CBSE Sample Papers for Class 10 Maths Paper 6 is part of CBSE Sample Papers for Class 10 Maths . Here we have given CBSE Sample Papers for Class 10 Maths Paper 6. According to new CBSE Exam Pattern, MCQ Questions for Class 10 Maths Carries 20 Marks.

## 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 Papers for Class 10 Maths is given below with free PDF download solutions.

Time: 3 Hours
Maximum Marks: 80

GENERAL INSTRUCTIONS:

• All questions are compulsory.
• This question paper consists of 30 questions divided into four sections A, B, C and D.
• Section A comprises of 6 questions of 1 mark each, Section B comprises of 6 questions of 2 marks each, Section C comprises of 10 questions of 3 marks each and Section D comprises of 8 questions 1 of 4 marks each.
• There is no overall choice. However, internal choice has been provided in one question of 2 marks, 1 three questions of 3 marks each and two questions of 4 marks each. You have to attempt only one of the alternatives in all such questions.
• In question of construction, drawings shall be neat and exactly as per the given measurements.
• Use of calculators is not permitted. However, you may ask for mathematical tables.

SECTION A

Question numbers 1 to 6 carry 1 mark each.

Question 1.
If $$\frac { p }{ q }$$ is a rational number (q ≠ 0), what is the condition on q so that the decimal representation of $$\frac { p }{ q }$$ is terminating?

Question 2.
Find the value of k so that the following system has no solution:
3x – y – 5 = 0; 6x – 2y – k = 0.

Question 3.
Find the 10th term of the AP √2, √8, √18 , … .

Question 4.
In the adjoining figure, if ∠ATO = 40°, find ∠AOB.

Question 5.
If cos A = $$\frac { 3 }{ 5 }$$, find 9 cot² A – 1.

Question 6.
In the adjoining figure, ∠M = ∠N = 46°. Express x in terms of a, b and c, where a, b and c are lengths of LM, MN and NK respectively.

SECTION B

Question numbers 7 to 12 carry 2 marks each.

Question 7.
For any natural number n check whether 6n end with digit 0.

Question 8.
Simplify:

Question 9.
If A, B and C are the interior angles of a triangle ABC, show that:

Question 10.
The angles of a cyclic quadrilateral ABCD are ∠A = (6x + 30)°, ∠B = (5x)°, ∠C = (x + 10)° and ∠D = (3y – 10)°. Find x and y.

Question 11.
Find the middle term(s) of the following AP:
213, 205, 197, …, 37.

Question 12.
The centre of a circle is (2a, a – 7). Find the values of a if the circle passes through the point (11, -9) and has diameter 10 √2 units.

SECTION C

Question numbers 13 to 22 carry 3 marks each.

Question 13.
If the roots of the equation (a – b) x² + (b – c) x + (c – a) = 0 are equal, prove that 2a = b + c.

Question 14.
The sum of first 7 terms of an AP is 63 and sum of its next 7 terms is 161. Find 28th term of AP.
OR
Find the sum of all multiples of 8 lying between 201 and 950.

Question 15.
Show that the square of any positive integer cannot be of the form 5m + 2 or 5m + 3 for some integer m.

Question 16.
In the adjoining figure, ∆ABC is right angled at C and DE ⊥ AB. Prove that ∆ABC ~ ∆ADE, and hence find the lengths of AE and DE.

Question 17.
In a quadrilateral ABCD, ∠A + ∠D = 90°. Prove that AC² + BD² = AD² + BC².

Question 18.
The average score of boys in the examination of a school is 71 and that of the girls is 73. The average score of the boys and girls in the examination is 71.8. Find the ratio of number of boys to the number of girls who appeared in the examination.
OR
Given below is the distribution of weekly pocket money received by students of a class. Calculate the pocket money that is received by most of the students.

Question 19.
If sec θ+ tan θ = p, prove that sin θ = $$\frac { { p }^{ 2 }-1 }{ { p }^{ 2 }+1 }$$
OR
Evaluate the following:
sec 41° sin 49° + cos 49° cosec 41° – $$\frac { 2 }{ \sqrt { 3 } }$$ tan 20° . tan 60° . tan 70° – 3(cos² 45° – sin² 90°)

Question 20.
In the adjoining figure, O is the centre of a circle such that diameter AB = 13 cm and AC = 12 cm. BC is joined. Find the area of the shaded region. (Take π = 3.14)

Question 21.
In the adjoining figure, find the area of the shaded region, . enclosed between two concentric circles of radii 7 cm and 14 cm, where ∠AOC = 40°.

