CBSE Sample Papers for Class 10 Maths Paper 10

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 10.

CBSE Sample Papers for Class 10 Maths Paper 10

Board CBSE
Class X
Subject Maths
Sample Paper Set Paper 10
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 10 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.
a and b are two positive integers such that the least prime factor of a is 3 and the least prime factor of his 5. Then calculate the least prime factor of (a + b). [2014]

Question 2.
Find the centroid of the triangle whose vertices are (-1,1), (-3,4) and (8,-11)

Question 3.
If sin A= [latex]\frac { 1 }{ 2 } [/latex] (0° < A < 90°) then find the value of cos3 A – 3 cos A [2014,2015]

Question 4.
Write the nature of roots of quadratic equation 4x2 + 4 √3x + 3 = 0

Question 5.
In ∆ABC, if X and Y are points on AB and AC respectively such that [latex]\frac { AX }{ XB } [/latex] = [latex]\frac { 3 }{ 4 } [/latex], AY = 5cm and YC = 9cm, then state whether XYand BC parallel or not. [2015,2016]

Question 6.
If an = [latex]\frac { n(n-3) }{ n+4 } [/latex] then find 18th term of this sequence.

Section – B

Question 7.
Express the number [latex]0.\overline { 3178 } [/latex] in the form of rational number [latex]\frac { a }{ b } [/latex].

Question 8.
A factory had 120 workers in January and 90 of them were female workers. In February, another 15 male workers are added. A worker is then picked at random. Calculate the probability of picking a female worker.

Question 9.
The sum of the digits of a two digit number is 8. The number obtained by reversing the digits exceeds the original number by 18. Find the given number.

Question 10.
If P (x,y) is any point on the line joining the points A (a, 0) andB (0, b), then show that [latex]\frac { x }{ a } [/latex] + [latex]\frac { y }{ b } [/latex] = 1

Question 11.
If p,q, r are in A.P. then find the value of p3 + r3 – 8q3.

Question 12.
Ajar contains only green, white and yellow marbles. The probability of selecting a green marble and white marble randomly from a jar is 1/4 and 1/3 respectively. If this jar contains 10 yellow marbles, what is the total number of marbles in the jar ?

Section – C

Question 13.
Prove that 3 + 2 √5 is irrational.

Question 14.
On dividing x– 3x2 + x + 2 by a polynomial g(x), the quotient and remainder were x – 2 and -2x + 4 respectively. Find g(x).

Question 15.
CBSE Sample Papers for Class 10 Maths Paper 10 1

Question 16.
CBSE Sample Papers for Class 10 Maths Paper 10 2

Question 17.
From given fig. express ‘x’ in terms of a, b, c.
CBSE Sample Papers for Class 10 Maths Paper 10 3
In figure, two line segments AC and BD intersect each other at the point P such that PA = 6 cm, PB = 3 cm, PC = 2.5 cm, PD = 5 cm, ∠APB = 50° and ∠CDP = 30° Then, find the value of ∠PBA
CBSE Sample Papers for Class 10 Maths Paper 10 4

Question 18.
In fig. A circle touches the side BC of ∆ABC at P and touches AB and AC produced at Q and R respectively. If AQ= 5 cm, find the perimeter of ∆ABC.
CBSE Sample Papers for Class 10 Maths Paper 10 5

Question 19.
Solve the equations :
CBSE Sample Papers for Class 10 Maths Paper 10 6
Question 20.
Find the area of a triangle with vertices (a, b + c), (b, c + a) and (c, a + b).

Question 21.
In the given figure, OACB is a quadrant of a circle with centre O and radius 3.5 an. If OD=2an, find the area of the shaded region. [2017]
CBSE Sample Papers for Class 10 Maths Paper 10 7

Question 22.
In figure, a tent is in the shape of a cylinder surmounted by a conical top of same diameter. If the height and diameter of cylindrical part are 2.1 m and 3 m respectively and the slant height of conical part is 2.8 m, find the cost of canvas needed to make the tent if the canvas is available at the rate of ₹ 500/sq. metre. ( Use π = [latex]\frac { 22 }{ 7 } [/latex] ) [2016]
CBSE Sample Papers for Class 10 Maths Paper 10 8

OR
Prove that the largest possible sphere is carved out from a wooden solid cube of side 7 cm. Find the volume of the wood left. ( Use π = [latex]\frac { 22 }{ 7 } [/latex] ) [2014]

Section – D

Question 23.
In a triangle, if the square of one side is equal to the sum of the squares of the other two sides, then prove that the angle opposite to the first side is a right angle.
OR
Prove that if a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.

Question 24.
Aboy onhorizontal plane finds bird flying at a distance of 100 m from him at an elevation of30°. Agirl standing on the roofof20 metre high building, finds the angle of elevation ofthe same bird to be 45°. Both the boy and the girl are on opposite sides of the bird. Find the distance of bird from the girl.
OR
A 1.2 m tall girl spots a balloon moving with the wind in a horizontal line at a height of 88.2 m from the ground. The angle of elevation of the balloon from the eyes of the girl at= any instant is 60°. After sometime, the angle of elevation reduces to 30°. Find the distance travelled by the balloon during the interval.

Question 25.
If the pth term of an A.P. is ( [latex]\frac { 1 }{ q } [/latex] ) and its qth term is ( [latex]\frac { 1 }{ p } [/latex] ). show that the sume of its first pq term is [latex]\frac { 1 }{ 2 } [/latex] (pq + 1).

Question 26.
Roots of the quadratic equation 36x2 – 12ax + (a2 – b2) = 0 are [latex]\frac { a+b }{ c } [/latex] and [latex]\frac { a-b }{ c } [/latex] then, find the value of c.
OR
Find the real roots of the equation x2/3 + x1/3 – 2 = 0

Question 27.
A farmer connects a pipe ofintemal diameter 20 cm from a canal into a cylindrical tank which is 10 m in
diameter and 2 m deep. If the water flows through the pipe at the rate of 4 km per hour, in how much time will the tank be filled completely? [2014]

Question 28.
The following table shows marks secures by 140 students in an examination :

Marks 0-10 10-20 20-30 30-40 40-50
No. of Student 20 24 40 36 20

Calculation of mean by Step-deviation method.

Question 29.
CBSE Sample Papers for Class 10 Maths Paper 10 9

Question 30.
Construct a tangent to a circle of radius 4 cm from a point on the concentric circle of radius 6 cm and measure its length.

Solutions

Solution 1.
Least prime factor of (a + b) is 2. Because a + b= 8 and least prime factor of 8 is 2. Since,

Solution 2.
CBSE Sample Papers for Class 10 Maths Paper 10 10

Solution 3.
CBSE Sample Papers for Class 10 Maths Paper 10 11

Solution 4.
∵ 4x2 + 4√3 x + 3 = 0
So,determinant = b2 – 4ac = (4√3)2 -4 × 4 × 3 = 0 Hence,
Hence, roots are real and equal

Solution 5.
CBSE Sample Papers for Class 10 Maths Paper 10 12

Solution 6.
CBSE Sample Papers for Class 10 Maths Paper 10 13

Solution 7.
Let x = 0.31783178 …(i)
Multiply by 10000
10000 x = 3178.31783178 …..(ii)
Subtracting (i) from (ii)
10000x = 3178.3178 …..
x = 0.3178 ….
CBSE Sample Papers for Class 10 Maths Paper 10 14

Solution 8.
Initial number of workers = 120
When 15 male workers are added, then the total number of workers = 120 + 15 = 135
Number of female workers = 90
Probability of female workers = [latex]\frac { 90 }{ 135 } [/latex] = [latex]\frac { 2 }{ 3 } [/latex]

Solution 9.
Let the unit digits be x
∴ Tens’ digits = 8 – x
∴ Required number = 10(8 – x) + x = 80 – 9x
Reverse number = 10x + (8 – x) = 9x + 8
∴ (9x + 8) – (80 – 9x) = 18
⇒ x = 5
∴ Tens digit = 8 – 5 = 3
∴ Required number = 35

Solution 10.
As the point P (x, y) lies on the line joining the points A (a, 0) and B (0, b), the points A, B and P are collinear
⇒ a(b – y) + 0 (y – 0) + x (0 – b) = 0
⇒ ab – ay – bx = 0 ⇒ bx + ay = ab
⇒ [latex]\frac { x }{ a } [/latex] + [latex]\frac { y }{ b } [/latex] = 1

Solution 11.
∵ 2q = p + r
∴ p + r – 2q = 0
so, p3 + r3 – 8q3 = 3 × p × r × (-2q) = -6 pqr

Solution 12.
Let the no. of green marbles = x
Let the no. of white marbles = y
∴ Total no. ofmarbles = x + y + 10
CBSE Sample Papers for Class 10 Maths Paper 10 15
CBSE Sample Papers for Class 10 Maths Paper 10 16

Solution 13.
CBSE Sample Papers for Class 10 Maths Paper 10 17

Solution 14.
We know that, if p(x) and g(x) are any two polynomials with g(x) ≠ 0, then we can find polynomials q (x) and r (x) such that
CBSE Sample Papers for Class 10 Maths Paper 10 18

Solution 15.
CBSE Sample Papers for Class 10 Maths Paper 10 19
CBSE Sample Papers for Class 10 Maths Paper 10 20

Solution 16.
CBSE Sample Papers for Class 10 Maths Paper 10 21

Solution 17.
CBSE Sample Papers for Class 10 Maths Paper 10 22

Solution 18.
To Find : Perimeter of ∆ABC
Let AQ = 5 cm
and
AQ = AR ….. (i)
BQ = BP ….. (ii)
CP = CR ….. (iii)
(Tangent drawn from an external points are equal)
CBSE Sample Papers for Class 10 Maths Paper 10 23
∴ Perimeter of ∆ABC = AB + BC + CA
= AB + BP + PC + CA [∵ BC = BP + PC]
= (AB + BQ) + (CR + CA) from (ii) and (iii)
= AQ + AR [∵ AQ = AR from (i)]
=AQ+AQ = 2 AQ =2 × 5 = 10cm
∴ Perimeter of ∆ABC = 10 cm.

