Author name: Raju

CBSE Sample Papers for Class 10 Maths Paper 3

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

Time Allowed : 3 hours
Maximum Marks : 80

General Instructions

• All questions are compulsory.
• The question paper consists of 30 questions divided into four sections — A, B, C and D.
• Section A contains 6 questions of 1 mark each. Section B contains 6 questions of 2 marks each. Section C contains 10 questions of 3 marks each. Section D contains 8 questions of 4 marks each.
• There is no overall choice. However, an internal choice has been provided in 4 questions of 3 marks each and 3 questions of 4 marks each. You have to attempt only one of the alternatives in all such questions.
• Use of calculator is not permitted.

SECTION-A

Question 1.
Vertices of a triangle are (- 4, 0), (4, 0), (0, 3). What type of triangle is it ? [1]

Question 2.
If two positive integers p and q can be expressed as p = ab² and q = a²b; a, b being prime
numbers, then find LCM (p, q).  [1]

Question 3.
D and E are respectively the points on the sides AB and AC of a triangle ABC such that
AD = 2 cm, BD = 4 cm, BC = 9 cm and DE || BC. Then, find the length of DE.  [1]

Question 4.
Find a quadratic polynomial, the sum and product of whose zeroes are 3 and – 2 respectively.  [1]

Question 5.
For what value of p are 2p + 1, 12, 5p – 3 are three consecutive terms of an A.P. ?  [1]

Question 6.

SECTION-B

Question 7.
The decimal expansion of the rational number \(\frac { 33 }{ { 2 }^{ 2 }.{ 5 }^{ n } } \) terminates after 3 places of decimal. Then find the value of n.  [2]

Question 8.
Find the value of a, if the distance between the points A(- 3, – 14) and B(a, – 5) is 9 units  [2]

Question 9.
17 cards numbered 1, 2, 3, …, 17 are put in a box and mixed thoroughly. One person draws a card from the box. Find the probability that the number on the card is :  [2]

• Odd
• A prime
• Divisible by 3
• Divisible by 3 and 2 bot

Question 10.
What is the probability that an ordinary year has 53 Sundays.  [2]

Question 11.
Determine the sum of first 100 odd natural number.  [2]

Question 12.  
Find the zeroes of the polynomial 2x² + x – 6.  [2]

SECTION-C

Question 13.
Find the largest number that divides 1251, 9377 and 15628 leaving remainders 1, 2 and 3 respectively.  [3]

Question 14.
The points A(2, 9), B(a, 5) and C(5, 5) are the vertices of a triangle ABC right angled at B. Find the values of a and hence the area of ∆ABC.  [3]
OR
A(6,1), B(8, 2) and C(9, 4) are three vertices of a parallelogram ∆BCD. If E is the mid-point of DC, find the area of ∆ADE.

Question 15.
Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals. [3]
OR
If BL and CM are median of a triangle ABC right angled at A then prove that:
4(BL² + CM²) = 5BC².

Question 16.
Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line segment joining the points of contact at the centre.  [3]

Question 17.
From a circular piece of cardboard of radius 3 cm, two sectors of 90° have been cut off. Find the perimeter of the remaining portion to nearest hundredth centimetres. (Take π = \(\frac { 22 }{ 7 } \))  [3]
OR
Find the area of the segment of a circle of radius 12 cm whose corresponding sector has a central angle of 60°.
(Use π = 3.14)

Question 18.
Water is flowing at the rate of 15 km/h through a pipe of diameter 14 cm into a cuboidal pond which is 50 m long and 44m wide. In what time will the level of water in pond
rise by 21 cm ?  [3]
OR
How many spherical lead shots each of diameter 4.2 cm can be obtained from a solid rectangular lead piece with dimensions 66 cm, 42 cm and 21 cm.

Question 19.
Find the zeroes of the polynomial 4x² – x – 3 and verify the relationship between the zeroes and the coefficients.  [3]

Question 20.
Solve graphically : 4x + 6y = 9; 7x + 5y = 2.  [3]
OR
Sum of a two-digit number and the number formed by reversing the order of digits is 88. If difference of digits is 2 and the unit digit is greater, determine the number.

Question 21.
If sec θ = x + \(\frac { 1 }{ 4x } \), prove that : sec θ + tan θ = 2x or \(\frac { 1 }{ 2x } \) [3]

Question 22.
Find the mean of the following data :  [3]

SECTION-D

Question 23.
If the angle of elevation of a cloud from a point h metre above a lake is α and the angle of depression of its reflection in the lake is β, prove that the distance of the cloud from the point of observation is   [4]

OR
A man is standing on the deck of a ship, which is 8 m above water level. He observes the angle of elevation of the top of a hill as 60° and the angle of depression of the base of the hill as 30°. Calculate the distance of the hill from the ship and the height of the hill.

Question 24.
Draw a line segment AB of length 8 cm. Taking A as centre, draw a circle of radius 4 cm and taking B as centre, draw another circle of radius 3 cm. Construct tangents of each circle from the centre of the other circle.  [4]

Question 25.
Show that the tangent at any point of a circle is perpendicular to the radius through the point of contact. In the given figure, AT is a tangent to the circle with centre O such that OT = 4 cm and ∠OTA = 30°. Find the length of AT. [4]

Question 26.
A solid right circular cone of height 120 cm and radius 60 cm is placed in a right circular cylinder full of water of height 180 cm such that it touches the bottom. Find the volume of water left in the cylinder, if the radius of the cylinder is equal to the radius of the cone.  [4]
OR
A well of diameter 3 m is dug 14 m deep. The earth taken out of it has been spread evenly all around it in the shape of a circular ring of width 4 m to form an embankment. Find the height of the embankment.

Question 27.
A boat can go 24 km downstream and return in 5 hours. If the speed of the stream is 2 km/hr, find the speed of the boat in still water.  [4]
OR
Product of digits of a two-digit number is 14. When 45 is added to the number then the digits interchanged their places. Find the number.

Question 28.
Shalini gets pocket money from her father every day. Out of the pocket money, she saves ₹30 on the first day and on each succeeding day, she increases her savings by 500 paise.  [4]

1. Find the amount saved by Shalini on 10th day.
2. Find the total amount saved by Shalini in 30 days.

Question 29.

Question 30.
Draw “more than ogive” for the following data :  [4]

SOLUTIONS
SECTION-A

Answer 1.
Given vertices of triangle are A(- 4, 0), B(4, 0) and C(0, 3).
We know that distance formula is,

Answer 2.
We have, p = ab² and q = a²b
Now, p = a × b × b
and q = a × a × b
∴ H.C.F. (p,q) = ab
We know,
Product of two numbers = H.C.F × L.C.M

Answer 3.
Given : AD = 2 cm, BD = 4 cm, BC = 9 cm
and DE || BC
In ΔABC and ΔADE,
DE || BC
∴ ∠B = ∠D [Corresponding angles]
and ∠C = ∠E [Corresponding angles]
∴ By AA similarly axiom

Answer 4.
Given : Sum of zeroes = 3 and Product of zeroes = – 2
We know that, equation of quadratic polyomial is given as
= k[x² – (Sum of zeroes)x + Product of zeroes]
= k[x² – 3x – 2] where k ≠ 0.

Answer 5.
Given : 2p + 1, 12, 5p – 3 are three consecutive terms of A.P.
12 – (2p + 1) = (5p – 3) – 12
12 – 2p – 1 = 5p – 3 – 12
11 – 2p = 5 p – 15
-2p – 5p = -15 – 11
– 7p = – 26
p = \(\frac { 26 }{ 7 } \).

Answer 6.

SECTION-B

Answer 7.
Given : Decimal expansion of rational number \frac { 33 }{ { 2 }^{ 2 }.{ 5 }^{ n } } terminate after 3 decimals.
∴ Denominator of decimal number will be in the form = (2 x 5)n and n = 3 to terminate after 3 decimals.

Answer 8.
Given : A(- 3, – 14), B(a, – 5) and AB = 9 units
By distance formula,

On squaring both sides, we get
81 = (a + 3)² + (9)²
81 – 81 = (a + 3)²
⇒ (a + 3)² = 0
On taking square root both side, we get
a + 3 = 0
a = – 3.

Answer 9.
Total number of cards = 1, 2, 3, …, 17.
n(S) = 17
(i) Probability that the number is odd.
Number of odd number cards = 1, 3, 5, 7, 9, 11, 13, 15, 17.
∴ n(E) = 9
Probability that number on the card is odd,
P(E) = \(\frac { n(E) }{ n(S) } \)
P(E) = \(\frac { 9 }{ 17 } \).

