grade ix mathematics...number systems_solutions solution 1 we know that, by definition, the decimal...

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Grade IXMathematics

Exam Preparation Booklet

Chapter WiseImportant Questions-Solutions

Number Systems_Questions

Q. No.1

The decimal representation of an irrational number is always1. terminating2. non-terminating3. recurring4. non-recurring

A. 1 and 3B. 1 and 4C. 2 and 3D. 2 and 4

Q. No.2

Which of the following is not a rational number?

A.

B.

C.

D.

Q. No.3

Find the value of , if the value of .

A. 1.2360B. 0.2360C. 0.23D. 0

Q. No.4

Which of the following is equal to 2?

A.

B.

C.

D.

1. 23417√15

√120

×2

√3

√6

√8

√√5−2

√5+2√5 = 2. 2360

√42

9√(16)5

2 − 49

525

(1)0 × (2)0

Q. No.5

The value of is

A.

B. 1

C.

D.

Q. No.6

The number obtained on rationalising the denominator of is . The difference between numbers a and b is _____.

Q. No.7

The value of the expression , upon simplification, is _____.

Q. No.8

If , then the value of is _____.

Q. No.9

The number of zeros at the end of the product is _____.

Q. No.10

Choose correct options from brackets to fill in the blank spaces.If n is a natural number then will either be a _____ (natural/whole/integer) number or _____ (rational/irrational) number.

0. 9 + 0. 1 3

29

30

31

3032

33

1

3−2√2a + b√2

2√3× 3√6× 6√93√2× 6√3

x = 1 + √2(x−1)(x+1)

x

23 × 34 × 45 × 56

√n

Number Systems_SolutionsSolution 1

We know that, by definition, the decimal representation of a rational number can be terminating, non-terminating and non-terminatingrecurring as well. But the decimal representation of an irrational number is always non-terminating and non-recurring.Hence, the correct answer is option D.

Solution 2

The decimal expansion of is terminating and recurring, therefore, it is a rational number.

The decimal expansion of is which is terminating and recurring, therefore, it is a rational number.

which is non-terminating and non-recurring, therefore, it is an irrational

number.

which is a rational number.

Hence, the correct answer is option C.

Solution 3

Hence, the correct answer is option B.

Solution 4


Solution 5


Solution 6

1. 2341

70. 142857

= √ = √ = = = 0. 353553391. . . . .√15

√120

15

120

1

8

1

2√2

√2

4

× = 2√ = 2√ = = 12

√3

√6

√8

624

14

22

√

= √ ×

=

⎷

=

= 2. 2360 − 2= 0. 2360

√5−2

√5+2

√5−2

√5+2

√5−2

√5−2

(√5−2)2

(√5)2−(2)2

√5−2

1

2 − 4

= 2 − ((2)2)

= 2 − 2

= 2

= 21

= 2

9

525

9

5

2

5

9

545

9−45

0. 9 + 0. 1 3

= +

=

=

=

9

10

12

90810+120

900

930

90031

30

Comparing it with , we get a = 3, b = 2.The required difference = 3 2 = 1

Solution 7

Solution 8

Given that

Therefore,

Solution 9

So, it is clear that the product will have 6 zeros at the end of the product .

Solution 10

If n is a natural numberCase I: n is a perfect square then will be a natural number.Case II: n is a non-perfect square then will be an irrational number.

= × = = 3 + 2√21

3−2√2

1

3−2√2

3+2√2

3+2√2

3+2√2

9−8

a + b√2−

= 2√3 × 3√ × 6√

= 2√3 × 3√3 × 6√3

= 3 × 3 × 3

= 3 + +

= 3= 3

2√3× 3√6× 6√93√2× 6√3

6

2

9

3

12

13

16

12

13

16

6

6

= = x −(x−1)(x+1)

x

x2−1x

1x

x = 1 + √2

⇒ = = × = = √2 − 11x

1

1+√2

1

1+√2

1−√2

1−√2

1−√2

1−2

x − = (1 + √2) − (√2 − 1) = 21x

23 × 34 × 45 × 56

= 23 × 34 × ((2)2)5

× 56

= 23 × 34 × (2)10 × 56

= 213 × 34 × 56

= 27 × 34 × (2 × 5)6

= 27 × 34 × 106

23 × 34 × 45 × 56

√n

√n

Polynomials_Questions

Q. No.1

The value of is

A. 2B. 2000C. 4000D. None of the above

Q. No.2

Which of the following is a factor of the polynomial ?

A. 3B.C.D. All of the above

Q. No.3

The values of k for which is a factor of the polynomial are

A. onlyB. 1 onlyC. For any real value of kD. For no value of k

Q. No.4

The value of the expression for a = 100 is

A. vanishesB.C.

D.

