CBSE Sample Papers for Class 10 Maths Paper 2

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.

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 andD.
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.
Identify: $\sqrt { \frac { 9 }{ \pi } }$ as rational or irrational.

Question 2.
Find an arithmetic mean between 10 $\frac { 1 }{ 2}$ and 25 $\frac { 1 }{ 2 }$ .

Question 3.
ABCD is a rectangle whose three vertices are B (4, 0), C (4, 3) and D (0, 3). What is the length of one of its diagonals ?

Question 4.
If cos 2A = sin(A – 15°), find A.

Question 5.
If x = – $\frac { 1 }{ 2}$ , is a solution of the quadratic equation 3x² + 2kx – 3 = 0, find the value of k.

Question 6.
In the figure of ∆ABC, the points D and E are on the sides CA, CB respectively such that DE || AB, AD = 2x, DC=x + 3,BE = 2x-1 and CE=x. Then, find x.

Section-B

Question 7.
Find the HCF of 1,656 and 4,025 by Euclid’s division algorithm.

Question 8.
If the point P (k – 1,2) is equidistant from the points A (3 ,k) and B (k, 5), find the values of k.

Question 9.
The probability of selecting a rotten apple randomly from a heap of900 apples is 0.18. What is the number of rotten apples in the heap?

Question 10.
The probability of selecting a red ball at random from ajar that contains only red, blue and orange balls
is $\frac { 1 }{ 4}$ . The probability of selecting a blue ball at random from the same jar is $\frac { 1 }{ 3}$ . If thejar contains 10 orange balls, find the total number of balls in the jar.

Question 11.
Sum of the ages of a father and the son is 40 years. If father’s age is three times that of his son, then find their respective ages.

Question 12.
If the ratio of the sum of first n terms of two A.P’s is (7n + 1): (4n + 27), find the ratio of their wth terms.

Section-C

Question 13.
Show that $\frac { 2 }{ 5\sqrt { 3 } }$ is irrational

Question 14.
Obtain all other zeroes of the polynomial 4x⁴ + x³ – 72x² – 18x, if two of its zeroes are 3 $\sqrt { 2 }$ and -3 $\sqrt { 2 }$ .

Question 15.
x-coordinate of a point P is twice its y-coordinate. If P is equidistant from Q (2, -5) and R (-3, 6), find the coordinates of P.
OR
The area of a triangle is 5 sq units. Two of its vertices are (2,1) and (3, -2). Ifthe third vertex is $\left( \frac { 7 }{ 2 } ,y \right)$ , find the value of y.

Question 16.
If $\frac { \cos { \alpha } }{ \cos { \beta } }$ = m and $\frac { \cos { \alpha } }{ \sin { \beta } }$ = n show that (m² + n²)cos² β n²
OR
Find the value of ‘x’ such that
2cos ec²30° + x sin² 60°- $\frac { 3 }{ 4}$ tan² 30° = 10

Question 17.
In figure, a quadrilateral ABCD is drawn to circumscribe a circle, with centre O, in such a way that the sides AB, BC, CD and DA touch the circle at the points P, Q, R and S respectively. Prove that AB + CD = BC + DA.

Question 18.
In the given figure, ABCD is a rectangle of dimensions 21 cm × 14 cm. A semicircle is drawn with BC as diameter. Find the area and the perimeter of the shaded region in the figure.

Question 19.
A cubical block of side 10 cm is surmounted by a hemisphere. What is the largest diameter that the hemisphere can have? Find the cost of painting the total surface area of the solid so formed, at the rate of ? 5 per 100 sq. cm [Use π=3.14].
OR
504 cones, each of diameter 3.5 cm and height 3 cm, are melted and recast into a metallic sphere. Find the diameter of the sphere and hence find its surface area. [Use π = $\frac { 22 }{ 7 }$ ]

Question 20.
Find the sum of values of a and b for which the following system of linear equations has infinite number of solutions:
2x + 3y = 7
(a+b+ 1)x + (a+2b + 2)y = 4(a+ b)+ 1

Question 21.
In the given figure, S and T trisect the side QR of a right triangle PQR. Prove that HPT² = 3 PR² + 5 PS²

OR
In the given figure PA, QB and RC, each are perpendicular to AC. Prove that $\frac { 1 }{ x } +\frac { 1 }{ z } =\frac { 1 }{ y }$

Question 22.
The mean and median of 100 observations are 50 and 52 respectively. The value of the largest observation is 100. It was later found that it is 110 not 100. Find the true mean and median.