Solution 19.
CBSE Sample Papers for Class 10 Maths Paper 10 24

Solution 20.
CBSE Sample Papers for Class 10 Maths Paper 10 25

Solution 21.
CBSE Sample Papers for Class 10 Maths Paper 10 26

Solution 22.
Canvas needed to make the tent = C.S.A of the conical part + C.S.A ofthe cylindrical part Given that
Radius of the conical part = Radius of the cylindrical part = [latex]\frac { 3 }{ 2 } [/latex] m
Slant height ofthe conical part = l = 2.8 m
Height of the cylindrical part = h = 2.1 m
CBSE Sample Papers for Class 10 Maths Paper 10 27
CBSE Sample Papers for Class 10 Maths Paper 10 28

Solution 23.
CBSE Sample Papers for Class 10 Maths Paper 10 29
CBSE Sample Papers for Class 10 Maths Paper 10 30
CBSE Sample Papers for Class 10 Maths Paper 10 31

Solution 24.
Let P, B and A be the positions of the bird, boy and girl respectively.
Given PB = 100m, AC = 20m, Let PM = xm.
In right-angled ∆PBO, we have,
CBSE Sample Papers for Class 10 Maths Paper 10 32
OR
Suppose P be the position of the balloon if its angle of elevation from the eyes of the girl is 60° and Q be the position if angle of elevation is 30°.
CBSE Sample Papers for Class 10 Maths Paper 10 33
CBSE Sample Papers for Class 10 Maths Paper 10 34

Solution 25.
CBSE Sample Papers for Class 10 Maths Paper 10 35
CBSE Sample Papers for Class 10 Maths Paper 10 36

Solution 26.
CBSE Sample Papers for Class 10 Maths Paper 10 37

Solution 27.
Internal diameter of pipe = 20 cm
Internal radius of pipe = 10 cm
Radius of cylindrical tank = 5 m = 500 cm
Height of cyclindrical tank = 200 cm
Water flows in 1 hour = 400000 cm
Let we assume the height of pipe = 400000 cm
Now, volume of tank = πr2 h = π (500)2 (200) = 50000000 π cm3
Volume of water flow in hour = π (20)2 (400000) = 160000000 π
Time taken = [latex]\frac { 5 }{ 16 } [/latex] hour = [latex]\frac { 5 }{ 16 } [/latex] × 60 min = [latex]\frac { 300 }{ 16 } [/latex] min = [latex]\frac { 75 }{ 4 } [/latex] × 60 sec = 1125 sec or 18.75 min

Solution 28.
CBSE Sample Papers for Class 10 Maths Paper 10 38

Solution 29.
CBSE Sample Papers for Class 10 Maths Paper 10 39

Solution 30.
CBSE Sample Papers for Class 10 Maths Paper 10 40
Steps of construction :

  1. Draw two circles with common centre O and of radii 4cm and 6cm.
  2. Take any point P on the outer circle.
  3. Joint the point P to the centre O.
  4. Draw perpendicular bisector of PO, which intersects PO at point Q.
  5. With centre Q and radius PQ or QO, draw a circle, which intersects the inner circle at points R and S.
  6. Draw rays PR and PS.
    Thus, PR and PS are the required tangents.
    By measurement PR = 4.5 cm = PS.

Verification by Actual Calculation :
CBSE Sample Papers for Class 10 Maths Paper 10 41

Join O to RandS.
Since, OR and OS are the radius through the point of contact of the tangent to the circle.
∴ OR ⊥PR and OS ⊥ PS
PR= √OP2 – OR2
= √36 – 16 = √20
= 4.5 cm (Approximately)
Since, PR and PS are two tangents from an exterior point to the same circle
∴ PR = PS = 4.5 cm (Approximately)
Hence verified.

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

CBSE Sample Papers for Class 10 Maths Paper 9

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 9.

CBSE Sample Papers for Class 10 Maths Paper 9

Board CBSE
Class X
Subject Maths
Sample Paper Set Paper 9
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 9 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.
What is the condition for the decimal expansion of a rational number to terminate? Explain with the help of an example.

Question 2.
In the given figure ∆ACB ~ ∆APQ. If BC = 8 cm, PQ = 4 cm, BA = 6.5 cm, AP = 2.8 cm, then find AC.
CBSE Sample Papers for Class 10 Maths Paper 9 1

Question 3.
An equation has been given as [latex]\frac { c }{ { x }^{ 2 } } +\frac { k }{ x }[/latex] = 1. Find the relation between c and k if x has real values.

Question 4.
If sec 2 A = cosec (A – 42°) where 2 A is acute angle, then find the value of A [NTSE 2015]

Question 5.
Find the next term of A.P. √2,√8,√18……..

Question 6.
Find the ratio in which the point P ([latex]\frac { 3 }{ 4 } [/latex],[latex]\frac { 5 }{ 12 } [/latex]) divides the line segment joining the points A ([latex]\frac { 1 }{ 2 } [/latex],[latex]\frac { 3 }{ 2 } [/latex]) and B (2, -5).

Section – B

Question 7.
How many three digit natural numbers are divisible by 7 ?

Question 8.
A certain class has ‘s’ students. If a student is picked at random, the probability of picking a boy is [latex]\frac { 8 }{ 13 } [/latex] If the class has 24 boys, what is the value of ‘s’?

Question 9.
A bag contains 40 coins, consisting of ₹ 2, ₹ 5 and ₹ 10 denominations. If a coin is drawn at random, the probability of drawing a ₹ 2 coin is [latex]\frac { 5 }{ 8 } [/latex] If x ₹ 2 coins are removed from the bag and then a coin is drawn at random, the probability of drawing a ₹ 2 coin is [latex]\frac { 1 }{ 2 } [/latex] . Find the value of x.

Question 10.
Solve the following system of equations :
[latex]\frac { 4 }{ X } [/latex] + 5y = 7 ; [latex]\frac { 3 }{ x } [/latex] + 4y = 5

Question 11.
If the pth term of an A.P. is [latex]\frac { 1 }{ Q } [/latex] and qth term is [latex]\frac { 1 }{ P } [/latex]. Prove that the sum of the first pq terms is [latex]\frac { 1 }{ 2 } [/latex] (pq +1)

Question 12.
If (3,0), (2, a) and (b, 6) are the vertices of a ∆ ABC, whose centroid is (2, 5). Find the values of a and b.

Section – C

Question 13.
Prove that √2 + √5 is irrational.

Question 14.
Quadratic polynomial 2x2 – 3x + 1 has zeroes as a and β Now form a quadratic polynomial whose zeroes are 3α and 3β.

Question 15.
Prove that the points (a, a), (-a,-a) and (-√3 a,√3 a) are the vertices of an equilateral triangle.|

Question 16.
CBSE Sample Papers for Class 10 Maths Paper 9 2

Question 17.
In a triangle ABC, AD ⊥ BC, If AD2 = BD.DC, prove that ∆ABC is right angle triangle.
OR
AD is the median of ∆ABC. The bisector of ∠ADB and ∠ADC meet AB and AC at points E and F. Prove that EF || BC.

Question 18.
In figure, PQRS is a square lawn with side PQ = 42 metres. Two circular flower beds are there on the sides PS and QR with centre at O, the intersection of its diagonals. F ind the total area of the two flower beds (shaded parts).
CBSE Sample Papers for Class 10 Maths Paper 9 3

Question 19.
The mean of x1, x2, x3, …., xn is [latex]\overline { x } [/latex] If (a – 2b) is added to each ofthe observations, show that the mean of the new observations is [latex]\overline { x } [/latex] + (a – 2b)

Question 20.
CBSE Sample Papers for Class 10 Maths Paper 9 4

Question 21.
If a circle touches the side BC of a triangle ABC at P and extended sides AB and AC at Q and R, respectively, prove that AQ = [latex]\frac { 1 }{ 2 } [/latex] (BC + CA + AB)

Question 22.
A toy is in the form of a cone mounted on a hemisphere of common base radius 7 cm. The total height of the toy is 31 cm. Find the total surface area ofthe toy. (use π = [latex]\frac { 22 }{ 7 } [/latex])
OR
The rain water from a roof 22 m × 20 m drains into a cylindrical vessel having diameter of base 2 m and height 3.5 m. Ifthe vessel is just lull, find the rainfall in cm.

Section – D

Question 23.
If the sum of p terms of an A.P. is q and the sum of q terms is p then, show that sum of (p – q) terms = (p – q) (1 + [latex]\frac { 2q }{ P } [/latex])

Question 24.
From a window (h meters high above the ground) of a house in a street, the angles of elevation and depression of the top and the foot of another house on the opposite side of the street are θ and Φ respectively, Show that the height of the opposite house is h (1 + tan θ cot Φ)) meters.
OR
A pole 5 m high is fixed on the top of the tower. The angles of elevation of the top of the pole observed from a point A on the ground is 60° and the angle of depression ofthe point ‘A’ from the top of the tower is 45°. Find the height of the tower.

Question 25.
Out of a certain number of Saras birds one-fourth the number arc moving in lotus plants, [latex]\frac { 1 }{ 9 } [/latex] th coupled with [latex]\frac { 1 }{ 4 } [/latex] th as well as 7 times the square root of the number move on a hill, 56 birds remain in Vacula tree. What is the total number of birds?

Question 26.
A bucket open at the top, and made up of a metal sheet is in the form of a frustum of a cone. The depth of the bucket is 24 cm and the diameters of its upper and lower circular ends are 30 cm and 10 cm respectively. Find the cost of metal sheet used in it at the rate of ₹ 10 per 100 cm2. [Use π = 3.14],

Question 27.
Find the value of
(1 + tan θ + sec θ) (1 + cot θ – cosec θ).

Question 28.
Draw a circle of radius 3 cm. Take two points P and Q on one of its extended diameter and on opposite sides of its centre, each at a distance of 7 cm from its centre. Draw tangents to the circle from these two points P and Q.