Q. No.5

If is a factor of the polynomial , then itmust be a factor of

A.

10012 − 9992

(a − b)3 − (a3 − b3)

ab

a − b

(x + k) 2x2 + kx − k2

−1

(a − 1)3

− ( )3

− ( − 1)33a

4

a

4

99 + 75 + 24

(99) (75) (24)

3 (99) (75) (24)

(x − k) q (x) = x3 − kx2 − x + k but not of the polynomial r (x) = x2 + x − 1

q (x) + r (x)

B.

C.

D.

Q. No.6

The degree of the polynomial is ______.

Q. No.7

For any two variables a, b and a polynomial , the value of is independent of the variable _____.

Q. No.8

For the polynomial , the value of is _____ the value of , where a is any even prime number.

Q. No.9

If is divided by , the remainder is _____.

Q. No.10

The total number of zeroes of a non-zero constant polynomial is _____.

q (x) − r (x)

q (x)r (x)r(x)

q(x)

0x5 + x4 + 2x3 + 3x2 + 5

p (x) = x2 + 3ax + 2b p (3) − p (2)

p (x) = x4 + x2 + 1 p (a) p (−a)

xn + n x − 1

Polynomials_SolutionsSolution 1


Solution 2

Hence, the correct answer is option D.

Solution 3

For to be a factor of the polynomial , we must have i.e.

This means that is a factor of the polynomial for any real value of k.Hence, the correct answer is option C.

Solution 4

We know that x3 + y3 + z3 – 3xyz = (x + y + z) (x2 + y2 + z2 – xy – yz – zx).If x + y + z = 0, then x3 + y3 + z3 – 3xyz = 0 i.e. x3 + y3 + z3 = 3xyz.

The given expression is for a = 100 i.e. i.e.

Clearly, Therefore, Hence, the correct answer is option D.

Solution 5

Clearly, will be a factor of the polynomial .Hence, the correct answer is option C.

Solution 6

Since the highest power of the variable x in above polynomial is 4Therefore, the degree of the polynomial is 4.

Solution 7

10012 − 9992

= 10012 − 10002 + 10002 − 9992 (Adding and subtracting 10002)

= (1001 − 1000) (1001 + 1000) + (1000 − 999) (1000 + 999) [Using a2 − b2 = (a − b) (a + b)]

= 1 (2001) + 1 (1999)

= 2001 + 1999

= 4000

(a − b)3 − (a3 − b3)

= (a − b)(a − b)2 − (a − b) (a2 + ab + b2)

= (a − b) [(a − b)2 − (a2 + ab + b2)]

= (a − b) [(a2 − 2ab + b2) − (a2 + ab + b2)]

= (a − b) [−3ab]

= −1 × 3 × ab × (a − b)

(x + k) p (x) = 2x2 + kx − k2 p (−k) = 0

2(−k)2

+ k (−k) − k2 = 0

2k2 − k2 − k2 = 00 = 0

(x + k) 2x2 + kx − k2

(a − 1)3 − ( )3

− ( − 1)3

3a

4

a

4(99)3 − (75)3 − (24)3 (99)3 + (−75)3 + (−24)3

99 − 75 − 24 = 0

(99)3 − (75)3 − (24)3 = 3 (99) (−75) (−24) = 3 (99) (75) (24)

q (x) = x3 − kx2 − x + k = (x − k) (x2 − 1) and r (x) = x2 + x − 1

(x − k) q (x)r (x)

0x5 + x4 + 2x3 + 3x2 + 5

= x4 + 2x3 + 3x2 + 5

0x5 + x4 + 2x3 + 3x2 + 5

and Therefore, which is independent of the variable b.

Solution 8

For the polynomial and a = 2, even prime

And

Thus, the value of is equal to the value of .

Solution 9

Let .Now, the remainder obtained when is divided by is .Therefore, the remainder is

Solution 10

Let be a non-zero constant polynomial where k is any non-zero constant.Its zero is given by But Therefore, a non-zero constant polynomial has no zeroes.The total number of zeroes of a non-zero constant polynomial is zero.

p (x) = x2 + 3ax + 2b

∴ p (3) = 9 + 9a + 2b p (2) = 4 + 6a + 2b

p (3) − p (2) = (9 + 9a + 2b) − (4 + 6a + 2b) = 5 + 3a

p (x) = x4 + x2 + 1

p (2) = 24 + 22 + 1 = 16 + 4 + 1 = 21

p (−2) = (−2)4 + (−2)2 + 1 = 16 + 4 + 1 = 21

p (a) p (−a)

p (x) = xn + n

xn + n x − 1 p (1)

p (1) = 1n + n = 1 + n

p (x) = k

p (x) = 0 ⇒ k = 0k ≠ 0

Coordinate Geometry_Questions

Q. No.1

How many points are there on the x-axis which are at the distance of 5 units from the point P(2, 0)?