Section-D

Question 23.
If cos θ + $\sqrt { 3 }$ sin θ = 2sin θ
Show that sin θ - $\sqrt { 3 }$ cos θ = 2cos θ.

Question 24.
A well of diameter 4 m is dug 21 m deep. The earth taken out of it has been spread evenly all around it in the shape of a circular ring of width 3 m to form an embankment. Find the height of the embankment.

Question 25.
Solve the following for x : $\frac { 1 }{ 2a+b+2x } =\frac { 1 }{ 2a } +\frac { 1 }{ b } +\frac { 1 }{ 2x }$
OR
Solve for x: $\frac { 1 }{ x+1 } +\frac { 1 }{ x+2 } =\frac { 4 }{ x+4 }$ , x ≠ -1,-2,-4

Question 26.
Prove that the ratio of the areas of two similar triangles is equal to the ratio of the square of their corresponding sides
OR
In a right triangle, prove that the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Question 27.
The angles of depression of the top and bottom of a 50 m high building from the top of a tower are 45° and 60° respectively. Find the height of the tower and the horizontal distance between the tower and the
building, (use $\sqrt { 3 }$ = 1.73)
OR
vertical tower stands on a horizontal plane and is surmounted by a vertical flag staff of height h. At a point on the plane, the angles of elevation of the bottom and the top of the flag staff are α and β respectively. Prove that the height of the tower is ( $\frac { h\tan { \alpha } }{ \tan { \beta } -\tan { \alpha } } \quad$ )

Question 28.
If $\frac { b+c-a }{ a } ,\frac { c+a-b }{ b } ,\frac { a+b-c }{ c } \quad$ are in A.P. then prove $\frac { 1 }{ a } ,\frac { 1 }{ b } ,\frac { 1 }{ c } \quad$ in A.P.

Question 29.
Draw a triangle ABC with sideBC = 7 cm. ∠B = 45°, ∠A = 105°. Construct a triangle whose sides are (4/3) times the corresponding side of ∆ ABC.

Question 30.
On the annual day of school, age-wise participation of students is given in the following frequency distribution table:

Solutions:
Section-A

Solution 1.
[latexs=2]\sqrt { \frac { 9 }{ \pi } } =(3)\left( \frac { 1 }{ \sqrt { \pi } } \right) [/latex] = (rational) (irrational) = irrational (1)

Solution 2.

Solution 3.

According to the figure BD is the diagonal of ABCD. By distance formula BD
BD = $\sqrt { { ({ x }_{ 2 }-{ x }_{ 1 }) }^{ 2 }+{ ({ y }_{ 2 }-{ y }_{ 1 }) }^{ 2 } } =\sqrt { { (4-0) }^{ 2 }+(0-3)^{ 2 } }$
= $\sqrt { 25 }$ = 5 units (1)

Solution 4.
sin(90° – 2A) = sin(A-15°)
⇒ 90°-2A = +4-15° ⇒ 3A = 105° (1/2)
∴ A = 35° (1/2)

Solution 5.
x = – $\frac { 1 }{ 2 }$ is a solution of 3x² + 2kx -3=0
Now, [latexs=2]{ 3\left( \frac { -1 }{ 2 } \right) }^{ 2 }+2k\left( \frac { -1 }{ 2 } \right) -3[/latex] = 0 (1/2)
$\frac { 3 }{ 4 }$ -k -3 =0 ⇒ k = $\frac { -9 }{ 4 }$ (1/2)

Solution 6.
$\frac { CD }{ AD } =\frac { CE }{ BE }$ ⇒ [latexs=2]\frac { x+3 }{ 2x } =\frac { x }{ 2x-1 } [/latex] (1/2)
⇒ 5x = 3 ⇒ x = $\frac { 3 }{ 5 }$ (1/2)

Section-B

Solution 7.
Since 4025 > 1656,
∴4025 =1656 × 2 + 713
1656 = 713 × 2 + 230
713 = 230 × 3+23 (1)
230 = 23 × 10+0
So, HCF (1,656,4,025) = 23 (1)

Solution 8.

Solution 9.
Since, the number of apples in the given heap = 900
And the probability of selecting a rotten apple randomly = 0.18

Hence, the number of rotten apples in the heap will be 900 × 0.18 i.e.,162 (1)

Solution 10.
∵P (Red) = $\frac { 1 }{ 4 }$ , and, P (Blue) = $\frac { 1 }{ 3 }$
So, P (Orange) = 1- $\frac { 1 }{ 4 } -\frac { 1 }{ 3 } =\frac { 5 }{ 12 }$ (1)
Therefore, $\frac { 5 }{ 12 }$ ×(Total no. of balls) = 10
Hence, total no. of balls = $\frac { 10\times 12 }{ 5 }$ = 24 (1)

Solution 11.
Let ages of father and son be x and y respectively.
x+y = 40 …(i)
x = 3y ….(ii) (1)
By solving eqs. (i) and (ii)
x =30 andy = 10
Ages are 30 years and 10 years. (1)

Solution 12.