Question 29.
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 30.
On the sports day of a school, 300 students participated. Their ages are given in the following distribution :

Age (in years) 5 – 7 7 – 9 9 – 11 11 – 13 13 – 15 15 – 17 17 – 19
Number of students 67 33 41 95 36 13 15

Find the mode of the data.
OR
Monthly expenditures of milk in 100 families of a housing society are given in the following frequency distribution :

Monthly expenditure (in ₹)

0 – 175 175 – 350 350 – 525 525 – 700 700 – 875 875 – 1050 1050 – 1225
Number of families 10 14 15 21 28 7 5

Find the mode for this distribution.

SOLUTIONS

Section – A

Solution: 1
The decimal expansion of a rational number terminates, if the denominator of rational number. [latex]\frac { p }{ q } [/latex], where p and q are co-prime can be expressed as 2m 5n, while m and n are non-negative integers.
e.g., [latex]\frac { 3 }{ 10 } [/latex] = [latex]\frac { 3 }{ { 2 }^{ 1 }\times { 5 }^{ 1 } } [/latex] = 0.3 (1)

Solution: 2
CBSE Sample Papers for Class 10 Maths Paper 9 5

Solution: 3
CBSE Sample Papers for Class 10 Maths Paper 9 6

Solution: 4
sec 2A = cosec (A – 42°)
cosec (90° – 2A) = cosec (A – 42°)
90° – 2A = A – 42°
3A = 132°
A = 44° (1)

Solution: 5
CBSE Sample Papers for Class 10 Maths Paper 9 7

Solution: 6
CBSE Sample Papers for Class 10 Maths Paper 9 8

Section – B

Solution: 7
Since, the numbers which are less than 1000 and divisible by 7 is [ [latex]\frac { 1000 }{ 7 } [/latex] ] i.e., 142. (1/2)
Now, the numbers which are less than 100 and divisible by 7 is [ [latex]\frac { 100 }{ 7 } [/latex] ] i.e., 14. (1/2)
So, the numbers which have three digits and divisible by 7 is 142 – 14 i.e., 128. (1)
Hence, there are total 128 three-digit natural numbers which are divisible by 7.

Solution: 8
Probability of picking a boy = [latex]\frac { Number of boys }{ Total students } [/latex] (1)
⇒ [latex]\frac { 8 }{ 13 } [/latex] = [latex]\frac { 24 }{ s } [/latex] ⇒ s = 24 × [latex]\frac { 13 }{ 8 } [/latex] ⇒ s = 3 × 13 ⇒ s = 39 (1)

Solution: 9
Total number of coins = 40
let number of ₹ 2 coins is C.
∵ P (drawing a ₹2 coins is c) = [latex]\frac { 5 }{ 8 } [/latex] ⇒ [latex]\frac { c }{ 40 } [/latex] = [latex]\frac { 5 }{ 8 } [/latex] ⇒ c = 25 (1)
Now, [latex]\frac { 25-x }{ 40-x } [/latex] = [latex]\frac { 1 }{ 2 } [/latex] ⇒ x = 10 (1)

Solution: 10
CBSE Sample Papers for Class 10 Maths Paper 9 9

Solution: 11
CBSE Sample Papers for Class 10 Maths Paper 9 10

Solution: 12
CBSE Sample Papers for Class 10 Maths Paper 9 11

Section – C

Solution: 13
Let us assume on the contrary that √2 + √5 is rational number. Then, there exist co-prime positive integers a and b such that
CBSE Sample Papers for Class 10 Maths Paper 9 12

Solution: 14
CBSE Sample Papers for Class 10 Maths Paper 9 13

Solution: 15
CBSE Sample Papers for Class 10 Maths Paper 9 14

Solution: 16
CBSE Sample Papers for Class 10 Maths Paper 9 15

Solution: 17
CBSE Sample Papers for Class 10 Maths Paper 9 16
CBSE Sample Papers for Class 10 Maths Paper 9 17

Solution: 18
CBSE Sample Papers for Class 10 Maths Paper 9 18

Solution: 19
CBSE Sample Papers for Class 10 Maths Paper 9 19

Solution: 20
CBSE Sample Papers for Class 10 Maths Paper 9 20
CBSE Sample Papers for Class 10 Maths Paper 9 21
CBSE Sample Papers for Class 10 Maths Paper 9 22

Solution: 21
CBSE Sample Papers for Class 10 Maths Paper 9 23

Solution: 22
CBSE Sample Papers for Class 10 Maths Paper 9 24
CBSE Sample Papers for Class 10 Maths Paper 9 25

Section – D

Solution: 23
CBSE Sample Papers for Class 10 Maths Paper 9 26

Solution: 24
CBSE Sample Papers for Class 10 Maths Paper 9 27
CBSE Sample Papers for Class 10 Maths Paper 9 28
CBSE Sample Papers for Class 10 Maths Paper 9 29

Solution: 25
Let the total number of birds be x.
∴ Numbe of birds moving in lotus plants = [latex]\frac { x }{ 4 } [/latex]
Number of birds moving on a hill = [latex]\frac { x }{ 9 } [/latex] + [latex]\frac { x }{ 4 } [/latex] + 7 √x (1)
Number of birds in Vacula tree = 56.
Using the given informations, we have
CBSE Sample Papers for Class 10 Maths Paper 9 30

Solution: 26
CBSE Sample Papers for Class 10 Maths Paper 9 31
Total surface area of bucket
= Curved surface area of frustum + base area of lower circular end ABA
= (π × 15 × 39 – π × 5 × 13) + π (5)2
= 520 π + 25 π = 54 π cm2 (1)
So, total cost of metal sheet = ₹ 545 π × 0.1
= ₹ 54.5 × 3.14 = ₹ 171.13
Hence, cost of required metal sheet is ₹ 171.13. (1)

Solution: 27
CBSE Sample Papers for Class 10 Maths Paper 9 32

Solution: 28
CBSE Sample Papers for Class 10 Maths Paper 9 33
Steps of construction :

  1. Draw a line segment PQ of length (2 × 7 = 14) cm. (1)
  2. Draw perpendicular bisector of PQ, which intersects PQ at O.
  3. With centre O and radius 3 cm, draw a circle,
  4. Draw perpendicular bisector ofPO, which intersects PO at R.
  5. With centre R and radius RP or RO, draw a circle, which intersects the circle drawn in step (iii), at S and T.
  6. Draw rays PS and PT
  7. With centre Q and radius equal to PS (or PT), draw arcs intersecting the circle drawn in step (iii) at U andV.
  8. Draw rays QU and QV.

Thus, PS, PT, QU and QV are required tangents.

Solution: 29
CBSE Sample Papers for Class 10 Maths Paper 9 34
CBSE Sample Papers for Class 10 Maths Paper 9 35
CBSE Sample Papers for Class 10 Maths Paper 9 36

Solution: 30
CBSE Sample Papers for Class 10 Maths Paper 9 37

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CBSE Sample Papers for Class 10 Maths Paper 8

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 8.

CBSE Sample Papers for Class 10 Maths Paper 8

Board CBSE
Class X
Subject Maths
Sample Paper Set Paper 8
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 8 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.
Find the smallest positive rational number by which [latex s=2]\frac { 1 }{ 7 } [/latex] should be multiplied so that its decimal expansion terminates after 2 places of decimal.

Question 2.
If the vertices ofa triangle are (1,2), (4, -6) and(3, 5) then, findthearea oftriangle (NTSE2014,2015)

Question 3.
If [latex s=2]\left( \tan { \theta } +\frac { 1 }{ \tan { \theta } } \right) [/latex] = 2, then find the value of [latex s=2]\tan ^{ 2 }{ \theta } +\frac { 1 }{ \tan ^{ 2 }{ \theta } } [/latex](NTSE 2015)

Question 4.
Evaluate: [latex]\sqrt { 6+\sqrt { 6 } +\sqrt { 6+ } } [/latex]…..

Question 5.
What is the common difference of an A. P. in which a21 -a7 = 84 ?

Question 6.
The areas of two similar triangles are 121 cm2 and 64 cm2 respectively. If the median of the first triangle is
12.1 cm, find the corresponding median of the other.

Section-B

Question 7.
Two bills of ₹ 6075 and ₹ 8505 respectively are to be paid separately by cheques of same amount. Find
the largest possible amount of each cheque.

Question 8.
Find the value of70 + 68 + 66 +……..+ 40.

Question 9.
Find the chance that a non-leap year contains 53 Saturdays.

Question 10.
Ajar contains 54 marbles each of which is blue, green or white. The probability of selecting a blue
marble and a green marble at random from the jar is [latex s=2]\frac { 1 }{ 3 } [/latex] and [latex s=2]\frac { 4 }{ 9 } [/latex] respectively. How many white marbles does the jar contain?

Question 11.
If the lines given by 3x + 2ky = 2 and 2x+5y= 1 are parallel, then find the value of k

Question 12.
Find ‘k’ so that the points (7, -2), (5,1) and (3, k) are collinear.

Section-C

Question 13.
If a2 – b2 is a prime number, show that a2 – b2 = a + b, where a, b are natural number.

Question 14.
If the polynomial 6x4 + 8x3 + 17x2 + 21x + 7 is divided by another polynomial 3x2 + 4x+ 1, the remainder comes out to be (ax + b ), find the values of a and b.

Question 15.
If tan θ + sin θ = m and tan θ – sin θ = n, then prove that m2 – n2 = 4[latex]\sqrt { mn } [/latex]
OR
CBSE Sample Papers for Class 10 Maths Paper 8 img 1

Question 16.
The following table gives the number of pages written by Sarika for completing her own book for 30 days:
CBSE Sample Papers for Class 10 Maths Paper 8 img 2
Find the mean number of pages written per day.

Question 17.
CBSE Sample Papers for Class 10 Maths Paper 8 img 3

Question 18.
A circle touches all the four sides of a quadrilateral ABCD. Prove that AB + CD = BC + DA.

Question 19.
In figure, two concentric circles with centre O, have radii 21 cm and 42 cm. If ZAOB = 60°, find the area of the shaded region [ Use π = [latex s=2]\frac { 22 }{ 7 } [/latex] ].
CBSE Sample Papers for Class 10 Maths Paper 8 img 4

Question 20.
If volumes of two spheres are in the ratio of64 : 27, then what is the ratio of their surface areas?
OR
A medicine-capsule is in the shape of a cylinder of diameter 0.5 cm with two hemispheres stuck to each of its ends. The length of entire capsule is 2 cm. What is the capacity of the capsule?