A. No pointB. One pointC. Two pointsD. Infinitely many points

Q. No.2

Which of the following points is collinear with the points and ?

A.

B.

C.D. A and C

Q. No.3

Two points which have different abscissae and the difference between their ordinates vanishes, lie on

A. x = y lineB. y-axisC. a line parallel to x-axisD. a line parallel to y-axis

Q. No.4

Which of the following statements is incorrect?

A. The point (7, 3) lies in the 1st quadrant.B. The point (0, −4) lies in the 4th quadrant.C. The point (−5, 5) lies in the 2nd quadrant.D. The point (−3, −2) lies in the 3rd quadrant.

Q. No.5

Which graph correctly represents the location of the point A (−2, 3) on the coordinate plane?

(−1, 2) (3, 0)

(1, 1)

(1, 2)

(−2, 1)

A.

B.

C.

D.

Q. No.6

The sum and the product of the abscissa and the ordinate of any point are always of same sign in _____ quadrant.

Q. No.7

The total number of points which are at a distance of 4 cm and 5 cm from x-axis and y-axis respectively are _____.

Q. No.8

The measure of the angle between positive x-axis and negative y-axis measured counterclockwise is _____.

Q. No.9

Let ΔABC is a triangle whose vertices have coordinates A (−1, 1), B (−1, 5) and C (4, 4). The area of ΔABC is _____.

Q. No.10

According to the map of a boarding school, the academic block is located at the origin. After attending classes in the academic block,Abhay went to his dormitory which is located at (−4, 0). After keeping his books, he went to the dining hall which is located at (−4, 2)for lunch. After having lunch, he goes straight to the recreation hall which is located at (3, 2) for playing chess. A distance of 1 unit onthe map represents an actual distance of 50 m. The total distance traveled by Abhay after class to play chess is _____.

Coordinate Geometry_SolutionsSolution 1

It is clear from the figure that there are exactly two points A and B(7, 0) which are at the distance of 5 units from the point P(2, 0).Hence, the correct answer is option C.

Solution 2

It is clear from the figure that the points and are collinear with the point Hence, the correct answer is option A.

Solution 3

(−3, 0)

(−1, 2) (3, 0) (1, 1)

Let and be two points which have different abscissae and the difference between their ordinates vanishes.

From the figure, it is clear that they lie on a line parallel to x-axis.Hence, the correct answer is option C.

Solution 4

If the x-coordinate and the y-coordinate of a point are positive and negative, respectively, then the point lies in the 4th quadrant.The x-coordinate of the point (0, −4) is 0, therefore, this point lies on the y-axis.Thus, the statement given in alternative B is incorrect.Hence, the correct answer is option B.

Solution 5

The point A (−2, 3) can be located on the coordinate plane by moving 2 units on the negative x-axis from the origin (0, 0), and then 3 unitson the positive y-axis.


Solution 6

In the I-quadrant, both the abscissa and the ordinate are of positive sign.Therefore, their sum and product will also be of same sign i.e. positive.Hence, the required answer is I-quadrant

Solution 7

(1, 2) (4, 2)

From the figure, it is clear that the total number of points which are at a distance of 4 cm and 5 cm from x-axis and y-axis respectively arefour.

Solution 8

The required angle is .

Solution 9

The points A (−1, 1), B (−1, 5) and C (4, 4) can be plotted on the coordinate plane as:

270°

Draw CM ⊥ AB.It is seen that, AB = 4 units and CM = 5 units.

Area of ΔABC = 10 square units

Thus, the area of ΔABC is 10 square units.

Solution 10

It is given that on the map, the academic block is located at the origin (0, 0).Then, the coordinates (−4, 0), (−4, 2), and (3, 2) can be located on the map as

From the map, it is clearly observed that the total distance travelled by Abhay after class to play chess is (4 + 2 + 4 + 3) units = 13 units.It is given that 1 unit on the map represents a distance of 50 m.Thus, total distance travelled by Abhay = (13 × 50) m = 650 m

Linear Equations in Two Variables_Questions

Q. No.1

Which of the following points doesn't lie on the line ?

A. (13, 8)B. (10, 10)C. (7, 12)D. (5, 14)

Q. No.2

Which of the following equations has the solution given by ?

A.B.C.D.

Q. No.3

How many linear equations can have as its solution?

A. NoneB. only one C. only two D. Infinitely many

Q. No.4

The number of solution(s) of the equation on the Number line is a and on the Cartesian plane is b, then which of thefollowing options is not correct?

A. The solution of given equation is

B. a = 1C. b = infinteD. a + b = finite

Q. No.5

At what point on the graph , the abscissa is equal to the ordinate?

