10th Standard Mathematics Questions -I- 2019-2020

Download Tamil Nadu 10th Standard Maths question papers, model tests, one-mark questions, important questions, and public exam papers in PDF format. Free study materials and answer keys for TN State Board students.

Questions:

Multiple Choice Question

40 x 1 = 40

The angle of elevation of the top of tree from a point at a distance of 250 m from its base is 60^o. The height of tree is ___________

(a)

250 m

(b)

\(250\sqrt { 3 } \)

(c)

\(\frac { 250 }{ 3 } m\)

(d)

\(200\sqrt { 3 } \)

If the smallest value and co-efficient of range a data are 25 and 0.5 respectively. Then the largest value is ___________

(a)

(b)

(c)

100

(d)

12.5

A purse contains 10 notes of Rs. 2000, 15 notes of Rs. 500, and 25 notes of Rs. 200.One note is drawn at random. What is the probability that the note is either a Rs. 500, note or Rs. 200 note?

(a)

\(\frac { 1 }{ 5 } \)

(b)

\(\frac { 3 }{ 10 } \)

(c)

\(\frac { 2 }{ 3 } \)

(d)

\(\frac { 4 }{ 5 } \)

A floating boat having a length 3m and breadth 2m is floating on a lake. The boat sinks by 1 cm when a man gets into it. The mass of the man is (density of water is 10000 kg/m³)

(a)

50 kg

(b)

60 kg

(c)

70 kg

(d)

80 kg

The curved surface area of a cylinder is 264 cm² and its volume is 924 cm². The ratio of diameter to its height is ___________

(a)

3:7

(b)

7:3

(c)

6:7

(d)

7:6

The material of a cone is converted into the shape of a cylinder of equal radius. If the height of the cylinder is 5 cm, then height of the cone is ___________

(a)

10 cm

(b)

15 cm

(c)

18 cm

(d)

24 cm

If the mid-point of the line segment joining \(A\left( \frac { x }{ 2 } ,\frac { y+1 }{ 2 } \right) \) and B(x + 1, y-3) is C(5, -2) then find the values of x, y ____________

(a)

(6, -1)

(b)

(-6, 1)

(c)

(-2, 1)

(d)

(3, 5)

Find the ratio in which the line segment joining the points (-3, 10) and (6,-8) is internally divided by (-1, 6) ____________

(a)

7:2

(b)

3:4

(c)

2:7

(d)

5:3

If the angle between two radio of a circle is ^o, the angle between the tangents at the end of the radii is ____________

(a)

50^o

(b)

90^o

(c)

40^o

(d)

70^o

10.

I the given figure DE||AC which of the following is true.

(a)

\(x=\frac { ay }{ b+a } \)

(b)

\(x=\frac { a+b }{ ay } \)

(c)

\(x=\frac { ay }{ b-a } \)

(d)

\(\frac { x }{ y } =\frac { a }{ b } \)

11.

If triangle PQR is similar to triangle LMN such that 4PQ = LM and QR = 6 cm then MN is equal to ____________

(a)

12 cm

(b)

24 cm

(c)

10 cm

(d)

36 cm

12.

If P and Q are matrices, then which of the following is true?

(a)

PQ ≠ QP

(b)

(P^T)^T≠ P

(c)

P + Q ≠ Q + P

(d)

All are true

13.

\(\frac { { x }^{ 2 }+7x12 }{ { x }^{ 2 }+8x+15 } \times \frac { { x }^{ 2 }+5x }{ { x }^{ 2 }+6x+8 } =\_ \_ \_ \_ \_ \_ \_ \_ \_ \)

(a)

x+2

(b)

\(\frac { x }{ x+2 } \)

(c)

\(\frac { 35{ x }^{ 2 }+60x }{ { 48x }^{ 2 }+120 } \)

(d)

\(\frac { 1 }{ x+2 } \)

14.

Which of the following is correct
(i) Every polynomial has finite number of multiples
(ii) LCM of two polynomials of degree 2 may be a constant
(iii) HCF of 2 polynomials may be constant
(iv) Degree of HCF of two polynomials is always less then degree of LCM

(a)

(i) and (ii)

(b)

(iii) and (iv)

(c)

(iii) only

(d)

(iv) only

15.

A boy saves Rs. 1 on the first day Rs. 2 on the second day, Rs. 4 on the third day and so on. How much did the boy will save upto 20 days?

(a)

2¹⁹+ 1

(b)

2¹⁹- 1

(c)

2²⁰- 1

(d)

2²¹- 1

16.

The difference between the remainders when 6002 and 601 are divided by 6 is ____________

(a)

(b)

(c)

(d)

17.

If f is constant function of value \(\frac { 1 }{ 10 } \), the value of f(1) + f(2) + ... + f(100) is _________

(a)

\(\frac { 1 }{ 100 } \)

(b)

100

(c)

\(\frac { 1 }{ 10 } \)

(d)

18.

If function f : N⟶N, f(x) = 2x then the function is, then the function is ___________

(a)

Not one - one and not onto

(b)

one-one and onto

(c)

Not one -one but not onto

(d)

one - one but not onto

19.

Let f(x) = x²- x, then f(x- 1) - (x + 1) is ___________

(a)

(b)

2-2x

(c)

2-4x

(d)

4x-2

20.

If f : R⟶R is defined by (x) = x²+ 2, then the preimage 27 are _________

(a)

0.5

(b)

5, -5

(c)

5, 0

(d)

\(\sqrt { 5 } ,-\sqrt { 5 } \)

21.

which of the following is true?

(a)

0 ≤ p(∈) ≤ 1

(b)

p(∈) > 1

(c)

p(∈) < 0

(d)

\(-\frac { 1 }{ 2 } \ge P(\in )\le \frac { 1 }{ 2 } \)

22.

A number x is chosen at random from -4, -3, -2, -1, 0, 1, 2, 3, 4 find the probability that |x| ≤ 4

(a)

(b)

(c)

\(\frac{1}{2}\)

(d)

\(\frac{1}{9}\)

23.

