9th Standard Mathematics creative test model

Download Tamil Nadu 9th 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:

Generic 3

15 x 3 = 45

In an office, where 42 staff members work, 7 staff members use cars, 20 staff members use two-wheelers and the remaining 15 staff members use cycles. Find the relative frequencies.

Find the TSA and LSA of a cuboid whose length, breadth and height are 7.5 m, 3 m and 5 m respectively.

Express
(i) sin 74° in terms of cosine
(ii) tan 12° in terms of cotangent
(iii) cosec 39° in terms of secant

The point (3, −4) is the centre of a circle. If AB is a diameter of the circle and B is (5, −6), find the coordinates of A.

Verify A -(BUC) = (A-B)∩(A-C) using Venn diagrams.

If A = {2,5,6,7} and B = {3,5,7,8}, then verify the commulative property of intersection of sets

Show that x+4 is a factor of x³+ 6x²- 7x - 60

Write in scientific notation: (500000)⁵\(\times\)(3000)³

The following table represents the marks obtained by a group of 12 students in a class test in Mathematics and Science

Marks (Mathematics)	52	55	32	30	60	44	28	25	50	75	33	62
Marks (Science)	54	42	48	49	27	25	24	19	28	58	42	69

Indicate in which subject, the level of achievement is higher?

10.

The average mark of 25 students was found to be 78.4. Later on, it was found that score of 96 was misread as 69. Find the correct mean of the marks

11.

In the figure find x⁰ and y⁰.

12.

Write the coordinates of quadrilateral PQRS as shown in the following figure.

13.

Find the quotient and the remainder when (5x²-7x+2) ÷ (x-1)

14.

Express the following decimal expression into rational numbers \(3.1\overline { 7 } \)

15.

Consider the given pairs of triangles and say whether each pair is that of congruent triangles. If the triangles are congruent, say ‘how’; if they are not congruent say ‘why’ and also say if a small modification would make them congruent:

Generic 2

25 x 2 = 50

16.

One card is drawn from a pack of 52 cards, each of the 52 cards being equally likely to be drawn. Find the probability that the card drawn is
(i) an ace,
(ii) either red card or king.

17.

Find the value of \(\frac{\tan 25^{\circ}}{\cot 65^{\circ}}+\frac{\sin 40^{\circ}}{\cos 50^{\circ}}\)

18.

A cuboid has total surface area of 40m² and its lateral surface area is 26 m², Find the area of its base.

19.

Using section formula, show that the points A (7, -5), B (9, -3) and C (13, 1), are collinear.

20.

The length, breadth and height of a chocolate box are in the ratio 5:4:3. If its volume is 7500 cm³, then find its dimensions.

21.

The probability of guessing the correct answer to a certain question is \(\frac { x }{ 3 } \). If the probability of not guessing the correct answer is \(\frac { x }{ 5 } \), then find the value of x.

22.

Find the coordinates of the point which divides the line segment joining the points A(4,−3) and B(9,7) in the ratio 3:2.

23.

From the given figure, find all the trigonometric ratios of angle B.

24.

Factorise the following : 25m^-2-16n²

25.

Evaluate : \(\left( \cfrac { 1 }{ 9 } \right) ^{ -3 }\)

26.

Find the value of X^o

27.

Find the value of X^o

28.

In a research laboratory scientists treated 6 mice with lung cancer using medicine. Ten days, they measured the volume of the tumor of the tumor in each mouse given the results in the table

Mouse marking	1	2	3	4	5	6
Tumor Volume(mm)³	145	148	142	141	139	140

Find the mean

29.

In a rice mill, seven labours are receiving the daily wages of Rs. 500, Rs. 600, Rs. 600, Rs. 800, Rs. 800, Rs. 800 and Rs. 1000, find the modal wage

30.

Factorise the following: l³- 8m³-27n³-18lmn

31.

Find the expansion of the following: (x + 4) (x + 5) (x + 6)

32.

Rationalise the denominator of (i) \(\frac { 7 }{ \sqrt { 14 } } \) (ii) \(\frac { 5+\sqrt { 3 } }{ 5-\sqrt { 3 } } \)

33.

Find the value of \({ 64 }^{ \frac { -2 }{ 3 } }\)

34.

Without actual division classify the decimal expansion of the following numbers as terminating or non-terminating and recurring.
\(-{11\over 75}\)

35.

IfA = {1,2,3} B = {3,5,6} C = {3,4,7}. Find
(i) A-(BUC)
(ii) BΔC
(iii) AΔB

36.

Write the following in "Roster" form?
(a) A = set of the months having 31 days.
(b) B = {x : x is a natural number of 2 digits divisible by 13}
(c) C = {set of vowels in the word "father"}
(d) D = {x : 5 < x < 10 ; x ∈ N}
(e) E = {x: x is a square natural number less than 16}

37.

Draw the following special quadrilaterals on a graph sheet, measure the sides and angles and complete the table to explore the properties of the quadrilaterals with respect to sides and angles.

38.