A.

2x + 3y = 50

x = , y = 3ππ

2

2x − 3y = 6

6x − y = 0

2x = 3y

3x = y

x = −1, y = 4

x + 3 =x

3

x =−9

2

4x + 7y = 33

(−7, 4)

B.

C.D.

Q. No.6

The graph of the linear equation meets a line that is perpendicular to the y-axis, at a distance 4 units from the origin andin the negative direction of y-axis. The abscissa of that meeting point is _____.

Q. No.7

The coordinate of a point on the graph of the equation for which = 4 is _____.

Q. No.8

The graph of lines meets at the point _____.

Q. No.9

The graph of the linear equation cuts the x-axis at _____ and the y-axis at _____.

Q. No.10

Let y varies indirectly as x. If y = 5 when x = 4, then the value of y when x = 15 is _____.

( , 5)−1

2

(−3, −3)

(3, 3)

2x + 3y = 0

9x − 4y + 4 = 0 x − y

x = y and x = −y

6x + 9y = 36

Linear Equations in Two Variables_SolutionsSolution 1

Let's consider the point (5, 14).Substitute in the given line , we get

Therefore, the point (5, 14) doesn't lie on the line .Hence, the correct answer is option D.

Solution 2

Consider the equation .Substitute to get


Solution 3

The required information represents a point on a plane.And we know that infinite number of lines can pass through a point.Hence, the correct answer is option D.

Solution 4

The given equation is

Therefore, the number of solution on the number line is 1 and on the cartesian plane is infinite.i.e. a = 1, b = infiniteNow a + b cannot be equal finite.Option D is incorrect.Hence, the correct answer is option D.

Solution 5

It is given that .Therefore, becomes Therefore, .The required point is (3, 3).Hence, the correct answer is option D.

Solution 6

The coordinates of the points lying on the line perpendicular to the y-axis, at a distance 4 units from the origin and in the negative directionof y-axis are of the form (x, ).Putting in the equation , we get

Thus, the abscissa of the required point is 6.

Solution 7

Given = 4

Put this value in , we get

x = 5, y = 14 2x + 3y = 50LHS : 2 (5) + 3 (14) = 10 + 42 = 52 ≠ 50 = RHS

2x + 3y = 50

6x − y = 0

x = , y = 3ππ

26x − y = 0

6( ) − (3π) = 0

3π − 3π = 0

0 = 0

π

2

x = −1, y = 4 (−1, 4)

x + 3 = x

3

x − + 3 = 0

+ 3 = 0

2x = −9

x =

x

3

2x

3

−9

2

x = y4x + 7y = 33 4x + 7x = 33 ⇒ 11x = 33 ⇒ x = 3y = x = 3

−4y = −4 2x + 3y = 0

2x + 3 (−4) = 0

2x = 12x = 6

x − y⇒ x = y + 4

9x − 4y + 4 = 0

Thus, the required point is .

Solution 8

The two lines are .Lets substitute , we will get

The required point is the origin.

Solution 9

The graph of the linear equation cuts the x-axis when y = 0 i.e.

And cuts the y-axis when x = 0 i.e.

Solution 10

Given that y varies indirectly as x

For x = 4 and y = 5, we have Now, when x = 15,

9 (y + 4) − 4y + 4 = 0

5y + 40 = 0

y = −8

∴ x = −4(−4, −8)

x = y and x = −yx = y in equation x = −y

y = −y

2y = 0

y = 0

⇒ x = 0

6x + 9y = 366x + 9 (0) = 36

6x = 36x = 6

6 (0) + 9y = 36

9y = 36

y = 4

⇒ xy = kk = 4 × 5 = 20

y = = =kx

20

15

4

3

Lines and Angles_Questions

Q. No.1

If one angle of a triangle is less and the other angle is more than the third angle, then the triangle is

A. an isosceles triangleB. an obtuse triangleC. an equilateral triangleD. a right triangle

Q. No.2

Which of the following statements is correct?Statement-I: The sum of two adjacent angles is 150°, so they can form a linear pair.Statement-II: A triangle can have all the angles less than 60°.Statement-III: A triangle can have a straight angle.

A. Statement-IIB. Both the statements-II and IIIC. All the above three statementsD. None of the above

Q. No.3

Which of the following is a line?

A. AOBB. CODC. EOFD. All three of them

Q. No.4

30° 30°

The value of is

A. 15B. 25C. 35D. 50

Q. No.5

The sum of all the exterior angles of a triangle is

A. 360B. 300C. 270D. 180

Q. No.6

In , OB and OC are the angle bisectors of respectively. The measure of y is _____.

Q. No.7

The maximum number of triangles that can be drawn having angles 55°, 63° and 61° is _____.