The value of sin²θ + \(\frac { 1 }{ 1+{ tan }^{ 2 }\theta } \) of ___________

(a)

sin²θ

(b)

cos²θ

(c)

secθ

(d)

24.

If sec θ + tan θ = n, and sec θ - tan θ = 0, then the value of mn is ___________

(a)

(b)

(c)

土1

(d)

土2

25.

If (sin α + cosec α)²+ (cos α + sec α)²= k + tan²α + cot²α, then the value of k is equal to

(a)

(b)

(c)

(d)

26.

The probability of getting a job for a person is \(\frac{x}{3}\). If the probability of not getting the job is \(\frac{2}{3}\) then the value of x is

(a)

(b)

(c)

(d)

1.5

27.

The range of the data 8, 8, 8, 8, 8. . . 8 is

(a)

(b)

(c)

(d)

28.

The electric pole subtends an angle of 30° at a point on the same level as its foot. At a second point ‘b’ metres above the first, the depression of the foot of the pole is 60°. The height of the pole (in metres) is equal to

(a)

\(\sqrt { 3 } \) b

(b)

\(\frac { b }{ 3 } \)

(c)

\(\frac { b }{ 2 } \)

(d)

\(\frac { b }{ \sqrt { 3 } } \)

29.

If the sequence t₁, t₂, t₃... are in A.P. then the sequence t₆, t₁₂, t₁₈,.... is

(a)

a Geometric Progression

(b)

an Arithmetic Progression

(c)

neither an Arithmetic Progression nor a Geometric Progression

(d)

a constant sequence

30.

The least number that is divisible by all the numbers from 1 to 10 (both inclusive) is

(a)

2025

(b)

5220

(c)

5025

(d)

2520

31.

The point of intersection of 3x − y = 4 and x + y = 8 is

(a)

(5, 3)

(b)

(2, 4)

(c)

(3, 5)

(d)

(4, 4)

32.

If (5, 7), (3, p) and (6, 6) are collinear, then the value of p is

(a)

(b)

(c)

(d)

33.

A tangent is perpendicular to the radius at the

(a)

centre

(b)

point of contact

(c)

infinity

(d)

chord

34.

If \(\triangle\)ABC is an isosceles triangle with \(\angle\)C = 90^o and AC = 5 cm, then AB is

(a)

2.5 cm

(b)

5 cm

(c)

10 cm

(d)

\(5\sqrt { 2 } \)cm

35.

The volume (in cm³) of the greatest sphere that can be cut off from a cylindrical log of wood of base radius 1 cm and height 5 cm is

(a)

\(\frac{4}{3}\pi\)

(b)

\(\frac{10}{3}\pi\)

(c)

\(5\pi\)

(d)

\(\frac{20}{3}\pi\)

36.

The height of a right circular cone whose radius is 5 cm and slant height is 13 cm will be

(a)

12 cm

(b)

10 cm

(c)

13 cm

(d)

5 cm

37.

If f(x) = 2x² and g(x) = \(\frac{1}{3x}\), then f o g is

(a)

\(\\ \frac { 3 }{ 2x^{ 2 } } \)

(b)

\(\\ \frac { 2 }{ 3x^{ 2 } } \)

(c)

\(\\ \frac { 2 }{ 9x^{ 2 } } \)

(d)

\(\\ \frac { 1 }{ 6x^{ 2 } } \)

38.

Let A = {1, 2, 3, 4} and B = {4, 8, 9, 10}. A function f: A ⟶ B given by f = {(1, 4), (2, 8), (3, 9), (4,10)} is a

(a)

Many-one function

(b)

Identity function

(c)

One-to-one function

(d)

Into function

39.

If the roots of the equation q²x² + p²x + r² = 0 are the squares of the roots of the equation qx² + px + r = 0, then q, p, r are in __________.

(a)

A.P

(b)

G.P

(c)

Both A.P and G.P

(d)

none of these

40.

y² + \(\frac {1}{y^{2}}\) is not equal to

(a)

\(\frac {y^{2} + 1}{y^{2}}\)

(b)

\({ \left( y+\frac { 1 }{ y } \right) }^{ 2 }\)

(c)

\({ \left( y-\frac { 1 }{ y } \right) }^{ 2 }+2\)

(d)

\({ \left( y+\frac { 1 }{ y } \right) }^{ 2 }-2\)

Generic 2

20 x 2 = 40

41.

Let A = {0, 1, 2, 3} and B = {1, 3, 5, 7, 9} be two sets. Let f : A \(\rightarrow\)B be a function given by f(x) = 2x + 1. Represent this function as an arrow .

42.

Let A = {0, 1, 2, 3} and B = {1, 3, 5, 7, 9} be two sets. Let f: A \(\rightarrow\)B be a function given by f(x) = 2x + 1. Represent this function as a set of ordered pairs.

43.

Let A = {1,2, 3, 4} and B = {-1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12} Let R = {(1, 3), (2, 6), (3, 10), (4, 9)} \(\subseteq \) A x B be a relation. Show that R is a function and find its domain, co-domain and the range of R.

44.

Find the sum and product of the roots for each of the following quadratic equations:
2x² + 5x + 7 = 0

45.

Find the slope of a line joining the points (sin θ, -cos θ) and (-sin θ, cos θ)

46.

In fig. if PQ || BC and PR || CD prove that

\(\frac { QB }{ AQ } =\frac { DR }{ AR } \)

47.

If A and B are two mutually exclusive events of a random experiment and P(not A) = 0.45, P(A U B) = 0.65, then find P(B).

48.

A die is rolled and a coin is tossed simultaneously. Find the probability that the die shows an odd number and the coin shows a head.

49.

Find the slope of the line which is Parallel to y = 0.7 x − 11

50.

Find the maximum volume of a cone that can be carved out of a solid hemisphere of radius r units.

51.

A tower stands vertically on the ground. from a point on the ground, which is 48m away from the foot of the tower, the angel of elevation of the top of the tower is 30°.find the height of the tower.

52.

The ratio of the volumes of two cones is 2 : 3. Find the ratio of their radii if the height of second cone is double the height of the first.

53.

Using the functions f and g given below, find f o g and g o f. Check whether f o g = g o f.
f(x) = x - 6, g(x) = x²

54.

Find the first term and common difference of the Arithmetic Progressions whose n^th terms are given below t_n = -3 + 2n

55.

Check whether the following sequences are in A.P. or not?
x + 2, 2x + 3, 3x + 4, ....

56.

Find the LCM of the following
8x⁴y², 48x²y⁴

57.

Determine whether the sets of points are collinear? \((-\frac12 ,3)\), (- 5, 6) and (-8, 8)

58.

\(\angle A=\angle CED\) prove that \(\Delta\ CAB \sim \Delta CED\) Also find the value of x.

59.

Let A = {1,2,3} and B = {x| x is a prime number less than 10}. Find A x B and B x A.

60.

prove that \(\frac { sinA }{ 1+cosA } =\frac { 1-cosA }{ sinA } \)

Generic 5

29 x 5 = 145

61.

The shadow of a tower, when the angle of elevation of the sum is 45^o is found to be 10 metres, longer than when it is 60^o. find the height of the tower

62.

From a point on a bridge across a river, the angles of depression of the banks on opposite sides at the river are 30° and 45°, respectively. If the bridge is at a height at 3 m from the banks, find the width at the river.

63.

A wooden article was made by scooping out a hemisphere from each end of a cylinder as shown in figure. If the height of the cylinder is 10 cm and its base is of radius 3.5 cm find the total surface area of the article.

64.

S.D. of a data is 2102, mean is 36.6, then find its C.V.

65.

Find two consecutive natural numbers whose product is 20.

66.

Seven years ago, Varun's age was five times the square of Swati's age. Three years hence Swati's age will be two fifth of Varun's age. Find their present ages.

67.

How many terms of the AP: 24, 21, 18, ... must be taken so that their sum is 78?

68.

If the sum of the first 14 terms of an AP is 1050 and its first term is 10, find the 20th term.

69.

Find the area of the triangle formed by the points P(-1, 5, 3), Q(6, -2) and R(-3, 4).

70.

f(x) = (1+ x)
g(x) = (2x - 1)
Show that fo(g(x)) = gof(x)

71.

Let f = {(2, 7); (3, 4), (7, 9), (-1, 6), (0, 2), (5,3)} be a function from A = {-1,0, 2, 3, 5, 7} to B = {2, 3, 4, 6, 7, 9}. Is this
(i) an one-one function
(ii) an onto function,
(iii) both one and onto function?

72.

If A = \(\left[ \begin{matrix} 1 & 8 & 3 \\ 3 & 5 & 0 \\ 8 & 7 & 6 \end{matrix} \right] \), B = \(\left[ \begin{matrix} 8 & -6 & -4 \\ 2 & 11 & -3 \\ 0 & 1 & 5 \end{matrix} \right] \), C = \(\left[ \begin{matrix} 5 & 3 & 0 \\ -1 & -7 & 2 \\ 1 & 4 & 3 \end{matrix} \right] \) compute the following
\(\\ \frac { 1 }{ 2 } A-\frac { 3 }{ 2 } B\)

73.

if sin\(\theta \) (1 + sin²\(\theta \)) = cos²\(\theta \), then prove that cos⁶\(\theta \) - 4cos⁴\(\theta \) + 8cos²\(\theta \) = 4

74.

The King, Queen and Jack of the suit spade are removed from a deck of 52 cards. One card is selected from the remaining cards. Find the probability of getting
(i) a diamond
(ii) a queen
(iii) a spade
(iv) a heart card bearing the number 5.

75.

If two dice are rolled, then find the probability of getting the product of face value 6 or the difference of face values 5.

76.

Find X and Y if X + Y = \(\left[ \begin{matrix} 7 & 0 \\ 3 & 5 \end{matrix} \right] \) and X - Y = \(\left[ \begin{matrix} 3 & 0 \\ 0 & 4 \end{matrix} \right] \)

77.

The mean and variance of seven observations are 8 and 16 respectively. If five of these are 2, 4, 10, 12 and 14, then find the remaining two observations.

78.

A vertical pole fixed to the ground is divided in the ratio 1:9 by a mark on it with lower part shorter than the upper part. If the two parts subtend equal angles at a place on the ground, 25 m away from the base of the pole, what is the height of the pole?

79.

A(-3, 0) B(10, - 2) and C(12, 3) are the vertices of ΔABC. Find the equation of the altitude through A and B.

80.

A kite is flying at a height of 75m above the ground, the string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is \(60°\).find the length of the string ,assuming that there is no slack in the string.

81.

An artist has created a triangular stained glass window and has one strip of small length left before completing the window. She needs to figure out the length of left out portion based on the lengths of the other sides as shown in the figure.

82.

In \(\triangle\)ABC , with \(\angle\)B=90° , BC = 6 cm and AB = 8 cm, D is a point on AC such that AD = 2 cm and E is the midpoint of AB. Join D to E and extend it to meet at F. Find BF.

83.

Find the equation of a line whose intercepts on the x and y axes are given below. 4, -6

84.

As shown in figure a cubical block of side 7 cm is surmounted by a hemisphere. Find the surface area of the solid.

85.

A hemispherical section is cut out from one face of a cubical block such that the diameter l of the hemisphere is equal to side length of the cube. Determine the surface area of the remaining solid.

86.

Let A = {1, 2} and B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}, Verify whether A x C is a subset of B x D?

87.

Iniya bought 50 kg of fruits consisting of apples and bananas. She paid twice as much per kg for the apple as she did for the banana. If Iniya bought Rs. 1800 worth of apples and Rs. 600 worth bananas, then how many kgs of each fruit did she buy?

88.

Find the volume of the iron used to make a hollow cylinder of height 9 cm and whose internal and external radii are 21 cm and 28 cm respectively

89.

Construct a triangle \(\triangle\)PQR such that QR = 5 cm, \(\angle\)P = 30^o and the altitude from P to QR is of length 4.2 cm.

Generic 8

1 x 8 = 8

90.

Construct a \(\triangle\)PQR such that QR = 6.5 cm,\(\angle\)P = 60^oand the altitude from P to QR is of length 4.5 cm.

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Answers Key:

Multiple Choice Question

40 x 1 = 40

(b)
\(250\sqrt { 3 } \)

(b)
75

(d)
\(\frac { 4 }{ 5 } \)

(b)
60 kg

(b)
7:3

(b)
15 cm

(a)
(6, -1)

(d)
70^o

10.

11.

(b)
24 cm

12.

(d)
All are true

13.

(b)
\(\frac { x }{ x+2 } \)

14.

15.

16.

(b)
1

17.

(d)
10

18.

(d)
one - one but not onto

19.

20.

(b)
5, -5

21.

(a)
0 ≤ p(∈) ≤ 1

22.

(b)
1

23.

(d)
1

24.

(b)
1

25.

(b)
7

26.

(b)
1

27.

(a)
0

28.

(b)
\(\frac { b }{ 3 } \)

29.

(b)
an Arithmetic Progression

30.

(d)
2520

31.

32.

33.

(b)
point of contact

34.

(d)
\(5\sqrt { 2 } \)cm

35.

(a)
\(\frac{4}{3}\pi\)

36.

(a)
12 cm

37.

38.

39.

(b)
G.P

40.

(a)
\(\frac {y^{2} + 1}{y^{2}}\)

Generic 2

20 x 2 = 40

41.

An arrow diagram

42.

A = {1,2,3},B = {1,3,5, 7,9}
f(x) = 2x + 1
f(0)) = 2(0) + 1 = 1
f(1) = 2(1) + 1=3
f(2) = 2(2) + 1 = 5
f(3) = 2(3) + 1 = 7
(i) A set of ordered pairs.
f = {(0, 1), (1, 3), (2, 5), (3, 7)}
(ii) A table

x	0	1	2	3
f(x)	1	3	5	7

(ii) An arrow diagram

(iv) A Graph f = \(\{(x, f(x) / x \in A\}\)
= {(0, 1), (1, 3), (2, 5), (3, 7)}

43.

Domain of R = {1,2,3,4}
Co-domain of R = B = {-1, 2, 3,4,5,6, 7, 9, 10, 11,12}
Range of R = {3, 6,10, 9}

44.

2x² + 5x + 7 = 0
a = 2. b = 5, c = 7
α + β = \(-\frac {b}{a}\) = \(-\frac {5}{2}\) and αβ = \(\frac {c}{a}=\frac {7}{2}\)
α + β = \(-\frac {5}{2}\); αβ = \(\frac {7}{2}\)

45.

Given points (sin θ, -cos θ) and (-sin θ, cos θ)
Slope of a line = \(\frac{y_{1}-y_{2}}{x_{1}-x_{2}}\)
\(=\frac{-\cos \theta-\cos \theta}{\sin \theta+\sin \theta}=-\frac{2 \cos \theta}{2 \sin \theta}=-\cot \theta\)

46.

From (1) and (2) we have
\(\frac{A Q}{A B} =\frac{A R}{A D} \)
\(\frac{A B}{A Q} =\frac{A D}{A R} \)
\(\frac{A Q+Q B}{A Q} =\frac{A R+R D}{A R} \)
\(1+\frac{Q B}{A Q} =1+\frac{R D}{A R} \)
\(\Rightarrow \frac{Q B}{A Q} =\frac{D R}{A R} \)

47.

Since A and B are mutually exclusive events
\(\mathrm{P}(A \cap B)=0\)
P(not A) = 0.45
P(A) = 1 - P(not A)
P(A) = 1 - 0.45 = 0.55
\(\mathrm{P}(A \cup B)=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(A \cap B)\)
0.65 = 0.55 + P(B) - 0
P(B) = 0.65 - 0.55 = 0.1

48.

Sample space
S = {1H,1T,2H,2T,3H,3T,4H,4T,5H,5T,6H,6T};
n(S) = 12
Let A be the event of getting an odd number and a head.
A = {1H, 3H, 5H}; n(A) = 3
P(A) = \(\frac { n(A) }{ n(S) } =\frac { 3 }{ 12 } =\frac { 1 }{ 4 } \)

49.

Given line is y = 0.7x - 11 is in the form
y = mx + c
Slope m = 0.7
slope of the line parallel to y = 0.7 x - 11 is
m = 0.7.
(Slopes are equal when the lines are parallel)

50.

Radius of hemisphere = r units
Radius of cone = Radius of hemisphere
Height of cone = Radius of hemisphere
Maximum volume of cone = \(\frac{1}{3} \pi r^{2} h \text { cu. units }\)
\(=\frac{1}{3} \pi\left(r^{2}\right) r=\frac{1}{3} \pi r^{3} \text { cu. units }\)

51.

Let PQ the height of the tower.
Take PQ = h and QR is the distance between the tower and the point R.in right triangle PQR,\(\angle \)PRQ=30°
tan\(\theta =\frac { PQ }{ QR } \)
tan30° = \(\frac { h }{ 48 } \) gives,\(\frac { 1 }{ \sqrt { 3 } } =\frac { h }{ 48 } \) so, h =\(16\sqrt { 3 } \)
Therefore the height of the tower \(16\sqrt { 3 } \) m

52.

Let r₁ and h₁ be the radius and height of the cone - I and let r₂ and h₂ be the radius and height of the cone-II.
Given h₂ = 2h₁ = 2 and \(\frac { Volume\ of\ the\ cone\ I }{ Volume\ of\ the\ cone\ II } =\frac { 2 }{ 3 } \)
\(\frac { \frac { 1 }{ 3 } { \pi r }_{ 1 }^{ 2 }{ h }_{ 1 } }{ \frac { 1 }{ 3 } { \pi r }_{ 2 }^{ 2 }{ h }_{ 2 } } =\frac { 2 }{ 3 } \)
\(\frac { { r }_{ 1 }^{ 2 } }{ { r }_{ 2 }^{ 2 } } \times \frac { { h }_{ 1 } }{ 2{ h }_{ 2 } } =\frac { 2 }{ 3 } \)
\(\frac { { r }_{ 1 }^{ 2 } }{ { r }_{ 2 }^{ 2 } } =\frac { 4 }{ 3 } \text {gives} \frac { { r }_{ 1 } }{ { r }_{ 2 } } =\frac { 2 }{ \sqrt { 3 } } \)
Therefore, ratio of their radii = 2 : \(\sqrt3\)

53.

f(x) = x - 6, g(x) = x²
fog(x) = f(g(x)) = f(x²) = x²- 6 ...(1)
gof(x) = g(f(x)) = g(f(x)) = g(x - 6) = (x - 6)²
= x²- 12x + 36
fog(x) ≠ gof

54.

Given the n^thterm of the A.P. is t_n = - 3 + 2n
Put n = 1
t₁ = -3 + 2(1) = - 3 + 2
a = t₁= -1
Put n = 2
t₂ = -3 + 2(2) = - 3 + 4
t₂ = 1
Common difference d = t₂ - t₁ = 1 - (-1)
= 1 + 1 = 2
First term a = - 1; Common difference d = 2

55.

To check that the given sequence is in A.P., it is enough to check if the differences between the consecutive terms are equal or not.
t₂ - t₁ = (2x + 3) - (x - 2) = x + 1
t₃ - t₁ = (3x +4) - (2x +3) = x +1
t₂ - t₁ = t₃ - t₂
Thus, the differences between consecutive terms are equal.
Hence the sequence x + 2, 2x + 3, 3x + 4,..... is in A.P

56.

8x⁴y², 48x²y⁴
First let us find the LCM of the numerical coefficients.
That is, LCM (8, 48) = 2 x 2 x 2 x 6 = 48
Then find the LCM of the terms involving variables.
That is, LCM (x⁴y², x²y⁴) = x⁴y⁴
Finally find the LCM of the given expression.
We conclude that the LCM of the given expression is the product of the LCM of the numerical coefficient and the LCM of the terms with variables.
Therefore, LCM (8x⁴y², 48x²y⁴) = 48 x⁴y⁴

57.

Given points are \((-\frac12 ,3)\), (- 5, 6) and (-8, 8)
Let us use area of triangle formula
Area of triangle = \(\frac{1}{2}\left[x_{1}\left(y_{2}-y_{3}\right)+\right. \left.\quad x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)\right] \)
\(=\frac{1}{2}\left[-\frac{1}{2}(6-8)-5(8-3)-8(3-6)\right] \)
\(=\frac{1}{2}\left[-\frac{1}{2}(-2)-5(5)-8(-3)\right] \)
\(=\frac{1}{2}[1-25+24]=\frac{1}{2}(0)=0 \)
Since, the area of triangle is zero, the given points are collinear.

58.

\(\Delta \ CAB\) and \(\Delta CED\),\(\angle C\) is common, \(\angle A=\angle CED\)
Therefore, \(\Delta CAB\sim \Delta CED\)
Hence, \(\frac { CA }{ CE } =\frac { AB }{ DE } =\frac { CB }{ CD } \)
\(\frac { AB }{ DE } =\frac { CB }{ CD } \quad \frac { 9 }{ x } =\frac { 10+2 }{ 8 } ,x=\frac { 8\times 9 }{ 12 } =6\) cm.

59.

A = {1,2,3} B = {2,3,5,7}
A x B = {(1,2),(1,3),(1,5),(1,7),(2,2),(2,3),(2,5),(2,7),(3,2),(3,3),(3,5),(3,7)}
B x A = {(2,1),(2,2),(2,3),(3,1),(3,2),(3,3),(5,1),(5,2),(5,3),(7,1),(7,2),(7,3)}

60.

\(\frac { sinA }{ 1+cosA } = \)\(\frac { sinA }{ 1+cosA } \)\(\times \frac { 1-cosA }{ 1-cosA } \) [ multiply numerator and denominator by the conjugate of 1+cosA]
= \(\frac { sinA(1-cosA) }{ (1+cosA)\quad (1-cosA) } =\frac { sinA(1-cosA) }{ 1-co{ s }^{ 2 }A } \)
= \(\frac { sinA(1-cosA) }{ si{ n }^{ 2 }A } =\frac { 1-cosA }{ sinA } \)

Generic 5

29 x 5 = 145

61.

62.

A and B represent points on the bank on opposite sides at the river, so that AB is the width of the river. P is a point on the bridge at a height of 3m i.e., DP = 3 m. We are interested to determine the width at the river which is the length at the side AB of the ΔAPB.

Now, AB = AD + DB
In right ΔAPD,
ㄥA = 30^o
So, tan 30° = \(\frac { 1 }{ \sqrt { 3 } } =\frac { 3 }{ AD } \)
i.e., \(\frac { 1 }{ \sqrt { 3 } } =\frac { 3 }{ AD } \) (or) AD = 3\(\sqrt { 3 } \)
Also, in right ΔPBD,
B = 45^o
So, BD = PD = 3m
Now, AB = BD + AD
= 3 + 3\(\sqrt { 3 } \) = 3(1+\(\sqrt { 3 } \))m
Therefore, the width at the river is 3(+\(\sqrt { 3 } \))m

63.

Radius of the cylinder be r
Height of the cylinder be h
Total surface area of the article
= CSA of cylinder + CSA of 2 hemispheres
= 2ㅠrh + 2πr²= 2πr(h + 2r)
\(=2\times \frac { 22 }{ 7 } \times 3.5\times (10+2\times 3.5)\)
= 22 x 17 = 374 cm²

64.

σ = 21.2, \(\bar { x } \) = 36.6
CV = \(\frac { \sigma }{ \bar { x } } \) x 100 = \(\frac { 21.2 }{ 36.6 } \) x 100 = 57.92%

65.

Let a natural number be x.
The next number = x + 1
x (x + 1) = 20
X²+x-20 = 0
(x + 5)(x - 4) = 0
x=-5,4
∴ x=4
(∵ x≠-5, x is a natural number
The next number = 4 + 1= 5
Two consecutive numbers are 4,5.

66.

Seven years ago, let Swathi's age be x years .
Seven years ago, let Varun's age was 5x² years.
Swathi's present age = x + 7 years
Varun's present age = (5x² + 7) years
3 years hence, we have
Swathi's age = x + 7 + 3 years
=x + 10 years
Varun's age = 5x²+ 7 + 3 years
= 5x²+ 10 years
It is given that 3 years hence Swathi's age will
be \(\frac{2}{5}\) of Varun's age.
∴ x+10=\(\frac{2}{5}\)(5x²+10)
⇒ x+10=2x²+4
⇒ 2x²-x-6=0
⇒ 2x(x-2)+3(x-2)=0
⇒(2x+3)(x-2)=0
⇒ x-2=0
⇒ x=2(∵2x+3≠0 as x>0)
Hence Swathi's present age = (2 + 7) years
= 9 years
Varun's present age = (5 x 2² + 7) years
= 27 years

67.

Here a = 24, d = 21-24 = -3, S_n = 78. We need to find n.
We know that,
S_n= \(\frac{n}{2}\) (2a + (n - 1)d)
78 = \(\frac{n}{2}\) 48 + (n -1)(-3))
78 = \(\frac{n}{2}\) 2(51-3n)
or 3n²- 51n + 156 = 0
n² -17n + 52 = 0
(n - 4)(n - 13) = 0
n = 4 or 13
The number of terms are 4 or 13.

68.

Here S₁₄ = 1050
n = 14
a = 10
S_n = \(\frac{n}{2}\) 2(2a + (n -1)d)
1050 = \(\frac{14}{2}\)(20 + 13d)
= 140 + 91d
910 = 91d
d = 10
a₂₀ = 10 + (20 - 1) x 10
= 20
∴ 20th term = 200.

69.

The area of the triangle formed by the given points is equal to
= \(\frac { 1 }{ 2 } \) [-1.5 (-2 - 4) + 6 (4 - 3) + (-3) (3 + 2)]
= \(\frac { 1 }{ 2 } \) [9 + 6 - 15] = 0
We can have a triangle at area 0 square units? What does this mean?
If the area of a triangle is 0 square units, then its vertices will be collinear.

70.

f(x) = 1 + x
g(x) = (2x - 1)
fog(x) = f(g(x) = f(2x - 1)
= 1 + 2x - 1 = 2x ...(1)
gof(x) = g(f(x) = g(1 + x) = 2(1 + x) - 1
= 2 + 2x - 1
= 2x + 1 ...(2)
(1) \(\neq \)(2)
\(\therefore\) fog(x) \(\neq \) gof(x)
It is verified

71.

It is both one-one and onto function

All the elements in A have their separate images in B. All the elements in B have their preimage in A. Therefore it is one-one and onto function.

72.

\(\\ \frac { 1 }{ 2 } A-\frac { 3 }{ 2 } B\) = \(\frac {1}{2}\)(A - 3B)
= \(\frac { 1 }{ 2 } \left( \left[ \begin{matrix} 1 & 8 & 3 \\ 3 & 5 & 0 \\ 8 & 7 & 6 \end{matrix} \right] -3\left[ \begin{matrix} 8 & -6 & -4 \\ 2 & 11 & -3 \\ 0 & 1 & 5 \end{matrix} \right] \right) \)
= \(\frac { 1 }{ 2 } \left[ \left[ \begin{matrix} 1 & 8 & 3 \\ 3 & 5 & 0 \\ 8 & 7 & 6 \end{matrix} \right] +\left[ \begin{matrix} -24 & 18 & 12 \\ -6 & -33 & 9 \\ 0 & -3 & -15 \end{matrix} \right] \right] =\frac { 1 }{ 2 } \left[ \begin{matrix} -23 & 26 & 15 \\ -3 & -28 & 9 \\ 8 & 4 & -9 \end{matrix} \right] \)
= \(\left[ \begin{matrix} -\frac { 23 }{ 2 } & 13 & \frac { 15 }{ 2 } \\ -\frac { 3 }{ 2 } & -14 & \frac { 9 }{ 2 } \\ 4 & 2 & -\frac { 9 }{ 2 } \end{matrix} \right] \)

73.

Given sin \(\theta \) (1+ sin² \(\theta \)) = cos² \(\theta \)
Squaring on both the sides.
sin² \(\theta \)(1+ sin² \(\theta \))² = cos⁴ \(\theta \)
\(\Rightarrow\left(1-\cos ^{2} \theta\right)\left\{1+\left(1-\cos ^{2} \theta\right)\right\}^{2}=\cos ^{4} \theta\)
\(\Rightarrow\left(1-\cos ^{2} \theta\right)\left\{1+\left(1-\cos ^{2} \theta\right)^{2}+2(1)\left(1-\cos ^{2} \theta\right)\right\} =\cos ^{4} \theta \)
\(\Rightarrow\left(1-\cos ^{2} \theta\right)\left(1+1+\cos ^{4} \theta-2(1) \cos ^{2} \theta\right. \left.+2-2 \cos ^{2} \theta\right)=\cos ^{4} \theta \)
\(\Rightarrow\left(1-\cos ^{2} \theta\right)\left(4-4 \cos ^{2} \theta+\cos ^{4} \theta\right)=\cos ^{4} \theta \)
\(\Rightarrow 4-4 \cos ^{2} \theta+\cos ^{4} \theta-4 \cos ^{2} \theta+4 \cos ^{4} \theta-\cos ^{6} \theta \ =\cos ^{4} \theta \)
\(\Rightarrow \ 4-8 \cos ^{2} \theta+5 \cos ^{4} \theta-\cos ^{6} \theta=\cos ^{4} \theta \)
\(4-8 \cos ^{2} \theta+4 \cos ^{4} \theta-\cos ^{6} \theta=0 \)
\(4=8 \cos ^{2} \theta-4 \cos ^{4} \theta+\cos ^{6} \theta \)
\(\Rightarrow \cos ^{6} \theta-4 \cos ^{4} \theta+8 \cos ^{2} \theta=4 \)

74.

