB’C’ is the required triangle, whose sides are $$\frac { 7 }{ 5 }$$ of the corresponding sides of ∆ABC.

Let price of a book be ₹ x and number of books be n.
Then nx = 300
Also (x – 5) (n + 5) = 300
=> nx + 5x -5n-25 = 300
5x – 5n – 25 =0
x – n – 5 = 0
We have,
n = $$\frac { 300 }{ x }$$
Equation (ii) becomes,
x – $$\frac { 300 }{ x }$$ – 5 = 0
x² – 5x – 300 = 0
x² – 20x + 15x – 300 = 0
x(x-20) + 15(x-20) = 0
(x-20) (x+15) = 0
x = 20, -15
x ≠ – 15, as price of a book cannot be negative.
∵ price of each book is ₹20

OR

We have,
$$\frac { 1 }{ (x-1)(x-2) } +\frac { 1 }{ (x-2)(x-3) } \frac { 2 }{ 3 }$$ x ≠ 1,2,3
$$\frac { (x+3)+(x-1) }{ (x-1)(x-2)(x-3) } =\frac { 2 }{ 3 }$$
3(x-3) + 3(x- 1) = 2(x – 1) (x – 2) (x-3)
3x – 9 + 3x – 3 = 2(x – 1) (x – 2) (x-3)
6x – 12 = 2(x – 1) (x – 2) (x-3)
6(x – 2) = 2(x – 1) (x – 2) (x-3)
3 =(x-1) (x-3)
3 = x2– 3x – x + 3
x² – 4x = 0
x(x – 4)= 0
x = 0 or 4

Statement : The line drawn from the mid-point of one side of a triangle parallel to another side bisects the third side.
Given : A ∆ABC is which D is the mid-point of side AB and the line DE is drawn parallel to BC,
meeting AC at E.
To prove : E is the mid-point of Ac.
Proof : In ∆ABC,
DE || BC
∴ $$\frac { AD }{ BD } =\frac { AE }{ EC }$$
Since, D is the mid-point of AB
Thus, from equation (i),
$$\frac { AE }{ EC } =1$$
⇒ AE = EC
Hence, E bisects AC.

OR

Given : Two triangles ABC and DEF such that ΔABC ~ ΔDEF
To prove :

$$\frac { Area(\triangle ABC) }{ Area(\triangle DEF) } =\frac { { AB }^{ 2 } }{ { DE }^{ 2 } } =\frac { { BC }^{ 2 } }{ { EF }^{ 2 } } =\frac { { AC }^{ 2 } }{ { DF }^{ 2 } }$$

Construction : Draw AL ⊥ BC and DM ⊥ EF
Proof : Since, similar triangles are equiangular and their corresponding sides proportional,
Therefore, ∠A = ∠D, ∠B = ∠E, ∠C = ∠F
and $$\frac { AB }{ DE } =\frac { BC }{ EF } =\frac { AC }{ DF }$$
Thus, in ∆ALB and ∆DME, we have
∠ALB = ∠DME = 90°
and ∠B = ∠E
So, by AA-criterion of similarity, we have
∆ALB ~ ∆DME
⇒ $$\frac { AL }{ DM } =\frac { AB }{ DE }$$
From (i) and (ii), we get
$$\frac { AB }{ DE } =\frac { BC }{ EF } =\frac { AC }{ DF } =\frac { AL }{ DM }$$

Let r be the radius of well, h be the height of the well, x be the width of embankment and H be the height of embankment.
Then, 2r = 3
=> r = $$\frac { 3 }{ 2 }$$ m;
x = 4 m and h = 14 m
R = r + x = (4 + $$\frac { 3 }{ 2 }$$ ) m = $$\frac { 11 }{ 2 }$$ m
Volume of embankment (hollow cylinder) = Volume of well cylinder

Given system of equation is
5x – y = 5
and 3x – y = 3

Triangle formed by these lines and Y-axis is ABC.
∴ Area of ∆ ABC = $$\frac { 1 }{ 2 }$$ × BC × OA
= $$\frac { 1 }{ 2 }$$ × 2 × 1 = 1 sq.unit
OR
Let the two digit number be l0x + y.
Then, according to the question,
Case 1:
l0x + y = 8(x + y) – 5
l0x + y = 8x + 8y – 5
2x – 7y = – 5
Case II:
16(x – y) + 3 = 10x + y
16x – 16y + 3 = 10x + y
6x – 17y = – 3
Multiplying equation (i) by 3 and then subtracting it with equation (ii), we get

Substituting the value of y in equation (i),
2x – 7 × 3 = – 5
⇒  2x – 21 = – 5
2x = – 5 + 21
⇒  2x =16
⇒  x = 8
Therefore, required number is 83.

 Age (yrs.) ƒ x ƒx 5-15 6 10 60 15-25 11 20 220 25-35 21 30 630 35-45 23 40 920 45-55 14 50 700 55-65 5 60 300 Total ∑ƒ = 80 ∑ƒx=2830

Mean, $$\bar { x }$$ =$$\frac { \Sigma fx }{ \Sigma f } =\frac { 2830 }{ 80 }$$ = 35.375 yrs.
Model class = group with highest frequency
= 35 – 45
So, Lower limit of the modal class, l = 35
Frequency of modal class, ƒ1 = 23
Frequency of class preceding modal class, ƒ0 = 21
Frequency of class successing model class, ƒ2 = 14

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