Q. No.8

A triangle can have two _____ (acute/right/obtuse) angles.

Q. No.9

α

°

°

°

°

°

°

°

°

△ ABC ∠B and ∠C

Two straight lines on a plane, when produced indefinitely, are either _____ or _____.

Q. No.10

If line l is perpendicular to the line m, and line m is perpendicular to another line n, then the lines l and n are _____ to each other.

Lines and Angles_SolutionsSolution 1

Let the three angles be .Using the angle sum property, we have

Therefore, the three angles are The given triangle is a right-angled triangle.Hence, the correct answer is option D.

Solution 2

For a linear pair, the sum of two adjacent angles must be 180°.A triangle can't have all the angles less than 60° as it would not satisfy the angle sum property of triangles.A triangle can't have a straight angle as it would become a straight line and not a triangle.All the above three statements are incorrect.Hence, the correct answer is option D.

Solution 3

Since, and Therefore, COD is a line.Hence, the correct answer is option B.

Solution 4

In , using the exterior angle property, we have

In , using the angle sum property, we get

In , using the exterior angle property, we will get

Hence, the correct answer is option A.

Solution 5

We know that the sum of all the exterior angles of any polygon is always 360 .Hence, the correct answer is option A.

Solution 6

Using angle sum property in , we get

Since, OB and OC are the angle bisectors of respectively, therefore

Now, using angle sum property in , we get

Solution 7

Since, 55° + 63° + 61° = 179°This triangle doesn't satisfy the angle sum property of triangles.Therefore, no such triangle exists.Hence, the required number of triangles is zero.

Solution 8

A triangle cannot have two right or obtuse angles. Therefore, a triangle can have two acute angles.

x − 30°, x, x + 30°

x − 30° + x + x + 30° = 180°

3x = 180°x = 60°

30°, 60°, 90°

∠COA + ∠AOF + ∠FOD = 70° + 50° + 60° = 180° ∠COE + ∠EOB + ∠BOD = 65° + 40° + 75° = 180°

△ ABC12y = 7x + 5z . . . . . (1)△ XYZ

5x + 12y + 7z = 180°

5x + (7x + 5z) + 7z = 180° [Using (1)]

12x + 12z = 180°

x + z = 15° . . . . . (2)

△ PQRα = x + z = 15°

°

△ OBCx + 10° + 150° = 180°

x = 20°

∠B and ∠C∠B = 2x = 40° and ∠C = 2 × 10° = 20°

△ ABCy + 20° + 40° = 180°

y = 120°

Solution 9

Two straight lines on a plane, when produced indefinitely, are either intersecting or parallel.

Solution 10

If line l is perpendicular to the line m, and line m is perpendicular to another line n, then the lines l and n are parallel to each other.

Triangles_Questions

Q. No.1

Which of the following statements is incorrect for a triangle?

A. the side opposite to the greater angle is longerB. The angle opposite to the longer side is greaterC. The sum of any two sides is always greater than the third sideD. The sum of any two angles is always greater than the third angle

Q. No.2

If , then which of the following options is correct?

A. AC = PRB. Perimeter of = Perimeter of C. Area of = Area of D. All of the above

Q. No.3

Two triangles ABC and PQR are such that AB = QR = a, PQ = BC = b and . These triangles are always

A. Congruent and equilateralB. Congruent and isoscelesC. Equilateral but not congruentD. Neither congruent nor equilateral

Q. No.4

The given two triangles ABC and EFG, in which , are congruent by AAS rule if

A.B. AB = EFC. BC = FGD. AC = FG

△ ABC ≅△ PQR

△ ABC △ PQR

△ ABC △ PQR

∠P = ∠C = 61°

∠A = ∠F and ∠E = ∠B

∠C = ∠G

Q. No.5

Which of the following triangles are always congruent?

A. All equilateral trianglesB. All right-angled trianglesC. All right-angled isosceles trianglesD. None of the above

Q. No.6

A point equidistant from two intersecting lines lies on the _____ of the angles formed by the two lines.

Q. No.7

A triangle ABC has two sides of measure 3 cm and 4 cm. The possible length of the third side of this triangle lies between _____ and_____.

Q. No.8

Choose the correct option from to fill in the blank.The relation between in given by _____ .

Q. No.9

The total number of triangles that are possible to construct with lengths 7 cm, 5 cm and 13 cm are _____.

Q. No.10

If one side of the triangle is extended, then the exterior angle so formed will be equal to the sum of the two _____.

(≤, <, =, >, ≥)α and β α β

Triangles_SolutionsSolution 1

Consider the statement-IV.Let's consider a triangle with angles .Here, we can see that Thus, this statement is incorrect.Hence, the correct answer is option D.