Question 21.
If A is the area of a right angled triangle and ‘b’ is one of the sides containing the right angle, prove that
the length of the altitude on the hypotenuse is [latex s=2]\frac { 2Ab }{ \sqrt { { b }^{ 4 }+4{ A }^{ 2 } } } [/latex]
OR
ABC is a right. A, right angled at B. AD and CE are the two medians drawn from A and C respectively. If
AC = 5 cm and AD = [latex s=2]\frac { 3\sqrt { 5 } }{ 2 } [/latex] cm, find CE

Question 22.
If the points A (k + 1,2k), B (3k, 2k+ 3) and C (5k- 1,5k) are collinear, then find the value of k.
OR
Show that the points A (1, 0), B (5,3), C (2,7) and D (-2,4) are the vertices of a parallelogram.

Section-D

Question 23.
An aeroplane left 30 minutes later than its scheduled time. The pilot decided to increase its speed. In order to reach its destination 1500 km away in time, pilot has to increase its speed by 250 km/hr from its usual speed. Determine the usual speed.

Question 24.
CBSE Sample Papers for Class 10 Maths Paper 8 img 5

Question 25.
Draw a line segment AB of length 8 cm. Taking A as centre, draw a circle of radius 4 cm and taking B as centre, draw another circle of radius 3 cm. Construct tangents to each circle from the centre of the other circle.

Question 26.
If h, c, v are respectively the height, the curved surface area and the volume of a cone, prove that 3πvh3– c2h2+ 9v2=0

Question 27.
CBSE Sample Papers for Class 10 Maths Paper 8 img 6

Question 28.
If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then prove that the other two sides are divided in the same ratio.
OR
If a line divides any two sides of a triangle in the same ratio, then prove that the line is parallel to the third side.

Question 29.
From the top of a tower, the angles of depression of two objects on the same side of the tower are found to be α and β ( α > β). If the distance between the objects is P; show that the height ‘h’ of the tower is given by h = [latex s=2]\frac { P\tan { \alpha } \tan { \beta } }{ \tan { \alpha } -\tan { \beta } } [/latex] .
OR
A vertical tower stands on a horizontal plane and is surmounted by 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 [latex s=2]\frac { h\tan { \alpha } }{ \tan { \beta -\tan { \alpha } } } [/latex]

Question 30.
The weekly expenditure of5000 families is tabulated below :
CBSE Sample Papers for Class 10 Maths Paper 8 img 7
Find the median expenditure.
OR
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.
CBSE Sample Papers for Class 10 Maths Paper 8 img 8

Solutions
Section-A

Solution 1.
Since: [latex s=2]\frac { 1 }{ 7 } [/latex] × [latex s=2]\frac { 7 }{ 100 } [/latex] = [latex s=2]\frac { 1 }{ 100 } [/latex] = 0.01
The smallest rational number is [latex s=2]\frac { 7 }{ 100 } [/latex] (1)

Solution 2.
CBSE Sample Papers for Class 10 Maths Paper 8 img 9

Solution 3.
CBSE Sample Papers for Class 10 Maths Paper 8 img 10

Solution 4.
Let x = [latex]\sqrt { 6+\sqrt { 6 } +\sqrt { 6+ } } [/latex] ….Then, x = [latex]\sqrt { 6+x } [/latex]
On squaring both the sides, we get
x2 = 6+ x ⇒x2-x-6 = 0
⇒ (x-3)(x + 2) = 0 ⇒ x = 3 or x = -2.
∵ x > 0
So, x = 3 (1)

Solution 5.
Since, a21 – a7 = 84 …(i)
Let the first term and common difference of
given A.P. be a and d respectively.
So, a21 = a+ (21 – 1)d = a + 20d
Now, a7= a + (7 – 1)d = a + 6d
From equation (i),
(a + 20d) – (a + 6d) = 84
⇒ (20-6)d= 84 ⇒ 14d= 84 ⇒d = 6
Hence, the common difference of given A.P. is 6. (1)

Solution 6.
Let ABC and DEFbe two triangles such that ∆ABC ~ ∆DEF
CBSE Sample Papers for Class 10 Maths Paper 8 img 11
Let AL, DM be their medians respectively
CBSE Sample Papers for Class 10 Maths Paper 8 img 12
Hence required corresponding median = 8.8 cm.

Section-B

Solution 7.
Largest possible amount of cheque will betheHCF (6075,8505).
Applying Euclid’s division lemma to 8505 and 6075, we have,
8505 = 6075 × 1 +2430 (1/2)
Since, remainder 2430 ≠ 0 again applying division lemma to 6075 and 2430
6075 = 2430 × 2+1215 (1/2)
Again remainder 1215 ≠ 0
So, again applying the division lemma to 2430 and 1215
2430=1215 × 2 +0 (1/2)
Here the remainder is zero
So,H.C.F= 1215 (1/2)
Therefore, the largest possible amount of each cheque will be 1215.

Solution 8.
Series: 70 + 68 + 66 + +40
Here a = 70, d = 68 – 70 = -2, tn = 1 = 40.
Now, tn = a + (n-1)d =70 + (n-1)(-2) ⇒ 40 = 70 – 2n + 2 ⇒n= 16 (1)
∴ Sn =[latex s=2]\frac { n }{ 2 } [/latex][a + l] ⇒Sn = [latex s=2]\frac { 16 }{ 2 } [/latex][70 + 40] = 880 (1)

Solution 9.
S = {S, M, T, W, Th, F, Sa} (1/2)
n(S)=7
A non-leap year contains 365 days,
i.e., 52 weeks + 1 day.
E={Sa}
n(E)=1 (1/2)
∴ P(E) = [latex s=2]\frac { n(E) }{ n(S) } [/latex] = [latex s=2]\frac { 1 }{ 7 } [/latex] (1)

Solution 10.
Let there be x blue, y green and z white marbles in the jar.
Then, x+y+ z=54 …(i) (1/2)
∴ P(selecting a blue marble) = [latex s=2]\frac { x }{ 54 } [/latex] ⇒[latex s=2]\frac { x }{ 54 } [/latex] = [latex s=2]\frac { 1 }{ 3 } [/latex] ⇒ x = 18 (1/2)
Similarly, /’(selecting a green marble) = [latex s=2]\frac { y }{ 54 } [/latex] ⇒ [latex s=2]\frac { y }{ 54 } [/latex]= [latex s=2]\frac { 4 }{ 9 } [/latex] ⇒y = 24 (1/2)
Substituting thevaluesofxandyin(i),z= 12
Hence, the jar contains 12 white marbles. (1/2)

Solution 11.
Given lines,
3x + 2ky- 2 = 0
and 2x+ 5y-1 =0
Here, a1 =3,b1 =2k, c1 =-2
a2 = 2, b2 = 5, c2 =-1 (1)
Condition for parallel lines is [latex s=2]\frac { 3 }{ 2 } [/latex] = [latex s=2]\frac { 2k }{ 5 } [/latex] ⇒k = [latex s=2]\frac { 15 }{ 4 } [/latex] (1)

Solution 12.
Since points are collinear,
so, x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2) = 0 (1/2)
Taking (7, -2) as (x1, y1), (5,1) as (x2, y2) and (3, k) as (x3, y3), we have
7(1 – k) + 5(k + 2) + 3 (-2 -1) = 0 (1/2)
or, 7 – 7k + 5k + 10 – 9 = 0 ⇒ 2k = 8 ⇒k = 4. (1)

Section-C

Solution 13.
a2 – b2 = (a-b)(a +b) ….(1)
∵ a2 -b2 is a prime number (1)
∴One of the two factors = 1
So, a — b= 1 [∵a – b< a+b]
∵ The only divisors of a prime number are 1 and it self. (1)
(1) become a2 -b2= 1 (a + b)
or a2 – b2 = a + b
e.g., 32 – 22 = 5 (which is prime)
⇒ 32 – 22 = 3 +2 (1)

Solution 14.
CBSE Sample Papers for Class 10 Maths Paper 8 img 13

Solution 15.
CBSE Sample Papers for Class 10 Maths Paper 8 img 14
OR
CBSE Sample Papers for Class 10 Maths Paper 8 img 15

Solution 16.
CBSE Sample Papers for Class 10 Maths Paper 8 img 16

Solution 17.
CBSE Sample Papers for Class 10 Maths Paper 8 img 17

Solution 18.
Let there be a circle with centre O whereas AB, BC, CD and DA are tangents at P, Q, R and S respectively.
CBSE Sample Papers for Class 10 Maths Paper 8 img 18
Here,
AP = AS …(i)
BP = BQ …(ii)
CR=CQ …(iii)
DR=DS …(iv)
[Tangents drawn from a point (outside the circle) on a given circle are equal in lengths]
From equations (i), (ii), (iii) and (iv), (1)
(AP +BP) + (CR + DR) = (BQ + CQ) + (DS + AS)
AB + CD = BC + DA (Hence proved.) (1)

Solution 19.
CBSE Sample Papers for Class 10 Maths Paper 8 img 19

Solution 20.
CBSE Sample Papers for Class 10 Maths Paper 8 img 20
CBSE Sample Papers for Class 10 Maths Paper 8 img 21
OR
CBSE Sample Papers for Class 10 Maths Paper 8 img 22

Solution 21.
CBSE Sample Papers for Class 10 Maths Paper 8 img 23
OR
CBSE Sample Papers for Class 10 Maths Paper 8 img 24

Solution 22.
CBSE Sample Papers for Class 10 Maths Paper 8 img 25
OR
CBSE Sample Papers for Class 10 Maths Paper 8 img 26
Since, the coordinates of the mid-points of diagonals AC and BD are same.
∴ They bisect each other.
Hence, ABCD is a parallelogram. (1)