OR
In the adjoining figure, APB and AQO are semicircles and OA = OB. If the perimeter of the figure is 40 cm, find the area of the shaded region.

Question 22.
Construct a triangle with sides 5 cm, 5.5 cm and 6.5 cm. Now construct another triangle whose sides are $$\frac { 3 }{ 5 }$$ times the corresponding sides of the given triangle. 5

SECTION D

Question numbers 23 to 30 carry 4 marks each.

Question 23.
In the adjoining figure, the vertices of ∆ABC are A(4, 6), B(1, 5) and C(7, 2). A line segment DE is drawn to intersect the sides AB and AC at D and E respectively such that $$\frac { AD }{ AB } =\frac { AE }{ AC } =\frac { 1 }{ 3 }$$. Calculate the area of ∆ADE and compare it with area of ∆ABC.

OR
Find the point on the x-axis which is equidistant from the points (5, 4) and (-2, 3). Also find the area of the triangle formed by these points.

Question 24.
If α and β are the zeroes of the polynomial p(x) = 2x² + 5x + k satisfying the relation α² + β² + αβ = $$\frac { 21 }{ 4 }$$, then find the value of k. Also find the zeroes of the polynomial p(x).
OR
If the remainder on division of x3 + 2x2 + kx + 3 by x – 3 is 21, find the quotient and the value of k. Hence, find the zeroes of the cubic polynomial x3 + 2x2 + kx – 18.

Question 25.
A number x is selected from the numbers 1, 2, 3 and 4. Another number y is selected at random from the numbers 1, 4, 9 and 16. Find the probability that product of x and y is
(i) less than 16
(ii) not less than 16.

Question 26.
Due to heavy floods in a State, thousands were rendered homeless. 50 schools collectively offered to the State Government to provide place and canvas for 1500 tents to be fixed by the Government and decided to share the whole expenditure equally. The lower part of each tent is cylindrical of base radius 2.8 m and height 3.5 m, with conical upper part of same base radius but of height 2.1 m. If the canvas used to make the tents costs Rs 120 per sq. m, find the amount shared by each school to set up the tents.
What value is generated by the above problem?

Question 27.
The median of the following data is 525. Find the values of x and y, if the total frequency is 100.

Question 28.
A rectangular park is to be designed whose breadth is 3 m less than its length. Its area is 4 square metres more than the area of a park that has already been made in the shape of an isosceles triangle with its base as the breadth of the rectangular park and of altitude 12 m (shown in the adjoining figure). Find the dimensions of the rectangular park.

Question 29.
In the adjoining figure, two equal circles with centres O and O’, touch each other at X. OO’ produced meets the circle with O’ at A. AC is tangent to the circle with centre O, at the point C. O’D is perpendicular to AC. Find the value of $$\frac { DO’ }{ CO }$$

OR
Prove that the tangent drawn at the mid-point of an arc of a circle is parallel to the chord joining the end points of the arc.

Question 30.
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 √3 = 1.732).

q is of the form 2m x 5n, where m, n are non-negative integers, where p, q are coprime.

Given: 3x – y – 5 = 0
6x – 2y – k – 0
For no solution, we have

Given AP is √2, √8, √18 ,……..
i.e. √2, 2√2, 3√2,……
∴ a = √2, d = √2
T10 = a + (10 – 1 )d
= √2 + 9√2 = 10√2 = √200
Hence, 10th term of given AP is √200 .

Given ∠ATO = 40°
and ∠OAT = 90° (tangent is perpendicular to the radius)
∴ ∠AOT = 180° – (90° + 40°)
= 50°
∠AOB = 2∠AOT = 2 x 50° = 100°

∵ cos A = $$\frac { 3 }{ 5 }$$

In ∆KPN and ∆KLM,
∠N = ∠M = 46° (given)
∠K = ∠K (common)
∴ ∆KPN ~ ∆KLM

6n can end with digit 0 only if 6n is divisible by 2 and 5 both. .
But prime factors of 6n = 2n x 3n, so 6n is not divisible by 5.
∴ By Fundamental Theorem of Arithmetic, there is no natural number n for which 6n ends with digit zero.
Hence, 6n does not end with digit zero

As A, B and C are the interior angles of ∆ABC, A + B + C = 180

In cyclic quadrilateral ABCD,
∠A + ∠C = 180°
=> (6x + 30)° + (x + 10)° = 180°
=> 7x = 140 => x = 20 …(i)
and ∠B + ∠D = 180°
=> (5x)° + (3y – 10)° = 180°
=> 5x + 3y = 190
=> 5 x 20 + 3y = 190 => 3y = 90 => y = 30 (using (i))
x = 20, y = 30.

