CBSE Sample Papers for Class 10 Maths Paper 5 is part of CBSE Sample Papers for Class 10 Maths . Here we have given CBSE Sample Papers for Class 10 Maths Paper 5. According to new CBSE Exam Pattern, MCQ Questions for Class 10 Maths Carries 20 Marks.

## CBSE Sample Papers for Class 10 Maths Paper 5

 Board CBSE Class X Subject Maths Sample Paper Set Paper 5 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 5 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.
Which term of the sequence 114, 109, 104,…… is the first negative term?

Question 2.
Find the value of k for which the following are the consecutive terms of an AP: k, 2k – 1, 2k + 1.

Question 3.
Find the distance between the points [$$\frac { -8 }{ 5 }$$,2] and [$$\frac { 2 }{ 5 }$$,2]

Question 4.
Which measure of central tendency is given by the x-coordinate of the point the of intersection of “more than ogive” and “less than ogive”?

Question 5.
What is the distance between two parallel tangents of a circle of radius 4 cm?

Question 6.
In the adjoining figure, O is the centre of a circle. The area of sector OAPB is $$\frac { 5 }{ 18 }$$ of the area of the circle. Find x.

SECTION B

Question numbers 7 to 12 carry 2 marks each.

Question 7.
Using Euclid’s algorithm, find the HCF of 1656 and 4025.

Question 8.
Find the two numbers whose sum is 75 and difference is 15.

Question 9.
If m and n are the zeroes of the polynomial 3x² + 11x – 4, find the value of $$\frac { m }{ n } +\frac { n }{ m }$$ .

Question 10.
If the distances of P(x, y) from the points A(3, 6) and B(-3, 4) are equal, prove that 3x + y = 5.

Question 11.

Question 12.
Three metallic solid cubes whose edges are 3 cm, 4 cm and 5 cm are melted and formed into a jingle cube. Find the edge of the cube so formed

SECTION C

Question numbers 13 to 22 carry 3 marks each.

Question 13.
In the adjoining figure, $$\frac { XP }{ PY } =\frac { QX }{ QZ }$$ = 3. If the area of ∆XYZ is 32 cm², then find the area of the quadrilateral PYZQ.

Question 14.
In an equilateral triangle ABC, a point D is taken on base BC such that BD : DC = 2:1. Prove that 9 AD² = 7AB².

Question 15.
Find the values of a and b so that 8x4 + 14x3 – 2x2 + ax + b is exactly divisible by 4x2 + 3x – 2.
OR
If one zero of the polynomial 3x2 – 8x – (2k + 1) is seven times the other, find both zeroes of the polynomial and the value of k.

Question 16.
Solve for x and y:
(a – b)x + (a + b)y = a² – 2ab – b²
(a + b) (x + y) = a² + b².

Question 17.
Find the area of the triangle formed by joining the mid points of the sides of the triangle whose vertices are A(2, 1), B (4, 3) and C(2, 5).
OR
Show that the points (1, 7), (4, 2), (-1, -1) and (-4, 4) are the vertices of a square.

Question 18.
There are 100 cards in a box on which numbers from 1 to 100 are written. A card is taken out from the box at random. Find the probability that the number on the selected card is
(i) divisible by 9 and is a perfect square
(ii) a prime number greater than 80.

Question 19.
In a single throw of a pair of different dice, what is the probability of getting
(i) a prime number of each dice?
(ii) a total of 9 or 11?

Question 20.
A bucket is in the form of a frustum of a cone and holds 28.490 litres of water. The radii of the top and bottom are 28 cm and 21 cm respectively. Find the height of the bucket.
OR
A solid right-circular cone of height 60 cm and radius 30 cm is dropped in a right-circular cylinder full of water of height 180 cm and radius 60 cm. Find the volume of water left in the cylinder in cubic metre.

Question 21.
Prove the following: sin6 θ + cos6 θ + 3 sin2 θ cos2 θ = 1.
OR
Prove that

Question 22.
If cos θ + sin θ = √2 cos θ, show that cos θ – sin θ = √2 .

SECTION D

Question numbers 23 to 30 carry 4 marks each.

Question 23.
If the roots of the quadratic equation x² + 2px + mn = 0 are real and equal, show that the roots of the quadratic equation x² – 2(m + n) x + (m² + n² + 2p²) = 0 are also real and equal.