This is a copy of the tangram puzzle. The tangram puzzle consists of 7 geometric pieces which are normally boxed in the shape·of a square. The pieces, called 'tans', are used to create different patterns including animals, people, numbers, geometric shapes and many more.
You can make several polygons using the pieces in different ways.

39.

Find the symmetric difference between the following sets. P = {2, 3, 5, 7, 11} and Q = {1, 3, 5, 11}

40.

Draw Venn diagram and shade the region representing the following sets
(A – B)′

Multiple Choice Question

40 x 1 = 40

41.

The total surface area of a cuboid is ______________

(a)

4a²sq. units

(b)

6a² sq. units

(c)

2(l + b)h sq. units

(d)

2(lb + bh + lh) sq. units

42.

The perimeter of an equilateral triangle is 30 cm. The area is _______.

(a)

\(10\sqrt { 3 } \) cm²

(b)

\(12\sqrt { 3 } \) cm²

(c)

\(15\sqrt { 3 } \) cm²

(d)

\(25\sqrt { 3 } \) cm ²

43.

The probability of all possible outcomes of a random experiment is always equal to _______.

(a)

One

(b)

Zero

(c)

Infinity

(d)

Less than one

44.

A number between 0 and 1 that is used to measure uncertainty is called _______.

(a)

Random variable

(b)

Trial

(c)

Simple event

(d)

Probability

45.

The value of 2tan30° tan60° is

(a)

(b)

(c)

\(2\sqrt { 3 } \)

(d)

46.

if sin 30⁰ = x and cos 60⁰ = y, then x² + y² is________.

(a)

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

(b)

(c)

sin90°

(d)

cos90°

47.

Which of the following is a solution of the equation 2x − y = 6.

(a)

(2,4)

(b)

(4,2)

(c)

(3, −1)

(d)

(0,6)

48.

If the coordinates of the mid-points of the sides AB, BC and CA of a triangle are (3, 4), (1, 1) and (2, −3) respectively, then the vertices A and B of the triangle are ______.

(a)

(3, 2), (2, 4)

(b)

(4, 0), (2, 8)

(c)

(3, 4), (2, 0)

(d)

(4, 3), (2, 4)

49.

If one of the factor of x^2-9x+18 is (x-3) then the other factor is_________

(a)

x-9

(b)

x+6

(c)

(x-6)

(d)

x-18

50.

Degree of the linear polynomial is ________

(a)

(b)

(c)

(d)

51.

(a-b-c)² is equal to________

(a)

(a+b+c)²

(b)

(a+b+c)²

(c)

(-a + b + c)²

(d)

(a+b+c)²

52.

Which is the best example of a number written in scientific notation?

(a)

2.71 \(\times\) 10⁵

(b)

0.6 \(\times\) 10⁴

(c)

0.9871

(d)

125.4 \(\times\) 10⁴

53.

The angle in a semi circle is_________

(a)

180°

(b)

90°

(c)

60°

(d)

45°

54.

The angle subtend by equal chords of a circle at the centre is________

(a)

Complementary

(b)

Supplementary

(c)

equal

(d)

unequal

55.

The mean of a, b, c, d and e is 28. If the mean of a, c and e is 24, then mean of b and d is _______________

(a)

(b)

(c)

(d)

56.

The algebraic sum of the deviations of a set of n values from their mean is ___________

(a)

(b)

n-1

(c)

(d)

n+1

57.

For which set of number do the mean,median and mode all have the same valuas?

(a)

2,2,2,4

(b)

1,3,3,3,5

(c)

1,1,2,5,6

(d)

1,1,2,1,5

58.

A particular observation which occurs maximum number of times in a given data is called is

(a)

Frequency

(b)

range

(c)

mode

(d)

median

59.

The mean of a set of numbers is \(\bar X\). If each number is multiplied by z, the mean is _______.

(a)

\(\bar X + z\)

(b)

\(\bar X-z\)

(c)

z\(\bar X\)

(d)

\(\bar X\)

60.

The mean of the square of first 11 natural numbers is _______.

(a)

(b)

(c)

(d)

61.

If \(\sqrt { { 9 }^{ x } } =\sqrt [ 3 ]{ 9^{ 2 } } \) , then x = _______.

(a)

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

(b)

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

(c)

\(\frac{1}{3}\)

(d)

\(\frac{5}{3}\)

62.

If n(A \(\cup \) B \(\cup \) C) = 40, n(A) = 30, n(B) = 25, n(C) = 20, n(A\(\cap \)B) = 12, n(B\(\cap \)C) = 18 and n(A\(\cap \)C) = 15 , then n(A\(\cap \)B\(\cap \)C) is ___________

(a)

(b)

(c)

(d)

63.

if a polynominal p(x) is divided by (ax+b), then the reminder is _______________________

(a)

\(p\left( \frac { b }{ a } \right) \)

(b)

\(p\left( -\frac { b }{ a } \right) \)

(c)

\(p\left( \frac { a }{ b } \right) \)

(d)

\(p\left( -\frac { a }{ b } \right) \)

64.

The point of concurrency of the medians of a triangle is known as __________

(a)

circumcentre

(b)

incentre

(c)

orthocentre

(d)

centroid

65.