Given AP is 213, 205, 197, … 37.
Here, a = 213, d = -8, l = 37
∵ l = a + (n – 1 )d
=> 37 = 213 + (n – 1) (-8) => n – 1 = $$\frac { -176 }{ -8 }$$ => n – 1 = 22
=> n = 23
Total number of terms = 23

Radius of circle = $$\frac { 1 }{ 2 }$$ x 10√2 = 5√2 .
The centre of the circle is C(2a, a – 7) and it passes through the point P(11, -9), so CP = radius of circle

The given equation is (a – b)x² + (b – c) x + (c – a) = 0.
Comparing it with Ax² + Bx + C = 0, we get
A = a – b, B = b – c, C = c – a.
Discriminant = B² – 4AC = (b – c)² – 4(a – b) (c – a).
For equal roots, discriminant = 0
=> (b – c)² – 4(a – b) (c – a) = 0
=> b² – 2bc + c² – 4(ca – a² – bc + ab) = 0
=> 4a² + b² + c² – 4ab – 4ca + 2bc = 0
=> (2a – b – c)² = 0
=>2a – b – c = 0
=> 2a = b + c.

Let the first term and common difference of AP be a and d respectively.
Given, S7 = 63 => $$\frac { 7 }{ 2 }$$ [2a + (7 – 1 )d] = 63
=> 2a + 6d = 18 …(i)
and sum of next 7 terms = 161
i.e. S14 – S7 = 161
=> S14 = 161 + 63 => S14 = 224
=> $$\frac { 14 }{ 2 }$$ [2a + (14 – 1 )d] = 224
=> 2 a + 13d = 32 … (ii)
Subtracting (i) from (ii), we get
2a + 13d – 2a – 6d = 32 – 18 => 7d = 14 => d = 2.
Substituting d = 2 in equation (i), we get
2A + 6 x 2 = 18 => 2a – 6 => a = 3.
Now, T28 = 3 + (28 – 1) x 2 = 3 + 54 = 57.
Hence, 28th term of AP is 57.
OR
The multiples of 8 lying between 201 and 950 are 208, 216, 224, …, 944.
These numbers form are AP with a = 208, d = 8 and l = 944.
Let the number of these numbers be n, then
944 = 208 + (n – 1) x 8 => 736 = 8 (n – 1)
=> 92 = n – 1 => n = 93.
Sum of these integers = $$\frac { 93 }{ 2 }$$ (208 + 944) = $$\frac { 93 }{ 2 }$$ x 1152
= 93 x 576 = 53568.

Let n be any positive integer. Applying Euclid’s division lemma with divisor = 5, we get
n = 5q, 5q + 1, 5q + 2, 5q + 3 or 5q + 4, where q is some whole number.
Now (5q)² – 25q² = 5 m, where m = 5q², which is an integer;
(5q + 1)² = 25q² + 10q + 1 = 5(5q² + 2q) + 1 = 5m + 1, where
m = 5q² + 2q, which is an integer;
(5q + 2)² = 25q² + 20q + 4 = 5(5q² + 4q) + 4 = 5m + 4, where
m = 5q² + 4q, which is an integer;
(5q + 3)² = 25q² + 30q + 9 = 5(5q² + 6q + 1) + 4 = 5m + 4, where
m = 5q² + 6q + 1, which is an integer;
(5q + 4)² = 25q² + 40q + 16 = 5(5q² + 8q + 3) + 1 = 5m + 1, where
m = 5q² + 8q + 3, which is an integer.
Thus, the square of any positive integer is of the form 5m, 5m + 1 or 5m + 4 for some integer m.
It follows that the square of any positive integer cannot be of the form 5m + 2 or 5m + 3 for some integer m.

In ∆ABC and ∆ADE
∠BAC = ∠EAD (same angle)
∠ACB = ∠AED (each 90°)
∴ ∆ABC ~ ∆ADE (AA Similarity criterion)

Given, in quadrilateral ABCD,
∠A + ∠D = 90° …(i)
Produce AB and DC to meet at the point E.
In ∆AED, ∠A + ∠D + ∠E = 180°
=> 90° + ∠E = 180° (using (i))
=> ∠E = 90°.
In ∆AED, ∠AED = 90°
∴ AD² = AE² + ED² …(ii)
In ∆BEC, ∠BEC = 90°
∴ BC² – BE² + EC² …(iii)
In ∆AEC, ∠AEC = 90°
∴ AC² = AE² + EC² …(iv)
In ∆BED, ∠BED = 90°
∴ BD² = BE² + ED² …(v)
On adding (iv) and (v), we get
AC² + BD² = (AE² + ED²) + (BE² + EC²)
=> AC² + BD² = AD² + BC² (using (ii) and (iii))