Solution 2

Given that in accordance with the correspondence .Threfore, AB = PQ, BC = QR, AC = PRAlso, Perimeter of = AB + BC + AC = PQ + QR + PR = Perimeter of And Area of = Area of (By CPCT)Hence, the correct answer is option D.

Solution 3

There is no criterion rule such as SSA

It is clear from the figure that the two triangles are not congruent and nor equilateral as each angle of an equilateral triangle is of measure .


Solution 4

The given two triangles ABC and EFG are congruent by AAS rule in accordance with the correspondence It is clear from the figure that AC = FGHence, the correct answer is option D.

Solution 5

All equilateral triangles/right-angled triangles / right-angled isosceles triangles are not congruent as they do not follow any of theSSS/SAS/ASA/AAS/RHS criteria.Hence, the correct answer is option D.

Solution 6

A point equidistant from two intersecting lines lies on the bisectors of the angles formed by the two lines.

Solution 7

Since triangle ABC has two sides of measure 3 cm and 4 cm.Therefore, the possible length of the third side of this triangle is given by

A triangle ABC has two sides of measure 3 cm and 4 cm. The possible length of the third side of this triangle lies between 1cm and 7cm.

Solution 8

The given triangle is a right-angled triangle.Therefore, using the Pythagoras theorem, we get BC = 8 cm.Now, we know that if two sides of a triangle are unequal, then the longer side has a greater angle opposite it.

30°, 60°, 90°30° + 60° = 90° ≯ 90°

△ ABC ≅△ PQR A ↔ P, B ↔ Q, C ↔ R

△ ABC △ PQR△ ABC △ PQR

60°

A ↔ F, B ↔ E, C ↔ G

Difference of the other two sides < The length of third side < Sum of the other two sides⇒ 4 − 3 < The length of the third side < 4 + 3

⇒ 1 < The length of the third side < 7

Since BC > AB.

Solution 9

We know that the sum of two sides of a triangle is always greater than the third side.But, in this case Therefore, a triangle with given measurements is not possible to construct.The required number of such triangles is zero.

Solution 10

If one side of the triangle is extended, then the exterior angle so formed will be equal to the sum of the two interior opposite angles.

∴ ∠A > ∠B i. e. α > β

5 + 7 = 12 ≯ 13

Quadrilaterals_Questions

Q. No.1

Diagonals of a quadrilateral ABCD bisect each other and are equal and vice-versa.Which of the following can not be the required quadrilateral?

A. TrapeziumB. RectangleC. RhombusD. Square

Q. No.2

ABCD is a parallelogram whose each side is equal, then its diagonals are always

A. equalB. perpendicularC. perpendicular bisectorD. equal and perpendicular bisector

Q. No.3

The diagonal of a parallelogram divides it into two triangles of

A. equal area and unequal perimeterB. equal perimeter and unequal areaC. equal area and perimeterD. equal area but unequal perimeter

Q. No.4

The perimeter of a rhombus with diagonals and is

A.

B. 2

C. 3

D. 4

Q. No.5

In the given figure, WXYZ is a parallelogram. Irrespective of the values of y and z, find the values of x.

d1 d2

√d12

+ d22

√d12

+ d22

√d12

+ d22

√d12

+ d22

A. 52°B. 50°C. 75°D. 80°

Q. No.6

The quadrilateral obtained by joining the mid-points of an irregular quadrilateral or a parallelogram will always be a _____.

Q. No.7

If PQRS is a rectangle as shown in the figure, then the value of is______.

Q. No.8

The quadrilateral formed by joining the mid-points of the sides of a _____ is a _____ and vice-versa.

Q. No.9

The angles of a quadrilateral, in order, are in the ratio 4 : 4 : 5 : 5. This type of quadrilateral is a _____.

Q. No.10

∠ SOR

ABCD is a kite in which ∠ACD = 33º. Then the measure of ∠DBC is _____.

Quadrilaterals_SolutionsSolution 1

We know that the diagonals of a rectangle/square/rhombus bisect each other and are equal and vice-versa.Hence, the correct answer is option A.

Solution 2

If ABCD is a parallelogram whose each side is equal, then it is either a square or a rhombus.The diagonals of both squares and rhombuses are perpendicular bisector.Hence, the correct answer is option C.

Solution 3

We know that a diagonal of a parallelogram divides it into two congruent triangles.And we also know that the corresponding sides and angles, perimeter and area of congruent triangles are also equal.Hence, the correct answer is option C.

Solution 4

Let AC = and BD =

Therefore, OA = and OB = Using Pythagoras theorem, we get

The perimeter of rhombus = 4 AB = Hence, the correct answer is option B.