Question 24.
The sum of three numbers in AP is 12 and sum of their cubes is 288. Find the numbers.
OR
The sums of first n terms of three arithmetic progressions are S1, S2 and S3 respectively. The first term of each AP is 1 and their common differences are 1, 2 and 3 respectively. Prove that S1 + S3 = 2S2.

Question 25.
Dudhnath has two vessels containing 720 mL and 405 mL of milk. Milk from these containers is poured into glasses of equal capacity to their brim. Find the minimum number of glasses that can be filled.
OR
Show that the cube of any positive integer is of the form 4m, 4m + 1 or 4m + 3 for some integer m.

Question 26.
The following table gives production yield per hectare of wheat of 100 farms of a village.

Change the distribution to more than type, and draw its ogive. Using the ogive, find the median of the given data.
What is the value of proper knowledge of farming?

Question 27.
In the adjoining figure, AB is a chord of length 16 cm of a circle with centre O and of radius 10 cm. The tangents at A and B intersect at the point P. Find the length of PA.

OR
Prove that the opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

Question 28.
The angle of elevation of a cloud from a point h metres above the surface of a lake is θ and the angle of depression of its reflection in the lake is φ. Prove that the height of the cloud above the

Question 29.
Draw a circle of radius 3.5 cm. From a point P Outside the circle at a distance of 6 cm from the centre of the circle, draw two tangents to the circle. Also measure their lengths.

Question 30.
In the adjoining figure, ABCD is a trapezium with AB || DC, AB = 18 cm, DC = 32 cm and distance between AB and DC = 14 cm. If the arcs of equal radii 7 cm with centres A, B, C and D have drawn, then find the area of the shaded region.

The given sequence is an AP with first term a = 114 and common difference d = 5.
∴ an = a + (n – 1)d < 0
⇒114 + (n – 1) (-5) < 0
⇒ -5n + 119 < 0 ⇒ 5n > 119

∴ n = 24
Hence, 24th term is 1st negative term.

Given k, 2k – 1, 2k + 1 are in AP
⇒ 2(2k – 1) = k + 2k + 1
⇒ 4k – 2 = 3k + 1
⇒ 4k – 3k = 2 + 1
⇒ k = 3.

Median

Distance between parallel tangents = (4 + 4) cm = 8 cm.

Area of sector OAPB = $$\frac { x }{ 360 } \times \pi { r }^{ 2 }$$, where r is the radius of the circle.

Hence x = 100

By Euclid’s division algorithm, we have
4025 = 1656 × 2 + 713; 1656 = 713 × 2 + 230
713 = 230 × 3 + 23; 230 = 23 × 10 + 0
Thus, last non-zero remainder = 23
Hence, HCF of 1656 and 4025 is 23.

Let the two numbers be x and y (x > y), then according to given
x + y = 75 …(i)
and x – y = 15 …(ii)
Adding (i) and (ii), we get
2x = 90 => x = 45
Putting x = 45 in (i), we get
45 + y = 75 => y = 30
Hence, two numbers are 45 and 30.

Given, m and n are zero’s of the polynomial 3x² + 11x – 4.

Given PA = PB => (PA)² = (PB)²
=> (x – 3)² + (y – 6)² = (x + 3)² + (y – 4)²
=> x² – 6x + 9 + y² – 12y + 36 = x² + 6x + 9 + y² – 8y + 16
=> 6x + 6x – 8y + 12y = 9 + 36 – 9 – 16
=> 12x + 4y = 20
=> 3x + y = 5

Let the edge of the cube formed by x cm, then
volume of the cube formed = sum of volumes of three cubes of edges 3 cm, 4 cm and 5 cm
=> x3 = 33 + 43 + 53
=> x3 = 27 + 64 + 125 = 216 = 63
=> x = 6
Hence, the edge of the cube so formed is 6 cm.

In ∆XYZ, $$\frac { XP }{ PY } =\frac { QX }{ QZ }$$ (given)
∴PQ || YZ (converse of Thales theorem)
⇒ ∠XPQ = ∠XYZ and ∠XQP = ∠XZY (corres. ∠s)
∆XPQ ~ ∆XYZ.

As ∆ABC is equilateral, BC = AB.
Given BD : DC = 2 : 1 => BD = $$\frac { 2 }{ 3 }$$BC = $$\frac { 2 }{ 3 }$$AB.
Draw AE ⊥ BC, then E is mid-point of BC,
so BE = $$\frac { 1 }{ 2 }$$ BC = $$\frac { 1 }{ 2 }$$ AB.
From fig., ED = BD – BE = $$\frac { 2 }{ 3 }$$AB – $$\frac { 1 }{ 2 }$$AB = $$\frac { 1 }{ 6 }$$AB.
In ∆ABE, ∠AEB = 90°
∴ AB² = AE² + BE² …(i)
In ∆AED, ∠AED = 90°
∴ AD² = AE² + ED² …(ii)
Subtracting (ii) from (i), we get

Dividing 8x4 + 14x3 – 2x2 + ax + b by 4x² + 3x – 2, we get

The given equations are:
(a – b)x + (a + b)y = a² – 2ab – b²
(a + b) (x + y) = a² + b²
Equation (ii) can be written as
(a + b)x + (a + b)y = a² + b²
Subtracting equation (i) from equation (iii), we get
2bx = 2ab + 2b² => x = a + b.
Substituting x = a + b in equation (iii), we get
(a + b) (a + b) + (a + b)y = a² + b²
=> a² + b² + 2ab + (a + b)y = a² + b²

Let D, E and F be the mid-points of the sides BC,
CA and AB respectively

Total number of cards in the bag = 100
By saying that a card is selected at random means all cards are equally likely to be selected.
So, the sample space of the experiment has 100 equally likely outcomes.
(i) Numbers divisible by 3 and perfect square are 9, 36, 81.
∴ The number of outcomes favourable to the event ‘divisible by 3 and is a perfect square’ is 3.
Required probability = $$\frac { 3 }{ 100 }$$.
(ii) Prime numbers greater than 80 but less than or equal to 100 are 83, 89, 97.
∴ The number of outcomes favourable to the event ‘prime number less than 80’ is 3.
∴Required probability = $$\frac { 3 }{ 100 }$$

When two different dice are thrown, the total number of outcomes is 36 and all the outcomes are equally likely.
(i) The outcomes favourable to the event ‘a prime number on each dice’ are (2, 2), (2, 3), (2, 5), (3, 2), (3, 3), (3, 5), (5, 2), (5, 3), (5, 5).
These are 9 in number.
∴ P (a prime number on each die) = $$\frac { 9 }{ 36 }$$ = $$\frac { 1 }{ 4 }$$.
(ii) The outcomes favourable to the event ‘a total of 9 or 11’ are (3, 6), (6, 3), (4, 5), (5, 4), (5, 6), (6, 5). These are 6 in number.
∴ P (a total of 9 or 11) = $$\frac { 6 }{ 36 }$$ = $$\frac { 1 }{ 6 }$$.

Let h cm be height of the bucket.
Here, radius of top = R = 28 cm,
radius of bottom = r = 21 cm.
Volume of bucket = 28.490 litres = (28.490 x 1000) cm3
= 28490 cm3

LHS = sin6 θ + cos6 θ + 3 sin2 θ cos2 θ
= (sin2 θ)3 + (cos2 θ)3 + 3 sin2 θ cos2 θ
= (sin2 θ + cos2 θ)3 – 3 sin2 θ cos2 θ (sin2 θ + cos2 θ) + 3 sin2 θ cos2 θ
(∵ a3 + b3 = (a + b)3 – 3ab (a + b))
= (1)3 – 3 sin2 θ cos2 θ (1) + 3 sin2 θ cos2 θ
= 1 – 3 sin2 θ cos2 θ + 3 sin2 θ cos2 θ = 1 = RHS
OR

Given cos θ + sin θ = √2cos θ ⇒sin θ = (√2 – 1) cos θ

Given that the roots of the quadratic equation x² + 2px + mn = 0 are real and equal
=> its discriminant = 0
=> (2p)² – 4 x 1 x mn = 0 => p² = mn …(i)
Now, the roots of the quadratic equation x² – 2(m + n) x + (m² + n² + 2p²) = 0 are real and equal if its discriminant = 0
i.e. if (-2 (m + n))² – 4 x 1 x (m² + n² + 2p²) = 0
i.e. if 4 (m² + n² + 2mn) – 4 (m² + n² + 2p²) = 0
i.e. if 8mn – 8p² = 0 i.e. if p² = mn, which is true from (i).

Let the three numbers are in AP be a – d, a, a + d, then according to given,
a – d + a + a + d = 12
3a = 12 => a = 4 …(i)
and (a – d)3 + a3 + (a + d)3 – 288
=> a3 – 3a²d + 3ad² – d3 + a3 + a3 + 3a²d + 3ad² + d3 = 288
=> 3a3 + 6ad² = 288
=> 3(4)3 + 6 x 4 x d² = 288 [using (i)]
=> 192 + 24d² = 288
=> 24d² = 96 => d² = 4
d = ±2
When a = 4, d = 2, the numbers are 4 – 2, 4, 4 + 2 i.e. 2, 4, 6
and when a = 4, d = -2, the numbers are 4 – (-2), 4, 4 + (-2) i.e. 6, 4, 2
Hence, the three numbers are 2, 4, 6.
OR

Two vessels contain 720 mL and 405 mL of milk respectively. Since we need the minimum number of glasses of equal capacity, so the capacity of each glass should be maximum. Therefore, we have to find the HCF of 720 mL and 405 mL.
720 = 2 x 2 x 2 x 2 x 3 x 3 x 5 = 24 x 32 x 51 and
405 = 3 x 3 x 3 x 3 x 5 = 34 x 51
∴ HCF (720, 405) = 32 x 51 = 9 x 5 = 45.
∴Maximum capacity of each glass = 45 mL.

Construct the table for cumulative frequency distribution of more than type as under:

Take 1 cm along x-axis 5 kg of production/ha and 1 cm along y-axis = 10 farms.
Plot the points (50, 100), (55, 98), (60, 90), (65, 78), (70, 54), (75, 16) and (80, 0). Join these points by a free hand drawing. The required ogive (more than type) is drawn on the graph sheet given below. Since the scale on x-axis starts at 50, a kink (break) is shown on the x-axis near the origin.

To find median:
Let A be the point on y-axis representing frequency = $$\frac { n }{ 2 }$$ = $$\frac { 100 }{ 2 }$$ = 50.
Through A, draw a horizontal line to meet the ogive at P. Through P, draw a vertical line to meet
the x-axis at M. The abscissa of the point M represents 70.5 kg.
Hence, the median = 70.5 kg of wheat/ha.
The values of proper knowledge of farming are:
Arrangement of irrigation, testing of soil, rotation of crops, quality and quantity of manure and
pesticides will help in increasing the yield per hectare.

Given, AB = 16 cm, so AL = BL = 8 cm
(∵ OP is the perpendicular bisector of AB)
In ∆OLB, ∠OLB = 90° (∵ OP ⊥ AB)
By Pythagoras theorem,
OL² = OB² – BL² = 10² – 8² = 100 – 64 = 36
=> OL = 6 cm.
Let LP = x cm and BP = y cm, then OP = OL + LP = (6 + x) cm.
Since tangent is perpendicular to radius, OB ⊥ PB.
In ∆OPB, ∠OBP = 90°. By Pythagoras theorem,
OP² = BP² + OB²
=> (x + 6)² = y² + 10²
=> x² + 12x + 36 = y² + 100

Let O be the point of observation h metres above the lake and let the cloud be at the point P. If Q is the reflection of the cloud in the lake then BQ = BP.
Let the height of the cloud above the lake be x metres. From O, draw OA ⊥ PQ. Then, AB = OM = h metres.
AP = BP – BA = (x – h) metres and
AQ = AB + BQ = (x + h) metres.
Let OA = d metres.
From right angled ∆OAP, we get

Steps of construction:
1. Mark a point O. With O as centre and radius 3.5 cm, draw a circle.
2. Take a point P at a distance of 6 cm from O. P lies outside the circle.
3. Join OP and draw its perpendicular bisector to meet OP at M.
4. Taking M as centre and OM (or MP) as radius, draw a circle. Let this circle intersect the previous circle at points A and B.
5. Join PA and PB. Then PA and PB are the required tangents. On measuring, we find that AP = BP = 4.9 cm (approximately).

Justification:
Join OA, then ∠OAP = 90° (angle in a semicircle = 90°)
As OA is radius and ∠OAP = 90°, so PA has to be tangent to the circle.