Section-D

Solution 23.
CBSE Sample Papers for Class 10 Maths Paper 8 img 27

Solution 24.
CBSE Sample Papers for Class 10 Maths Paper 8 img 28

Solution 25.
CBSE Sample Papers for Class 10 Maths Paper 8 img 29
Steps of construction:
(i) Drawa line AB of length 8 cm.
(ii) With centre A and radius 4 cm, draw a circle, which intersect AB at point P.
(iii) With centre B and radius 3 cm, draw the second circle.
(iv) With centre P and radius PA or PB draw third circle, which intersects the circle drawn in step (ii) at C & D and the circle drawn in step (iii) at E & F.
(v) Draw rays AE,AF,BC and BD.
(vi) Thus AE,AF,BC and BD are required tangents. (1)

Solution 26.
CBSE Sample Papers for Class 10 Maths Paper 8 img 30

Solution 27.
CBSE Sample Papers for Class 10 Maths Paper 8 img 31
CBSE Sample Papers for Class 10 Maths Paper 8 img 32

Solution 28.
CBSE Sample Papers for Class 10 Maths Paper 8 img 33
OR
CBSE Sample Papers for Class 10 Maths Paper 8 img 34
CBSE Sample Papers for Class 10 Maths Paper 8 img 35

Solution 29.
CBSE Sample Papers for Class 10 Maths Paper 8 img 36
OR
Let AB be the tower and BC be the flag.
LetAB = H
BC = h
CBSE Sample Papers for Class 10 Maths Paper 8 img 37

Solution 30.
CBSE Sample Papers for Class 10 Maths Paper 8 img 38
CBSE Sample Papers for Class 10 Maths Paper 8 img 39
OR
CBSE Sample Papers for Class 10 Maths Paper 8 img 40

We hope the CBSE Sample Papers for Class 10 Maths paper 8 help you. If you have any query regarding CBSE Sample Papers for Class 10 Maths paper 8, drop a comment below and we will get back to you at the earliest.

CBSE Sample Papers for Class 10 Maths Paper 7

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 7.

CBSE Sample Papers for Class 10 Maths Paper 7

Board CBSE
Class X
Subject Maths
Sample Paper Set Paper 7
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 7 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.
What type of decimal expansion does a rational number has? How can you distinguish it from decimal expansion of irrational numbers?

Question 2.
Coordinates of P and Q are (4,-3) and (-1, 7). What is the abscissa of a point R on the line segment PQ such that [latex]\frac { PR }{ PQ } [/latex] = [latex]\frac { 3 }{ 5 } [/latex] ? (NTSE 2012)

Question 3.
For what value of k will k + 9, 2k – 1 and 2k + 7 are the consecutive terms of an A.P. ?

Question 4.
For the equation 3×2 + px + 3 = 0, if one of the roots is the square of the other then find p. (NTSE 2014-2015)

Question 5.
Find the value of sin (45° + θ) – cos (45° – θ)

Question 6.
In Fig., if PQ || RS, prove that ∆ POQ ~ ∆ SOR
CBSE Sample Papers for Class 10 Maths Paper 7 img 1

Section-B

Question 7.
When 2256 is divided by 17 then find the remainder.

Question 8.
In what ratio, the line segment joining the points (3,5) & (-4,2) is divided byy-axis?

Question 9.
Find the solution of the pair equations [latex s=2]\frac { x }{ 10 } [/latex] + [latex s=2]\frac { y }{ 5 } [/latex] – 1= 0 and [latex s=2]\frac { x }{ 8 } [/latex] + [latex s=2]\frac { y }{ 6 } [/latex] = 15. Hence, find λ, if y = λx + 5.

Question 10.
Find the sum of first 24 terms of the sequence whose nth term is an = 3+[latex s=2]\frac { 2n }{ 3 } [/latex]

Question 11.
A girl calculates that the probability of her winning the first prize in a lottery is 0.08. If6000 tickets are sold,

Question 12.
A book containing 100 pages is opened at random. Find the probability that a doublet page is found.

Section-C

Question 13.
Show that any positive odd integer is of the form 8q ± 1 and 8q ± 3, where q is some integer.

Question 14.
If a and 8 are the zeroes of the quadratic polynomial f(x) = ax2 + bx + c then evaluate [latex]\frac { { \alpha }^{ 2 } }{ { \beta }^{ 2 } } +\frac { { \beta }^{ 2 } }{ { \alpha }^{ 2 } } [/latex]

Question 15.
In the given fig, BD ⊥ AC and CE ⊥ AB. Prove that
(i) ∆AEC ~ ∆ADB
(ii) [latex s=2]\frac { CA }{ AB } [/latex] = [latex s=2]\frac { CE }{ DB } [/latex]
CBSE Sample Papers for Class 10 Maths Paper 7 img 2
OR
In the given fig, [latex s=2]\frac { OA }{ OC } [/latex] = [latex s=2]\frac { OD }{ OB } [/latex] prove that ∠A = ∠C and ∠B = ∠D
CBSE Sample Papers for Class 10 Maths Paper 7 img 3

Question 16.
Prove that the area of a triangle with vertices (t, t – 2), (t + 2, t + 2) and (t + 3, t) is independent of t.
OR
The three vertices of a parallelogram, taken in order, are (1, -2), (3,6) and (5,10). Find the coordinates of its fourth vertex .

Question 17.
In the given figure, PA and PB are tangents to the circle from an external point P. CD is another tangent touching the circle at Q. If PA = 12 cm, QC = QD = 3 cm, then find PC + PD.
CBSE Sample Papers for Class 10 Maths Paper 7 img 4

Question 18.
Calculate the mean of the following frequency distribution :
CBSE Sample Papers for Class 10 Maths Paper 7 img 5

Question 19.
In figure, ABCD is a square of side 10 cm and semicircles are drawn with each side of the square as diameter. Find area of the shaded region.
CBSE Sample Papers for Class 10 Maths Paper 7 img 6

Question 20.
A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Roohi paid ₹ 27 for a book kept for seven days, while Shushama paid₹ 21 for the book she kept for five days. What is the fixed charge?

Question 21.
Without using trigonometric tables, prove that:- tan7° tan 23° tan 60° tan 67° tan 83° = [latex]\sqrt { 3 } [/latex]
OR
If 5 tan θ = 4, then find the value of [latex s=2]\frac { 5\sin { \theta } -3\cos { \theta } }{ 5\sin { \theta } +2\cos { \theta } } [/latex]

Question 22.
A copper rod of diameter 1cm and length 8 cm, is drawn into a wire of length 18 m of uniform thickness. Find the thickness of wire.
OR
If a solid piece of iron in the form of a cuboid of dimensions 49 cm × 33 cm × 24 cm, is moulded to form a solid sphere. Find the radius of the sphere

Section-D

Question 23.
In a right triangle, prove that the square of the hypoten’use is equal to the sum of the squares of the other two sides.
OR
In a triangle, if the square of one side is equal to the sum of the squares of the other two sides, then prove that the angle opposite to the first side is a right angle.

Question 24.
What is the value of x if
cot x = [latex s=2]\frac { 5 }{ 3 } [/latex] tan 130 tan 37° tan 45 ° tan 53° tan 77° -[latex s=2]\frac { 2 }{ 3 } [/latex] cosec2 58° + [latex s=2]\frac { 2 }{ 3 } [/latex] cot 58° tan 32°.

Question 25.
Solve the equation: [latex s=2]\left( \frac { 2x-3 }{ x-1 } \right) -4\left( \frac { x-1 }{ 2x-3 } \right) [/latex] = 3 x ≠ 1, 3/2

Question 26.
Find the median of the following data :
CBSE Sample Papers for Class 10 Maths Paper 7 img 7

OR
The median ofthe following data is 525. Find the values of* and y if the total frequency is 100.
CBSE Sample Papers for Class 10 Maths Paper 7 img 8

Question 27.
Find the volume and total surface area of a tumbler in the form of a frustum of a cone, if the diameter of the ends are 6.50 cm and 3.50 cm and the perpendicular height ofthe tumbler is 7.80 cm.

Question 28.
Draw a circle of radius 4 cm. Draw two tangents to the circle inclined at an angle of 60° to each other.

Question 29.
If the ratio ofthe sum ofthe first n terms oftwoA.Ps is (7n+ 1): (4n + 27), then find the ratio of their 9th terms.

Question 30.
An aeroplane is flying at a height of 300 m above the ground. Flying at this height, the angles of depression from the aeroplane of two points on both banks of a river in opposite directions are 45° and
60° respectively. Find the width ofthe river. [Use [latex]\sqrt { 3 } [/latex] = 1.732]
OR
The angle of elevation of a cloud from a point 60 m above the surface ofthe water of a lake is 30° and the angle of depression of its shadow in water of lake is 60°. Find the height ofthe cloud from the surface of water.

Solutions
Section-A

Solution 1.
A rational number is either terminating or non-terminating repeating.
An irrational number is non-terminating and non-repeating. (1)

Solution 2.
P (4, -3) and Q (-1,7)
PR: RQ = 3:2
4 × 2 + (-1) × 3
Abscissa = [latex s=2]\frac { 4\times 2+(-1)\times 3 }{ 3+2 } [/latex]= 1 (1)

Solution 3.
Given k + 9,2k – 1 and 2k + 7 are in AP.
∴ 2(2k – 1) = (k + 9)+(2k + 7) [By Arithmetic mean] (1/2)
⇒ 4k-2 = 3k+16 ⇒ k=18 (1/2)

Solution 4.
α + α2 = [latex s=2]\frac { -p }{ 3 } [/latex] ………..(i)
α3 = [latex s=2]\frac { 3 }{ 3 } [/latex] = 1 =>(α- l)(α3+ α+ 1) = 0 ⇒ α = 1
∴ p= – 6 (∵From (i)) (1)

Solution 5.
Given that
sin (45° + θ) – cos (45° – θ) = cos [90° – (45° + θ)] – cos (45° – θ) [∵ cos (90° – θ) = sin θ]
= cos (45° – θ) – cos (45° – θ) = 0 (1)

Solution 6.
PQ || RS (Given)
So, ∠P = ∠S (Alternate angles)
and ∠Q=∠R (1/2)
Also, ∠POQ = ∠ SOR (Vertically opposite angles)
Therefore, ∆POQ ~ ∆SOR (AAA similarity criterion) (1/2)

Section-B

Solution 7.
CBSE Sample Papers for Class 10 Maths Paper 7 img 9

Solution 8.
CBSE Sample Papers for Class 10 Maths Paper 7 img 10

Solution 9.
CBSE Sample Papers for Class 10 Maths Paper 7 img 11

Solution 10.
CBSE Sample Papers for Class 10 Maths Paper 7 img 12

Solution 11.
P (winning) = 0.08
Total tickets sold = 6000
Let the number of tickets she bought be x, then probability of winning =[latex s=2]\frac { x }{ 6000 } [/latex] (1)
⇒[latex s=2]\frac { x }{ 6000 } [/latex] = 0.08 ⇒ x = 6000 × 0.08 ⇒ x = 480 (1)
Thus the girl bought 480 tickets.