(ii) Probability that the number is a prime.
Number of prime number cards = 2, 3, 5, 7, 11, 13, 17
∴ n(E) =7
Probability that a card is prime number,
n(E) = \(\frac { n(E) }{ n(S) } \) = \(\frac { 7 }{ 17 } \)

(iii) Probability that the number is divisible by 3.
Number of cards divisible by 3 = 3, 6, 9, 12, 15
∴ n(E) = 5
Probability that the number is divisible by 3,
P(E) = \(\frac { n(E) }{ n(S) } \) = \(\frac { 5 }{ 17 } \)

(iv) Probability the number is divisible by 3 and 2 both.
Number of cards are divisible by 3 and 2 both = Number of cards are divisible by 6 = 6,12
∴ n(E) = 2
Probability that the number is divisible by 3 and 2 both,
P(E) = \(\frac { n(E) }{ n(S) } \) = \(\frac { 2 }{ 17 } \)

Answer 10.
1 year has 365 days = (52 x 7 + 1) days
Ordinary year has 52 Sunday + 1 day
1 day may be {Sun, Mon, Tue, Wed, Thu, Fri, Sat}
n(S) = 7
That 1 day may be Sun = {Sun}
∴ n(E) = 1
Probability of 53 Sundays in ordinary year
P(E) = \(\frac { n(E) }{ n(S) } =\frac { 1 }{ 7 } \)

Answer 11.
Sum of first 100 odd natural numbers i.e.,
1 + 3 + 5 + 7 + 9+ …
These are in arithmetic progression, with
a = 1, d = 3 – 1 = 2, n = 100
We know that

Answer 12.
Given polynomial is,
2x² + x – 6 = 0
2x² + 3x – 3x – 6 = 0
2x(x + 2) – 3(x + 2) = 0
(x + 2) (2x – 3) = 0
Either x + 2 = 0 ⇒ x = -2 or 2x – 3 = 0 ⇒ x = \(\frac { 3 }{ 2 }\)
∴ Zeroes of the polynomial are – 2 and \(\frac { 3 }{ 2 }\)

SECTION-C

Answer 13.
Given : 1251, 9377 and 15628 leave remainders 1, 2 and 3 respectively.
Required number = H.C.F. of (1251 – 1), (9377 – 2) and (15628 – 3)
= H.C.F. of 1250, 9375 and 15625
By Euclid’s lemma,
15625 = 1 x 9375 + 6250
9375 = 1 x 6250 + 3125
6250 = 2 x 3125 + 0
∴ H.C.F. of 15625 and 9375 = 3125.
and H.C.F. of 3125 and 1250 = 625.
Again by Euclid’s lemma
3125 = 2 x 1250 + 625
1250 = 2 x 625 + 0
∴ H.C.F. of 1250, 9375 and 15625 = 625
Hence, 625 is largest number that divides 1251, 9377 and 15628 leaving remainders 1, 2 and 3 respectively.

Answer 14.
Given : Vertices of triangle are A(2, 9), B(a, 5) and C(5, 5).
By distance formula,

Answer 15.

Answer 16.
Given : PA and PB are tangents to a circle having centre O.
To prove : ∠AOB + ∠APB = 180°.

Proof : OA is the radius of circle and PA is the tangent to the circle.
∴ ∠OAP = 90° …(i)
Similarly, OB is the radius of circle and PB is the tangent to the circle.
∴ ∠OBP = 90°
In quadrilateral APBO,
∠OAP + ∠APB + ∠PBO + ∠BOA = 360°
90° + ∠APB + 90° + ∠BOA = 360°
∠APB + ∠BOA = 360° – 180°
∠APB + ∠BOA = 180°.

Answer 17.
We have, radius of cardboard, r = 3 cm.
Two sectors AOB and COD have been cut off in the circular cardboard.
∴ Perimeter of remaining cardboard = OA + length of arc APD + OD + OB + Length of arc BQC + OC
= r + Length of arc APD + r + r + Length of arc BQC + r
= 4 x r + length of arc APD + length of arc BQC
|
OR
Given : Radius of circle, r =12 cm
Central angle of segment = 60°
Area of segment = Area of sector OACD – Area of ∆AOB

Answer 18.
Given : Diameter of cylindrical pipe = 14 cm


OR
Given : Diameter of lead ball = 4.2 cm
∴ Radius, r = \(\frac { 4.2 }{ 2 }\) = 2.1 cm
Dimensions of lead piece = 66 cm × 42 cm × 21 cm
Volume of rectangular lead piece = l × b × h = 66 × 42 × 21 cm³

Answer 19.
Given polynomial is
4x² – x – 3 = 0
⇒ 4x² – 4x + 3x – 3 =0
⇒ 4x(x – 1) + 3(x – 1) =0
⇒ (x – 1) (4x + 3) = 0
Either x – 1 = 0 ⇒ x = 1 or x + 3 = 0 ⇒ x = – \(\frac { 1 }{ 4 }\)
∴ Zeroes of the polynomial are 1 and – \(\frac { 3 }{ 4 }\)
Now,

Answer 20.



The point of intersection of these lines is (- 1.5, 2.5).
∴ x = -1.5 and y = 2.5
OR
Let the ten’s place of digit be x.
Therefore, unit place of digit will be (x + 2)
Two-digit number = 10 × x + (x + 2) = 11x + 2
Reverse of the number = 10(x + 2) + x = 11x + 20
According to the question,
11x + 2 + 11x + 20 = 88
22x = 88 – 22
x = \(\frac { 66 }{ 22 }\) = 3
∴ Two-digit number = 11x + 2 = 11 × 3 + 2 = 35.

Answer 21.

Answer 22.

SECTION-D

Answer 23.




Putting the value of x in equation (i), we get
h = 8 + √3 . 8√3
h = 8 + 24 = 32 m
and x = 8 × 1.732 = 13.856 m
Thus, height of the hill = 32 m
and distance between hill and ship = 13.856 m.

Answer 24.
Steps of construction :

• Draw a line segment AB = 8 cm.
• With A as centre and radius 4 cm, draw a circle.
• With B as centre and radius 3 cm, draw another circle.
• Draw perpendicular bisector of AB which meets AB at M.
• With M as centre and MA as radius, draw a circle intersecting the circle with centre A at P and Q and the circle with centre B at R and S.
• Join AR, AS, BP, BQ which are the required tangents.

Answer 25.
Given : A circle C(0, r) and a tangent l at point A.
To prove : OA ⊥ l.
Construction : Take a point B, other than A, on the tangent l. Join OB. Suppose OB meets the circle at C.

Proof : We know that, among all line segment joining the point O to a point on l, the perpendicular is shortest to l.
Clearly, OA = OC
Now, OB = OC + BC
∴ OB > OC
⇒ OB > OA
⇒ OA < OB
Thus, OA is shortest than any other line segment joining O to any point on l.
Thus, OA ⊥ l.
Given : OT = 4 cm
∠OTA = 30°
Construction : Join OA.
Now, OA is the radius of circle and AT is the tangent to the circle.

∴ OA ⊥ AT
In ΔOAT, ∠A = 90°
∴ cos T = \(\frac { AT }{ OT } \)
cos 30° = \(\frac { AT }{ 4 } \)
\(\frac { \sqrt { 3 } }{ 2 } \) = \(\frac { AT }{ 4 } \)
∴ AT = 2√3 = 2 × 1.732 = 3.464 cm

Answer 26.
Given :
Radius of cylinder = Radius of cone = r = 60 cm
Height of cylinder (h) =180 cm
and Height of cone (H) =120 cm

Now,

OR
Given :
Diameter of well = 3 m
∴ Radius, r = \(\frac { 3 }{ 2 } \)
Depth of well, h = 14 m
and Width of embankment = 4 m
Volume of sand dug out = πr²h

Answer 27.
Given : Speed of stream = 2 Km/h
Let the speed of the boat in still water = x km/hr
∴ Speed of boat in downstream = (x + 2) km/hr
and speed of boat in upstream = (x – 2) Km/hr
We know
Time = \(\frac { Distance }{ Speed } \)
∴ Time taken to go 24 Km Up stream = \(\frac { 24 }{ x-2 } \)
and time taken to go 24 Km downstream = \(\frac { 24 }{ x+2 } \)
According to the question,



OR
Let the unit digit be x and tens digit be y
Number = 10y + x
and, reverse of the number = 10x + y
According to the question,
xy = 14
and, 10y + x + 45 = 10x + y
⇒ 9x – 9y = 45
⇒ x-y =5
⇒ x = 5 + y
Substituting the value of x in equation (i), we get
(5 + y)y = 14
y² + 5y – 14 = 0
y² + 7y – 2y – 14 = 0
y(y + 7)-2(y + 7) =0
(y + 7) (y-2) =0
y + 7 = 0 ⇒ y = – 7 (digit is not negative)
y – 2 = 0 ⇒ y = 2
x =5 + y = 7
When y = 2, then
∴ Required two-digit number = 10y + x = 10 × 2 + 7 = 27

Answer 28.
Money saved on 1st day = ₹ 30
Money saved on IInd day = ₹ 30 + 500 p = ₹ 35
Money saved on IIIrd day = ₹ 30 + 500 p + 500 p = ₹ 35
and so on.
Amount of money saved on successive days is an AP with
First term, a = 30 and common difference, d = 5.
(i) Money saved on 10th day,
We know that
∴ a_n = a + (n – 1 )d
a₁₀ = a + 9d = 30 + 9 x 5 = ₹ 75
(ii) Money saved in 30 days

Answer 29.

Answer 30.