King spade, Queen spade, Jack spade are removed
∴ total number of cards = 52 - 3 = 49.
(i) Probability (diamond)= \(\frac { 13 }{ 49 } \)
(ii) Probability (queen) \(\frac { 4-1 }{ 49 } =\frac { 3 }{ 49 } \)
(iii) Probability (spade) = \(\frac { 13-3 }{ 49 } =\frac { 10 }{ 49 } \)
(iv) Probability (heart bearing number 5) = \(\frac { 13-5 }{ 49 } =\frac { 8 }{ 49 } \)

75.

Product of face values 6: { (1, 6), (2, 3), (6, 1), (3,2)}
Difference of face value 5: {(1, 6), (6, 1)}
P(product 6) = \(\frac { 4 }{ 6\times 6 } =\frac { 4 }{ 36 } =\frac { 1 }{ 9 } \)
p( difference 5) = \(\frac { 2 }{ 2\times 6 } =\frac { 1 }{ 18 } \)

76.

X + Y = \(\left[ \begin{matrix} 7 & 0 \\ 3 & 5 \end{matrix} \right] \) ...(1)
X - Y= \(\left[ \begin{matrix} 3 & 0 \\ 0 & 4 \end{matrix} \right] \) ...(2)
______________
\((1)+(2)\Rightarrow 2x=\left[ \begin{matrix} 10 & 0 \\ 3 & 9 \end{matrix} \right] \)
\(=\left[ \begin{matrix} 5 & 0 \\ \frac { 3 }{ 2 } & \frac { 9 }{ 2 } \end{matrix} \right] \)
\((1)-(2)\Rightarrow X+Y=\left[ \begin{matrix} 7 & 0 \\ 3 & 5 \end{matrix} \right] \)

\(2Y=\left[ \begin{matrix} 4 & 0 \\ 3 & 1 \end{matrix} \right] \Rightarrow Y=\frac { 1 }{ 2 } \left[ \begin{matrix} 4 & 0 \\ 3 & 1 \end{matrix} \right] \)
\(\therefore Y=\left[ \begin{matrix} 2 & 0 \\ \frac { 3 }{ 2 } & \frac { 1 }{ 2 } \end{matrix} \right] \)
\(X=\left[ \begin{matrix} 5 & 0 \\ \frac { 3 }{ 2 } & \frac { 9 }{ 2 } \end{matrix} \right] , Y=\left[ \begin {matrix} 2 & 0 \\ \frac { 3 }{ 2 } & \frac { 1 }{ 2 } \end{matrix} \right] \)

77.

Mean = 8
Variance = 16.
Let a and b be the missing observations
\(\frac{\text { Sum of observations }}{\text { Number of observations }}=\text { Mean }\)
\(\frac{2+4+10+12+14+a+b}{7}=8\)
42 + a + b = 56
a + b = 56 - 42 = 14
b = 14 - a
Variance \(=\frac{\sum_{i=1}^{n}\left(x_{i}-\bar{x}\right)^{2}}{n} \)
\(= \frac{\sum_{i=1}^{7}\left(x_{i}-\bar{x}\right)^{2}}{7} \)
\(=\frac{(2-8)^{2}+(4-8)^{2}+(10-8)^{2}+(12-8)^{2}+(14-8)^{2}+(a-8)^{2}+(b-8)^{2}}{7}\)
\(16=\frac{36+16+4+16+36+(a-8)^{2}+(b-8)^{2}}{7}\)
108 + (a - 8)² + (b - 8)² = 112
(a - 8)² + (b - 8)² = 4 ..(2)
Put b = 14 - a in (2)
(a - 8)² + (b - 8)² = 4
(a - 8)² + (14 - a ) - 8)² = 4
(a - 8 )² + (6 - a)² = 4
a² + 64 - 16a + 36 + a² - 12a - 4 = 0
2a² - 28a + 96 = 0
Divided by 2,
a² - 14a + 48 = 0
(a - 6) (a - 8) = 0
a = 6 or a = 8
a = 6 , b = 14, a = 14 - 6 = 8
a = 6 and b = 8
Remaining numbers are 6 and 8.

78.

Let AC be the pole and let point 'B' divide it in the ratio
\(\not x: 9 \not x=1: 9\)
Let 'D' be the point 25 m.
\(tan\alpha =\frac { x }{ 25 } \) \(tan2\alpha =\frac { 10x }{ 25 } \)
\(tan2\alpha =\frac { 2tan\alpha }{ 1-{ tan }^{ 2 }\alpha } \)
\(\frac{\not 10 x^{5}}{\not 25}=\frac{2 \times \frac{x}{25}}{1-\frac{x^{2}}{625}}\)

Height of pole = 10x
= \(100\sqrt { 5 } \)
x = \(10\sqrt { 5 } \) m

79.

Given vertices are A(- 3, 0), B(10, - 2) and, C(12, 3).
Slope of BC = \(\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=\frac{-2-3}{10-12}=\frac{-5}{-2}=\frac{5}{2}\)
Altitude AD is perpendicular to BC and passing through A(- 3, 0)
Slope of AD = \(-\frac{2}{5}\)
Equation of AD y - y₁ = m(x - x₁)
\(y-0=-\frac{2}{5}(x+3)\)
5y = -2x - 6
2x + 5y + 6 = 0
Slope of AC \(=\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=\frac{0-3}{-3-12}=\frac{-3}{-15}=\frac{1}{5}\)
Altitude BE is perpendicular to AC and passing through B(10, - 2). Slope of BE = 5
Equation of BE y - y₁= m(x - x₁)
y + 2 = - 5(x - 10)
y + 2 = -5x + 50
5x + y - 48 = 0

80.

Let AB be the height of the kite above the ground. Then, AB = 75.
Let AC be the length of the string.
In right triangle ABC,\(\angle \)ACB = \(60°\)
\(sin\theta =\frac { AB }{ AC } \)
\(sin60°=\frac { 75 }{ AC } \)
gives \(\frac { \sqrt { 3 } }{ 2 } =\frac { 75 }{ AC } \) so, AC = \(\frac { 150 }{ \sqrt { 3 } } =50\sqrt { 3 } \)
Hence, the length of the string is 50\(\sqrt { 3 } m\)

81.