Solution 12.
S= {1,2,3, ,100} (1/2)
n(S)= 100
E ={ 11,22,33,44, 55,66,77, 88,99} (1)
n(E) = 9
∴P(E) = [latex s=2]\frac { 9 }{ 100 } [/latex] (1/2)

Section-C

Solution 13.
Let a and b be two positive integers where a is odd.
Applying division lemma a = 8q + r where 0 < r < 8
So, r can take any of the values 0, 1,2,3,4,5,6,7
Therefore, a= 8q, 8q + 1, 8q + 2, 8q + 3, 8q + 4,
8q + 5, 8q + 6, 8q + 7 (1)
Since, a is odd.
Therefore, a cannot take values 8q, 8q + 2, 8q + 4, 8q + 6 since they can expressed as multiples of 2.
So, a will take values 8q+ 1,8q + 3, 8q + 5, 8q + 7. (1)
Also, 8q + 5 = 8q + 8 – 3 = 8 (q + 1) – 3 = 8q’ — 3
where q’ = q + 1, 8q + 7 = 8q + 8 – 1 =8q’- 1
So, every positive odd integer is of the form 8q ± 1, 8q ± 3. (1)

Solution 14.
CBSE Sample Papers for Class 10 Maths Paper 7 img 13

Solution 15.
Given: In the fig., BD ⊥ AC and CE ⊥AB
To prove: (i) ∆AEC ~ ∆ADB
(ii) [latex s=2]\frac { CA }{ AB } [/latex] = [latex s=2]\frac { CE }{ DB } [/latex] (1)
CBSE Sample Papers for Class 10 Maths Paper 7 img 14
Proof (i) In ∆AEC and ∆ADB
∠1 = ∠2 (each 90°)
∠A = ∠A (common)
∴ ∆AEC ~ ∆ADB (by AArule)
(ii) ∆AEC ~ ∆ADB
[latex s=2]\frac { CA }{ AB } [/latex] = [latex s=2]\frac { CE }{ DB } [/latex] (∵ Angles are similar ∴ corresponding sides are proportional) (1)
Hence proved
OR
[latex s=2]\frac { OA }{ OC } [/latex] = [latex s=2]\frac { OD }{ OB } [/latex]
CBSE Sample Papers for Class 10 Maths Paper 7 img 15
To prove: ∠A = ∠C and ∠B = ∠D (1)
Proof: In ∆AOD and ∆BOC
[latex s=2]\frac { OA }{ OC } [/latex] = [latex s=2]\frac { OD }{ OB } [/latex] (Given)
and ∠AOD = ∠BOC (Vertically opposite angles)
∴ ∆AOD ~ ∆BOC (by SAS) (1)
∴ ∠A = ∠C and ∠B = ∠D (C.P.C.T.) (1)

Solution 16.
Let A (t, t – 2), B (t + 2, t + 2) and C (t + 3, t) be the vertices of the given triangle. (1/2)
Since the area of the triangle having vertices (x1, y1), (x2, y2) and(x3, y3)
CBSE Sample Papers for Class 10 Maths Paper 7 img 16
Therefore, the area of the triangle with specified vertices is independent of t. (1/2)
OR
Let fourth vertex be D (a, b)
Let the diagonals AC and BD intersect at E,
CBSE Sample Papers for Class 10 Maths Paper 7 img 17
We know that the diagonals of a parallelogram bisect each other.
∴ E is the mid-point of AC as well as that ofBD.
mid-point of AC is
CBSE Sample Papers for Class 10 Maths Paper 7 img 18
Hence, The fourth vertex of the given parallelogram is D (3,2). (1)

Solution 17.
Given : PA and PB are tangents to the circle from an external point P. CD is another tangent at Q. PA= 12 cm, QC = QD = 3 cm
CBSE Sample Papers for Class 10 Maths Paper 7 img 19
To find: PC + PD (1/2)
Proof: PA = PC + AC
12 = PC+ 3
[∵ QC = AC = 3 cm, tangents from external point to a circle are equal in length]
PC = 9 cm …(i) (1)
Similarly, BD = QD = 3 cm
and PB = PA =12 cm (1/2)
PB = PD + BD
12 = PD +3
PD = 9 cm
Now, PC + PD = 9 + 9 = 18 cm (1)

Solution 18.
CBSE Sample Papers for Class 10 Maths Paper 7 img 20

Solution 19.
We mark the area of unshaded regions as A1,A2A3 and A4 as shown in the figure.
Area of the shaded region = Area of the square ABCD – Area of the unshaded portion (A1+A2+A3+ A4).
CBSE Sample Papers for Class 10 Maths Paper 7 img 21
Area of unshaded region A1+A3= Area of square ABQD – Area of semi circle on BC as diameter- Area of semi-circle on AD as diameter.
CBSE Sample Papers for Class 10 Maths Paper 7 img 22
Similarly, area of unshaded region A2 – A4 = Area of square ABCD – Area of the semi-circles on diameter AB and DC
So, Area A2+A4 = 102 -2× [latex s=2]\frac { \pi \times { 5 }^{ 2 } }{ 2 } [/latex] =100-25π.
So,Area A1 + A2 + A3 + A4 = 200-50π.
Area of the shaded region = Area of square ABCD – Area (A1 + A2 + A3 + A4) (1)
= 102-(200-50π)
= (100-200 + 50 π)sq. cm.
= 50 π -100 = 50 (π – 2) sq. cm.
= 50 (3.14-2)= 50 × 1.14 = 57 cm2. (1)

Solution 20.
x = F ixed charge
y = Additional charge for each day
x + 4y = 27 …(i)
x + 2y = 21 …(ii) (1)
⇒ (x + 4y)-(x+ 2y) = 27-21 => 2y = 6 ⇒y = 3 (1)
∵ x + 2y=21 =>orx + 2(3) = 21 ⇒ x= ₹ 15 (1)

Solution 21.
L.H.S.
= tan 7° tan 23° tan 60° tan 67° tan 83°= tan 7° tan 83° tan 23° tan 67° tan 60° (1)
= tan 7° cot 7° tan 23° cot 23° tan 60°(1)
CBSE Sample Papers for Class 10 Maths Paper 7 img 23
OR
CBSE Sample Papers for Class 10 Maths Paper 7 img 24
CBSE Sample Papers for Class 10 Maths Paper 7 img 25

Solution 22.
Volume of rod = πr2h = π × (1/2)2 × (8) = 2πcm3 ….(i) (1)
Length of wire of same volume = 18m = 1800 cm.
Let ‘r’ be the radius of cross-section of wire (1/2)
∴ Volume = πr2h
Volume of wire = πr2 × 1800 ….(ii)
From (i) & (ii), πr2 × 1800 = 2π
r2 = 2/1800 =>r2= 1/900 =>r=1/30cm. (1)
Thickness of wire = diameter of cross-section = 2r = 2 × [latex s=2]\frac { 1 }{ 30 } [/latex] = [latex s=2]\frac { 1 }{ 15 } [/latex] = 0.067 cm (approximately) (1/2)
OR
Given, dimensions of the cuboid
= 49 cm x 33 cm x 24 cm
Now, volume of the cuboid
= 49x33x24 = 38808 cm3
Let the radius of the sphere is r, then (1/2)
Volume of the sphere = [latex s=2]\frac { 4 }{ 3 } [/latex]πr3 (1/2)
According to the given condition
Volume of the sphere = Volume of the cuboid (1/2)
CBSE Sample Papers for Class 10 Maths Paper 7 img 26

Section-D

Solution 23.
Given: A triangle ABC in which ZB = 90°.
CBSE Sample Papers for Class 10 Maths Paper 7 img 27
To prove : ACsup>2 = ABsup>2 + BCsup>2 or (Hypotenuse)sup>2 = (Base)sup>2 + (Perpendicular)sup>2
Construction : From B, drawBD⊥ AC. (1)
Proof: Since BD ⊥AC.
∴∆ADB ~ ∆ABC
∴[latex s=2]\frac { AD }{ AB } [/latex] = [latex s=2]\frac { AB }{ AC } [/latex] ⇒AB2 =AC ×AD …(i) (1)
CBSE Sample Papers for Class 10 Maths Paper 7 img 28
OR
CBSE Sample Papers for Class 10 Maths Paper 7 img 29

Solution 24.
CBSE Sample Papers for Class 10 Maths Paper 7 img 30

Solution 25.
CBSE Sample Papers for Class 10 Maths Paper 7 img 31

Solution 26.
CBSE Sample Papers for Class 10 Maths Paper 7 img 32
OR
CBSE Sample Papers for Class 10 Maths Paper 7 img 33
CBSE Sample Papers for Class 10 Maths Paper 7 img 34

Solution 27.
CBSE Sample Papers for Class 10 Maths Paper 7 img 35
CBSE Sample Papers for Class 10 Maths Paper 7 img 36
CBSE Sample Papers for Class 10 Maths Paper 7 img 37

Solution 28.
Steps of construction:
1. Draw a circle of radius 4 cm, with O as centre.
2. Take a pointAon the circumference ofthe circle andjoin OA. Draw a perpendicular to OA at A.
3. Draw a radius OB, making an angle of 120° with OA.
4. Draw a perpendicular to OB at B. Suppose these perpendiculars intersect at P.
CBSE Sample Papers for Class 10 Maths Paper 7 img 38
Here, PA and PB are two tangents drawn to the circle inclined at an angle of 60° to each other.
Justification
The construction can be justified by proving that
∠APB = 60°.
∠OAP = 90° (By Construction)
∠OBP = 90° (By Construction)
∠AOB=120° (By Construction)
The sum of all interior angles of a quadrilateral is 360°,.
∠OAP + ∠AOB + ∠OBP + ∠APB = 360° .
⇒ 90°+120°+ 90° +∠APB = 360° (1)
⇒ ∠APB = 60°

Solution 29.
CBSE Sample Papers for Class 10 Maths Paper 7 img 39

Solution 30.
CBSE Sample Papers for Class 10 Maths Paper 7 img 40
OR
CBSE Sample Papers for Class 10 Maths Paper 7 img 41

We hope the CBSE Sample Papers for Class 10 Maths paper 7 help you. If you have any query regarding CBSE Sample Papers for Class 10 Maths paper 7, drop a comment below and we will get back to you at the earliest.