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 5

These Sample Papers are part of CBSE Sample Papers for Class 10 Maths. Here we have given CBSE Sample Papers for Class 10 Maths Paper 5

Time Allowed : 3 hours
Maximum Marks : 80

General Instructions

• All questions are compulsory.
• The question paper consists of 30 questions divided into four sections – A, B, C and D.
• Section A contains 6 questions of 1 mark each. Section B contains 6 questions of 2 marks each. Section C contains 10 questions of 3 marks each. Section D contains 8 questions of 4 marks each.
• There is no overall choice. However, an internal choice has been provided in 4 questions of 3 marks each and 3 questions of 4 marks each. You have to attempt only one of the alternatives in all such questions.
• Use of calculator is not permitted.

SECTION-A

Question 1.
AOBC is a rectangle whose three vertices are A(0, 3), 0(0, 0) and B(5, 0). Find the length of its diagonal CO.

Question 2.
In the given figure, if ∠AOB = 125°, then find the measure of ∠DOC.

Question 3.
A vessel has 136 lit. of oil. Another vessel has 92 lit. of oil. What is the capacity of the largest possible ladle which can measure out all the oil in each vessel in exact number of times ?

Question 4.
For what value of k, the system : kx + 3y = 11; 2x + 5y = 3, has unique solution.

Question 5.
If one root of 2x² + kx – 6 = 0 is 2, find the value of k.

Question 6.
Evaluate : sin 30° cos 60° + sin 60° cos 30°.

SECTION-B

Question 7.
If the HCF of 85 and 153 is expressible in the form 85m – 153, find the value of m.

Question 8.
Check whether (5, – 2), (6, 4) and (7, – 2) are the vertices of an isosceles triangle or not.

Question 9.
A bag contains 5 red balls, 8 white balls, 4 green balls and 7 black balls. If one ball is drawn at random, find the probability that it is :
(i) Black ball
(ii) Red ball
(iii) Not a green ball
(iv) Green or white ball

Question 10.
A two digit number is selected at random. What is the chance that will be :
(i) A composite number which is odd
(ii) Successor of a prime number

Question 11.
Can x – 1 be the remainder on division of a polynomial p(x) by 2x + 3 ? Justify your answer.

Question 12.
Find k if the sum of roots of equation x² – x + k(2x – 1) is Zero.

SECTION-C

Question 13.
Show that exactly one of the numbers n, n + 2 or n + 4 is divisible by 3.

Question 14.
If P(9a – 2, – b) divides line segment joining A(3a + 1,- 3) and B(8a, 5) in the ratio 3 : 1, find the value of a and b.

Question 15.
In the given figure, D and E trisect BC. Prove that 8AE² = 3AC² + 5AD²

Question 16.
In the given figure, AB is a chord of the circle and AOC is its diameter such that ∠ACB = 50°. If AT is the tangent to the circle at the point A, then find the measure of ∠BAT.

A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively. Find the sides AB and AC.

Question 17.
A path of 4 m width runs round a semi-circular grassy plot whose circumference is \(163 \frac { 3 }{ 7 } \) m.
Find :
(i) The area of the path.
(ii) The cost of gravelling the path at the rate of Rs 1.50 per m².
(iii) The cost of turfing the plot at the rate of 45 paise per m².

OR

All the vertices of a square lie on a circle. Find the area of the square, if area of the circle is 1256 cm². (Use π = 3.14)

Question 18.
500 persons are taking a dip into a cuboidal pond which is 80 m long and 50 m broad. What is the rise of water level in the pond, if the average displacement of the water by a person is 0.04 m3 ?

OR

How many spherical lead shots of diameter 4 cm can be made out of a solid cube of lead whose edge measures 44 cm ?

Question 19.
Show that the sum of all odd integers between 1 and 1000 which are divisible by 3 is 83667.

OR

If sum of first 6 terms of an A.P. is 36 and that of the first 16 terms is 256, find the sum of first 10 terms.

Question 20.
Solve for x :\(2\left( { x }^{ 2 }+\frac { 1 }{ { x }^{ 2 } } \right) -9\left( x+\frac { 1 }{ x } \right) +14=0\)

Question 21.
If tan (A + B) = √3 and tan (A – B) = \(\frac { 1 }{ \sqrt { 3 } } \);0° < A + B ≤ 90°; A > B, find sin 2A and cos 6B.

Question 22.
Find the mean of the following data :

SECTION-D

Question 23.
The angle of elevation of a stationary cloud from a point 2500 m above a lake is 15° and the angle of depression of its reflection in the lake is 45°. What is the height of the cloud above the lake level ?

OR

The angle of elevation of the top of a vertical tower PQ from a point X on the ground is 60°. At a point Y, 40 m vertically above X, the angle of elevation of the top is 45°. Calculate the height of the tower.

Question 24.
Construct a right triangle in which the sides (other than hypotenuse) are of lengths 5 cm and 4 cm. Then construct another triangle whose sides are 5/3 times the corresponding sides of the given triangle.

Question 25.
In a class test, the sum of Kamal’s marks in Maths and English is 40. Had he got 3 marks more in Maths and 4 marks less in English, the product of their marks would have been 360. Find his marks in two subjects separately.

Question 26.
A metallic right circular cone 20 cm high and whose vertical angle is 60° is cut into two parts at the middle of its height by a plane parallel of its base. If the frustum so obtained be drawn into a wire of diameter 1/16 cm, find the length of the wire.

OR

A pen stand made of wood is in the shape of a cuboid with four conical depressions and a cuboidal depression to hold the pens and pins, respectively. The dimensions of the cuboid are 10 cm, 5 cm and 4 cm. The radius of each of the conical depressions is 0.5 cm and the depth is 2.1 cm. The edge of the cubical depression is 3 cm. Find the volume of the wood in the entire stand.

Question 27.
Solve graphically x – y = 0, 2x + 3y – 30 = 0. Also find coordinates of points where 2x + 3y – 30 = 0 meets axis of X and Y.

OR

Students of a class are made to stand in rows. If 4 students are extra in each row, there would be two rows less. If 4 students are less in each row, there would be 4 more rows. Find the number of students in the class.

Question 28.
Two pipes running together can fill a tank in \(5 \frac { 5 }{ 21 } \) minutes. If one pipe takes 1 minute more than the other, then find the time in which each pipe would individually fill the tank.

Question 29.
If sin α = a sin β and tan α = b tan β, then prove that cos² α = \(\frac { { a }^{ 2 }-1 }{ { b }^{ 2 }-1 } \)

Question 30.
Draw more than ogive’ and ‘less than ogive’ for the following distribution on the same graph. Also find the median from the graph.

SOLUTIONS

SECTION-A

Solution 1:
We know, diagonals of rectangle are equal.
OC = AB
=> \(OC=\left| \sqrt { { (5-0) }^{ 2 }+{ (0-3) }^{ 2 } } \right| \)
=> \(OC=\left| \sqrt { 25+9 } \right| =\left| \sqrt { 34 } \right| \)
=> \(OC=\sqrt { 34 } \) units

Solution 2:
Given
∠AOB = 125°
We know, opposite sides of a quadrilateral circumscribing a circle subtend supplements angles at the centre of a circle.
∴∠AOB +∠DOC = 180°
=> 125° + ∠DOC = 180°
=> ∠DOC = 180° – 125°
= 55° Ans.

Solution 3:
Largest capacity of the ladle which can measure out all the oil
= H.C.F. of 136 and 92
Now, 136 = 2 x 2 x 2 x 17
and 92 = 2 x 2 x 23
H.C.F. (136, 92) = 4
∴ 4 litre capacity ladle can measure out all the oil in each vessel in exact number of times. Ans.

Solution 4:
Given equations are,
kx + 3y – 11 = 0
and 2x + 5y – 3 =0

Solution 5:
Given equation is,
2x² + kx – 6 = 0
One root of the equation is 2.
x = 2
2(2)² + k(2) – 6 = 0
8 + 2k – 6 = 0
2k = – 2
k = – 1.

Solution 6:
We have,
sin 30° cos 60° + sin 60° cos 30°
=> \(\frac { 1 }{ 2 } .\frac { 1 }{ 2 } +\frac { \sqrt { 3 } }{ 2 } .\frac { \sqrt { 3 } }{ 2 } \)
=> \(\frac { 1 }{ 4 } +\frac { 3 }{ 4 } \)
=> 1

SECTION-B

Solution 7:
85 = 5 x 17
153 = 3 x 3 x 17
∴H.C.F. (85, 153) = 17
But
∴H.C.F. = 85m – 153
17 = 85m – 153
17 + 153 = 85m
\(m= \frac { 170 }{ 85 } \)
= 2

Solution 8:
Let the given points be A(5, – 2), B(6, 4), C(7, – 2).
Now,

Here, we observe that
AB = BC
and AB+BC≠AC
∴Points A,B,C forms an isosceles triangle with AB = BC

Solution 9:
We have,
Number of red balls = 5
Number of white balls = 8
Number of green balls = 4
Number of black balls = 7
Total balls = 24

Solution 10:
Total two-digit numbers 10 to 99 = 90
∴n(S) = 90
(i) Composite numbers which is odd = {15, 21, 25, 27, 33, 35, 39, 45, 49, 51, 55, 57, 63, 65, 69, 75, 77, 81, 85, 87, 91, 93, 95, 99}
∴ n(E) = 24
Probability of odd composite number
= \(\frac { n(E) }{ n(S) } =\frac { 24 }{ 90 } =\frac { 4 }{ 15 } \)
(ii) Successor of a prime number = {12, 14, 18, 20, 24, 30, 32, 38, 42, 44, 48, 54, 60, 62, 68, 72, 74, 80, 84, 90, 98}
∴ n(E) = 21
Probability of a successor of a prime number
= \(\frac { n(E) }{ n(S) } =\frac { 21 }{ 90 } =\frac { 7 }{ 30 } \)

Solution 11:
No, degree of the remainder is always less than degree of the divisor.