One angle of a parallelogram is a right angle. The name of the quadrilateral is ______

(a)

square

(b)

rectangle

(c)

rhombus

(d)

kite

66.

The set of (A U B) - (A ก B) is ____________

(a)

(AUB)'

(b)

AΔB

(c)

(A∩B)'

(d)

A'UB'

67.

Sets having the same number of elements are called ___________

(a)

overlapping sets

(b)

disjoints sets

(c)

equivalent sets

(d)

equal sets

68.

A = {set of odd natural numbers}, B = {set of even natural numbers}, then A and B are ___________

(a)

equal set

(b)

equivalent sets

(c)

overlapping sets

(d)

disjoint sets

69.

Which of the following are irrational numbers?
\(\sqrt { 2+\sqrt { 3 } } \)
\(\sqrt [ 3 ]{ 5+\sqrt { 7 } } \)
\(\sqrt { 8-\sqrt [ 3 ]{ 8 } } \)
\(\sqrt { 4+\sqrt { 25 } } \)

(a)

(ii), (iii) and (iv)

(b)

(i), (ii) and (iv)

(c)

(i), (ii) and (iii)

(d)

(i), (iii) and (iv)

70.

The decimal form of -\(\frac { 3 }{ 4 } \) is_______________

(a)

- 0.75

(b)

- 0.50

(c)

-0.25

(d)

- 0.125

71.

The point which is on y-axis with ordinate - 5 is _____________

(a)

(0, - 5)

(b)

(-5,0)

(c)

(5,0)

(d)

(0,5)

72.

The distance between the points (a, 0) and (0, b) is____________

(a)

a unit

(b)

b unit

(c)

\(\sqrt{a^2+{b^2}}\ unit\)

(d)

\(\sqrt{a^2-{b^2}}\ unit\)

73.

The distance between the points (4, -1) and the origin is___________

(a)

\(\sqrt{24}\)

(b)

\(\sqrt{37}\)

(c)

\(\sqrt{26}\)

(d)

\(\sqrt{17}\)

74.

The point whose abscissa is 5 and lies on the x-axis is__________

(a)

(-5, 0)

(b)

(5,5)

(c)

(0,5)

(d)

(5,0)

75.

Which one of the following, regarding sum of two irrational numbers, is true?

(a)

always an irrational number

(b)

may be a rational or irrational number

(c)

always a rational number

(d)

always an integer.

76.

If x³ + 6x² + kx + 6 is exactly divisible by (x + 2), then k= ?

(a)

-6

(b)

-7

(c)

-8

(d)

77.

Point (–10, 0) lies ___________

(a)

on the negative direction of x-axis

(b)

on the negative direction of y-axis

(c)

in the III quadrant

(d)

in the IV quadrant

78.

If X = {x : x = 4(n – 1), n ∈ N} and Y = {y : y = 3n – 2n – 1, n ∈ N}, then X∪Y is ______________

(a)

(b)

(c)

(d)

79.

The interior angle made by the side in a parallelogram is 90° then the parallelogram is a ________.

(a)

rhombus

(b)

rectangle

(c)

trapezium

(d)

kite

80.

The correct statement out of the following is ________.

(a)

ΔABC ≅ ΔDEF

(b)

ΔABC ≅ ΔDEF

(c)

ΔABC ≅ ΔFDE

(d)

ΔABC ≅ ΔFED

Generic 5

8 x 5 = 40

81.

Find the area of a quadrilateral ABCD whose sides are AB = 8cm, BC = 15 cm, CD = 12 cm, AD = 25 cm and = 90°.

82.

If A =\(\left\{ \cfrac { 1 }{ 2 } ,1\cfrac { 5 }{ 4 } ,\cfrac { 7 }{ 4 } ,3 \right\} \) ,B =\(\left\{ 1\cfrac { 5 }{ 4 } ,\cfrac { 7 }{ 4 } ,3\cfrac { 7 }{ 2 } \right\} \) Now and C =\(\left\{ \cfrac { 1 }{ 2 } ,\cfrac { 5 }{ 4 } ,2,3\cfrac { 7 }{ 2 } \right\} \), then verify \(A\cap \left( B\cap C \right) \)= \(\left( A\cap B \right) \cap C\)

83.

Prove that x -1 is a factor x⁵ - 45x⁴ + 36x³ + 45x²- 36x^-1

84.

Find the angle of the given cyclic quadrilateral ABCD in the figure.

85.

Find the Arithmetic Mean of the following data using Step Deviation Method

Age	15-19	20-24	25-29	30-34	35-39	40-44
No.of persons	4	20	38	24	10	9

86.

Represent \(-\frac { 2 }{ 11 } ,-\frac { 5 }{ 11 } and-\frac { 9 }{ 11 } \)on the number line.

87.

Find any seven rational numbers between \(\frac { 5 }{ 8 } \) and \(\frac { 5 }{ 6 } \)

88.

Find x such that PQ = QR where P(6, -1) Q (1, 3) and R (x, 8) respectively.

Generic 8

1 x 8 = 8

89.

Construct the centroid of \(\triangle\)PQR such that PQ = 9 cm, PQ = 7cm, RP = 8 cm.

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Answers Key:

Generic 3

15 x 3 = 45

Total number of staff members = 42
The relative frequencies:
Car users \(=\frac { 7 }{ 42 } =\frac { 1 }{ 6 } \)
Two-wheeler users \(=\frac { 20 }{ 42 } =\frac { 10 }{ 21 } \)
Cycle users \(=\frac { 15 }{ 42 } =\frac { 5 }{ 14 } \)

Given the dimensions of the cuboid;
that is length (l) = 7.5 m, breadth (b) = 3 m and height (h) = 5 m.
TSA = 2(lb + bh + lh)

= 2[(7.5 × 3) + (3 × 5) + (7.5 × 5)]
= 2(22.5 +15 + 37.5)
= 2 × 75
= 150 m²
LSA = 2(l + b) × h
= 2(7.5 + 3) × 5
= 2 ×10.5 × 5
= 105 m²

(i) sin74° = sin(90⁰ -16⁰) (since, 90⁰ -16⁰ = 74⁰ )
RHS is of the form sin(90⁰ - \(\theta\)) = cos\(\theta\)
Therefore sin74° = cos16⁰
(ii) tan12° = tan(90⁰ - 78⁰) (since, 12⁰= 90⁰ = 78⁰ )
RHS is of the form tan(90⁰- \(\theta\)) = cot \(\theta\)
Therefore tan12° = cot 78⁰
(iii) cosec 39° = cosec(90⁰ - 51⁰) (since, 39⁰ = 90⁰ - 51⁰ )
RHS is of the form cosec(90⁰ - \(\theta\)) = sec\(\theta\)
Therefore cosec39° = sec51⁰

Let the coordinates of A be (x₁, y₁) and the given point is B(5,−6). Since the centre is the mid-point of the diameter AB, we have
\(\frac { { x }_{ 1 }+{ x }_{ 2 } }{ 2 } \) = 3
x₁ + 5 = 6
x₁ = 6 − 5
x₁ = 1
\(\frac { { y}_{ 1 }+{ y }_{ 2 } }{ 2 } \) = -4
y₁ − 6 = −8
y₁ = −8 + 6
y₁ = −2

Therefore, the coordinates of A is (1, −2) .

From (1) and (2), we get A -(BUC) = (A-B)∩(A-C). Hence it is verified.

\(A\cap B\) = {5,7}
\(B\cap A\)= {5,7}
From (3) and (4) we get, \(A\cap B=B\cap C\)
It is verifield that insersection of sets is commutative.

Let p (x) = x³+ 6x²- 7x - 60
By factor theorem (x+4) is a factor of p(-4) = 0
p(-4) = (-4)³+ 6(-4)²-7(-4) -60 = -64 + 96 + 28 - 60 = 0
Therefore,(x+4) is a factor of x³+ 6x²+ (-7x) -60

(500000)⁵\(\times\) (3000)³
\( =\left(5.0 \times 10^{5}\right)^{3} \times\left(3.0 \times 10^{3}\right)^{3} \\ =(5.0)^{2} \times\left(10^{5}\right)^{2} \times(3.0)^{3} \times\left(10^{3}\right)^{3} \\ =25 \times 10^{10} \times 27 \times 10^{9}=675 \times 10^{19} \\ =675.0 \times 1019=6.75 \times 102 \times 1019=6.75 \times 10^{21} \)

Let us arrange the marks in the two subjects in ascending order.

Marks (Mathematics)	52	55	32	30	60	44	28	25	50	75	33	62
Marks (Science)	54	42	48	49	27	25	24	19	28	58	42	69

Since the number of students is 12, the marks of the middle-most student would be the mean mark of 6^th and 7^th students.
Therefore, Median mark in Mathematics \(={44+50\over2}=47\)
Median mark in Science \(={42+42\over2}=42\)
Here the median mark in Mathematics is greater than the median mark in Science. Therefore, the level of achievement of the students is higher in Mathematics than Science.

10.

Given that the total number of students n = 25, \(\bar X\)= 784
So, Incorrect \(\sum x=\bar X\times n=78.4\times25=1960\)
Correct Σx = incorrect Σx - wrong entry + correct entry
= 1960 - 69 + 96 = 1987
Correct \(\bar X={correct\sum x\over n}={1987\over 25}=79.48\)

11.

ㄥACD = ㄥA + ㄥB
(An exterior angle of a triangle is sum of its interior opposite angles)
120° = 50° + x⁰
x⁰ = 120° - 50°
= 70°
In the the triangle ABC
ㄥA + ㄥB + ㄥACB = 180° (Sum of the angles of a Δ)
50° +x + ㄥACB = 180°
50° + 70° + ㄥACB = 180°
ㄥACB = 180° - 120°
y = 60° (OR)
ㄥACD + ㄥACB = 180⁰(Angles of a linear pair)
ㄥACB = 180° - 120°
= 60°
The value ofx = 70° andy = 60°.

12.