Clearly In \(\triangle\)ABC, D, E, F are points on lines BC, CA, AB respectively using Ceva's theorem, we have
\(\frac{A E}{E C} \times \frac{C D}{D B} \times \frac{B F}{F A}=1\) ...(1)
From the diagram it is clear that
AE = 3, EC = 4, CD = 10, DB = 3, FA = 5
Substituting these values in (1)
\( \frac{3}{4} \times \frac{10}{3} \times \frac{B F}{5} =1 \)
\(B F =\frac{1 \times 4 \times 3 \times 5}{3 \times 10}=2 \mathrm{~cm}\)

82.

Consider DABC, Then D, E, F are respective points on the sides CA, AB and Be. By construction D, E, F are collinear
By Menelaus' Theorem, \(\frac { AE }{ EB } \times \frac { BF }{ FC } \times \frac { CD }{ DA } =1\)
FC = FB + BC = BF + 6
By Pythagoras Theorem AC² = AB² + BC²
=64+36=100
\(\therefore \) AC = 10, CD = 8

\(4BF=BF+6\Rightarrow BF=2cm\)

83.

Given intercepts are 4, - 6
a = 4, b = - 6
Equation of the line in the intercepts form is
\(\frac{x}{a}+\frac{y}{b}=1 \)
\(\frac{x}{4}+\frac{y}{-6}=1 \)
3x - 2y - 12 = 0

84.

Edge of cube = 7 cm
surface area of a cube = 6a² sq. units
= 6(7)²
= 294 cm²
radius of hemisphere = \(\frac{7}{2} \mathrm{~cm}\)
[Only C.S.A is considered as the hemisphere surmounted]
C.S.A of hemisphere \(=2 \pi r^{2} \text { sq. units } \)
\(=2 \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \)
= 77 Cm²
Surface area of = T.S.A of cube + C.S.A the solid of hemisphere area of circular region (bottom of hemisphere)

\(=294+77-\left(\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2}\right)\)
= 371 - 38.5
= 332.5 cm²

85.

Let r be the radius of the hemisphere.
Given that, diameter of the hemisphere = side of the cube = l
Radius of the hemisphere = \(\frac{l}{2}\)
TSA of the remaining solid = Surface area of the cubical part + C.S.A. of the hemispherical part − Area of the base of the hemispherical part
= 6 x (Edge)²+ 2\(\pi\)r²−\(\pi\)r²
= 6 x (Edge)²+ \(\pi\)r²
\(=6{ \times (l) }^{ 2 }+\pi { \left( \frac { l }{ 2 } \right) }^{ 2 }=\frac { 1 }{ 4 } (24+\pi ){ l }^{ 2 }\)
Total surface area of the remaining solid \(=\frac { 1 }{ 4 } (24+\pi ){ l }^{ 2 }\)sq. units

86.

A = {1, 2),B = {1, 2, 3, 4}, C = {5, 6},D = {5, 6, 7, 8}
A x C = {1, 2} x {5, 6}
\(\mathrm{A} \times \mathrm{C}=\{(1,5),(1,6),(2,5),(2,6)\}\)
B x D = { 1, 2, 3, 4} x { 5, 6, 7, 8}
\(\begin{array}{r} \mathrm{B} \times \mathrm{D}=\{({1,5}),(1,6),(1,7),(1,8) ({2,5}),(2,6),(2,7),(2,8) (3,5),(3,6),(3,7),(3,8) (4,5),(4,6),(4,7),(4,8)\} \end{array}\)
(A x C) is a subset of (B x D)

87.

Let the weight of applies be a kg.
Let the weight of bananas be b kg.
a+b = 50
ax = Rs. 1800...(1)
by = Rs. 600 ..(2)
x = 2y ....(3)
Use (3) in (1) ⇒
∵ 3b = 2a ....(5)
∵ a + b = 50
\(a+\frac { 2a }{ 3 } =50\Rightarrow \frac { 5a }{ 3 } =50\)
\(\Rightarrow a=50\times \frac { 3 }{ 5 } \)
= 30
∴ b = 20
∴ Iniya bought 30kg of applies and 20 kg of bananas.

88.

Let r, R and h be the internal radius, external radius and height of the hollow cylinder respectively.
Given that, r = 21cm, R = 28 cm, h = 9 cm
Now, volume of hollow cylinder = \(\pi\)(R²− r²)h cu. units
\(=\frac { 22 }{ 7 } \left( { 28 }^{ 2 }-21^{ 2 } \right) \times 9\)
\(=\frac { 22 }{ 7 } (784-441)\times 9=9702\)
Therefore, volume of iron used = 9702 cm³

89.

Construction
Step 1 : Draw a line segment QR = 5 cm.
Step 2 : At Q draw QE such that \(\angle\)RQE = 30^o.
Step 3 : At Q draw QF such that \(\angle EQF\) = 90^o
Step 4 : Draw the perpendicular bisector XY to QR which intersects QF at O and QR at G.
Step 5 : With O as centre and OQ as radius draw a circle.
Step 6: From G mark an arc in the line XY at M, such that GM = 42. cm.
Step 7 : Draw AB through M which is parallel to QR.
Step 8 : AB meets the circle at P and S.
Step 9 : Join QP and RP. Then\(\triangle\)PQR is the required triangle

Generic 8

1 x 8 = 8

90.

Construction:
Steps (1) Draw QR = 6.5 cm.
Steps (2) Draw \(\angle RQE={ 60 }^{ 0 }\)
Steps (3) Draw \(\angle FQE={ 90 }^{ 0 }\)
Steps (4) Drawn the perpendicular bisector XY to ER which intersects QF at O and ER at G.
Steps (5) With O as center and OQ as radius drawn a circle
Steps (6) XY intersects QR at G. On XY, from G marked an arc at M, such that GM = 4.5 cm
Steps (7) Drawn AB through M which is parallel to QR
Steps (8) AB meets the circle at P and S
Steps (9) Joined QP and RP Then \(\triangle\)PQR is the required triangle.
Steps (10) Here \(\triangle\)SQR is also another required triangle.