Solution 12:
Given equation is,
x² – x + k(2x – 1) = 0
=> x² – x + 2kx – k = 0
=> x² + (2k – 1)x – k = 0

SECTION-C

Solution 13:
Let q be the quotient and r be the remainder when n is divisible by 3.
Therefore, n = 3q + r, where r = 0, 1, 2
=> n = 3q or n = 3q + 1 or n = 3q + 2
Case I: If n = 3q, then n is divisible by 3 but n + 2 = 3q + 2 and n + 4 = 3a + 4 are not divisible by 3.
Case II: If n = 3q + 1, then n + 2 = 3q + 3 = 3(q + 1), which is divisible by 3 and n + 4 = 3q + 5, which is not divisible by 3.
So, only (n + 2) is divisible by 3.
Case III: If n = 3q + 2, then n + 2 = 3q + 4, which is not divisible by 3 and n + 4 = 3q + 6 = 3(q + 2) is divisible by 3.
So, only (n + 4) is divisible by 3.
Hence, one and only one out of n, (n + 2), (n + 4) is divisible by 3.

Solution 14:
By section formula,

Solution 15:
Here, D and E trisect BC.
=>BD = DE = EC = \(\\ \frac { 1 }{ 3 } \) BC
or BE = 2 BD,
BC = 3BD
In right ∆ABD,
we have
AD² = AB² + BD²….(i)

Solution 16:
In the given circle,
∠ABC = 90°
∠ACB + ∠CAB = 90°
50° + ∠CAB = 90°
∠CAB = 90° – 50°
∠CAB = 40°
Now, OA is radius of the circle and AT is the tangent of circle
∠OAT = 90°
∠OAB + ∠BAT = 90°
40° + ∠BAT = 90°
∠BAT = 90° – 40°
∠BAT = 50°

Solution 17:
We have

Solution 18:
Given
length of pond, l = 80m
breadth of pond, b = 50m

Solution 19:
Odd integers between 1 and 1000 divisible by 3 are 3, 9,15, 21, … 999
The above series form an A.P.
where a = 3, d = 9 – 3 = 6 and an = 999
We know

Solution 20:
We have
\(2\left( { x }^{ 2 }+\frac { 1 }{ { x }^{ 2 } } \right) -9\left( x+\frac { 1 }{ x } \right) +14=0\)
\(2\left( { x }^{ 2 }+\frac { 1 }{ { x }^{ 2 } } +2 \right) -9\left( x+\frac { 1 }{ x } \right) +14-4=0\)

Solution 21:
We have
tan(A+B) = √3
=> tan(A+B) = tan 60°

Solution 22:

SECTION-D

Solution 23:
Let B be the point of oservation, P be the cloud and P’ be its reflection in the take.
AB = 2500 m
∠PBC = 15°
and ∠CBP = 45°
Let the height of cloud from lake be h m.
OP = OP’= h m

Solution 24:
Steps, of construction :
(1) Draw a line segment AB = 4 cm.
(2) Construct ∠XAB = 90°.
(3) With A as centre, draw an arc of radius 3 cm intersecting AX at C.
(4) Join BC. Thus, ∆ABC is obtained.
(5) Draw an acute angle ∠BAY below of AB.
(6) Mark points A1, A2, A3, A4, A5 on AY such that
AA1 = A1A2 = A2A3 = A3A4 = A4A5.
(7) Join A3B.
(8) Draw a line through A5, parallel to A3B intersecting extended line segment AB at B’.
(9) Through B’, draw a line parallel to BC intersecting AX at C’.
(10) Thus, ∆AB’C’ is obtained whose sides are \(\\ \frac { 5 }{ 3 } \) times the corresponding sides of ∆ABC.

Solution 25:
Let the marks obtained in Maths be x, then marks obtained in English = 40 – x
According to the question,
(x + 3) (40 – x – 4) = 360
=> (x + 3) (36 – x) = 360
=> – x² + 36x + 108 – 3x = 360
=> – x² + 33x – 252 = 0
=> x² – 21x – 12x + 252 = 0
=> x(x – 21) – 12(x – 21) = 0
(x – 21) (x – 12) = 0
Either x – 21 = 0 or x – 12 = 0
=> x = 21 or x = 12
If x = 21, then marks obtained in Maths = 21
and marks obtained in English = 40 – 21 = 19
If x = 12, then marks obtained in Maths = 12
and marks obtained in English = 40 – 12 = 28

Solution 26:
Given
Height of the cone = 20 cm
PQ (h) = 10 cm

Solution 27:
Given equations are
x – y = 0
y = x
=> x – y = 0

Solution 28:
Let one pipe fill the tank in x minute
then other will fill the tank in (x + 1) minutes.
According to the question,

Solution 29:
Given : sin α = a sin β
sin² α = > a² sin² β

Solution 30:

We hope the CBSE Sample Papers for Class 10 Maths Paper 5 help you. If you have any query regarding CBSE Sample Papers for Class 10 Maths Paper 5, drop a comment below and we will get back to you at the earliest.

CBSE Sample Papers for Class 10 Maths Paper 8

These Sample Papers are part of CBSE Sample Papers for Class 10 Maths. Here we have given CBSE Sample Papers for Class 10 Maths Paper 8

Time Allowed : 3 hours
Maximum Marks : 80

General Instructions

• All questions are compulsory.
• The question paper consists of 30 questions divided into four sections – A, B, C and D.
• Section A contains 6 questions of 1 mark each. Section B contains 6 questions of 2 marks each. Section C contains 10 questions of 3 marks each. Section D contains 8 questions of 4 marks each.
• There is no overall choice. However, an internal choice has been provided in 4 questions of 3 marks each and 3 questions of 4 marks each. You have to attempt only one of the alternatives in all such questions.
• Use of calculator is not permitted.

SECTION-A

Question 1.
If two integers a and b are written as a = x³y² and b = xy⁴; x, y are prime numbers, then find H.C.F. (a, b).

Question 2.
Find the perimeter of a triangle with vertices (0, 4), (0, 0) and (3, 0).

Question 3.
If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of 80°, then find the measure of ∠POA.

Question 4.
Determine the pair of linear equations from 2x + 3y = 5, y + \(\\ \frac { 2 }{ 3 } \) x = 5, 4x = 6y + 10, which has infinite solution.

Question 5.
The first term of an A.P. is p and its common difference is q. Find its 10th term.

Question 6.
If 9 sec A = 41, find cos A and cot A.

SECTION-B

Question 7.
Find a point which is equidistant from the points A(- 5, 4) and B(- 1, 6). How many such points are there ?

Question 8.
Which term of A.P. 129, 125, 121, … is its first negative term ?

Question 9.
Write the denominator of the rational number \(\\ \frac { 257 }{ 5000 } \) in the form 2^m x 5ⁿ, where m, n are non – negatives. Hence, write its decimal expansion without actual division.

Question 10.
It is known that a box of 600 electric bulbs contains 12 defective bulbs. One bulb is taken out at random from this box. What is the probability that it is a non-defective bulb ?

Question 11.
In a lottery of 50 tickets numbered 1 to 50, one ticket is drawn. Find the probability that the drawn ticket bears a prime number.

Question 12.
For what value of k the quadratic equation (2k + 3) x² + 2x – 5 = 0 has equal roots ?

SECTION-C

Question 13.
Find the smallest number which when divided by 161, 207 and 184 leaves remainder 21 in each case.

Question 14.
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 the points A( 1, – 2), B(2, 3), C(a, 2) and D(- 4, – 3) form a parallelogram, find the value of a and height of the parallelogram taking AB as base.

Question 15.
O is the point of intersection of the diagonals AC and BD of a trapezium ABCD with AB || DC. Through O, a line segment PQ is drawn parallel to AB meeting AD in P and BC in Q. Prove that PO = QO.

Question 16.
Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle.

Question 17.
The diameter of coin is 1 cm (fig.). If four such coins be placed on a table so that the rim of each touches that of the other two, find the area of the shaded region (Take π = 3.1416).

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

Question 18.
A hemispherical tank of diameter 3 m, is full of water. It is being emptied by a pipe at the rate of \(3 \frac { 4 }{ 7 } \) litres per second. How much time will it take to make the tank half empty ?

OR

A wall 24 m long, 0.4 m thick and 6 m high is constructed with the bricks each of dimensions 25 cm x 16 cm x 10 cm. If the mortar occupies 1/10 of the volume of the wall, then find the number of bricks used in constructing the wall.