Section-C

Solution 13.
Let us assume $\frac { 2 }{ 5\sqrt { 3 } }$ , to the contrary, that $\frac { 2 }{ 5\sqrt { 3 } }$ is rational.
∴ $\frac { 2 }{ 5\sqrt { 3 } }$ = $\frac { p }{ q }$ where p, q (q ≠ 0) are integers and p, q are co-prime.  (1)
⇒ $\frac { 2q }{ 5p }$ = $\sqrt { 3 }$   (1/2)
Since, 2, 5, p, q are integers
∴ $\frac { 2q }{ 5p }$ is rational, and so $\sqrt { 3 }$ is rational.  (1)
But this contradicts the fact that $\sqrt { 3 }$ is irrational
Hence, $\frac { 2 }{ 5\sqrt { 3 } }$ is irrational.  (1/2)

Solution 14.

Solution 15.
Let the y-coordinate of the point P be a.
And, its x-coordinate will be 2a.
So, the coordinates of the point P are (2a, a).
Since, the point P (2a, a) is equidistant from Q (2, -5) and R(-3,6).
∴ $\sqrt { { (2a-2) }^{ 2 }+{ (a-(-5)) }^{ 2 } } =\sqrt { { (2a-(-3)) }^{ 2 }+{ (a-6) }^{ 2 } }$ (1)
$\sqrt { { (2a-2) }^{ 2 }+{ (a+5) }^{ 2 } } =\sqrt { { (2a+3) }^{ 2 }+{ (a-6) }^{ 2 } }$
So, $\sqrt { { 5a }^{ 2 }+2a+29 } =\sqrt { { 5a }^{ 2 }+45 }$ (1)
After squaring both sides, we get
5a² + 2a + 29 = 5a² + 45
5a² + 2a – 5a² = 45 – 29
Thus, a = 8
Hence, the coordinates of the point P are (16, 8).
OR

Solution 16.

OR

Solution 17.
As, the tangents drawn from the exterior point to a circle are equal in length. (1/2)
So, DR=DS ….(i)
AP= AS …(ii)
BP = BQ …(iii)
CR=CQ ….(iv) (1)

Adding(i), (ii), (iii) and (iv), we get
DR + AP + BP + CR = DS + AS + BQ + CQ (1/2)
⇒ (DR + CR) + (AP + BP) = (DS + AS) + (BQ + CQ)
⇒ CD + AB = DA + BC
⇒ AB+CD = BC + DA (Hence Proved). (1)

Solution 18.
Area of shaded region
= Area of rectangle – Area of semi-circle (1/2)
(14×21) – $\frac { 1 }{ 2 } \times \frac { 22 }{ 7 } \times { 7 }^{ 2 }$ (1/2)
= 294 – 77 = 217 cm² (1/2)
Perimeter of shaded region = (2 × 21) +14 + $\left( \frac { 22 }{ 7 } \times { 7 } \right)$ (1)
= 42+14 + 22 = 78 cm (1/2)

Solution 19.
∵ Largest possible diameter ofhemisphere = 10 cm
So, radius ofhemisphere = 5 cm (1/2)
Now, total surface area = 5(10)² + 3.14 × (5)² × 2 (1)
= 657 cm² (1/2)
Hence, cost of painting = $\frac { 657\times 5 }{ 100 }$ = ₹ 32.85
OR

Solution 20.

Solution 21.
S and T trisect the side QR.
Let QS = ST = TR = x units
Let PQ = y units (1/2)
In right ∆PQS, PS² = PQ² + QS² (By Pythagoras Theorem)
= y² + x² …(i) (1/2)
In right ∆PQR, PT2 = (By Pythagoras Theorem)
= y² + (2x)² = y² + 4x² ….(ii) (1/2)
In right ∆PQR, PR²2 = PQ² + QR²(By Pythagoras Theorem)
=y² + (3x)² = y² + 9x² …(iii) (1/2)
R.H.S. = 3 PR² + 5 PS²
= 3(y² + 9x²)+ 5(y²+x²) [From (i) and (iii)]
= 3y² + 27x² + 5y² + 5x² = 8y² + 32x²
= 8(y² + 4x²) = 8PT² = L.H.S. [From(ii)] (1)
Thus 8PT² = 3PR² + 5PS²2
OR
In APAC and AQBC, We have
∠PAC = ∠QBC [Each = 90°]
∠PCA = ∠QCB [Common] (1)
∴∆PAC ~ ∆QBC
∴ $\frac { x }{ y } =\frac { AC }{ BC }$ i.e. [latexs=2]\frac { y }{ x } =\frac { BC }{ AC } [/latex] ..(i) (1/2)

Solution 22.