Let the number of boys who appeared in the examination be x and that of the girls be y, so the total number of students who appeared in the examination = x + y.
Since the average score is boys in the examination is 71 and that of the girls is 73, therefore, the sum of scores of boys = 71x and the sum of scores of girls = 73y.
∴ The total sum of scores of boys and girls = 71x + 73y.
The average score of the boys and girls

Hence, the pocket money received by most of the students is Rs 86.32

Given sec θ + tan θ = p

In ∆ABC, ∠ACB = 90°. By Pythagoras theorem,
AB² = AC² + BC²
=> 13² = 12² + BC² (given AB = 13 cm and AC = 12 cm)
=> BC² = 25 => BC = 5 cm

Let R and r be the radii of the outer and inner concentric circles then R = 14 cm and r = 7 cm.
The central angle made by the major sector of the outer and inner circles θ = 360° – 40° = 320°.
∴ Area of shaded region = area of major sector OAC – area of major sector OBD

Steps of construction:
1. Draw AB = 6.5 cm.
2. With A as centre and radius 5.5 cm draw an arc.
3. With B as centre and radius 5 cm draw an arc to meet the previous arc at C.
4. Join AC and BC, then ABC is a triangle with AB = 6.5 cm, AC = 5.5 cm and BC = 5 cm.
5. Draw any ray AX making an acute angle with AB on the side opposite to the vertex C.
6. Locate 5 points A1, A2, A3, A4 and A5 on AX such that AA1 = A1 A2 = A2A3 = A3A4 = A4A5.
7. Join A5B. Through A3 draw a line parallel to A5B to intersect AB at B’.
8. Through B’ draw a line parallel to BC to intersect the AC at C’. Then AB’C’ is the required triangle.

Given the vertices of ∆ABC are A(4, 6), B(1, 5) and C(7, 2).

Given α and β are the zeroes of the polynomial p(x) = 2x² + 5x + k.
∴ Sum of the zeroes = α + β = $$\frac { -5 }{ 2 }$$ …(i)
and product of the zeroes = αβ = $$\frac { k }{ 2 }$$ …(ii)

For the product of outcomes, we construct a table as under:

Total number of outcomes are 4 x 4 i.e. 16 which are all equally likely.
(i) Table shows that the outcomes favourable to the event ‘product xy less than 16’ are (1, 1), (1, 4), (1, 9), (2, 1), (2, 4), (3, 1), (3, 4) and (4, 1) i.e. products as 1, 2, 3, 4, 4, 8, 9, 12 which are 8 in numbers.
∴ Required probability = $$\frac { 8 }{ 16 }$$ = $$\frac { 1 }{ 2 }$$
(ii) P(Product of x and y is not less than 16)
= 1 – P(Product of x and y is less than 16).
= $$1-\frac { 1 }{ 2 }$$ = $$\frac { 1 }{ 2 }$$.

Let r and h be the radius and height of the cylindrical
part and l be the slant height of conical part.
Given r = 2.8 m, l = 3.5 m, height of cone = 2.1 m

Construct the cumulative frequency distribution table as under:

It is given n = 100, so 76 + x + y = 100 => x + y = 24 …(i)
As the median is 525, which lies in the class 500 – 600, so the median class is 500 – 600.
∴I = 500, f = 20, c.f. = 36 + x and h = 100

=> 5(14 – x) = 25 => 14 – x = 5 =>x = 9
∴ From (i), 9 + y = 24 => y = 15.
Hence, x = 9 and y = 15.

Let the breadth of the rectangular park be x metres, then its length = (x + 3) metres.
∴ Area of rectangular park = (x + 3) x m²
Base of isosceles triangle = x metres and its height = 12 m.
∴ Area of isosceles triangle = $$\left( \frac { 1 }{ 2 } \times x\times 12 \right)$$ m² = 6x m².
According to given, (x + 3) x = 6x + 4
=> x² + 3x = 6x + 4 => x² – 3x – 4 = 0
=> x² – 4x + x – 4 = 0 => x(x – 4) + 1 (x – 4) = 0
=> (x – 4) (x + 1) = 0 =>x – 4 = 0 or x + 1 = 0
=> x = 4 or x = -1.
Since x is the breadth of the rectangular park, it cannot be negative.
Hence, the breadth of the park is 4 metres and its length = (x + 3) metres
= (4 + 3) metres = 7 metres.