Question 19.
The first term of an A.P. is 2 and the last term is 50. The sum of all these terms is 442. Find the common difference.

OR

Find the sum of first 17 terms of an A.P. whose 4th and 9th terms are – 15 and – 30 respectively.

Question 20.
Solve for x :
\({ \left( \frac { 2x+1 }{ x+1 } \right) }^{ 4 }-13{ \left( \frac { 2x+1 }{ x+1 } \right) }^{ 2 }+36=0\)

Question 21.
If cosec A + cot A = p, then prove that sec A = \(\frac { { p }^{ 2 }+1 }{ { p }^{ 2 }-1 } \)

Question 22.
Find mode of the following data :

SECTION-D

Question 23.
From the top of a lighthouse, the angles of depression of two ships on the opposite sides of it are observed to be a and p. If the height of the lighthouse is h metres and the line joining the ships passes through the foot of the lighthouse, show that the distance between the ships is \(\frac { h\left( tan\alpha +tan\beta \right) }{ tan\alpha \quad tan\beta } \) meters

OR

A boy is standing on the ground and flying a kite with 100 m of string at an elevation of 30°. Another boy is standing on the roof of a 10 m high building and is flying his kite at an elevation of 45°. Both the boys are on opposite sides of both the kites. Find the length of the string that the second boy must have so that the two kites meet.

Question 24.
Construct a right triangle in which sides (other than the hypotenuse) are of lengths 8 cm and 6 cm. Then construct another triangle whose sides are 3/4 times the corresponding sides of the first triangle.

Question 25.
State and prove the converse of Pythagoras theorem.

Question 26.
A hemispherical bowl of internal radius 9 cm is full of liquid. The liquid is to be filled into cylindrical shaped bottles each of radius 1.5 cm and height 4 cm. How many bottles are needed to empty the bowl ?

OR

A gulab jamun, contains sugar syrup about 30% of its volume. Find approximately how much syrup would be found in 45 gulab jamuns, each shaped like a cylinder with two hemispherical ends with length 5 cm and diameter 2.8 cm.

Question 27.
A boat goes 12 km downstream and 26 km upstream in 8 hours. It can go 16 km upstream and 32 km downstream in same time. Find the speed of boat in still water and the speed of stream.

OR

Solve graphically x – 2y = 1; 2x + y = 7. Also find the coordinates of points where the lines meet X-axis.

Question 28.
A sum of Rs 700 is to be used to give seven 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 prize.

Question 29.
Show that :
\(\frac { sinA-cosA+1 }{ sinA+cosA-1 } =\frac { 1 }{ secA-tanA } \)

Question 30.
The median of the following data is 20.75. Find the missing frequencies ‘x’ and ‘y’, if the total frequency is 100.

SOLUTIONS

SECTION-A

Solution 1.
Given : a = x³y²
b = xy⁴
H.C.F.(a, b) = xy².

Solution 2.
Given vertices are A(3, 0), B(0, 4) and 0(0, 0).
OA = 3 units
OB = 4 units
In ∆AOB, ∠O = 90°
∴From Pythagoras theorem,
AB² = OA² + OB² = 3² + 4²
AB = √9+16 = 5 unit
Perimeter of ∆OAB = OA + OB + AB
= 3 + 4 + 5 = 12 unit.

Solution 3.
Given :
∠APB = 80°
In ∆OAP and ∆OBP,
∠B = ∠A [each 90°]
OA = OB [radii]
OP = OP [common]
By RHS congruency rule

Solution 4.
Given equations are,
2x + 3y – 5 = 0
y + \(\\ \frac { 2 }{ 3 } \)x – 5=0
and 4x + 6y = 10
=> 2x + 3y – 5 = 0
2x + 3y – 15 = 0
and 4x + 6y – 10 =0

Solution 5.
Given : The first term of A.P. = a
and common difference = q
we know, a_n = a + (n – 1)d
∴ a₁₀ = p + (10 – 1)d
=> a₁₀ = P + 9d.

Solution 6.
Given : 9 sec A = 41
=> sec A = \(\\ \frac { 41 }{ 9 } \)
cos A = \(\\ \frac { 1 }{ sec A } \)
cos A = \(\\ \frac { 9 }{ 41 } \)

SECTION-B

Solution 7:
Let the point P(x, y) be equidistant from the points A (- 5, 4) and B(- 1, 6).

\(\left| \sqrt { { (x+5) }^{ 2 }+{ (y-4) }^{ 2 } } \right| =\left| \sqrt { { (x+1) }^{ 2 }+{ (y-6) }^{ 2 } } \right| \)
Squaring both sides
(x + 5)² + (y – 4)² = (x + 1)² + (y – 6)²
x² + 10x + 25 + y² – 8y + 16 = x² + 2x + 1 + y² – 12y + 36
10x – 2x – 8y + 12y + 41 – 37 = 0
8x + 4y + 4 = 0
2x + y + 1 =0
The points lying on 2x + y + 1 = 0 are equidistant from A and B.

Solution 8:
Given A.P. is 129, 125, 121,…
Here, a = 129, d = 125 – 129 = 121 – 125 = – 4
Let nth term be the first negative term.
Then T_n < 0.
We know, nth term, T_n = a + (n – 1)d
= 129 + (n -1) (- 4)
= 129 – 4n + 4
= 133 – 4n
∴T_n< 0
∵133 – 4n < 0
=> 133 < 4n
=> 4n > 133
=> n =\(\\ \frac { 133 }{ 4 } \)
= \(33 \frac { 1 }{ 4 } \)
Hence, 34th term will be the first negative term.

Solution 9:
We have,

Solution 10:
Total bulbs = 600
Number of defective bulbs = 12
Number of non-defective bulbs = 600 – 12 = 588
probability of a non defective bulb = \(\frac { Number\quad of\quad non\quad defective\quad bulbs }{ total\quad bulbs } \)
= \(\\ \frac { 588 }{ 600 } \)
= \(\\ \frac { 147 }{ 150 } \)

Solution 11:
Total tickets 1, 2, 3, 4, …, 50
=> n(S) = 50
Prime number tickets = 2, 3, 5, 7,11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47.
n(E) = 15
Probability that the ticket bears a prime number
P(E) = \(\\ \frac { n(E) }{ n(S) } \)
= \(\\ \frac { 15 }{ 50 } \)
= \(\\ \frac { 3 }{ 10 } \)
= 0.3

Solution 12:
Given quadratic equation is,
(2k + 3)x² + 2x – 5 = 0
On comparing the equation with ax² + bx + c = 0, we get
a = 2k + 3, b = 2, c = – 5
For equal roots,
b² – 4ac = 0
(2)² – 4(2k + 3) ( – 5) = 0
4 + 20(2k + 3) = 0
4 + 40k + 60 = 0
40k = – 64
k = \(\\ \frac { -64 }{ 40 } \)
k = \(\\ \frac { -8 }{ 5 } \)

SECTION-D

Solution 13:
Given numbers are 161, 207 and 184.
161 = 7 x 23
207 = 3 x 3 x 23
∴ 184 = 2 x 2 x 2 x 23
L.C.M. (161, 207, 1841) = 2 x 2 x 2 x 3 x 3 x 23 x 7
= 11592
Smallest number which when divided by 161, 207 and 184, leaves remainder 21 in each case
= 11592 + 21
= 11613. Ans.

Solution 14:
Given points are A(k + 1, 2k), B(3k, 2k + 3), C(5k – 1, 5k)
If the points A, B and C are collinear, then
ar (∆ABC) = 0

Solution 15:
Given, ABCD is trapezium in which
AB || CD || PQ

Solution 16:
Let O be the common centre of two concentric circles and let AB be a chord of large circle touching the smaller circle at P.
Construction : Join OP
Since OP is the radius of the smaller circle and AB is tangent to this circle at P,
∴ OP ⊥ AB
We know that the perpendicualr drawn from the centre of a circle to any chord of the circle bisects the chord.