Section-D

Solution 23.
(cos θ + $\sqrt { 3 }$ sin θ)² + (sin θ – $\sqrt { 3 }$ cos θ)²
= cos² θ + 3sin² θ + 2 $\sqrt { 3 }$ cos θ sin θ + sin² θ + 3cos² θ – 2 $\sqrt { 3 }$ cos θ sin θ (1)
⇒ (2sin θ)² + (sin θ – $\sqrt { 3 }$ cos θ)² = (cos² θ + sin² θ) + 3 (sin² θ + cos² θ) (1)
⇒ (sin θ – $\sqrt { 3 }$ cos θ)² = 4 – 4 sin² θ (1/2)
⇒ (sin θ- $\sqrt { 3 }$ cos θ)² = 4 cos² θ (1)
⇒ sin θ – $\sqrt { 3 }$ cos θ = 2 cos θ (Hence proved) (1/2)

Solution 24.
Let r and h be the radius and depth of the well, respectively.
Here, r = $\frac { 4 }{ 2 }$ = 2 m and h = 21 m (1/2)
Let R and H be the outer radius and height of the embankment, respectively.
So, R = r + 3= 2 + 3 = 5m (1)
As, volume of the earth to form the embankment = Volume of the earth dug out
π(R²-r²)H = πr²h (1)
∴ H = $\frac { { r }^{ 2 }h }{ { R }^{ 2 }-{ r }^{ 2 } }$ (1/2)

∴ H = $\frac { { 2 }^{ 2 }\times 21 }{ { 5 }^{ 2 }-{ 2 }^{ 2 } }$ = 4 m (1)

Solution 25.
Since, $\frac { 1 }{ 2a+b+2x } =\frac { 1 }{ 2a } +\frac { 1 }{ b } +\frac { 1 }{ 2x }$
$\frac { 1 }{ 2a+b+2x } =\frac { 2xb+4ax+2ab }{ (2a)(b)(2x) }$ (1)
4abx = (2xb + 4ax + 2ab) (2a + b + 2x)
4abx = 12axb + 8a²x + 4a²b + 2xb² + 2ab² + 4x²b + 8ax²
x² (4b + 8a) + x(8ab + 8a² + 2b2) + (2ab² + 4a²b) = 0 (1)
4x² (2a + b) + x [4a (2a + b) + 2b (2a+ b)] + 2ab (b + 2a) = 0
4(2a + b) x[x + a] + 2b (2a + b) [x + a] = 0 (1)
(x + a) (2x + b) = 0 [ ∵ 2a + b ≠ 0]
x = -a, $\frac { -b }{ 2 }$
Hence, the value of x is -a and $\frac { -b }{ 2 }$ . (1)
OR

Solution 26.

OR

Solution 27.

Solution 28.

Solution 29.
Steps of construction:
1.DrawBC = 7cm.
2. Draw a ray BX and CY such that ∠CBX = 45° and ∠BCY =180°- (45° + 105°) = 30°
Suppose BX and CY intersect each other at A. ∆ ABC so obtained is the given triangle.
3. Draw a ray BZ making a suitable acute angle with BC on opposite side of vertex A with respect to B BC.
4. Draw four (greater of 4 and 3 in 4/3) arcs intersecting the ray BZ at B₁, B₂, B₃, B₄ such that BB₁ = B₁B₂ = B₂B₃ = B₃B₄.
5. Join B₃ to C and draw a line through B₄ parallel to B₃C, intersecting the extended line segment BC atC’.
6. Draw a line through C’ parallel to CA intersecting the extended line segment BA at A’. Triangle A’BC’ so obtained is the required triangle. (1)

Solution 30.

We hope the CBSE Sample Papers for Class 10 Maths paper 2 help you. If you have any query regarding CBSE Sample Papers for Class 10 Maths paper 2, drop a comment below and we will get back to you at the earliest.

CBSE Sample Papers for Class 10 Maths Paper 2

Share this: