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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

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 [,2] and [,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 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 .

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, = 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.

CBSE Sample Papers for Class 10 Maths Paper 4 is part of CBSE Sample Papers for Class 10 Maths . Here we have given CBSE Sample Papers for Class 10 Maths Paper 4

CBSE Sample Papers for Class 10 Maths Paper 4

Board	CBSE
Class	X
Subject	Maths
Sample Paper Set	Paper 4
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 4 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.
The decimal expansion of the rational number will terminate after how many places of decimal?

Question 2.
Is x = -2 a solution of the equation x² – 2x + 8 = 0?

Question 3.
In the adjoining figure, P and Q are points on the sides of AB and AC respectively of ∆ABC such that AP = 3.5 cm, PB = 7 cm, QC = 6 cm. Prove that PQ || BC.a

Question 4.
If tan θ = cot (30° + θ), find the value of θ.

Question 5.
A cylinder and a cone are of same base radius and c the ratio of volume of cylinder to that of the cone.

Question 6.
Two coins are tossed simultaneously. Find the probability of getting exactly one head.

SECTION B

Question numbers 7 to 12 carry 2 marks each.

Question 7.
Complete the following factor tree and find the numbers x, y and z.

Question 8.
If α and β are zeroes of the polynomial f(x) = x² – x – k such that α – β = 9, find k.

Question 9.
The coordinates of the vertices of ∆ABC are A(4, 1), B(-3, 2) and C(0, k). Given that the area of ∆ABC is 12 unit², find the value of k. .

Question 10.
Without using trigonometric tables, evaluate the following:

Question 11.
Prove that

Question 12.
Area of a sector of a circle of radius 36 cm is 54π cm². Find the length of the corresponding arc of the sector.

SECTION C

Question numbers 13 to 22 carry 3 marks each.

Question 13.
The HCF of 65 and 117 is expressible in the form 65m – 117. Find the value of m. Also find the LCM of 65 and 117 using prime factorization

Question 14.
Solve the following pair of linear equations: px + qy = p – q;qx – py = p + q.
OR
In the given figure, ABCDE is a pentagon with BE || CD and BC || ED. BC is perpendicular to CD. If the perimeter of pentagon ABCDE is 21 cm, find the values of x and y.

Question 15.
Find the roots of the following equation

Question 16.
A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows: Rs 200 for the first day, Rs 250 for the second day, Rs 300 for the third day, etc., the penalty for each succeeding day being Rs 50 more than for the preceding day. How much money the contractor has to pay as penalty, if he has delayed the work for 30 days?
OR
Jaspal Singh repays his total loan of Rs 118000 by paying every month starting with the first instalment of Rs 1000. If he increases the instalment by Rs 100 every month, what amount will be paid by him in the 30th instalment? What amount of loan does he still to pay after the 30th instalment?

Question 17.
Prove that : sin θ(1 + tan θ) + cos θ(1 + cot θ) = sec θ + cosec θ.

Question 18.
D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C. Prove that AE² + BD² = AB² + DE².
OR
In the adjoining figure, DB ⊥ BC, DE ⊥ AB
and AC ⊥ BC. Prove that

Question 19.
In the adjoining figure, a triangle ABC is drawn to circumscribe a circle of radius 2 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 4 cm and 3 cm respectively. If area of ∆ABC = 21 cm², then find the lengths of sides AB and AC.

Question 20.
In the adjoining figure, APB and CQD are semicircles of diameter 7 cm each, while ARC and BSD are semicircles of diameter 14 cm each. Find the perimeter of the shaded region.

Question 21.
Two different dice are rolled together. Find the probability of getting:
(i) the sum of numbers on two dice as 5.
(ii) even numbers on both dice.
OR
The probability of selecting a red ball at random from a jar that contains only red, blue and orange balls is . The probability of selecting a blue ball at random from the same jar is . If the jar contains 10 orange balls, find the total number of balls in the jar.

Question 22.
Find the value of p for the following distribution whose mean is 10:

SECTION D

Question numbers 23 to 30 carry 4 marks each.

Question 23.
If pth, qth and rth terms of an AP are a, b and c respectively, then show that (a – b) r + (b – c) p + (c – a) q = 0.

Question 24.
An icecream seller sells his icecreams in two ways:
(A) In a cone of r = 5 cm, h – 8 cm with hemispherical top.
(B) In a cup in shape of cylinder with r – 5 cm, h – 8 cm.
He charges the same price both but prefers to sell his icecream in a cone.
(a) Find the volume of the cone and the cup.
(b) Which out of the two has more capacity?
(c) By choosing a cone, which value is not followed by the icecream seller?

Question 25.
Solve the following pair of linear equations graphically:
x + 3y = 6; 2x – 3y = 12
Also find the area of the triangle formed by the lines representing the given equations with y-axis.

Question 26.
If the centroid of ∆ABC, in which A(a, b), B(b, c), C(c, a) is at the origin, then calculate the value of (a³ + b³ + c³).
OR
The three vertices of a parallelogram ABCD are A(3, -4), B(-1, -3) and C(-6, 2). Find the coordinates of vertex D and find the area of parallelogram ABCD.

Question 27.
From the top of a tower 100 m high, a man observes two cars on the opposite sides of the tower with angles of depression 30° and 45° respectively. Find the distance between the cars. (Use √3 = 1.73)
OR
A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car at an angle of depression of 30°, which is approaching the foot of the tower with a uniform speed. Six seconds later, the angle of depression of the car is found to be 60°. Find the time taken by the car to reach the foot of the tower from this point.

Question 28.
Prove that the lengths of tangents drawn from an external point to a circle are equal.
OR
Two circles with centres O and O’ of radii 3 cm and 4 cm, respectively intersect at two points P and Q such that OP and O’P are tangents to the two circles. Find the length of the common chord PQ.

Question 29.
Draw an isosceles triangle ABC in which AB = AC = 6 cm and BC = 5 cm. Construct a triangle PQR similar to ∆ABC in which PQ = 8 cm. Also justify the construction.

Question 30.
Size of agriculture holding in a survey of 200 families is given in the following table:

Compute the median and mode size of holding.

Answers

Answer 1.

Hence, the decimal expansion of given rational number will terminate after 3 places of decimal.

Answer 2.
(-2)² – 2 x (-2) + 8 = 4 + 4 + 8 = 16≠0
∴ x = -2 is not a solution of the equation x² – 2x + 8 = 0.

Answer 3.

∴ By the converse of B.P.T., PQ || BC.

Answer 4.
tan θ = cot (30° + θ)
= tan [90° – (30° + θ)]
= tan (60° – θ)
θ = 60° – θ => 2θ = 60° => θ = 30°.

Answer 5.
Let r be the base radius of cylinder and cone and h be their heights.

Hence, the ratio of volume of cylinder to that of cone = 3:1.

Answer 6.
When two coins are tossed simultaneously, then the outcomes are HH, HT, TH, TT.
So total number of outcomes = 4.
The outcomes favourable to the event ‘getting exactly one head’ are HT and TH.
∴The number of outcomes favourable to the event = 2.
∴ P(exactly one head) =

Answer 7.
x = 3381 x 2 = 6762
y = 161 x 7 = 1127
and z = = 23.

Answer 8.
Given f(x) = x² – x – k

Answer 9.
Area of ∆ABC = | 14(2 – k) – 3(k – 1) + 0(1 – 2) |
=> 12 = |8 – 4k – 3k + 3|
=> 24 = |11 – 7k|
=> 11 – 7k = ±24
=> 11 – 7k = 24 or 11 – 7k = -24
=> 7k = -13 or 7k = 35
=> k = or k = 5
Hence, k = or k = 5

Answer 10.

Answer 11.

Answer 12.
Let the central angle (in degrees) be θ, then

Hence, the length of the corresponding arc of the sector = 3π cm.

Answer 13.
By Euclid’s division algorithm, we have
117 = 65 x 1 + 52; 65 = 52 x 1 + 13; 52 = 13 x 4 + 0
∴ HCF of 65 and 117 = 13
∴ 65m – 117 = 13 => 65m = 130 => m = 2.
Prime factorisation of 65 and 117 are as follows:
65 = 5 x 13 and 117 = 3 x 3 x 13
∴ LCM of 65 and 117 = 3 x 3 x 5 x 13 = 585.

Answer 14.
Given equations are
px + qy = p – q …(i)
qx – py = p + q …(ii)
Multiplying (i) by p and (ii) by q, we get
p²x + pqy = p² – pq …(iii)
and q²x – pqy – pq + q² …(iv)
On adding (iii) and (iv), we have

Putting x = 1 in equation (i), we have
p x 1 + qy = p – q => qy = -q => y = -1
∴ x = 1 ,y = -1.
OR
Given BE || CD and BC || ED, also BC ⊥ CD
=> BCDE is a rectangle
CD = BE
=> x + y = 5
Again, given perimeter of pentagon ABCDE = 21 cm
=> AB + BC + CD + DE + EA = 21 cm
=> 3 + x – y + x + y + BC + 3 = 21cm
=> 2x + x – y = 21 – 6
=> 3x – y = 15
Adding (i) and (ii), we get
4x = 20 => x = => x = 5
Putting x = 5 in (i), we get
5 + y = 5 =>y = 0
Hence, x = 5, y = 0.

Answer 15.
Given

(x + 4) (x – 7) = -30 => x² – 7x + 4x – 28 = -30
=> x² – 3x + 2 = 0 => x² – 2x – x + 2 = 0
=> x(x – 2) – 1(x – 2) = 0 => (x – 2) (x – 1) = 0
=> x – 2 = 0 or x – 1 = 0 => x = 2 or x = 1.
Hence, the roots of the given equation are 2 and 1.

Answer 16.
200, 250, 300, … is in AP.
whose first term = 200 and common difference = 50.
Let the number of days for which work is delayed be n, then

=> n[200 + 25 (n- 1)] = 27750
=> 25n² + 175n = 27750
=> n² + 7n – 1110 = 0
=> n² + 37n – 30n – 1110 = 0
=> n(n + 37) – 30(n + 37) = 0
=> (n + 37) (n – 30) = 0
=> n = -37 or n = 30
But n cannot be negative
∴ n = 30
Hence, the work was delayed by 30 days.
OR
Jaspal Singh starts repaying loan with first instalment of Rs 1000 and increases the instalment every month by Rs 100, so the numbers involved in instalments are 1000, 1100, 1200, …
These numbers form an AP with a = 1000 and d = 100.
30th term = a + 29d = 1000 + 29 x 100 = 3900.
Hence, his 30th instalment = Rs 3900
Sum of 30 terms = (100 + 3900)
= 15 x 4900 = 73500.
∴ The total amount of loan repaid = Rs 73500.
∴ Amount of loan still to be paid = Amount of total loan – amount of loan repaid
= Rs 118000 – Rs 73500 = Rs 44500.

Answer 17.

Answer 18.

In ∆ABC, ∠C = 90°
∴ AB² = AC² + BC²
In AECD, ∠ECD = 90°
∴ DE² = CD² + EC²
In ∆AEC, ∠ACE = 90°
∴ AE² = AC² + EC²
In ∆BCD, ∠BCD = 90°
∴ BD² = BC² + CD²
Adding (iii) and (iv), we get
AE² + BD² = (AC² + BC²) + (CD² + EC²)
= AB² + DE²
OR

Given, DB ⊥ BC, AC ⊥ BC
=> ∠DBC = ∠ACB = 90°
∴ AC || DB(sum of interior ∠s on the same side of BC is 180°)
∠ABD = ∠BAC (alt ∠s)
Now, in ABED and AACB,
∠EBD = ∠ABD = ∠BAC
∠C = ∠E (each = 90°)
∴ ∆BED ~ ∆ACB (AA similarity)

Answer 19.
Since tangents drawn from an external point to a circle are equal.
∴BF = BD = 4 cm (given)
and CE = CD = 3 cm (given)
Let AF = AE = x cm

Now, area (∆ABC) = area (∆AOB) + area (∆BOC) + area (∆AOC)
21 = AB x OF + BC x OD + AC x OE
=> 21 = (AF + BF) x 2 + (BD + CD) x 2 + (AE + CE) x 2
=> 21 = (x + 4) + (4 + 3) + (x + 3)
=> 21 = 2x + 14 => 2x = 7 => x = =>x = 3.5
AB = x + 4 = (3.5 + 4) cm = 7.5 cm
and AC = x + 3 = (3.5 + 3) cm = 6.5 cm.

Answer 20.
Given BC = R = 7 cm
and AE = CF = cm = r (say)
The perimeter of the shaded region
= perimeter of semicircle APB + perimeter of semicircle BSD + perimeter of semicircle DQC + perimeter of semicircle CRA

Answer 21.
Total possible outcomes = 6 x 6 = 36.
(i) The possible outcomes favourable to the event ‘sum of numbers on two dice is 5’ are (2, 3), (3, 2), (1, 4), (4, 1) i.e. 4 in number.
∴ Required probability =
(ii) The possible outcomes favourable to the event ‘even numbers on both dice’ are (2, 2), (2, 4), – (2, 6), (4, 2), (4, 4), (4, 6), (6, 2), (6, 4), (6, 6) i.e. 9 in number.
∴ Required probability =
We know that the sum of probabilities of all elementary events = 1
=> P(red ball) + P(blue ball) + P(orange ball) = 1

=> x = 24.
Hence, the total number of balls in the jar = 24.

Answer 22.

Answer 23.
Let A be the first term and D be the common difference of the AP.
Given,
a = pth term => a = A + (p – 1)D ….(i)
b = qth term => b = A +(q – 1)D ….(ii)
c = rth term => c = A + (r – 1)D ….(iii)
Subtracting (ii) from (i), we get
a – b = (p – q) D ….(iv)
Similarly, b – c = (q – r) D ….(v)
and c – a = (r – p) D ….(vi)
∴ (a – b) r + (b – c) p + (c – a) q = [(p – q) r + (q – r) p + (r – p)q] D
= (pr – qr + pq + qr – pq) D
= 0 x D = 0.

Answer 24.
Volume of type A = Volume of cone + Volume of hemisphere

Answer 25.
The given equations can be written as

Select the coordinate axes. Take 1 cm = 1 unit on both axes.
Plot the points (0, 2) and (3, 1) on the graph paper and join these points by a straight line. Plot the points (3, -2) and (6, 0) on the same graph paper and join these points by a straight line.
The given lines intersect at the point P(6, 0). Therefore, the solution of the given pair of linear equations is x = 6, y = 0.
The area bounded by the given lines and y-axis has been shaded.

From the figure AB = 6 units and OP = 6 units
The area of shaded region = area of ∆ABP
= x AB x OP = x 6 x 6
= 18 sq. units

Answer 26.
Given vertices of ∆ABC as A (a, b), B (b, c), C (c, a) and the centroid of ∆ABC is G(0, 0).
As we know that centroid of A whose vertices are (x1 y1), (x2, y2) and (x3, y3)

Answer 27.
Let AB be a tower 100 m high. P and Q be the position of two cars.
Angles of depressions are shown in adjoining figure.
∴∠APB = 90° and ∠AQB = 45°.
In ∆APB, ∠ABP = 90°
30° =

Answer 28.
Given: P is an external point to a circle with centre O. PA and PB are two tangents drawn from P to the circle, A and B being points of contact.
To prove: PA = PB
Construction: Join OA, OB and OP.
Proof: Since the tangent at any point of a circle is perpendicular to the radius through the point of contact.
∠OAP = 90° and ∠OBP = 90°
Now, in ∆OAP and ∆OBP,
OA = OB (radii of same circle)
OP = OP (common)
∠OAP = ∠OBP (each 90°)

Answer 29.
As ∆PQR ~ ∆ABC, so the side PQ of ∆PQR is the corresponding side AB of ∆ABC.
Given AB = 6 cm and we need PQ to be 8 cm, so
Thus, we are required to construct ∆PQR similar to ∆ABC such that sides of ∆PQR are times of the corresponding sides of ∆ABC.
Steps of construction:
1. Draw AB = 6 cm.
2. With A as centre and radius 6 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 an isosceles triangle with AB = AC = 6 cm and BC = 5 cm.
5. Take points A and P same.
6. Draw any ray AX making an acute angle with AB on the side opposite to the vertex C.

7. Locate 4 (the greater of 4 and 3) points A1, A2, A3 and A4 on AX such that AA1 = A1A2 = A2A3 = A3A4.
8. Join A3B.
9. Through A4, draw a line parallel to A3B (making an angle equal to ∠AA3B) to intersect the extended line segment AB at Q.
10. Through Q, draw a line parallel to BC (making an angle equal to ∠ABC) to intersect the extended line segment AC at R.
Then AQR i.e. PQR is the required triangle.
Justification:

Answer 30.
Construct the cumulative frequency distribution table as under:

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CBSE Sample Papers for Class 10 Maths Paper 3 is part of CBSE Sample Papers for Class 10 Maths . Here we have given CBSE Sample Papers for Class 10 Maths Paper 3

CBSE Sample Papers for Class 10 Maths Paper 3

Board	CBSE
Class	X
Subject	Maths
Sample Paper Set	Paper 3
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 3 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.
Write the polynomial whose zeroes are 2 + √3 and 2 – √3.

Question 2.
If the quadratic equation px² – 2√5 px + 15 = 0 has two equal real roots, then find the value of p.

Question 3.
Which term of the arithmetic progression 4, 9, 14, 19……is 109?

Question 4.
Explain, why 17 x 5 x 11 x 3 x 2 + 2 x 11 is a composite number?

Question 5.
If the adjoining figure is a sector of a circle of radius 10.5 cm, find the perimeter of the sector. [Take π=]

Question 6
Find the value of a, if the distance between the points A(-3, -14) and B (a, -5) is 9 units.

SECTION B

Question numbers 7 to 12 carry 2 marks each.

Question 7. Show that any positive odd integer is of the form 4q + 1 or 4q + 3, where q is some integer.

Question 8.
The coordinates of the mid-point of the line joining the points (3p, 4) and (-2, 2q) are (5, p). Find the values of p and q.

Question 9.
If the total surface area of a solid hemisphere is 462 cm², find its volume.[Take π=]

Question 10.
If 4 cos θ = 11 sin θ, find the value of

Question 11.
In an equilateral triangle ABC, AD is drawn perpendicular to BC meeting BC in D. Prove that AD² = 3BD².

Question 12.
The incircle of an isosceles triangle ABC, in which AB = AC, touches the sides BC, CA and AB at D, E and F respectively. Prove that BD = DC.

SECTION C

Question numbers 13 to 22 carry 3 marks each.

Question 13.
Find the largest number that divides 2053 and 967 and leaves a remainder of 5 and 7 respectively.

Question 14.
Solve graphically the pair of linear equations 2x + 3y = 11 and 2x - 4y = - 24. Hence, find the value of m, given that the line represented by y = mx + 3 passes through the intersection of the given pair.

Question 15.
Find the value of k for which the following pair of linear equations has no solution : x + 2y = 3; (k - 1)x + (k + 1)y = (k + 2).

Question 16.
Find the sum of all two digit numbers which when divided by 3 yield 1 as remainder.
OR
The sum of 4th and 8th terms of an AP is 24 and the sum of 6th and 10th terms is 44. Find the AP.

Question 17.
Find the mean of the following frequency distribution :

Question 18.
Find the area of the quadrilateral whose vertices taken in order are A(-5, -3), B(-4, -6), C(2, -1) and D(1, 2).
OR
Prove that the points A(2, 3), B(-2, 2), C(-1, -2) and D(3, -1) are the vertices of a square ABCD.

Question 19.
Draw a line segment of length 7.8 cm and divide it in the ratio 5 : 8. Measure the two parts. Also, justify your construction.

Question 20.
Prove the following identities:

OR
Prove that

Question 21.
In ∆PQR, right angled at Q, PQ = 5 cm and PR + QR = 25 cm. Find the values of sin P, cos P and tan P.

Question 22.
A toy is in the form of a cone of base radius 3.5 cm mounted on a hemisphere of base diameter 7 cm. If the total height of the toy is 15.5 cm, find the total surface area of the toy.
OR
A hemispherical bowl of internal diameter 36 cm contains liquid. This liquid is filled into 72 cylindrical bottles of diameter 6 cm. Find the height of each bottle, if 10% liquid is wasted in this transfer.

SECTION D

Question numbers 23 to 30 carry 4 marks each.

Question 23.
Ram's mother has given him money to buy some boxes from the market at the rate of 4x² + 3x - 2. The total amount of money given by his mother is represented by 8x⁴ + 14x³ - 2x² + 7x - 8. Out of this money he donated some amount of money to a child who was studying in the light of street lamp. Find how much amount of money he donated after purchasing the maximum number of boxes from the market? Write the values shown by Ram.

Question 24.
A student scored a total of 32 marks in class tests in mathematics and science. Had he scored 2 marks less in science and 4 more in mathematics, the product of his marks would have been 253. Find his marks in two subjects.
OR
If -4 is a root of the equation x² + 2x + 4p = 0, find the values of k for which the equation x² + p(1 + 3k)x + 7(3 + 2k) = 0 has equal roots.

Question 25.
Through the mid-point M of the side CD of a parallelogram ABCD, the line BM is drawn intersecting AC at L and AD produced in E. Prove that EL = 2BL.
OR
In a ∆PQR, N is a point on PR such that QN ⊥ PR. If PN x NR = QN², prove that ∠PQR = 90°.

Question 26.
In the adjoining figure, O is the centre of a circle of radius 5 cm. T is a point such that OT = 13 cm, PT and QT are tangents to the circle at points P and Q respectively. OT intersects the circle at the point E. If AB is tangent to the circle at E, find the length of AB.

Question 27.
A vertical tower stands on a horizontal plane and is surmounted by a vertical flag staff of height h. At a point on the plane, the angles of elevation of the bottom and the top of the flagstaff are α and β respectively. Prove that the height of the tower is

Question 28.
PS is a diameter of a circle of radius 6 cm. Q and R are points on the diameter that PQ, QR and RS are equal. Semicircles are drawn with PQ and QS as diameters, as shown in the figure. Find the perimeter of the shaded region.
Also find the area of the shaded region. (Use π = 3.14)

Question 29.
A die has six faces marked 0, 1, 1, 1, 6, 6. Two such dice are thrown together and the total score is recorded.
(i) How many different scores are possible?
(ii) What is the probability of getting a total of 7?
OR
Two customers Shyam and Ekta visit a particular shop in the same week from Tuesday to Saturday. Each is equally likely to visit the shop on any day. Find the probability that they visit the shop
(i) on the same day
(ii) on consecutive day
(iii) on different days
(iv) neither on same day nor on consecutive days.

Question 30.
The following distribution gives the daily wages of 50 workers of a factory:

Convert the distribution to a less than type cumulative frequency distribution, and draw its ogive. Hence, obtain the median wages from the graph.

Answers

Answer 1.
Required polynomial = x² - (Sum of the zeroes) x + Product of the zeroes
= x² - (2 + √3 + 2 - √3)x + (2 + √3)(2 - √3)
= x² - 4x + 1.

Answer 2.
Given px² - 2√5 px + 15 = 0
For equal roots, D = (-2√5 p)² - 4 x p x 15 = 0
=> 20p² - 60p = 0 => 20p(p - 3) = 0
=> p = 0 or p = 3 but p ≠ 0 => p = 3.

Answer 3.
Given 4, 9, 14, 19 ... 109
Here, a = 4, d = 5, l = 109
Now, l = a + (n - 1)d
109 = 4 + (n - 1) x 5 => 5(n - 1) = 105 => n - 1 = 21 => n = 22.
Hence, 22nd term is 109.

Answer 4.
Given number = 17 x 5 x 11 x 3 x 2 + 2 x 11 = 11 x 2 x (17 x 5 x 3 + 1)
which is product of more than one prime number, therefore, the given number is prime.

Answer 5.
Radius = 10.5 cm = cm

Perimeter = AC + BC + AB = (10.5 + 10.5 + 11) cm = 32 cm.

Answer 6.
AB = 9

Answer 7.
Let a be any positive odd integer.
Applying Euclid's division lemma with divisor 4, we get
a = 4q, 4q + 1, 4q + 2 or 4q + 3 where q is some whole number.
But 4q and 4q + 2 are even integers and a is odd integer, therefore, a cannot be of the form 4q or 4q + 2.
Hence, a is of the form 4q + 1 or 4q + 3, where q is some whole number.

Answer 8.
The mid-point of the line segment joining the points (3p, 4) and (-2, 2q) is

and 2 + q = p => 2 + q = 4 => q = 2.
Hence, p = 4 and q = 2.

Answer 9.
Given, total surface area of hemisphere = 462 cm²
=> 3πr² = 462

Answer 10.
Given 4 cos θ = 11 sin θ
⇒ ....(i)

Answer 11.
In ∆ABD right angled at D.
By Pythagoras theorem,
AB² = AD² + BD²
=> BC² = AD² + BD² (∵ AB = BC = CA)
=> (2BD)² = AD² + BD²
(∵in an equilateral ∆ perpendicular is the median , ∴)
=> 4BD² = AD² + BD²
=> AD² = 3BD²

Answer 12.
Given: A circle inscribed in an isosceles ∆ABC with AB = AC, touching the sides BC, CA and AB at D, E and F respectively.
To prove: BD = DC
Proof: ∵ Length of tangents from an external point to the circle are equal

∴ AF = AE ...(i)
BF = BD ...(ii)
and CD = CE ...(iii)
∴ AB = AC
=> AF + BF = AE + EC
=> BF = EC (using (i))
=> BD = CD (using (ii) and.(iii))

Answer 13.
Given that required number when divides 2053 and 967 leaves a remainder 5 and 7 respectively.
∴ The required number is HCF of 2053 - 5 = 2048 and 967 - 7 = 960

∴HCF of 2048 and 960 = 64
Hence, the required largest number is 64.

Answer 14.
The given equations can be written as

The given lines intersect at the point P(-2, 5). Therefore, the solution of the given pair of linear equation is x = -2, y = 5.
Given that the line y = mx + 3 passes through the intersection of the given pair i.e. (-2, 5)
∴5 = m x (-2) + 3 => -2m - 2 => m = -1.

Answer 15.
The given pair of linear equations can be written as
x + 2y - 3 = 0 and (k - 1)x + (k + 1)y - (k + 2) = 0
Here, a1 = 1,b1 = 2, c1 = -3 and a2 = k - 1, b2 = k + 1, c2 = -(k + 2).
Now, the given pair of linear equations has no solution, if

=> 2k - 2 = k + 1 and 2k + 4 ≠ 3k + 3
=> k = 3 and k ≠ 1.
Hence, the given pair of linear equations will have no solution when k = 3.

Answer 16.
Two digit numbers which leaves remainder 1 when divided by 3 are 10, 13, 16, 19 ..., 97.
These numbers form an AP with a = 10, d = 3 and l = 97.
∵ l = a + (n - 1 )d
=> 97 = 10 + (n - 1) x 3 => 3(n - 1) = 87 => n - 1 = 29
=> n = 30
Now, the sum of all such numbers

= 15 x 107 = 1605.
Hence, the required sum is 1605.
OR
Let the first term and common difference of AP be a and d respectively.
Given: T4 + T8 = 24 and T6 + T10 = 44
=> a + 3d + a + 7d = 24 and a + 5d + a + 9d = 44
=> 2a + 10d = 24 ...(i)
and 2a + 14d = 44 ...(ii)
Subtracting (i) from (ii), we get
4d = 20 => d = 5.
Putting d = 5 in (i), we get
2a + 10 x 5 = 24
=> 2a = -26 => a = -13.
Hence, the required AP is
-13, -13 + 5, -13 + 2 x 5, -13 + 3 x 5, ......
i.e. -13, -8, -3, 2, ...

Answer 17.
We shall use step-deviation method. Taking assumed mean a = 25, construct the table as under. Here, h = 10.

Answer 18.
Join AC.
Now,

∵AB = BC = CD = DA i.e. all 4 sides are equal
and AC = BD i.e. diagonals are equal.
∴ ABCD is a square.

Answer 19.
Steps of construction:
1. Draw a line segment AB of length 7.8 cm by using ruler.
2. Draw any ray AX making an acute angle with AB.
3. Locate 13 (5 + 8) points, A1, A2, A3, ..., A13 on AX such that AA1 = A1A2 = A2A3 = ...... = A12A13.
4. Join A13B.
5. Through A5, draw a line parallel to A13B (making an angle equal to ∠AA13B) to intersect AB at C. Then C is the required point which divides AB in the ratio 5 : 8.
On measuring, we find that AC = 3 cm and CB = 4.8 cm.

Answer 20.

Answer 21.
Given PR + QR = 25 cm => QR = (25 - PR) cm.
In ∆PQR, ∠Q = 90°. By Pythagoras Theorem,
PR² = PQ² + QR²
=> PR² = 5² + (25 - PR)²
=> PR² = 25 + 625 - 50 PR + PR²
=> 50 PR = 650 => PR = 13 cm
∴ QR = (25 - 13) cm = 12 cm.

Answer 22.
Given radius of cone = 3.5 cm
Diameter of hemisphere = 7 cm

Answer 23.
Total amount of money p(x) = 8x⁴ + 14x³ - 2x² + 7x - 8.
Cost of each box g(x) = 4x² + 3x - 2
Maximum number of boxes purchased = q(x)
Amount of money donated to the child = r(x)
We need to find q(x) and r(x).
On dividing p(x) by g(x), we have

Hence, the amount of money he donated to the child = 14x - 10
and maximum number of boxes, he purchased = 2x² + 2x - 1.
The values shown by Ram are kindness, helpfulness, well wisher of society, care for children and promoting education.

Answer 24.
Let the student's marks in mathematics be x.
As the total marks in mathematics and science is 32,
so the student's marks in science = 32 - x.
Given if he scores 2 less marks in science and 4 more marks in mathematics, then his new marks in science = 32 - x - 2 = 30 - x
and his new marks in mathematics = x + 4.
According to given, (x + 4) (30 - x) = 253
=> 30x + 120 - x² - 4x = 253
=> x² - 26x + 133 = 0
=> x² - 19x - 7x + 133 = 0
=> (x - 7) (x - 19) = 0 => x = 7 or x = 19.
When x = 7, 32 - x = 25; when x = 19, 32 - x = 13.
Hence, student's marks in mathematics = 7 and in science 25
or student's marks in mathematics = 19 and in science = 13.
OR
Given x = -4 is a root of the equation x² + 2x + 4p = 0
(-4)² + 2 x (-4) + 4p = 0
=> 16 - 8 + 4p = 0 => 4p = -8 => p = -2.
Now, the equation x² + p(1 + 3k)x + 7(3 + 2k) = 0 has equal roots
D = 0 => {p( 1 + 3k)}² - 4 x 1 x 7(3 + 2k) = 0
=> p²(1 + 6k + 9k²) - 28(3 + 2k) = 0
=> (-2)² (1 + 6k + 9k²) - 84 - 56k = 0 (∵ p = -2)
=> 4 + 24k + 36k² - 84 - 56k = 0
=> 36k² - 32k - 80 = 0
=> 9k² - 8k - 20 = 0
=> (9k + 10) (k - 2) = 0
=> or k = 2.

Answer 25.
In ∆BMC and ∆EMD,

MC = MD (∵ M is mid-point of CD)
∠CBM = ∠DEM (alt. ∠s, BC || AE)
∠BMC = ∠EMD (vert. opp. ∠s)
∴ ∆BMC ≅ ∆EMD (AAS criterion of congruence)
=> BC = DE (c.p.c.t.)
Also AD = BC (opp sides of ||gm ABCD)
∴ AE = AD + DE = BC + BC => AE = 2BC ...(i)
In AAEL and ACBL,
∠AEL = ∠CBL
∠ALE = ∠CLB
∴ ∆AEL ~ ∆CBL

Answer 26.
Since tangent is perpendicular to radius, OP ⊥ PT.
In ∆OPT, ∠OPT = 90°. By Pythagoras theorem,
PT² = OA² - OP² = 13² - 5² = 144 => PT = 12 cm
As the lengths of tangent drawn from a point to a circle are equal,
AP = AE = x cm (say)
=> AT = PT - AP = (12 - x) cm
ET = OT - OE = 13 cm - 5 cm = 8 cm.
Given, AB is tangent to the circle at E, OE ⊥ AB => ∠AET = 90°.
In ∆AET, ∠AET = 90°. By Pythagoras theorem,
AT² = AE² + ET² => (12 - x)² = x² + 8²
=> 144 - 24x + x² = x² + 64

Answer 27.
Let AB be a vertical tower and BP be a flagstaff of height h surmounted on the tower AB, and O be the point of observation on the plane
(ground), then BP = h.
Let OA = d and AB = H.
Given, ∠BOA = α and ∠POA = β.

Answer 28.
PS = 2 x radius = (2x6) cm = 12 cm.
Given PQ = QR = RS
=> PQ = of PS = of 12 cm = 4 cm
and QS = 8 cm.

Answer 29.
(i) For the sum of outcomes on the top faces of two dice i.e. the total score obtained, we construct the table as under:

The different scores obtained are 0, 1, 2, 6, 7, 12. They are six in number.
Hence, the number of different possible scores are 6 (in number).
(ii) Total number of outcomes are 6x6 i.e. 36, which are all equally likely.
The table shows that we get score 7 twelve times.
So, the number of outcomes favourable to the event 'getting a total of 7' is 12.
∴ P(getting a total of 7) =
OR
Let T, W, Th, F and S represent Tuesday, Wednesday, Thursday, Friday and Saturday respectively. Since Shyam and Ekta visit a particular shop in the same week from Tuesday to Saturday, the outcomes are:
(T, T), (T, W), (T, Th), (T, F), (T, S)
(W, T), (W, W), (W, Th), (W, F), (W, S)
(Th, T), (Th, W), (Th, Th), (Th, F), (Th, S)
(F, T), (F, W), (F, Th), (F, F), (F, S)
(S, T), (S, W), (S, Th), (S, F), (S, S)
Total number of outcomes = 5 x 5 = 25 and all outcomes are equally likely.
(i) The outcomes favourable to the event 'visit on the same day' are (T, T), (W, W), (Th, Th), (F, F), (S, S); which are 5 in number.
∴ P (visit on the same day) =
(ii) The outcomes favourable to the event 'visit on consecutive days' are (T, W), (W, T), (W, Th), (Th, W), (Th, F), (F, Th), (F, S), (S, F); which are 8 in number.
∴ P (visit on consecutive days) = .
(iii) P (visit on different days) = 1 - P (visit on same day)
=
(iv) The outcomes favourable to the event 'visit neither on same day nor on consecutive days' are (T, Th), (T, F), (T, S), (W, F), (W, S), (Th, T), (Th, S), (F, T), (F, W), (S, T), (S, W), (S, Th); which are 12 in number.
∴P (visit neither on same day nor on consecutive days) = 

Answer 30.
From the given data, we construct the table for cumulative frequency distribution as under:

Take 1 cm along x-axis = Rs 20 and 1 cm along y-axis = 10 workers.
Plot the points (120, 12), (140, 26), (160, 34), (180, 40), (200, 50) and (100, 0). Join these points by a free hand drawing. The required ogive (less than type) is drawn on the graph sheet given below.

Since the scale on x-axis starts at 100, 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 = = 25.
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 138.5
Hence, the median daily wages = Rs 138.50

We hope the CBSE Sample Papers for Class 10 Maths Paper 3 help you. If you have any query regarding CBSE Sample Papers for Class 10 Maths Paper 3, drop a comment below and we will get back to you at the earliest.

CBSE Sample Papers for Class 10 Maths Paper 2 is part of CBSE Sample Papers for Class 10 Maths Here we have given CBSE Sample Papers for Class 10 Maths Paper 2

CBSE Sample Papers for Class 10 Maths Paper 2

Board	CBSE
Class	X
Subject	Maths
Sample Paper Set	Paper 2
Category	CBSE Sample Papers

Students who are going to appear for CBSE Class 10 Examinations are advised to practice the CBSE sample papers given here which is designed as per the latest Syllabus and marking scheme as prescribed by the CBSE is given here. Paper 2 of Solved CBSE Sample 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.
Find the value of k for which the following pair of linear equations has a unique solution:
2x + 3y = 7; (k – 1)x + (k + 2)y = 3k.

Question 2.
Find the nature of the roots of quadratic equation 2x² – √5 x + 1 = 0.

Question 3.
What is the probability that a non-leap year has 53 Mondays?

Question 4.
A die is thrown once. Find the probability of getting a prime number.

Question 5.
Find the mode of the data, whose mean and median are given by 10.5 and 11.5 respectively.

Question 6.
In the adjoining figure, DE || BC. If AD = x, DB = x – 2, AE = x + 2 and EC = x – 1, find the value of x.

SECTION B

Question numbers 7 to 12 carry 2 marks each.

Question 7.
Find HCF and LCM of 90 and 144 by method of prime factorisation.

Question 8.
Find the values of a and b for which the following pair of linear equations has infinitely many solutions:
3x – (a + 1)y = 2b – 1; 5x + (1 – 2a)y = 3b.

Question 9.
Without using trigonometric tables, evaluate the following:
(cos² 25° + cos² 65°) + cosec θ . sec (90° – θ) – cot θ tan (90° – θ).

Question 10.
ABC is a triangle and G (4, 3) is the centroid of the triangle. If A, B and C are the points (1, 3), (4, b) and (a, 1) respectively, find the values of a and b. Also find the length of side BC.

Question 11.
In the adjoining figure, DE || AC and . Prove that DC || AP.

Question 12.
In the adjoining figure, a circle is inscribed in a quadrilateral ABCD in which ∠B = 90°. If AD = 23 cm, AB = 29 cm and DS = 5 cm, find the radius (r) of the circle.

SECTION C

Question numbers 13 to 22 carry 3 marks each.

Question 13.
If two zeroes of the polynomial x⁴ + 3x³ – 20x² – 6x + 36 are √2 and – √2 , find the other zeroes of the polynomial.

Question 14.
If α and β are zeroes of the polynomial 6x² – 7x – 3, then form a quadratic polynomial whose zeroes are and .

Question 15.
How many terms of the A.P. -6, , -5,……. double answer.are needed to give the sum – 25? Explain the
OR
The 19th term of an AP is equal to three times its 6th term. If its 9th term is 19, find the AP.

Question 16.
The father’s present age is six times his son’s ages. Four years hence the age of the father will be four times his son’s age. Find the present ages of the father and son.
OR
The ratio of incomes of two persons is 9 : 7 and the ratio of their expenditures is 4 : 3. If each of them manages to save Rs 2000 per month, find their monthly incomes.

Question 17.
ABC is a right triangle, right angled at C. If p is the length of perpendicular from C to AB and a, b, c have usual meanings, then prove that
OR
If the diagonals of a quadrilateral divide each other proportionally, prove that it is a trapezium.

Question 18.
PQ is a tangent to a circle with centre O at the point Q: A chord QA is ‘drawn parallel to PO. If AOB is a diameter of the circle, prove that PB is tangent to the circle at the point B.

Question 19.
The radii of the circular ends of a bucket of height 15 cm are 14 cm and r cm (r < 14 cm). If the volume of bucket is 5390 cm³, then find the value of r.

Question 20.
Find the area of the major segment APB in adjoining figure, of a circle of radius 35 cm and ∠AOB = 90°.

OR
In adjoining figure, a semicircle is drawn with O as centre and AB as diameter. Semicircles are drawn with AO and OB as diameters. If AB = 28 m, find the perimeter of the shaded region.

Question 21.
Prove that :

Question 22.
Prove that :

SECTION D

Question numbers 23 to 30 carry 4 marks each.

Question 23.
Prove that √5 is an irrational number and hence show that 2 + √5 is also an irrational number.

Question 24.
If two vertices of an equilateral triangle are (3, 0) and (6, 0), find the third vertex.
OR
The mid-points D, E and F of the sides AB, BC and CA of a triangle are (3, 4), (8, 9) and (6, 7) respectively. Find the coordinates of the vertices of the triangle.

Question 25.
Water is flowing at the rate of 15 km/hour through a pipe of diameter 14 cm into a cuboidal pond which is 50 m long and 44 m wide. In what time will the level of water in the pond rise by 21 cm?

Question 26.
While boarding an aeroplane, a passenger got hurt. The pilot showing promptness and concern, made arrangements to hospitalise the injured and so the plane started late by 30 minutes. To reach the destination, 1500 km away in time, the pilot increased the speed by 100 km/h. Find the original speed/hour of the plane.
Do you appreciate the values shown by pilot, namely promptness in providing help to the injured and his efforts to reach in time.

Question 27.
Draw a pair of tangents to a circle of radius 3 cm which are inclined at an angle of 60° to each other.

Question 28.
The angle of elevation of the top of a building from the foot of a tower is 30° aid the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building.
OR
The angles of depression of two ships from the top of a lighthouse and on the same side of it are found to be 45° and 30°. If the ships are 200 m apart, find the height of the lighthouse.

Question 29.
Three coins are tossed simultaneously, find the probability of getting:
(i) atleast one head
(ii) atmost two heads
(iii) exactly 2 heads
(iv) no head.

Question 30.
The mean of the following frequency distribution is 62.8 and the sum Of all the frequencies is 50. Compute the missing frequencies f1 and f2:

OR
The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mode and mean of the data:

Answers

Answer 1.
For unique solution, ⇒ 2k + 4 ≠ 3k – 3
⇒ k ≠ 7
Hence, the given pair of linear equations will have unique solution for all real values of k except 7.

Answer 2.
Given 2x² – √5x + 1 = 0
D = (-√5)² – 4 x 2 x 1 = 5 – 8 = – 3
∵D < 0, therefore, given equation has no real roots.

Answer 3.
There are 365 days in a non-leap year.
365 days = 52 weeks + 1 day
∴ One day can be M, T, W, Th, F, S, Su = 7 ways
∴ P(53 Mondays in non-leap year) =

Answer 4.
Total number of outcomes = 6(1, 2, 3, 4, 5 or 6)
Favourable number of outcomes = 3(2, 3, 5)
∴ P(prime number) =

Answer 5.
Mode = 3 Median – 2 Mean .
= 3 x 11.5 – 2 x 10.5 = 34.5 – 21 – 13.5
Hence, mode = 13.5

Answer 6.
∵ DE || BC
∴ By Basic Proportionality Theorem, we have


⇒
⇒ x (x – 1) = (x – 2) (x + 2)
⇒ x² – x – x² – 4
⇒ -x = -4
⇒ x = 4.

Answer 7.
90 = 2 x 3 x 3 x 5
and 144 = 2 x 2 x 2 x 2 x 3 x 3
∴ HCF = 2 x 3 x 3 = 18
and LCM = 2 x 2 x 2 x 2 x 3 x 3 x 5 = 720
Hence, HCF = 18 and LCM = 720.

Answer 8.
Given 3x – (a + 1 )y – (2b – 1) = 0
and 5x + (1 – 2a)y – 3b = 0

Hence, a = 8 and b = 5.

Answer 9.
(cos² 25° + cos² 65°) + cosec θ . sec (90° – θ) – cot θ . tan (90° – θ)
= cos² 25° + cos² (90° – 25°) + cosec θ . cosec θ – cot θ . cot θ
= cos² 25° + sin² 25° + cosec² θ – cot² θ
= 1 + 1
= 2.

Answer 10.
Since G (4, 3) is the centroid of ∆ABC, we have

Answer 11.
In ∆ABC, DE || AC

Answer 12.
AR = AQ, DR = DS, BP = BQ (lengths of tangents)
but DS = 5 cm => DR = 5 cm.
AR = AD – DR = 23 cm – 5 cm = 18 cm => AQ = 18 cm.
BQ = AB – AQ = 29 cm – 18 cm = 11 cm.
As AB is tangent to the circle at Q, OQ ⊥ AB
=> ∠OQB = 90°.
Also ∠B = 90° (given) => OQBP is a rectangle.
But BP = BQ => OQBP is a square.
∴ Radius = r = OQ = BQ = 11 cm.

Answer 13.
∴ √2 and -√2 are the zero’s of the given polynomial.
∴ (x – √2) (x + √2) i.e. (x² – 2) is a factor of the given polynomial.

To find the other two zero’s, we proceed as follows:
x² + 3x – 18 = 0
⇒ (x + 6) (x – 3) = 0
⇒ x + 6 = 0 or x – 3 = 0
⇒ x = -6 or x = 3
Hence, other zero’s are -6 and 3.

Answer 14.
Given α and β are zero’s of quadratic polynomial 6x² – 7x – 3,

Answer 15.
Here, a = -6, d = = = ,S_n = -25.
We are required to find n.

Answer 16.
Let the father’s present age be x years
and the son’s present age be y years.
According to given, x – 6y …(i)
4 years later,
father’s age = (x + 4) years
and son’s age = (y + 4) years
∴ (x + 4) = 4(y + 4)
=> x – 4y = 12 …(ii)
Putting the value of x from (i) in (ii), we get
6y – 4 y = 12 => 2y = 12 => y = 6
∴ x = 6 x 6 = 36
Hence, father’s present age = 36 years and son’s present age = 6 years
OR
Let the incomes per month of two persons be Rs x and Rs y respectively. As each person saves Rs 2000 per month, so their expenditures are Rs (x – 2000) and Rs (y – 2000) respectively.
According to given, we have
i.e- 7x – 9y = 0 …(i)
and i.e. 3x – 4y + 2000 = 0 …(ii)
Multiplying equation (i) by 3 and equation (ii) by 7, we get
21x – 27y = 0 …(iii) and 21x – 28y + 14000 = 0 …(iv)
Subtracting equation (iv) from equation (iii), we get
y – 14000 = 0 => y = 14000.
Substituting this value of y in (i), we get
7x – 9 x 14000 = 0 =>x = 18000.
Hence, the monthly incomes of the two persons are Rs 18000 and Rs 14000 respectively.

Answer 17.
In ∆ACB and ∆CDB,
∠ACB = ∠CDB (both 90°)
∠B = ∠B (common)
∴ ∆ACB ~ ∆CDB (by AA similarity criterion)

Answer 18.
Given a circle with centre O and PQ is tangent to the circle at the point Q from an external point P. Chord QA is parallel to PO and AOB is a diameter.
We need to prove that PB is tangent to the circle at the point B.
Join OQ and mark the angles as shown in the adjoining figure.

As QA || PO,
∠1 = ∠2 (alt. ∠s)
and ∠4 = ∠3 (corres. ∠s)
.But ∠2 = ∠3 (∵ in ∆OAQ, OA = OQ being radii)
∴ ∠1 = ∠4.
In ∆OPB and ∆OPQ,
OB = OQ (radii of same circle)
∠1 = ∠4 (proved above)
OP = OP (common)
∴ ∆OPB ≅ ∆OPQ (SAS congruence rule)
∴ ∠OBP = ∠OQP (c.p.c.t)
=> ∠OBP = 90° (tangent is ⊥ to radius ,OQ⊥PQ)
=> OB ⊥ PB i.e. radius is perpendicular to PB at point B.
Therefore, PB is tangent to the circle at the point B.

Answer 19.
Given h = 15 cm, R = 14 cm, ‘r’ = r cm and volume of bucket = 5390 cm³
∵Volume of bucket = volume of frustum of cone

∴r cannot be negative
∵Radius = r = 7 cm.

Answer 20.
Given r = 35 cm and ∠AOB = 90°
Area of minor segment = area of minor sector – area (∆OAB)

Answer 21.

Answer 22.

Answer 23.
Let √5 be a rational number, then
√5 = , where p, q are integers, q ≠ 0 and p, q have no common factors (except 1)
=> => p² = 5q²
As 5 divides 5q², so 5 divides p², but 5 is prime.
=> 5 divides p
Let p = 5m, where m is an integer.
Substituting this value of p in (i), we get
(5m)² = 5q² => 25m² = 5q² => 5m² = q²
As 5 divides 5m², so 5 divides q², but 5 is prime
=> 5 divides q
Thus p and q have a common factor 5. This contradicts that p and q have no common factors (except 1)
Hence, √5 is not a rational number.
So, we conclude that √5 is an irrational number.
Let 2 + √5 be a rational number, say r
Then, 2 + √5 = r => √5 = r – 2
As r is rational, r – 2 is rational => √5 is rational
But this contradicts the fact that √5 is irrational.
Hence, our assumption is wrong. Therefore, 2 + √5 is an irrational number.

Answer 24.
Given vertices are A(3, 0) and B(6, 0) and let third vertex be C(x, y), then

OR
Let the vertices A, B and C of the triangle ABC be (x1 y1), (x2, y2) and (x3 y3) respectively.
Since points D and F are mid-points of the sides AB and
AC respectively, by mid-point theorem, DF || BC and
DF = but E is mid-point of BC, so DF || BE and
DF = BE.
Therefore, DBEF is a parallelogram.
Similarly, DECF and DEFA are parallelograms.
Since the diagonals of a parallelogram bisect each other, mid-points of diagonals BF and DE are same.

Answer 25.
Radius of pipe = cm = 7 cm = m = 0.07 m
As the water is flowing at the rate of 15 km per hour,
the length of water delivered by the circular pipe in 1 hour
= 15 km = 15000 m
Volume of water delivered by the circular pipe in 1 hour

Hence, the level of water in the pond rise by 21 cm in 2 hours.

Answer 26.
Let the original speed of the aeroplane be x km/h.
Time taken to cover the distance of 1500 km = hours
New speed of the aeroplane = (x + 100) km/h.
Time taken to cover the distance of 1500 km at new speed = hours

=> x² + 100x – 300000 = 0
=> x² + 600x – 500x – 300000 = 0 => (x – 500) (x + 600) = 0
=> x = 500 or x = -600
But speed cannot be negative.
Hence, the original speed of the aeroplane = 500 km/h.
Yes, I appreciate the values shown by the pilot. Along with showing concern for the injured passenger he did not fail to perform his duty, by increasing the speed of the plane, he reached the destination on time.

Answer 27.
Steps of construction:
1. Draw a circle of radius 3 cm with O as its centre.
2. Draw any radius OA.
3. At O, construct ∠AOC = 120° to meet the circle at B.
4. At A, construct ∠OAR = 90°.
5. At B, construct ∠OBQ = 90° to meet AR at P.

Then PA and PB are tangents to the circle inclined at an angle of 60° to each other.
Justification:
As ∠APB and ∠AOB are supplementary, so ∠APB = 60°.

Answer 28.
Let CD = h metres be the height of the building and AB be the tower, then AB = 50 m.
Let BD = d metres be the distance between the foot of the tower and the foot of the building.
Given, ∠CBD = 30° and ∠ADB = 60°.
From right angled ∆CBD, we get

OR
Let the height of the lighthouse AB be h metres and C, D be the positions of two ships. The angles of depressions are shown in the adjoining figure.
Then ∠ACB = 45° and ∠ADB = 30°
Given CD = 200 m, let BC = x metres.
From right angled ∆ABC, we get

Answer 29.
When three coins are tossed simultaneously, the outcomes of the random experiment are:
HHH, HHT, HTH, THH, HTT, THT, TTH, TIT
It has 8 equally likely outcomes.
(i) The outcomes favourable to the event ‘atleast one head’ are
HHH, HHT, HTH, THH, HTT, THT, TTH; which are 7 in number.
∴ P(atleast one head) =
(ii) The outcomes favourable to the event ‘atmost two heads’ are
HHT, HTH, THH, THT, HTT, TTH, TTT; which are 7 in number.
∴P(atmost two heads) =
(iii) The outcomes favourable to the event ‘exactly 2 heads’ are
HHT, HTH, THH; which are 3 in number.
∴ P(exactly two heads) =
(iv) The only outcome favourable to the event ‘no head’ is TTT.
∴P(no head) =

Answer 30.
Given, sum of all frequencies = 50

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CBSE Sample Papers for Class 10 Maths Paper 1 is part of CBSE Sample Papers for Class 10 Maths Here we have given CBSE Sample Papers for Class 10 Maths Paper 1

CBSE Sample Papers for Class 10 Maths Paper 1

Board	CBSE
Class	X
Subject	Maths
Sample Paper Set	Paper 1
Category	CBSE Sample Papers

Students who are going to appear for CBSE Class 10 Examinations are advised to practice the CBSE sample papers given here which is designed as per the latest Syllabus and marking scheme as prescribed by the CBSE is given here. Paper 1 of Solved CBSE Sample 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.
Given that LCM (91, 26) = 182, find HCF (91, 26).

Question 2.
Does the rational number has a terminating or a non-terminating decimal representation?

Question 3.
If 1 is a zero of the polynomial p(x) = ax² – 3 (a – 1) x – 1, then find the value of a.

Question 4.
The nth turn of an AP is 7 – 4n. Find its common difference.

Question 5.
If adjoining figure, DE || BC and AD = 1 cm, BD = 2 cm. What is the ratio of the area of ∆ABC to the area of ∆ADE?

Question 6.
In adjoining figure, PA and PB are tangents to the circle drawn an external point P. CD is a third tangent touching the circle at PB = 10 cm and CQ = 2 cm, what is the length of PC?

SECTION B

Question numbers 7 to 12 carry 2 marks each.

Question 7.
Determine the values of m and n, so that the following system of linear equations has infinite number of solutions:
(2m – 1)x + 3y – 5; 3x + (n – 1)y – 2 = 0.

Question 8.
Find the roots of the quadratic equation: 4 √3 x² + 5x – 2 √3 =0.

Question 9.
A bag contains 5 red, 8 green and 7 white balls. One ball is drawn at random from the bag. Find the probability of getting:
(i) a white ball or a green ball.
(ii) neither a green ball nor a red ball.

Question 10.
Without using trigonometric tables, find the value of
+ cos 57° . cosec 33° – 2 cos 60°.

Question 11.
In the adjoining figure, E is a point on the side CB produced of an isosceles triangle ABC with AB = AC. If AD ⊥ BC and EF ⊥ AC,
prove that: AB x EF = AD x EC.

Question 12.
Two circles touch externally. The sum of their areas is 58π cm² and the distance between their centres is 10 cm. Find the radii of the two circles.

SECTION C

Question numbers 13 to 22 carry 3 marks each.

Question 13.
Prove that

Question 14.
Prove that: (1 + cot A – cosec A) (1 + tan A + sec A) = 2.

Question 15.
Solve the following pair of linear equations:

OR
The sum of the numerator and denominator of a fraction is 4 more than twice the numerator. If the numerator and denominator both increased by 3, they are in the ratio 2 : 3. Determine the fraction.

Question 16.
Show that and are the zeroes of the polynomial 4x² + 4x – 3 and verify the relationship between zeroes and coefficients of the polynomial.
OR
Quadratic polynomial 2x² – 3x + 1 has zeroes as α and β. Form a quadratic polynomial whose zeroes are 3α and 3β.

Question 17.
Construct a ∆ABC in which CA = 6 cm, AB = 5 cm and ∠BAC = 45°, then construct a triangle similar to the given triangle whose sides are of the corresponding sides of the ∆ABC.

Question 18.
Find the values of k if the points A(k + 1, 2k), B(3k, 2k + 3) and C(5k – 1, 5k) are collinear. .
OR
If P(9a – 2, -b) divides the line segment joining A(3a + 1, -3) and B(8a, 5) in the ratio 3 : 1, find the values of a and b.

Question 19.
Prove that the points A (-3, 0), B (1, – 3) and C(4, 1) are the vertices of an isosceles right triangle.

Question 20.
Cards marked with all 2-digit numbers are placed in a box and are mixed thoroughly. One card is drawn at random. Find the probability that the number on the card is
(i) divisible by 10
(ii) a perfect square number
(iii) a prime number less than 25.
OR
The king, queen and jack of clubs are removed from a deck of 52 playing cards and the remaining cards are shuffled. A card is drawn from the remaining cards. Find the probability of getting a card of
(i) hearts
(ii) queen
(iii) clubs.

Question 21.
Find the median of the following data:

Question 22.
The following frequency distribution shows the number of rims scored by some batsmen of India in one-day cricket matches:

Find the mode for the above data.

SECTION D

Question numbers 23 to 30 carry 4 marks each.

Question 23.
Prove that n³ – n is divisible by 6 for every positive integer n.

Question 24.
A sum of Rs 1600 is to be used to give 10 cash prizes to students of a school for their overall academic performance. If each prize is Rs 20 less than its preceding prize, find the value of each of the prizes.
What is the importance of an academic prize in student’s life?

Question 25.
A motor boat whose speed is 20 km/h in still water takes 1 hour more to go 48 km upstream than to return downstream to the same spot. Find the speed of the stream.
OR
A shopkeeper buys some books for Rs 80. If he had bought 4 more books for the same amount, each book would have cost Rs 1 less. Find the number o.f books he bought.

Question 26.
Prove that in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Question 27.
In the adjoining figure, XY and X’Y’ are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersects XY at A and X’Y’ at B. Prove that ∠AOB = 90°.

OR
The radii of two concentric circles are 13 cm and 8 cm. AB is a diameter of the bigger circle. BD is a tangent to the smaller circle touching it at D. Find the length AD.

Question 28.
The angles of elevation of the top of a tower from two points P and Q at distances of a and b respectively from the base and in the same straight line with it are complementary. Prove that the height of the tower is √ab .
OR
The angle of elevation of the top of a vertical tower from a point on the ground is 60°. From another point 10 m vertically above the first, its angle of elevation is 45°. Find the height of the tower.

Question 29.
In the adjoining figure, OACB is a quadrant of a circle with centre O and radius 3.5 cm. If OD = 2 cm, find
(i) the area of the shaded region.
(ii) the perimeter of the shaded region.

Question 30.
A building is in the form of cylinder surmounted by a hemispherical dome (shown in the adjoining figure). The base diameter of the dome is equal to of the total height of the building. Find the height of the building if it contains m³ of air.

Answers

Answer 1.
∵ HCF x LCM = 91 x 26
⇒HCF x 182 = 91 x 26
⇒HCF = = 13.

Answer 2.

Now numerator and denominator are co-prime and denominator is of the form 2^m x 5ⁿ. Hence, given rational number is terminating.

Answer 3.
Given 1 is a zero of p(x) = ax² – 3 (a – 1) x – 1
⇒ p(1) = 0 ⇒ a x 1² – 3(a – 1) x 1 – 1 = 0
⇒ a – 3a + 3 – 1 = 0
⇒ – 2a = – 2 ⇒ a = 1

Answer 4.
Given a_n = 7 – 4n ⇒a₂ = 7 – 4 x 2 = -1 and a₁ = 7 – 4 x 1 = 3
∴ common difference d = a₂ – a₁ = -1 – 3 = -4

Answer 5.

Area of ∆ABC : Area of ∆ADE = 9:1

Answer 6.
PC = PA – CA = PB – CQ = (10 – 2) cm = 8 cm (∵ PA = PB and CA = CQ)

Answer 7.
For infinite number of solutions

Answer 8.
4√3 x² + 5x – 2√3 = 0
=> 4√3 x² + 8x – 3x – 2√3 = 0
=> 4x(√3x + 2) – √3(√3x + 2) = 0
=> (√3x + 2) (4x – √3 ) = 0
=> √3x + 2 = 0 or 4x – √3 = 0

Answer 9.
Total number of possible outcomes = 5 + 8 + 7 = 20
(i) Number of favourable outcomes = number of white balls or number of green balls
= 7 + 8 = 15
∴ P (a white or a green ball) =
(ii) Number of favourable outcomes = number of white balls = 7
∴ P (neither green nor red ball) =

Answer 10.
+ cos 57° . cosec 33° – 2 cos 60°.

Answer 11.
Given, ∆ABC is an isosceles triangle with AB = AC.
⇒ ∠C = ∠B (angles opp. equal sides of a ∆ are equal)
In ∆ABD and ∆ECF,
∠ABD = ∠ECF (∵ ∠B = ∠C, proved above)
and ∠ADB = ∠EFC (each = 90°)
∴ ∆ABD ~ ∆ECF (by AA similarity criterion)
∴ ⇒ AB x EF = AD x EC, as required.

Answer 12.
Let r₁ cm and r₂ cm be the radii of two circles, then
r₁ + r₂ = 10 ….(i)


If r₁ = 7, then r₂ = 3 and if r₁ = 3, then r₂ = 7.
Hence, the radii of two circles are 7 cm and 3 cm.

Answer 13.

Answer 14.
(1 + cot A – cosec A) (1 + tan A + sec A)

Answer 15.
Given

Answer 16.
Let f(x) = 4x² + 4x – 3, then

Answer 17.
Steps of construction:
1. Draw AB = 5 cm.
2. At the point A draw ∠BAX = 45°.
3. From AX cut off AC = 6 cm.
4. Join BC. ∆ABC is formed with given data.
5. Draw making acute angle with AB as shown in the figure.
6. Draw 6 arcs at P1 P2, P3, P4, P5 and P6 such that AP1 = P1P2 = P2P3 = P3P4 = P4P5 = P5P6.
7. Join BP5.
8. Draw B’P6 || BP5 meeting AB produced at B’.
9. From B’, draw B’C’ || BC meeting AX at C’.
∆AB’C’ ~ ∆ABC.

Answer 18.
Since the given points A(k + 1, 2k), B(3k, 2k + 3) and C(5k – 1, 5k) are collinear, the area of triangle formed by them is zero

Answer 19.

∴AB² + BC² = AC² and AB = BC
Hence ABC is a right angled isosceles triangle , right angled at B

Answer 20.
Cards have numbers 10 to 99 (both inclusive) i.e. 10, 11, 12, … 99.
∴ Total number of cards = 90 (99 – 9 = 90)
(i) The numbers from 10 to 99 which are divisible by 10 are 10, 20, 30, …, 90
The number of’such numbers = 9.
∴ Required probability =
(ii) The numbers from 10 to 99 which are perfect squares are 16, 25, 36, 49, 64, 81.
The number of such numbers = 6.
∴ Required probability =
(iii) The numbers from 10 to 99 which are prime numbers less than 25 are 11, 13, 17, 19, 23.
The number of such numbers = 5.
∴ Required probability =
OR
After removing king, queen and jack of clubs, the number of cards left = 49
(i) The number of cards of hearts left in the remaining cards = 13
∴ P (a card of hearts) =
(ii) The number of cards of queens left in the remaining cards = 13
∴ P (a card of queen) =
(iii) The number of cards of clubs left in the remaining cards = 10
∴ P (a card of clubs) =

Answer 21.

Answer 22.
The maximum class frequency is 10 and the class corresponding to this frequency is 6000 – 8000. So, the modal class is 6000 – 8000.
Here, l = 6000, h = 2000, f₁ = 10, f₀ = 8 and f₂ = 2

∴ Mode = 6400

Answer 23.
Given n is any positive integer. Applying Euclid’s lemma with divisor = 6, we get
n = 6q, 6q + 1, 6q + 2, 6q + 3, 6q + 4 or 6q + 5, where q is some whole number.
Six cases arise.
Case I.
If n = 6q, then
n³ – n = n(n – 1) (n + 1) = 6q(6q – 1 )(6q + 1), which is divisible by 6.

Case II.
If n = 6q + 1, then
n³ – n = n(n – 1 )(n + 1) = (6q + 1)(6q)(6q + 2)
= 12q(6q + 1)(3q + 1), which is divisible by 6.

Case III.
If n = 6q + 2, then
n³ – n – n(n – 1 )(n + 1) = (6q + 2)(6q + 1)(6q + 3)
= 6(3q + 1)(6q + 1)(2q + 1), which is divisible by 6.

Case IV.
If n – 6q + 3, then
n³ – n = n(n – 1 )(n + 1) = (6q + 3)(6q + 2)(6q + 4)
= 12(2q + 1)(3q + 1)(3q + 2), which is divisible by 6.

Case V.
If n = 6q + 4, then
n³ – n = n(n – 1 )(n + 1) = (6q + 4)(6q + 3)(6q + 5)
= 6(3q + 2)(2q + 1)(6q + 5), which is divisible by 6.

Case VI.
If n = 6q + 5, then
n³ – n = n(n – 1 )(n + 1) = (6q + 5)(6q + 4)(6q + 6)
= 12(69 + 5)(3q + 2)(q + 1), which is divisible by 6.
Thus, in all cases, n³ – n is divisible by 6.
Hence, for every positive integer n, n³ – n is divisible by 6.

Answer 24.
Let the first cash prize be Rs x.
It is given that each prize is Rs 20 less than its preceding prize.
These cash prizes form an AP with a = x, d = – 20, S_n = 1600 and n = 10.

⇒ 320 = 2x – 180
⇒ 2x = 500
⇒ x = 250.
Therefore, the prizes are 250, 250 – 20, 250 – 2 × 20, 250 – 3 × 20, 250 – 4 × 20, 250 – 5 × 20,
i.e., 250, 230, 210, 190, 170, 150, 130, 110, 90, 70
Hence, the values of cash prizes are
Rs 250, Rs 230, Rs 210, Rs 190, Rs 170, Rs 150, Rs 130, Rs 110, Rs 90 and Rs 70.
Academic prize generates spirit of competition. One starts doing hard work which leads to success.

Answer 25.
Let the speed of the stream be x km/h.
Given, speed of the motor boat in still water = 20 km/h.
Then, the speed of the boat upstream = (20 – x) km/h
and the speed of the bpat downstream = (20 + x) km/h
Time taken by the motor boat to cover 48 km upstream = hours,
time taken by the motor boat to cover 48 km downstream = hours.
According to given, – = 1
⇒ 48(20 + x) – 48(20 – x) = (20 – x) (20 + x) ⇒ 48(2x) = 20² – x²
⇒ x² + 96x – 400 = 0
⇒ (x – 4) (x + 100) = 0
⇒ x = A or x = – 100 but x being the speed of the stream cannot be negative
⇒ x = 4
Hence, the speed of the stream is 4 km/h.
OR
Let the number of books the shopkeeper bought be x.
Given, total cost of books = Rs 80
Then cost of one book =
When he had bought 4 more books for same amount i.e. Rs 80, then cost of one book =
According to given – = 1
⇒ 80(x + 4) – 80x = x(x + 4)
⇒ 320 = x² + 4x
⇒ x² + 4x – 320 = 0
⇒(x + 20) (x- 16) = 0
⇒ x = 16 or x = – 20 but x being the number of books cannot be negative
⇒ x = 16
Hence, the number of books he bought = 16.

Answer 26.
Given. ABC is a right triangle right angled at A so that BC is its hypotenuse.
To prove. BC² = AB² + AC².
Construction. From A, draw AD ⊥ BC.
Proof. In ∆DBA and ∆ABC,

∠ABD = ∠ABC (same angle)
and ∠ADB = ∠BAC (each = 90°)
∴ ∆DBA ~ ∆ABC (AA similarity criterion)
=> AB² = BD x BC …(i)
In ∆DAC and ∆ABC,
∠ACD = ∠ACB (same angle)
and ∠ADC = ∠BAC (each = 90°)
∴ ∆DAC ~ ∆ABC (AA similarity criterion)
=> AC² = DC x BC …(ii)
On adding (i) and (ii), we get
AB² + AC² = BD x BC + DC x BC
= (BD + DC) x BC = BC x BC
=> AB² + AC² = BC².
Hence, BC² = AB² + AC².

Answer 27.
Join OP, OQ and OC. Mark the angles as shown in the figure.

In ∆OAC and ∆OAP,
OA = OA (common)
OC = OP (radii of some circle)
AC = AP (tangent drawn from A)
∴ ∆OAC ≅ ∆OAP (by SSS rule of congruency)
∴∠1 = ∠2 => ∠PAC = 2∠1 …(i)
Similarly,
∴∆OBC ≅ ∆OBQ,
∴ ∠3 = ∠4 => ∠QBC = 2∠3 …(ii)
As XY || X’Y’ and AB is a transversal,
∠PAC + ∠QBC = 180° (sum of co-int ∠S)
=> 2∠1 + 2∠3 = 180° [using (i) and (ii)]
=> ∠1 + ∠3 = 90° …(iii)
In ∆OAB, ∠AOB + ∠1 + ∠3 = 180° (sum of angles of a ∆)
=> ∠AOB + 90° = 180° (using (iii))
=> ∠AOB = 90°.
OR

Let the line BD meet the bigger circle at E. Join AE. Let O be the centre of two concentric circles.
As AB is a diameter of the bigger circle, O is mid-point of AB.
BD is tangent to the smaller circle and OD is radius of smaller circle through the point of contact D, OD ⊥ BE.
Since BE is a chord of the bigger circle and OD ⊥ BE,
BD = DE
(∵ Perpendicular from the centre to a chord bisects it)
=> D is mid-point of BE
∴ OD =
(∵ Segment joining the mid-points of any two sides of a triangle is half of the third side)
=> AE = 2 OD => AE = (2 x 8) cm = 16 cm (∵ OD = 8 cm)
In ∆OBD; ∠ODB = 90°, by Pythagoras theorem, we get
OB² = BD² + OD² => 13² = BD² + 8² (∵ OB is radius of bigger circle, so OB = 13 cm)
=> BD² = 169 – 64 = 105 => BD = √105 cm
But DE = BD => DE = √105 cm
Now ∠AEB = 90° (∵ angle in a semicircle = 90°)
=> ∠AED = 90°
In ∆ADE; ∠AED = 90°, by Pythagoras theorem, we get
AD² = AE² + DE² => AD² = 16² + (√105)² = 256 + 105 = 361
=> AD = √361 cm => AD = 19 cm

Answer 28.
Let AB be the tower of height h (units).
AP = a, QA = b.
As the angles of elevation are complementary, if ∠APB = θ° then ∠AQB = 90° – θ°

Let CD be a vertical tower of height h metres. From a point A on the ground, the angle of elevation of the top is 60°. B is another point 10 m vertically above A and angle of elevation of the top of the tower from B is 45°.
Let AC = d metres
From B, draw BE ⊥ CD.
Then BE = AC and EC = BA = 10 m.
From right angled ∆ADC, we get

Answer 29.
(i) Radius of the circle = 3.5 cm =
∴ Area of the quadrant OACB = x πr²

Answer 30.
Let the radius of the spherical dome be r metres and the total height of the building be h metres.
Since the base diameter of the dome is of the total height of the building, therefore,

Hence , the height of the building = 6 m

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