Solution 17:
Given : Diameter of coin = 1 cm
Radius of coin, r = 0.5 cm
PQ = QR = RS = PS
= 2 x 0.5 = 1 cm

Solution 18:
Given : Diameter of hemispherical tank = 3 cm
Radius, r = \(\\ \frac { 3 }{ 2 } \)
Now, Volume of hemispherical tank = \(\frac { 2 }{ 3 } { \pi r }^{ 3 } \)

Solution 19:
Given, a = 2, a_n = 50, S_n = 442
We know that

Solution 20:
We have
\({ \left( \frac { 2x+1 }{ x+1 } \right) }^{ 4 }-13{ \left( \frac { 2x+1 }{ x+1 } \right) }^{ 2 }+36=0\)
let
\({ \left( \frac { 2x+1 }{ x+1 } \right) }^{ 2 }=a\)

Solution 21:
Given : cosec A + cot A = p ….(i)
We know,
cosec² A – cot² A = 1
=> (cosec A + cot A) (cosec A – cot A) = 1 [∵ a² – b² = (a+b)(a-b)]
=> (p) (cosec A – cot A) = 1

Solution 22:

SECTION-D

Solution 23:
Let PO be the lighthouse and A, B be the position of the two ships.
PO = h m
Angle of depression of ship at the point A = a°.
Angle of depression of ship at the point B = P°.
In ∆OPA, ∠P = 90°

Solution 24:
Steps of construction :
(1) Draw a line segment BC = 8 cm and make a right angle (∠DBC) at the point B.
(2) Taking B as centre and radius 6 cm, draw an arc which intersect BD at the point A.
(3) Join AC. Thus, ∆ABC is obtained.
(4) Draw a ray BX such that ∠CBX is an acute angle.
(5) Make four points B₁ B₂, B₃ and B₄ such that
BB₁ = B₁B₂ = B₂B₃ = B₃B₄
(6) Join B₄C.
(7) Through the point B₃, draw a line B₃C’ parallel to B₄C which intersect the line segment BC at the point C’.
(8) Through the point C’, draw a parallel line C’A’ parallel to CA which intersect the line segment BA at the point A’.
(9) Thus, ∆A B’C’ is the required triangle whose sides are \(\\ \frac { 3 }{ 4 } \) of corresponding sides of ∆ABC.

Solution 25:
Converse of Pythagoras Theorem : In a triangle, if the square of one side is equal to the sum of the squares of the other two sides, then the angle opposite to the side is a right angle.
Given : A triangle ABC such that AC² = AB² + BC²
Construction : Construct a triangle DEF such that
DE = AB, EF = BC and ∠E = 90°.
Proof : Since, ∆DEF is a right angled triangle with right angle at E, therefore, by Pythagoras theorem, we have

Solution 26:
Given :
Radius of bowl, r = 9 cm
For cylindrical bottle :
Radius, R = 1.5 cm
Height, H = 4 cm

Solution 27:
Let the speed of boat in still water be x km/h and speed of the stream be y km/hr.
.’. Speed of the boat in upstream = x – y
Speed of the boat in downstream = x + y

Solution 28:
Given :
Total amount = Rs 700
No. of cash prizes = 7
Each prize is Rs 20 less than preceding prize.
Therefore, the list of value of prizes form an A.P.
Let the first cash prize = Rs a
Difference between two consecutive prize = – Rs 20

Solution 29:
Given
L.H.S = \(\frac { sinA-cosA+1 }{ sinA+cosA-1 } \)

Solution 30:

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

CBSE Sample Papers for Class 10 Maths Paper 2

These Sample papers are part of CBSE Sample Papers for Class 10 Maths. Here we have given CBSE Sample Papers for Class 10 Maths Paper 2.

Time Allowed : 3 hours
Maximum Marks : 80

General Instructions

• All questions are compulsory.
• The question paper consists of 30 questions divided into four sections – A, B, C and D.
• Section A contains 6 questions of 1 mark each. Section B contains 6 questions of 2 marks each. Section C contains 10 questions of 3 marks each. Section D contains 8 questions of 4 marks each.
• There is no overall choice. However, an internal choice has been provided in 4 questions of 3 marks each and 3 questions of 4 marks each. You have to attempt only one of the alternatives in all such questions.
• Use of calculator is not permitted.

SECTION-A

Question 1.
What are the coordinates of the centroid of the triangle ABC if A(2,3), B(5, 8) and C(2, 1) ? [1]

Question 2.
From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. Find the radius of the circle. [1]

Question 3.
Express 4 as a sum of two irrational numbers.  [1]

Question 4.
Check if the system 3x – y – 8 = 0, 4x – 12y + 16 = 0 has infinite solution.  [1]

Question 5.
Check if x = 1 is a solution of 3x² – 2x – 1 = 0.  [1]

Question 6.
∆PQR is right angled at Q. If tan R = \(\frac { 24 }{ 7 } \), find sec R and sin R.  [1]

SECTION-B

Question 7.
If the points A(1, 2), O(0, 0) and C(a, b) are collinear, then find the relationship between
‘a’ and ‘b’.  [2]

Question 8.
There is a circular path around a sports field. Sania takes 24 minutes to drive one round of the field, while Ravi takes 16 minutes for the same. Suppose they both start at the same point and at the same time, and go in the same direction. After how many minutes will they meet again at the starting point ?  [2]

Question 9.
What is the probability that a number selected from the numbers 1, 2, 3,…., 25 is a prime number, when each of the given numbers is equally likely to be selected ?  [2]

Question 10.
If a number x is chosen at random from the numbers -2,-1, 0, 1, 2, what is the probability that x² < 2 ? [2]

Question 11.
Find the values of for which x² – 2x + k = 0 has real roots.  [2]

Question 12.
Write the value of a₃₀ – a₁₀ for the A.P. 4, 9, 14, 19, 24,….  [2]

SECTION-C

Question 13.
If the point A(2, – 4) is equidistant from P(3, 8) and Q(- 10, y), find the value of y. Also find the distance PQ. [3]

Question 14.
If ∆ABC ~ ∆DEF and AL and DM are their corresponding angle bisector segments then show that
\(\frac { AL }{ DM } =\frac { AB }{ DE } =\frac { BC }{ EF } =\frac { AC }{ DF } \)  [3]

Question 15.
Prove that the tangents drawn at the ends of a diameter of a circle are parallel.  [3]

Question 16.
Find the LCM of 396, 429 and 561 by prime factorization method.  [3]

Question 17.
The diameters of the front and rear wheels of a tractor are 80 cm and 2 m respectively. Find the number of revolutions that rear wheel will make to cover the distance which the front wheel covers in 140 revolutions.  [3]
OR
In the given figure, OACB is a quadrant of a circle with centre O and radius 3.5 cm. If OD = 2 cm, find the area of the shaded region.

Question 18.
A rectangular water tank of base 11 m x 6 m contains water up to a height of 5 m. If the water in the tank is transferred to a cylindrical tank of radius 3.5 m, find the height of the water level in the tank.  [3]
OR
A copper rod of diameter 1 cm and length 8 cm is drawn into a wire of length 18 m of uniform thickness. Find the thickness of the wire.  [3]

Question 19. 
Find the zeroes of the quadratic polynomial x² + 13x + 30, and verify the relationship between the zeroes and the coefficients.  [3]

Question 20.
The first and the last terms of an A.P. are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum ?  [3]
OR
How many numbers lie between 10 and 300, which when divided by 4 leave a remainder 3 ?

Question 21.

Question 22.
Find median for the following data :  [3]

SECTION-D

Question 23.
Construct a triangle ABC in which AC = AB = 4.5 cm and ∠A = 90°, then constuct a triangle similar to ∆ABC with its sides equal to (\(\frac { 5 }{ 4 } \))th of the corresponding sides of ∆ABC.  [4]

Question 24.
The angle of elevation of the top of a tower from a point A on the ground is 30°. On moving a distance of 20 metre towards the foot of the tower to a point B, the angle of elevation increases to 60°. Find the height of the tower and the distance of the tower from the
point A.  [4]
OR
As observed from the top of a lighthouse, 100 m above sea level, the angle of depression of a ship, sailing directly towards it, changes from 30° to 45°. Determine the distance travelled by the ship during the period of observation. [4]

Question 25.
In figure, P is the mid-point of BC and Q is the mid-point of AP. If BQ when produced meets AC at R, prove that RA = \(\frac { 1 }{ 3 } \)CA  [4]

Question 26.
A milk container of height 16 cm is made of metal sheet in the form of a frustum of a cone with radii of its lower and upper ends as 8 cm and 20 cm respectively. Find the cost of milk at the rate of ? 22 per litre which the container can hold.  [4]
OR
From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest cm².

Question 27.
In a rectangle if length is increased and breadth is reduced by 2 metre, the area is reduced by 28 sq. m. If length is reduced by 1 metre and breadth is increased by 2 metre, the area increased by 33 sq. m. Find the length and breadth of the rectangle.  [4]
OR
Out of group of swans, \(\frac { 7 }{ 2 } \) times the square root of the total number are playing on the shore of a pond. The remaining two are swimming in water. Find the total number of swans.  [4]

Question 28.
The students of a school decided to beautify the school on the Annual Day by fixing colourful flags on the straight passage of the school. They have 27 flags to be fixed at intervals of every 2 m. The flags are stored at the position of the middle most flag. Ruchi was given the responsibility of placing the flags. Ruchi kept her books where the flags were stored. She could carry only one flag at a time. How much distance did she cover in completing this job and returning back to collect her books ?  [4]

Question 29.  
Show \(\frac { { cot }^{ 2 }A }{ 1+cosecA } +1=cosec A\)  [4]

Question 30.
Find the value of p, if the mean of the following distribution is 7.5.  [4]

SOLUTIONS
Section – A

Answer 1.
Here A(2,3) B(5,8) and C(2,1).