P (- 4,4), Q (8, 2), R (6, -3) and S (-2, -2)

13.

(5x²-7x+2) ÷ (x-1)

∴ Quotient = 5x–2 Remainder = 0
(i) \(\frac{5x^{2}}{x}=5x\)
(ii) 5x(x-1)=5x²-5x
(iii) \(-\frac{2x}{x}=-2\)
(iv) -2(x-1)=(-2x+2)

14.

Let x = \(3.1\overline { 7 } \) = 3.1777 ....... (1)
Here period of decimal is 2, multiply equation (1) by 100
10x = 31.777 ......... (2)
(2)-(1)

\( =\frac{143}{45} \)

15.

In the given figure
AB = AC (Given)
BM = MC (AM is the median of the \(\triangle\)ABC)
AM is common (By SSS congruency)
\(\therefore\triangle ABM\cong\triangle ACM\)

Generic 2

25 x 2 = 50

16.

17.

\( \frac{\tan 25^{\circ}}{\cot 65^{\circ}}+\frac{\sin 40^{\circ}}{\cos 50^{\circ}} \)
\(= \frac{\tan \left(90^{\circ}-65^{\circ}\right)}{\cot 65^{\circ}}+\frac{\sin \left(90^{\circ}-50^{\circ}\right)}{\cos 50^{\circ}} \)
\(= \frac{\cot 65^{\circ}}{\cot 65^{\circ}}+\frac{\cos 50^{\circ}}{\cos 50^{\circ}}=1+1=2 \)

18.

Area of its base = \(\frac { TSA-LSA }{ 2 } \)
= \(\frac { 40-26 }{ 2 } \) m²
= 7 m²

19.

\((9,-3)=\left( \frac { m(12+n(7) }{ m+n } ,\frac { m(1)+n(-5) }{ m+n } \right) \)
\(\frac {13m+37n}{ m+n } =9\)
13m + 7n = 9m + 9n
4m = 2n
\(\frac { m }{ n } =\frac { 2 }{ 4 } =\frac { 1 }{ 2 } \)
\(\frac{m-5n}{m+nn}=-3\)
m -5n = -3m -3n
m + 3m= 5n-3n
4m=2n
\(\frac { m }{ n } =\frac { 2 }{ 4 } =\frac { 1 }{ 2 } \)

20.

Let 1 ratio = x then 5:4:3
⇒ 5x: 4x: 3x
5x \(\times\) 4x \(\times\) 3x = 60x³ = 7500 cm³

x = 5 cm
∴ 5x = 5 \(\times\) 5 = 25 cm
4x = 4 \(\times\) 5 = 20 cm
3x = 3 \(\times\) 5 = 15 cm
∴ Dimensions of a chocolate box are 25 cm \(\times\) 20 cm \(\times\) 15 cm.

21.

\(\frac { x }{ 3 } +\frac { x }{ 5 } \) = 1
\(\frac { 5x+3x }{ 15 } \) = 1
\(\frac { 8x }{ 15 } \) = 1
8x = 15
x = \(\frac { 15 }{ 8 } \)

22.

x₁ y₁ x₂ y₂ m : n
A (4, -3), B (9, 7), 3 : 2
By section formula \(P\left( \frac { { mx }_{ 2 }+{ nx }_{ 1 } }{ m+n } ,\frac { { my }_{ 2 }+{ ny }_{ 1 } }{ m+n } \right) =P(x,y) \)

\(P(x,y)=\left( \frac { 3(9)+2(4) }{ 3+2 } ,\frac { 3(7)+2(-3) }{ 3+2 } \right) \)
\(=\left( \frac { 27+8 }{ 5 } ,\frac { 21-6 }{ 5 } \right) =\left( \frac { 35 }{ 5 } ,\frac { 15 }{ 5 } \right) =(7,3)\)

23.

sin B = \(\frac { 9 }{ 41 } \);
cos B = \(\frac { 40 }{ 41 } \);
tan B =\(\frac { 9 }{ 40 } \) ;
cosec B = \(\frac { 1 }{ sinB } \) = \(\frac { 41 }{ 9 } \);
sec B =\(\frac { 1 }{ cotB } \) = \(\frac { 41 }{ 40 } \);
cot B =\(\frac { 1 }{ tanB } \) = \(\frac { 40 }{ 9 } \)

24.

25m^-2- 16n²= (5m)² - (4n)² [ \(\because \) a²-b²= (a - b)(a + b)]
= (5m - 4n) (5m+ 4n)

25.

\(\left( \cfrac { 1 }{ 9 } \right) ^{ -3 }\)=\(\cfrac { 1 }{ \left( \cfrac { 1 }{ 9 } \right) ^{ 3 } } =\cfrac { 1 }{ \left( \cfrac { 1 }{ 729 } \right) } =729\)

26.

MPN = 90^o(Angle subtended by the diameter is 90°)
\(\angle\)OMP +\(\angle\) OPM = 180^o-120^o= 60^o
\(\angle\)MPO = \(\cfrac { 60^{ o } }{ 2 } \) = 30^o
\(\therefore\)x = \(\angle \)OPN = 90^o- 30^o= 60^o

27.

\(\angle \)QPR = \(\cfrac { 1 }{ 2 } \) \(\angle \)QPR =\(\cfrac { 1 }{ 2 } \) X 70^o= 35^o
In\(\triangle\)QPR
\(\angle \)R + \(\angle \)P + \(\angle \)Q = 180^o
75^o+ 35^o+\(\angle \)Q =180^o
\(\angle \)Q =180^o-110^o= 70^o
\(\angle \)Q = x^o+ 55^o= 70^o
x^o= 70^o- 55^o= 15^o

28.

\(\bar { x } =\frac { \Sigma x }{ n } =\frac { 145+148+142+141+139+140 }{ 6 } =\frac { 855 }{ 6 } \)
x = 142.5 mm²

29.

In the given data Rs. 800 occurs thrice. Hence the mode is Rs. 800

30.

= l³+(-2m)³+(-3n)³-3(l)(-2m)(-3n)
= (l-2m-3n)(l²+(-2m)²+(-3n)²-l-2m-(-2mx-3n)-(-3nxl))
= (l-2m-3n)(l²+4m²+9n²+2lm-6mn+3ln)

31.

(x+4) (x+5) (x+6) = x³+(4 + 5 + 6)x²+(4\(\times\)5 + 5\(\times\)6 + 6\(\times\)4)x + 4\(\times\)5\(\times\)6
= x³+15x²+ (20 + 30 + 24)x + 120
(x+4) (x+5) (x+6) = x³+15x²+74x+120

32.

(i) Multiply both numerator and denominator by the rationalising factor \(\sqrt{14}\).
\(\frac { 7 }{ \sqrt { 14 } } =\frac { 7 }{ \sqrt { 14 } } \times \frac { \sqrt { 14 } }{ \sqrt { 14 } } =\frac { 7\sqrt { 14 } }{ 14 } =\frac { \sqrt { 14 } }{ 2 } \)
(ii) \(\frac { 5+\sqrt { 3 } }{ 5-\sqrt { 3 } } =\frac { (5+\sqrt { 3 } ) }{ (5-\sqrt { 3 } ) } \times \frac { (5+\sqrt { 3 } ) }{ (5+\sqrt { 3 } ) } =\frac { (5+\sqrt { 3 } )^{ 2 } }{ { 5 }^{ 2 }-(\sqrt { 3 } )^{ 2 } } \)
=\(\frac { { 5 }^{ 2 }+(\sqrt { 3 } )^{ 2 }+2\times 5\times \sqrt { 3 } }{ 25-3 } \)
=\(\frac { 25+3+10\sqrt { 3 } }{ 22 } =\frac { 28+10\sqrt { 3 } }{ 22 } =\frac { 2\times [14+5\sqrt { 3 } ] }{ 22 } \)
=\(\frac { 14+5\sqrt { 3 } }{ 11 } \).

33.

\({ 64 }^{ \frac { -2 }{ 3 } }=\frac { 1 }{ 64 ^{ \frac { 2 }{ 3 } }} = \frac { 1 }{ \left( \sqrt [ 3 ]{ 64 } \right) ^{ 2 } } =\frac { 1 }{ { 4 }^{ 2 } } \) (How?) = \(\frac{1}{16}\)

34.

\(-{11\over 75}=-{11\over 3\times 5^2}\)
Since it is not in the form of \(P\over 2^m\times 5^n\)
\(\therefore -{11\over 75}\) has non-terminating and recurring decimal expansion.

35.

(i) {1 ,2}
(ii) {4,5,6,7}
(iii) {1,2,5,6}

36.

(a) A = {Jan, March, May, July, Aug, Oct, Dec}
(b) B = {13,26,39,52,65,78,91}
(c) C = {a,e}
(d) D = {6,7,8,9,10}
(e) E = {1,4,9}

37.

38.

By using the given shapes, students has to make different shapes like birds, house, numbers, animals, rocket etc.
These pictures are called tangram pictures.

39.

Use anyone of the formula to find A & B
A\(\Delta\)B = (A-B) \(\cup \)(B-A) or A \(\Delta\)B = (A \(\cup \)B) - (A \(\cap \)B)
P\(\cup \)Q = {2, 3, 5, 7, 11} \(\cup \) {1, 3, 5, 11}
= {1, 2, 3, 5, 7, 11}
P\(\cap \)Q = {2, 3, 5, 7, 11} \(\cap \) {1, 3, 5, 11}
= {3, 5, 11}
P\(\Delta\)Q = (P\(\cup \)Q) - (P\(\cap \)Q)
= { 1, 2, 3, 5, 7, 11} - {3, 5, 11}
= {1, 2, 7}
P-Q = {2, 3, 5, 7, 11} - {1, 3, 5, 11}
= {2, 7}
Q-P = {1, 3, 5, 11} - {2, 3, 5, 7, 11}
= {1}
P\(\Delta\)Q = (P-Q) \(\cup \) (Q-P)
= {2, 7} \(\cup \) {1}
= {1, 2, 7}

40.