Solution 5

∠XYZ + ∠XYP = 180°(Linear pair angles)⇒ ∠XYZ + 84°= 180°⇒ ∠XYZ = 96°Now, applying angle sum property in ΔXYZ, we obtain∠XYZ + ∠YZX + ∠ZXY = 180°⇒ 96°+ 32°+ ∠ZXY = 180°⇒ ∠ZXY = 180°− 128°= 52°∴ ∠ZXY = ∠XZW (Alternate interior angles)∴ x = 52°Hence, the correct answer is option A.

Solution 6

The quadrilateral obtained by joining the mid-points of an irregular quadrilateral or a parallelogram will always be a parallelogram.

d1 d2d1

2

d2

2

AB2 = OA2 + OB2

⇒ AB2 = ( )2

+ ( )2

⇒ AB =

d1

2

d1

2

√d12+d2

2

2

2√d12 + d2

2

Solution 7

It is given that PQRS is a rectangle.We know that diagonals of a rectangle bisect each other

Using the angle sum property of triangle, we will get

(Vertically opposite angles)If PQRS is a rectangle as shown in the figure, then the value of is .

Solution 8

The quadrilateral formed by joining the mid-points of the sides of a rhombus is a rectangle and vice-versa.

Solution 9

Clearly, the adjacent angles of the quadrilaterals are equal.This is only possible in case of an isosceles trapezium.Hence, The angles of a quadrilateral, in order, are in the ratio 4 : 4 : 5 : 5. This type of quadrilateral is a isosceles trapezium.

Solution 10

We know that the diagonals of a kite are perpendicular to each other.

In , using the angle sum property, we will get In , BC = DC

∴ OP = OQIn △ OPQ,

OP = OQ

⇒ ∠OQP = ∠OPQ = 41° (Angles opposite to equal sides)

∠POQ = 180° − (41° + 41°) = 98°⇒ ∠SOR = ∠POQ = 98°

∠ SOR 98°

∴ ∠COD = 90°△ COD ∠ODC = 57°△ BCD

⇒ ∠DBC = ∠ BDC (Angles opposite to equal sides)∠DBC = 57°

Areas of Parallelograms and Triangles_Questions

Q. No.1

A parallelogram ABCD is such that the diagonal AC triangulates the given parallelogram into two congruent equilateral triangles. Thearea of the parallelogram if BD = 4 units and OC = units is

A. square units.

B. square units.

C. square units.

D. square units.

Q. No.2

PQRS is a parallelogram. A and B are any points on PQ and SR respectively. if ar( ) = 6 cm2, ar( ) = 10 cm2 and ar() = 4 cm2, then ar(PQRS) is

A. 10 cm2

B. 15 cm2

C. 20 cm2

D. 25 cm2

Q. No.3

The given parallelogram in the figure is divided into 12 congruent rhombi by equidistant parallel lines. The area of the given figure is144 cm2 . The area of the shaded region is given by

2

√3

1

√3

2

√3

4

√3

8

√3

△ PBR △ PBQ

△ SAB

A. 50 cm2

B. 60 cm2

C. 72 cm2

D. 90 cm2

Q. No.4

In the given figure, ABDE and CDEF are parallelograms. Then ar(ABDE) equals

A. ar(ΔFDC)B. 2 ar(ΔFDC)C. ar(ΔADC)D. 2 ar(ΔADC)

Q. No.5

ABCD is a parallelogram. E is the mid-point of side DC. The ratio of the area of triangle DBE to that of the parallelogram ABCD is

A. 1 : 1B. 1 : 2C. 1 : 3D. 1 : 4

Q. No.6

ABC is a triangle in which D is the mid-point of BC and E is the mid-point of AD. Also, ED = 6 cm and CD = 8 cm as shown in thefigure.

The area of is _____.△ ADB

Q. No.7

In the given figure, ABGE is a parallelogram. F is the midpoint of AC. Then ar(ΔBCF) : ar(ACDE) = _____.

Q. No.8

The area of the parallelogram ABEF is 180 cm2. If AB = 12 cm, then the height of the triangle ADF is _____.

Q. No.9

ABCD is a rectangle with length and breadth 7 cm and 5 cm respectively. ACEF is a parallelogram whose area is _____.

Q. No.10

Fill in the blanks.The diagonals of a _____ intersect to divide it into four triangles of equal area.

Areas of Parallelograms and Triangles_SolutionsSolution 1

Since the diagonal AC triangulate the parallelogram ABCD into two congruent equilateral triangles, this implies that AB = BC = CD = DA.Therefore, ABCD is a rhombus.And we know that the diagonals of rhombus bisect each other.Therefore, AC = 2(OC) = .

We also know that the area of a rhombus = square units.