Answer 2.
Let, radius of circle OP = r
∵ Tangent to a circle is perpendicular to the radius
∴ In ∆POQ, using Pythagoras theorem

OQ² = PQ² + OP²
25² = 24² + r²
⇒ r² = 25² – 24²
⇒ r² = (25 + 24)(25 – 24)
⇒ r² = 49
⇒ r = 7 cm

Answer 3.
We know that (2 + √3) and (2 – √3) are irrational numbers
(2 + √3) + (2 – √3)
2 + √3 + 2 – √3 = 4

Answer 4.
Given system of equations is,
3x – y – 8 =0
and 4x – 12y + 16 = 0
For infinite solution

Answer 5.
We have
3x² – 2x – 1 = 0
If x = 1,
Then,
3 × (1)² – 2 × 1 – 1 = 0
3 – 2 – 1 = 0
3 – 3 = 0
0 = 0
L.H.S = R.H.S
Hence, x = 1 is a solution of the given equation.

Answer 6.
We have,
tan R =\(\frac { 24 }{ 7 } \)
From ∆PQR, tan R = \(\frac { PQ }{ RQ } \)
In ∆PQR, PR² = PQ² + RQ²
PR² = 24² + 7²
PR² = 576+ 49
PR² = 625
PR = √625
PR 25

sec R = \(\frac { PR }{ RQ } =\frac { 25 }{ 7 }\)
sin R = \(\frac { PQ }{ PR } =\frac { 24 }{ 25 } \)

SECTION-B

Answer 7.
We have, A( 1, 2), O(0, 0) and C(a, b) are collinear.
Since, the given points are collinear, therefore, area of triangle formed by them is zero.
i.e., ar (∆AOB) = 0
⇒ \(\frac { 1 }{ 2 } \) x1(y2 – y3) + x2(y3 – yx) + x3{yx – y2) 1=0
⇒ \(\frac { 1 }{ 2 } \) 1(0 – b) + 0(b – 2) + a(2 – 0) i =0
⇒ – b + 2a = 0
⇒ b = 2a.

Answer 8.
Given, Sania takes 24 minutes for one round and Ravi takes 16 minutes for one round. It means that Ravi takes lesser time than Sonia to drive one round of field.
Now,
Required time = L.C.M. of the time taken by Sonia and Ravi for completing one round
= L.C.M. of 24 and 16
∵ 24 = 2³ x 3 and 16 = 2⁴
∴ L.C.M. = 2⁴ x 3 = 16 x 3 = 48
Hence, Ravi and Sonia will meet each other at the starting point after, 48 minutes.

Answer 9.
Numbers are 1, 2, 3,…., 25.
∴ Total numbers n(S) = 25
Prime numbers between 1 to 25 are 2, 3, 5, 7, 11, 13, 17, 19, 23
⇒ Total prime numbers = n(E) = 9
∴ P(Prime Number) = \(\frac { n(E) }{ n(S) } \quad \frac { 9 }{ 25 }\)

Answer 10.
Given numbers are -2,-1, 0, 1, 2
⇒ Total numbers = 5
Here, (-2)² = 4, (-1)² = 1, 0² = 0, 1² = 1, 2² = 4
For x = – 1, 0, 1  x² < 2
⇒ Favourable outcomes = 3
p(x² < 2) = \(\frac { Favourable\quad outcomes }{ Total\quad outcomes } = \frac { 3 }{ 5 } \)

Answer 11.
Given equation is,
x² – 2x + k = 0
Here a = 1, b = -2, c = k
We know, D = b² – 4ac
⇒ D = (-2)² – 4 × 1 × k
⇒ D = 4 – 4k

For real roots,
D =0
⇒ 4 – 4 k =0
⇒ – 4k = – 4
⇒ k = \(\frac { -4 }{ -4 } \)
⇒ k = 1

Answer 12.
Given A.P. is, 4, 9, 14, 19, 24
Here, a = 4
d = 9 – 4=5
We know
a_n = a + (n – 1) d
a₃₀ = 4 + (30 – 1) 5
⇒ a₃₀ = 4 + 29 x 5
⇒ a₃₀ = 4 + 145
⇒ a₃₀ = 149
And a_n = a + (10 – 1) d = 4 + (30 -1 )5
= 4 + (9 x 5) = 4 + 45
⇒ a₃₀ = 49
∴ a₃₀ – a₁₀ = 149 – 49
= 100

SECTION-C

Answer 13.

AP =AQ
⇒ AP² = AQ²
Let (x₁, y₁) = (2, – 4)
(x₂, y₂) = (3, 8)
(x₃, y₃) = (-10, y)
Then,
(x₂ – x₁)² + (y₂ – y₁)² = (x₃ – x₁ )² + (y₃ – x₁)²
(3 – 2)² + (8 + 4)² = (-10 – 2)² + (y + 4)²
(1)² + (12)² = (12)² + (y + 4)²
1 + 144 = 144 + (y + 4)²
1 = (y + 4)²
±1 = y+4
⇒ y = -1 – 4 or y = 1 – 4
⇒ y=-5 or y = -3

Answer 14.

Answer 15.
Given : A circle with centre at O. AB and CD are tangents drawn at the end of diameter.
To prove : AB || CD
Proof : We know that a tangent at any point of a circle is perpendicular to the radius throught the point of contact.
∴ OM ⊥ AB and ON ⊥ CD

⇒ ∠OMB = 90°
∠OMB = 90°
∠OND = 90°
∠ONC = 90°
∠OMA = ∠OND
∠OMB = ∠ONC
These are the pair of alternate interior angles.
Since, alternate angles are equal, the line AB and CD are parallel to each other.
i.e, AB || CD

Answer 16.

Prime factorisation of the given numbers are,
396 = 2 × 2 × 3 × 3 × 11 = 2² × 3² × 11
429 = 3 × 143
561 = 3 × 187
∴ LCM (396, 429, 561) = 2² × 3² × 143 × 187 × 11 = 10589436

Answer 17.
Given, Diameter of front wheel = 80 cm
∴ Radius, r = \(\frac { 80 }{ 2 }\) = 40 cm
and Diameter of rear wheel = 2 m
∴ Radius, R = \(\frac { 2 }{ 2 }\) = 1 m = 100 cm
Number of revolutions covered by front wheel = 140.
Let the number of revolutions covered by rear wheel be n.
According to the question,
Distance covered by real wheel in n revolutions = Distance covered by front wheel in 140 revolutions
⇒ n x 2πR = 140 x 2πr
n × R = 140 x r
n x 100 = 140 x 40
n = \(\frac { 140\times 40 }{ 100 }\) = 56
Thus, the number of revolutions covered by rear wheel is 56.
OR
Given : Radius, OA = 3.5 cm
and OD = 2 cm
Area of shaded region = Area of quadrant OACB – Area of ∆DOB

Hence, area of shaded region is 6-125 cm².

Answer 18.
Given : Length of cuboid, l = 6 m
Breadth of cuboid, b = 11 m
Height of cuboid, h = 5 m
Radius of cylinder, r = \(\frac { 3.5 }{ 10 } \) m = \(\frac { 7 }{ 2 } \) m
Let height of water level in the cylindrical tank be H m
Since, water in rectangular tank is transferred to cylindrical tank.
∴ Volume of cuboid = Volume of cylinder of height H
l ×b × h =πr²H
6 x 11 x 5 = \(\frac { 22 }{ 7 } \times \frac { 7 }{ 2 } \times \frac { 7 }{ 2 }\) × H
\(\frac { 30\times 2 }{ 7 } \) = H
H = \(\frac { 60 }{ 7 }\) m = 8.57 m.
OR
Given : Diameter of the copper rod = 1 cm
So, Radius (r) of the copper rod = 1/2 cm
Length of copper rod, h = 8 cm
Length of wire, H = 18 m = 18 x 100 cm
Let R be the radius of the cross-section (circular) of the wire.
We know that volume of cylinder = πr²h Now, according to the question
Now, according to the question,

Answer 19.
Let p(x) = x² + 13x + 30
= x² + 10x + 3x + 30
= x(x + 10) + 3(x + 10)
=(x + 3) (x + 10)
⇒ x + 3 = 0; x + 10 = 0
⇒ x = – 3, x = – 10
Let α = – 3
and β = – 10
Now, α + β = \(-\frac { b }{ a }\)
– 3 – 10 = \(-\frac { (-13) }{ 1 } \)
– 13 = – 13
and αβ = \(\frac { c }{ a } \)
(-3) (-10) = \(\frac { 30 }{ 1 }\)
30 = 30
Thus, the releationship between the zeroes and coefficients is verified.

Answer 20.
Given : First term of A.P., a = 17
Last term, l = 350
Common difference, d = 9
We know,
a_n = a + (n – 1) d
∵ 350 is the nth term of given A.P.,
∵ 350 = 17 + (n – 1) 9
350 – 17 = (n – 1) 9
333 = (n – 1) 9
\(\frac { 333 }{ 9 } \) = n – 1
37 = n – 1
37 + 1 =n
38 = n
We know that sum of n terms of A.P is given as,
S_n = \(\frac { n }{ 2 } \)[a + l]
⇒ S₃₈ = \(\frac { 38 }{ 2 }\)[17 + 350]
= 19 [367] = 6973
n = 38 and S_n = 6973.
OR
The first number greater than 10 which when divided by 4 leaves a remainder 3 is 11.
So, the next number will be 11 + 4 = 15
The other numbers will be 15 + 4 = 19, 19 + 4 = 23….
Thus, A.P. = 11, 15, 19, 23 …..
The last term of this A.P. will be 299.
We have to find n. i.e., number of terms in the A.P.
We know,
a + (n – 1 )d = a_n
⇒ 11 + (n – 1)4 =299
⇒ 11 + 4n – 4 = 299
⇒ 7 + 4n =299
⇒ 4n = 299 – 7
⇒ 4n = 292
⇒ n = \(\frac { 292 }{ 4 } \)
⇒ n = 73
∴ numbers lie between 10 and 300 which divided by 4 leaves a remainder 3.