(A – B)′

Multiple Choice Question

40 x 1 = 40

41.

(d)
2(lb + bh + lh) sq. units

42.

(d)
\(25\sqrt { 3 } \) cm ²

43.

(a)
One

44.

(d)
Probability

45.

(b)
2

46.

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

47.

(b)
(4,2)

48.

(b)
(4, 0), (2, 8)

49.

50.

(a)
1

51.

(d)
(a+b+c)²

52.

(a)
2.71 \(\times\) 10⁵

53.

(b)
90°

54.

55.

(d)
34

56.

(a)
0

57.

(b)
1,3,3,3,5

58.

59.

60.

(b)
46

61.

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

62.

(b)
10

63.

(b)
\(p\left( -\frac { b }{ a } \right) \)

64.

(d)
centroid

65.

(b)
rectangle

66.

(b)
AΔB

67.

68.

(d)
disjoint sets

69.

(d)
(i), (iii) and (iv)

70.

(a)
- 0.75

71.

(a)
(0, - 5)

72.

73.

(d)
\(\sqrt{17}\)

74.

(d)
(5,0)

75.

(b)
may be a rational or irrational number

76.

(d)
11

77.

(a)
on the negative direction of x-axis

78.

(b)
X

79.

(b)
rectangle

80.

(d)
ΔABC ≅ ΔFED

Generic 5

8 x 5 = 40

81.

In the quadrilateral ABCD, join one of the diagonals, say AC.
Area of \(\triangle\)ABC = \(\frac{1}{2}\)\(\times\) base \(\times\) height
=\(\frac{1}{2}\)\(\times\)8\(\times\)15\(\times\) 60 cm²
By Pythagoras theorem, in right angled triangle ABC,
AC² = AB² + BC²
= 8² +15² = 64 + 225 = 289 cm
Therefore, AC =\(\sqrt{289}\) =17cm
Now, for\(\triangle\)ACD, let us consider a = 17 cm, b =12 cm, c =25 cm
then, s = \(\frac{a+b+c}{2}=\frac{17+12+25}{2}=\frac{54}{2}\) = 27cm

Area of \(\triangle\)ACD =\(\sqrt { s(s-a)(s-b)(s-c) } \)
=\(\sqrt{27(27-17)(27-12)(27-25)}\)
=\(\sqrt{27\times10\times15\times2}\)
=\(\sqrt{3\times3\times3\times2\times5\times5\times3\times2}\)
= 3 × 3 × 2 × 5 = 90cm²
Therefore, Area of quadrilateral ABCD
=Area of \(\triangle\)ABC + Area of \(\triangle\)ACD
= 60 + 90 = 150 cm²

82.

Now \(B\cap C\) =\(\left\{ 1,\cfrac { 5 }{ \underline { 4 } } ,\cfrac { 7 }{ 4 } ,\underline { 3 } ,\cfrac { 7 }{ \underline { 2 } } \right\} \cap \)= \(\left\{ \cfrac { 1 }{ 2 } ,\cfrac { 5 }{ \underline { 4 } } ,2,\underline { 3, } \cfrac { 7 }{ \underline { 2 } } \right\} \)=\(\left\{ 3,\cfrac { 7 }{ 2 } ,\cfrac { 5 }{ 4 } \right\} \)
\(A\cap \left( B\cap C \right) \)=\(\left\{ \cfrac { 1 }{ 2 } ,1,\cfrac { 5 }{ 4 } ,\cfrac { 7 }{ 4 } ,\underline { 3 } \right\} \cap \left( \underline { 3 } ,\cfrac { 7 }{ 2 } ,\cfrac { 5 }{ 4 } \right) \)=\(\left\{ \cfrac { 5 }{ 3 } ,3 \right\} \)
Then \(A\cap B\)=\(\left\{ \cfrac { 1 }{ 2 } ,\underline { 1 } ,\cfrac { 5 }{ 4 } ,\cfrac { 7 }{ 4 } ,\underline { 3 } \right\} \cap \left[ \underline { 1 } ,\cfrac { 5 }{ \underline { 4 } } ,\cfrac { 7 }{ \underline { 4 } } ,\underline { 3 } ,\cfrac { 7 }{ 2 } \right] \)=\(\left\{ 1,\cfrac { 5 }{ 4 } ,\cfrac { 7 }{ 4 } ,3 \right\} \)
\(A\cap B\cap C\)=\(\left\{ 1,\cfrac { 5 }{ \underline { 4 } } ,\cfrac { 7 }{ 3 } ,\underline { 3 } \right\} \cap \left\{ \cfrac { 1 }{ 2 } ,\cfrac { 5 }{ 4 } ,2,\underline { 3 } ,\cfrac { 7 }{ 2 } \right\} \) =\(\left\{ \cfrac { 5 }{ 4 } ,3 \right\} \)
From(1) and (2) it is property of sets
\(A\cap \left( B\cap C \right) \)=\(\left( A\cap B \right) \cap C\)

83.