CBSE Sample Papers for Class 10 Maths Paper 6

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 6.

CBSE Sample Papers for Class 10 Maths Paper 6

Board CBSE
Class X
Subject Maths
Sample Paper Set Paper 6
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 6 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.
Explain why 3 × 5 × 7 + 7 is composite number.

Question 2.
If sin A = [latex]\frac { 1 }{ \sqrt { 2 } } [/latex] and tan B = 1. Prove that sin (A+ B) = 1, where A, B are acute angles.

Question 3.
In Fig. DE || BC. HAD = x, DB = x – 2,AE=x +2 and EC = x – 1, find the value ofx.
CBSE Sample Papers for Class 10 Maths Paper 6 img 1

Question 4.
Find the distance of origin from the point P(3, -2). (NTSE2012)

Question 5.
Find the 9th term from the end (towards the first term) ofthe AP. 5,9,13,……, 185

Question 6.
Find the discriminant ofthe quadratic equation 3[latex]\sqrt { 3 } [/latex]x2 +10x + [latex]\sqrt { 3 } [/latex] = 0

Section-B

Question 7.
Show that 5- [latex]\sqrt { 3 } [/latex] is not irrational.

Question 8.
Cards marked with numbers 13,14,15,…….., 60 are placed in a box and mixed thoroughly. One card is
drawn at random from the box. Find the probability that the number on drawn card is
(i) divisible by 5
(ii) a number which is a perfect square

Question 9.
If p, q, r are in A.P. then find the value of p3 +r3 – 8q3.

Question 10.
For all real values of c, the pair of equations
x – 2y = 8
5x- 10y = c
have a unique solution. Justify whether it is true or false.

Question 11.
If the coordinates of points A and B are (-2, -2) and (2, -4) respectively, find the coordinates of P such that AP= [latex s=2]\frac { 3 }{ 7 } [/latex]AB, where Plies on the line segment AB.

Question 12.
A piggy bank contains hundred 50p coins, fifty ₹ 1 coins, twenty ₹ 2 coins and ten ₹ 5 coins. If it is equally likely that one of the coins will fall out when the bank is turned upside down, what is the probability that the coin (i) will be a 50 p coin ? (ii) will not be a ₹ 5 coin?

Section-C

Question 13.
Show that any positive odd integer is of the form 6q – 1 or 6q + 5, where q is some integer.

Question 14.
What must be subtracted from x3 – 6x2 – 15x + 80, so that the result is exactly divisible by x2 + x— 12.

Question 15.
If a cos θ + b sinθ = m and a sin θ – b cosθ = n, prove that a2 + b2 = m2 + n2
OR
If x = r sin A cos C, y = r sin A sin C, z = r cos A, Prove that : r2 = x2+ y2+ z2

Question 16.
Calculate the area of the shaded portion.
The quadrants shown in the figure are each of radius 7 cm. [Take π = [latex s=2]\frac { 22 }{ 7 } [/latex]]
CBSE Sample Papers for Class 10 Maths Paper 6 img 2

Question 17.
If mean of the following frequency distribution is 7.5, then find the value of p.
CBSE Sample Papers for Class 10 Maths Paper 6 img 3

Question 18.
An iron pillar has some part in the form of a right circular cylinder and remaining in the form of a right circular cone. The radius of the base of each 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 one cubic cm. of iron weights 7.8 gms.
OR
A cylinder whose height is two-third of its diameter has the same volume as a sphere of radius 4 cm. Calculate the radius of the base of the cylinder.

Question 19.
The numerator of a fraction is 4 less than the denominator. If the numerator is decreased by 2 and the denominator is increased by 1, then the denominator is eight times the numerator. The fraction is [latex s=2]\frac { a }{ b } [/latex] . Find [latex s=2]\frac { a+b }{ 2 } [/latex]

Question 20.
If the diagonal BD of a quadrilateral ABCD bisects both ∠B and ∠D then show that [latex s=2]\frac { AB }{ BC } [/latex] = [latex s=2]\frac { AD }{ CD } [/latex]
OR
In an equilateral triangle ABC, AD is drawn perpendicular to BC. Prove that 4 AD2 = 3BD2

Question 21.
Find the values ofk so that the area of the triangle with vertices (1, -1), (-4,2k) and (-k, -5) is 24 sq. units.
OR
If the point P (x, y) is equidistant from the points A (a + b, b – a) and B (a – b, a + b). Prove that bx = ay.

Question 22.
If from an external point P of a circle with centre 0, two tangents PQ and PR are drawn such that ∠QPR =120°, prove that 2PQ = PO.

Section-D

Question 23.
The ratio of the sums of first m and first n terms of an A.P. is m2: n2. Show that the ratio of its mth and nth terms is (2m -1): (2n – 1).

Question 24.
If the areas of two similar triangles are equal, prove that they are congruent.
OR
If the corresponding sides of two triangles are proportional (i.e., in the same ratio), then prove that their corresponding angles are equal and hence the two triangles are similar.

Question 25.
At a point A, 20 metres above the level of water in a lake, the angle of elevation of a cloud is 30°. The angle of depression of the reflection of the cloud in the lake, at A is 60°. Find the distance of the cloud from A.
OR
A bird is sitting on the top of a 80 m high tree. From a point on the ground, the angle of elevation of the bird is 45°. The bird flies away horizontally in such a way that it remained at a constant height from the ground. After 2 seconds, the angle of elevation of the bird from the same point is 30°. Find the speed of flying of the bird.
(Take [latex]\sqrt { 3 } [/latex] =1.732)

Question 26.
A semicircular sheet of metal of diameter 28 cm is bent into an open conical cup. Find the depth and capacity of the cup.

Question 27.
If θ be an acute angle such that tan2 θ = 1 – a2, then find the value of {sec θ + tan3 θ cosec θ}2

Question 28.
Solve the quadratic equation (x2 – 5x)2 – 7(x2 – 5x) + 6 = 0.

Question 29.
Construct a triangle with sides 5 cm, 5.5 cm and 6.5 cm. Now construct another triangle, whose sides are [latex s=2]\frac { 3 }{ 5 } [/latex] times the corresponding sides of the given triangle.

Question 30.
Following table gives the ages in years of militants operating in a certain area of a country:
CBSE Sample Papers for Class 10 Maths Paper 6 img 4
If mean of the above distribution is 47.2, find how many militants in the age groups 49-52 are active in the area?
OR
One hundred students took an examination. Compute values ofx and y from the following data if mean is 42.
CBSE Sample Papers for Class 10 Maths Paper 6 img 5

Solutions
Section-A

Solution 1.
3 × 5 ×7+7 = 7 (3 × 5 + 1) = 7 (16), which has more than two factors.
∴ It is a composite number. (1)

Solution 2.
sinA= [latex s=2]\frac { 1 }{ \sqrt { 2 } } [/latex] ⇒ A=45°

and tan B = 1 ⇒ B = 45° (1/2)
∴ sin (A + B) = sin (45° + 45°) ⇒ sin 90° = 1 Hence Proved. (1/2)

Solution 3.
In ∆ABC, we have DE || BC
∴ [latex s=2]\frac { AD }{ DB } [/latex] = [latex s=2]\frac { AE }{ EC } [/latex] By Thale’s Theorem] (1/2)

⇒ [latex s=2]\frac { x }{ x-2 } [/latex] = [latex s=2]\frac { x+2 }{ x-1 } [/latex]
⇒x(x-1) = (x-2)(x +2) ⇒ x2 – x = x2 – 4 ⇒x = 4 (1/2)

Solution 4.
Distance = [latex]\sqrt { { (0-3) }^{ 2 }+{ (0+2) }^{ 2 } } =\sqrt { 9+4 } =\sqrt { 13 } [/latex] (1)

Solution 5.
Common difference of the AP = d = 9 – 5 = 4
Last term of the AP = l= 185 (1/2)
Since the nth term from the end of an AP is given by l – (n – 1 )d.
So, the 9th term from the end is = 185 – (9 – 1)4 = 185 – 32= 153 (1/2)

Solution 6.
D = b2 – 4ac = (10)2 – 4(3[latex]\sqrt { 3 } [/latex] )([latex]\sqrt { 3 } [/latex]) = 100 – 36 = 64 (1)

Section-B

Solution 7.
If 5 – [latex]\sqrt { 3 } [/latex] is not irrational.
Let 5 – [latex]\sqrt { 3 } [/latex] = [latex s=2]\frac { p }{ q } [/latex] [p and q are integers, q ≠ 0] (1/2)
⇒[latex s=2]\frac { 5q-p }{ q } [/latex] = [latex]\sqrt { 3 } [/latex] ⇒ [latex s=2]\frac { 5q-p }{ 3 } [/latex] is rational number  (1)
So, [latex]\sqrt { 3 } [/latex] should be rational.
But [latex]\sqrt { 3 } [/latex] is irrational, it cannot be equal to rational number, Hence 5 -[latex]\sqrt { 3 } [/latex] is irrational. (1/2)