Answer 21.

Answer 22.

N = 40 ⇒ \(\frac { N }{ 2 } \) = 20.
Frequency just greater than 20 is 22.
So, 30 – 49 will be median class.
Here, l = 30
ƒ = 4
c ƒ = 18
h = 9

SECTION-D

Answer 23.
Steps of construction :

• Draw a line segment AB = 4-5 cm
• Construct ∠YAB = 90°
• Taking A as centre and radius 4-5 cm, draw an arc intersecting AY at C.
• Join BC. Thus, ∆ABC is obtained. Y
• Draw any ray AX making acute angle with AB on the side opposite to the vertex A.
• Mark 5 points namely A_1, A₂, A_3, A_4, A₅ on AX such that, AA₁ = A₁A₂ = A₂A₃ = A₃ A₄ = A₄ A₅
• Join A₄ to B and draw a line through A₅ parallel to A₄B intersecting extended AB at B’
• Draw a line B’C’ parallel to BC to intersect AY at C’.

Thus, ∆AB’C’ is the required triangle whose sides are \(\frac { 5 }{ 4 }\) times of the corresponding sides of ∆ABC.

Answer 24.
Let CD be the tower of height y m.
Let BD = x.
In ∆CAD, we have

tan 30° = \(\frac { CD }{ AD } \)
\(\frac { 1 }{ \sqrt { 3 } } \) = \(\frac { y }{ 20+x } \)
20 + x = y√3
x = y√3 – 20
In ∆CBD, we have
tan 60°= \(\frac { CD }{ BD } \)
√3 = \(\frac { y }{ x } \)
⇒ x√3 = y
Putting the value of y in equation (i), we get
x = (x√3) (√3) – 20
x = 3x – 20
20 = 3x – x
20 = 2x
\(\frac { 20 }{ 2 } \) = x
10 = x
Putting the value of x in equation (ii), we get
10√3 = y
⇒ y = 10 × 1.732 = 1.732
Hence, distamce of the foot of tower from point A
= 20 + x = 20 + 10 = 30 m
and Height of tower = 17.32 m

OR

Let A and B be the two positions of the ship. Let d be the distance travelled by ship during the period of observation i.e. AB = d metres.
Period of observation i.e. AB = d meters.
Let the observer be at O, the top of the light house PO.
It is given that PO = 100 m and the angles of depression from O of A and B are 30° and 45° respectively.
∴ ∠OAP = 30° and ∠OBP = 45°

In ∆OPA, we have
tan 45° = \(\frac { OP }{ BP } \)
⇒ 1 = \(\frac { 100 }{ BP }\)
⇒ BP = 100 m
In ΔOPA, we have
⇒ tan 30° = \(\frac { OP }{ AP } \)
⇒ \(\frac { 1 }{ \sqrt { 3 } } \) = \(\frac { 100 }{ d+BP }\)
⇒ d + BP = 100√3
⇒d + 100 = 100√3 [∵ BP = 100 m]
⇒ d = 100√3 – 100
⇒ d = 100(√3 – 1)
= 100(1.732 – 1) = 73.2 m
Hence, the distance travelled by the ship from A to B is 73.2 m

Answer 25.
Given : ∆ABC in which P is the mid-point of BC, Q is the mid-point of AP, such that BQ produced meets AC at R.
To prove : RA = \(\frac { 1 }{ 3 } \) CA.

Construction : Draw PS || BR, meeting AC at S.
Proof : In ∆BCR, P is the mid-point of BC and PS || BR.
∴ By basic proportionality theorem, S is the mid-point of CR.
⇒ CS = SR
Similarly, in ∆APS, Q is the mid-point of AP and QR || PS.
∴ R is the mid-point of AS.
⇒ AR=RS
From equations (i) and (ii), we get
AR =RS = SC .
⇒ AC =AR +RS +SC = 3 AR
AR = \(\frac { 1 }{ 3 } \) AC = \(\frac { 1 }{ 3 } \) CA

Answer 26.
Given :
Radii of upper end of frustum, R = 20 cm
Radii of lower end of frustum, r = 8 cm
Height of frustum, h = 16 cm

We know that
1 cm³ = 0.001 litre
⇒ Volume of container = 10.46 litre
Cost of 1 litre milk = ₹ 22
⇒ Total cost = 22 × 10.46 = ₹ 230.12
OR
Let r cm be the radius and h cm be height of the cone .
Then r = 0.7 cm and h = 2.4 cm
let r₁ cm be the radius, l cm be the slant height and h₁ cm be the height of the cone.
Then r₁ = 0.7 cm and h₁ = 2.4 cm


Now, total surface area of the remaining solid
= C.S.A. of cylinder + C.S.A. of cone + Area of base of the cylinder
= 2πrh + πr₁l + πr₂
= 2πrh + πr₁ + πr²   [∵ r = r1]
= [ 2 × \(\frac { 22 }{ 7 } \) × 0.7 × 2.4 + \(\frac { 22 }{ 7 } \) × 0.7 × 2.5 + \(\frac { 22 }{ 7 } \) × 0.7 × 0.7 ] cm²
= ( 10.56 + 5.5 + 1.54 )cm2 = 17.6 cm²

Answer 27.
Let Length of rectangle = x
and Breadth of rectangle = y
∴ Area of rectangle = xy
According to question,
(x + 2)(y – 2) = xy – 28
and (x – 1) (y + 2) = xy + 33
From equation (i), we have
xy – 2x + 2y – 4 = xy – 28
– 2x + 2y = – 28 + 4
– 2x + 2y = – 24
2x – 2y = 24
x – y =12
x =12 + y
From equation (ii), we have
(x -1) (y + 2) = xy + 33
xy + 2x – y – 2 = xy + 33
2x – y =33 + 2
2x – y =35
Putting the value of x from equation (iii) to equation (iv)
2(12 + y) – y =35
24 + 2y – y =35
24 + y = 35
y = 35 – 24 =11
Putting the value y in equation (i), we get
x = 12 + 11 = 23
Hence, Length of rectangle = 19 m
Breadth of rectangle = 11 m
OR
Let the total number of swans be x².
Then, number of swans playing at the pond shore
= \(\frac { 7 }{ 2 } \sqrt { { x }^{ 2 } } =\frac { 7 }{ 2 } x\)
Number of swans playing in water = 2
Total number of swans = No. of swans playing on the shore + No.of swans playing in water
⇒ x² = \(\frac { 7 }{ 2 } \)x + 2
⇒ 2x² = 7x + 4
⇒ 2x² – 7x – 4 = 0
⇒ 2x² – 8x + x – 4 = 0
⇒ 2x(x – 4) + (x – 4) = 0
⇒ (2x + 1) + (x – 4) = 0
⇒2x + 1 = 0 or x – 4 = 0
⇒ x = \(\frac { -1 }{ 2 } \) or x = 4
But x = \(\frac { -1 }{ 2 } \) is not feasible. Thus, x = 4
Hence, number of swans = x² = 4² = 16.

Answer 28.

Let A be the middle-most point and 13 flags are fixed at points A₁,A₂,A₃,…. A₁₃, to the left of A and remaining to the right of A.
Distance travelled for fixing and coming back to A for
(i) first flag = (2 + 2)m = 4m = a
(ii) second flag = (4 + 4) m = 8 m = a₂
(iii) third flag = (6 + 6) m = 12 m = a₃
∴ a_1, a₂, a₃…., form an AP in which
a = 4, a₂ = 8, a₃ = 12……n = 13
and common difference, d =8 – 4 = 4
S₁₃ = \(\frac { 13 }{ 2 } \)[2 x 4 + (13-1) 4]
= \(\frac { 13 }{ 2 } \)(8 + 48) = \(\frac { 13 }{ 2 } \) × 56 = 364
∴ Distance travelled to fix 13 flags to the left of A = 364 m.
Similarly, the distance travelled to fix remaining 13 flags to right of A = 364 m
∴ Total distance travelled = 2 × 364 m = 728 m.

Answer 29.

Answer 30.

We have, ∑f_i = 41 + p, ∑f_ix_i = 303 + 9p
But Mean = 7.5
∴ Mean = \(\frac { \Sigma { f }_{ i }{ x }_{ i } }{ \Sigma { f }_{ i } }\)
⇒ 7.5 = \(\frac { 303+9p }{ 41+p } \)
⇒ 7.5 × (41 + p) = 303 + 9p
⇒ 307.5 + 7.5p = 303 + 9p
⇒ 9p – 7.5p = 307.5 – 303
⇒ 1.5p = 4.5
⇒ p = 3

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