Letp (x) x⁵ - 45x⁴ + 36x³ + 45x² - 36x- 1
Sum of co-efficients = 1 - 45 + 36 + 45 - 36 - 1 = 0
Thus x -1 is a factor of p (x)

84.

\(\angle\)In the cyclic quadrilateral \(\angle\) A+\(\angle\)C = 180^o
y + 4^o + 3y^o+ 8^o = 180^o
4y^o + 12^o = 80^o
4y^o = 180^o -12^o = 168
y =\(\cfrac { 168 }{ 4 } \) = 42
\(\angle\)B + \(\angle\)D = 180^o
8x + 12 = 180^o
8x = 180^o- 12^o = 168
x = \(\cfrac { 168 }{ 8 } \) = 21^o
\(\therefore\) \(\angle\)A = y + 4^o = 42^o + 4^o = 46^o
\(\angle\)C = 3y + 8 = 3 \(\times\) 42 + 8 = 126 + 8 = 134^o
\(\angle\)B = 3x + 6 = 3 \(\times\) 21 + 6 = 63 + 6 = 69^o
\(\angle\)D = 5x + 6 = 5 \(\times\) 21 + 6
= 105 + 6 = 111^o

85.

Step Deviation Method
\(\bar { x } =A+\left[ \frac { \Sigma fd }{ \Sigma f } \times c \right] \) Where \(d=\frac { x-A }{ c } \)
Let Assumed mean A = 32
Class Interval c = 45

Age Class Interval	x	Mid x	No.of person f	d=\(\cfrac{x-A}{c}\)	fd
15-19	14.5-19.5	17	4	-3	-12
20-24	19.5-24.5	22	20	-2	-40
25-29	24.5-29.5	27	38	-1	-38
30-34	29.5-34.5	32	24	0	0
35-39	34.5-39.5	37	10	1	10
40-44	39.5-44.5	42	9	2	18
			\(\Sigma f=105\)		\(\Sigma fd=-62\)

Mean \(\bar { x } =A+\left[ \cfrac { \Sigma fd }{ \Sigma f } \times c \right] \)
\(=32+\left[ \cfrac { -62 }{ 105 } \times 5 \right] \)
= 32 + (-2.952)
= 29.05

86.

To represent \(-\frac { 2 }{ 11 } ,-\frac { 5 }{ 11 } and-\frac { 9 }{ 11 } \)on the number line we make 11 markings each being equal distance \(\frac { 1 }{ 11 } \) on the left of 0.
The point A represents \(\left( -\frac { 2 }{ 11 } \right) \) , the point B represents\(\left( -\frac { 2 }{ 11 } \right) \) and the point C represents \(\left( -\frac { 9 }{ 11 } \right) \)

87.

Let us convert the given rational numbers having the same denominators.
L.C.M of 8 and 6 is 24.
\(\frac { 5 }{ 8 } \times \frac { 5\times 3 }{ 8\times 3 } =\frac { 15 }{ 24 } \)
\(\frac { 5 }{ 8 } \times \frac { 5\times 3 }{ 8\times 3 } =\frac { 15 }{ 24 } \)
Now the rational numbers between \(-\frac { 202 }{ 24 } and\frac { 15 }{ 24 } are-\frac { 19 }{ 24 } ,-\frac { 18 }{ 24 } ,-\frac { 17 }{ 24 } ,....,\frac { 0 }{ 24 } ,\frac { 1 }{ 24 } ,\frac { 22 }{ 24 } ,...,\frac { 14 }{ 24 } \)
we can take any seven of them \(\frac { 1 }{ 24 } ,\frac { 2 }{ 24 } ,\frac { 3 }{ 24 } ,\frac { 4 }{ 24 } ,\frac { 5 }{ 24 } ,\frac { 6 }{ 24 } ,\frac { 7 }{ 24 } \)

88.

Distance = \(\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\)
\(PQ=\sqrt{(1-6)^2(3+1)^2}\)
\(=\sqrt{(-5)^2+4^2}=\sqrt{25+16}\)
\(=\sqrt{41}\)
\(QR=\sqrt{(x-1)^2+(8-3)^2}\)
\(=\sqrt{(x-1)^2+5^2}\)
\(=\sqrt{(x-1)^2+25}\)
But PQ = QR
\(\sqrt{(x-1)^2+25}=\sqrt{41}\)
Squaring on both sides
(x - 1)²- 25 = 41
(x-1)² = 41-25 = 16
x-1 =\(\sqrt{16}\) = ±4
x-1 = 4 (or) x-1 = - 4
x = 5 (or) x = - 4 + 1 = - 3
The value of x = 5 or - 3

Generic 8

1 x 8 = 8

89.

In \(\triangle\)PQR, PQ = 5 cm, PR = 6 cm, \(\angle\)QPR = 60°

Construction:
Step 1: Draw \(\triangle\) PQR using the given measurements PQ = 9 cm, QR = 7 cm and RP = 8 cm and construct the perpendicular bisector of any two sides (PQ and QR) to find the mid-points M and N of PQ and QR respectively.
Step 2: Draw the medians PN and RM and let them meet at G. The point G is the centroid of the given \(\triangle\)PQR.