Solution 8.
Total no. of cards = 60-12 = 48
Numbers are 13,14,15,16,……….,60.
(i) Numbers divisible by 5 are 15, 20,25, 30, 35, 40, 45, 50, 55, 60. (1/2)
∴ Favourable outcomes = 10
∴ P(no. is divisible by 5) = [latex s=2]\frac { 10 }{ 48 } [/latex] = [latex s=2]\frac { 5 }{ 24 } [/latex] (1/2)
(ii) Perfect square numbers are 16,25,36,49 (1/2)
∴ Ptperfect square) = [latex s=2]\frac { 4 }{ 48 } [/latex] = [latex s=2]\frac { 1 }{ 12 } [/latex] (1/2)

Solution 9.
CBSE Sample Papers for Class 10 Maths Paper 6 img 6

Solution 10.
CBSE Sample Papers for Class 10 Maths Paper 6 img 7

Solution 11.
CBSE Sample Papers for Class 10 Maths Paper 6 img 8

Solution 12.
Number of 50 p coins = 100
Number of ₹ 1 coins = 50
Number of ₹ 2 coins = 20
Number of ₹ 5 coins = 10
Total number of coins = 180 (1/2)
Then (i) P (50 p coin) = [latex s=2]\frac { 100 }{ 180 } [/latex] = [latex s=2]\frac { 5 }{ 9 } [/latex]
(ii) P (not a ₹ 5 coin) = 1 – P (a ₹ 5 coin) (1/2)
= 1- [latex s=2]\frac { 10 }{ 180 } [/latex] = 1- [latex s=2]\frac { 1 }{ 18 } [/latex] = [latex s=2]\frac { 17 }{ 18 } [/latex] (1)

Section-C

Solution 13.
Let a be any positive integer and b = 6
Then by Euclid’s algorithm, a = bq + r, o < r < b (1/2)
Wehavea = 6q+r   …(1)
Since 0 ≤ r < 6, So, r = 0,1,2,3,4,5
∴ From(1), forr = 0, a = 6q
forr = 1, a = 6q + 1
forr = 2, a = 6q + 2
forr = 3, a = 6q + 3
forr = 4, a = 6q + 4
forr = 5, a = 6q + 5   (1/2)
Since 6q is divisible by 2,
∴ 6q is even. (1)
6q + 1 is not divisible by 2. So, 6q + 1 l is odd.
6q + 2 is divisible by 2
∴ 6q + 2 is even. (1)
6q + 3 is not divisible by 2. So, 6q + 3 is odd.
6q + 4 is divisible by 2.
∴ 6q + 4 is even.
6q + 5 is not divisible by 2. So, 6q + 5 is odd. (1/2)
So, we see that 6q, 6q + 2, 6q + 4 are even. Since the number which are not divisible by 2 are odd integer.
∴ 6q +1, 6q + 3, 6q + 5 are odd integer. Hence any positive odd integer is ofthe form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer. (1/2)

Solution 14.
Let (ax + b) be subtracted from given polynomial p(x), so that it is exactly divisible by x2 + x – 12 Then, .
q (x) = x3-6x2 – 15x + 80 – (ax + b)
= x3 -6x2 – (15 + a)x + (80 -b) (1/2)
∵ Dividend = Divisor × quotient + remainder (A)
But remainder will be zero.
∴ Dividend = Divisor x quotient
q (x) = (x2 + x – 12) x quotient
∴q(x) = x3-6x2-(15+a)x + (80 – b) (1/2)
Also, x(x2 +x- 12)-7(x2 +x- 12)
= x3 + x2 – 7x2 – 12x – 7x + 84
= x3 – 6x2 – 19x + 84 (1)
Hence, x3 – 6x2 – 19x + 84 = x3 – 6x2 – (15 + a)x + (80 – b)
– 15 – a = – 19 ⇒ a = + 4 and 80 – b = 84 ⇒ b = – 4 (1)
Hence, ifinp (x) we subtracted 4x – 4 = (ax + b), then it is exactly divisible by x2 + x – 12

Solution 15.
m = a cos θ + b sin θ
(m)2= (a cos θ + b sin θ)2
m2 = a2 cos2 θ + bs2 sin2θ + 2ab cos θ sin θ …(i) (1)
also, n = a sin θ – b cos θ ;
(n)2 = (a sin θ – b cos θ)2
n2 = a2sin2 θ + b2 cos2 θ – 2ab cos θ sin θ …(ii) (1)
On adding equation (i) and (ii), we get
m2 + n2 = a2 cos2 θ + b2 sin2 θ + a2 sin2 θ + b2 cos2 θ
= 2(cos2 θ + sin2 θ) + b2(sin2θ + cos2 θ) (1)
m2 + n2 = a2 + b2
Hence proved.
OR
x = r sin A cos C, y = r sin A sin C and z = r cos A
x2+ y2+ z2 = r2sin2A cos2C + =(r2 sin2A ) (cos2C + sin2C) +r2 cos2A  (1)
= r2 sin2 A(1) + r2 cos2(1/2)
= r2(sin2 A + cos2 A) =r2 Hence Proved. (1/2)

Solution 16.
Draw square PQRS as shown in the figure given below.
Here PQ= QR = PS= RS = 14 cm (1)
Area of the shaded portion = Area of the square PQRS – Area of four equal quadrants (1)
= 14 × 14 – 4 × [latex s=2]\frac { 1 }{ 4 } [/latex]π × (7)2 =196 – [latex s=2]\frac { 22 }{ 7 } [/latex] 7 ×7
= 196-154 =42 cm2
CBSE Sample Papers for Class 10 Maths Paper 6 img 9

Solution 17.
CBSE Sample Papers for Class 10 Maths Paper 6 img 10

Solution 18.
V of cylindrical part = π(8)2 (240) = 64π × 240 (1/2)
V of conical part = [latex s=2]\frac { 1 }{ 3 } [/latex]π × 64 × 36 =64π(12) (1/2)
CBSE Sample Papers for Class 10 Maths Paper 6 img 11
Total volume of iron = 64π(240 + 12) = 64×[latex s=2]\frac { 22 }{ 7 } [/latex] ×252 = 64 × 22× 36 cm3 (1)
∴ Total weight ofpillar = (64) (22) (36) × 7.8 = 395366.4 gms = 395.3664 kg (1)
OR
CBSE Sample Papers for Class 10 Maths Paper 6 img 12

Solution 19.
CBSE Sample Papers for Class 10 Maths Paper 6 img 13

Solution 20.
CBSE Sample Papers for Class 10 Maths Paper 6 img 14
CBSE Sample Papers for Class 10 Maths Paper 6 img 15
OR
CBSE Sample Papers for Class 10 Maths Paper 6 img 16

Solution 21.
∵ Area of the triangle = 24 sq. units (1)
∴[latex s=2]\frac { 1 }{ 2 } [/latex][1(2k + 5)-4(-5 + 1)-k(-1-2k)] = 24
So, 2k2 + 3k- 27 = 0 ⇒ (2k + 9) (k – 3) = 0
After solving we get k = 3, -[latex s=2]\frac { 9 }{ 2 } [/latex] (1)
OR
The point P (x, y) is equidistant form the points A (a + b, b – a) and B (a – b, a + b)
∴ PA= PB
⇒ [latex]\sqrt { { (a+b-x) }^{ 2 }+{ (b-a-y) }^{ 2 } } =\sqrt { { (a-b-x) }^{ 2 }+{ (a+b-y) }^{ 2 } } [/latex] (1/2)
⇒ (a + b – x)2 + (b – a – y)2 = (a – b – x)2 + (a + b – y)2 (1/2)
⇒ (a + b – x)2 – (a – b – x)2 = (a + b – y)2 – (b – a – y)2 (1/2)
⇒ (a + b-x + a-b-x) (a + b-x-a + b + x) = (a + b-y + b-a-y) (a + b-y-b+ a + y) (1/2)
⇒ (2a-2x)(2b) = (2b-2y)(2a)
⇒ (a – x) b = (b – y) a
⇒ ab – bx = ab – ay ⇒ bx = ay (1)

Solution 22.
CBSE Sample Papers for Class 10 Maths Paper 6 img 17

Section-D

Solution 23.
CBSE Sample Papers for Class 10 Maths Paper 6 img 18

Solution 24.
Let ∆ABC ~ ∆PQR and .
area (∆ABC) = area (∆PQR) (Given) (1/2)
CBSE Sample Papers for Class 10 Maths Paper 6 img 19
OR
CBSE Sample Papers for Class 10 Maths Paper 6 img 20
CBSE Sample Papers for Class 10 Maths Paper 6 img 21

Solution 25.
CBSE Sample Papers for Class 10 Maths Paper 6 img 22
OR
CBSE Sample Papers for Class 10 Maths Paper 6 img 23
CBSE Sample Papers for Class 10 Maths Paper 6 img 24

Solution 26.
CBSE Sample Papers for Class 10 Maths Paper 6 img 25

Solution 27.
CBSE Sample Papers for Class 10 Maths Paper 6 img 26

Solution 28.
CBSE Sample Papers for Class 10 Maths Paper 6 img 27
CBSE Sample Papers for Class 10 Maths Paper 6 img 28

Solution 29.
CBSE Sample Papers for Class 10 Maths Paper 6 img 29
Steps of construction
1. Draw a line segment BC = 6.5 cm.
2. Take B as centre and mark an arc of length 5.5 cm above BC.
3. Similarly, take C as centre and cut an arc above BC of length 5 cm which cuts previous arc at point A.
4. Join AB and AC. Therequired triangle is formed i.e., ∆ABC
5. Draw a ray BX below BC.
6. Mark equal arcs B1, B2, B3, B4 and B5 on BX such that BB1 = B1B2 = B2B3 = B3B4 = B4B5.
7. Join B5C and draw a line || to B5C from point B3 which cuts BC at C’.
8. Drawaline||toACfromC’whichcutsAB at A’. (1)
9. Now, ∆A’BC’ is the required triangle.

Solution 30.
CBSE Sample Papers for Class 10 Maths Paper 6 img 30
CBSE Sample Papers for Class 10 Maths Paper 6 img 31
OR
CBSE Sample Papers for Class 10 Maths Paper 6 img 32
CBSE Sample Papers for Class 10 Maths Paper 6 img 33

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