Solution 2

We have,ar(PQRS) = ar( ) + ar( ) + ar( ) = ar( ) + ar( ) + ar( )

Similarly,

ar(PQRS) = 6 +10 + 4 = 20 cm2.Hence, the correct answer is option C.

Solution 3

The given parallelogram is divided into 12 congruent small rhombi.We can clearly observe from the figure that all the 10 shaded triangles are of equal areas as they line on equal bases and between twoparallel lines.Since, ar(ABCD) = 144 cm2

The area of each smaller rhombus = cm2

Also, the area of each shaded triangle = cm2

So, the area of 10 shaded triangles = cm2.Hence, the correct answer is option B.

Solution 4

Clearly ABDE and CDEF are two parallelograms on same base DE and between same parallels.Therefore, ar(ABDE) = ar(CDEF)Also, CDEF and ΔFDC lie on same base FC and between same parallels.Therefore, ar(CDEF) = ar(ABDE) = 2 ar(ΔFDC)Hence, the correct answer is option B.

Solution 5

Clearly, and the parallelogram ABCD lie on same base DC and between two parallel sides. ar( ) = ar(ABCD) .....(1)

Also, E is the mid-point of side DC.So, BE is the median of .

ar( ) = ar( ) = ar(ABCD) [Using (1)]

Thus, .

The required ratio is 1 : 4.Hence, the correct answer is option D.

Solution 6

2 × =2

√3

4

√3

d1d2 = × 4 × =12

12

4

√3

8

√3

△ PBR △ PRQ △ PSB△ PBR △ PBQ △ SAB

[∵ ar (PRQ) = ar (PBQ) as they both lie on same base PQ and between two parallel lines of parallelogram]

ar (PSB) = ar (SAB) as they both lie on same base PQ and between two parallel lines of parallelogram

⇒

⇒ = 1214412

= 6122

10 × 6 = 60

△ BDC∴ △ BDC 1

2

△ BDC⇒ △ BDE 1

2△ BDC 1

4

=ar(△DBE)

ar(ABCD)

14

Since D is the mid-point of BC, so AD is the median. ar( ACD) = ar( ABD) .....(1)

Also, E is the mid-point of AD, so CE acts as the median for ar( ACE) = ar( ECD) = ar( ACD) = ar( ABD) [Using (1)]

ECD is a right-angled triangle, ar( ECD) = cm2.

ABD = 2 ar( ECD) = 2 24 = 48 cm2

Solution 7

Given ABGE is a parallelogram.The parallelograms on the same base and between the same parallels are equal in area.∴ ar(ABGE) = ar(ACDE) ...(1)If a triangle and a parallelogram lie on the same base and are between the same parallels, then the area of the triangle is half the area ofthe parallelogram.

∴ ar(ΔABF) = ar(ABGE)

⇒ ar(ΔABF) = ar(ACDE) ...(2) [Using (1)]

It is known that the median of a triangle divides it into triangles with equal areas.∴ ar(ΔABF) = ar(ΔBCF)

ar(ΔBCF) = ar(ACDE) [Using (2)]

∴ ar(ΔBCF) : ar(ACDE) = 1 : 2

Solution 8

The parallelograms ABEF and triangle ADF lie between the same parallels AB and DE.Given, ar(parallelogram ABEF) = 180 cm2

⇒ AB × Height = 180 cm2

⇒ 12 cm × Height = 180 cm2

Thus, the height of the triangle ADF = the parallelogram ABEF = 15 cm.

Solution 9

Area of rectangle ABCD = Length × Breadth = 7 cm × 5 cm = 35 cm2

Now, the diagonal of a rectangle divides it into two triangles of equal areas.∴Area (ΔABC) = Area (ΔADC) Here, ΔABC and parallelogram ACEF lie on the same base AC and between the same parallels.

Solution 10

The diagonals of a parallelogram intersect to divide it into four triangles of equal area.

∴ △ △

△ ACD∴ △ △

12

△12

△

△

∴ △ × CD × ED = × 8 × 6 = 2412

12

∴ △ × △ ×

Height = = 15 cm180

12

= ar (ABCD) = × 35 = 17. 5 cm212

12

∴ ar (△ ABC) = ar (ACEF)

⇒ ar (ACEF) = 2 × ar (△ ABC) = 2 × 17. 5 = 35 cm2

12

Circles_Questions

Q. No.1

In the figure, AB, CD and EF are equal chords. Find the angle subtended by the arc AD at point G.

A.B.C.D.

Q. No.2

If the sum of a pair of opposite angles of a quadrilateral is 180º, the quadrilateral is

A. RectangleB. SquareC. CyclicD. All of the above

Q. No.3

AB and CD are equal chords of the given circle with centre O. Find the measure of ∠BCO.

A. 50°B. 80°C. 100°D. Data is inadequate

35°

45°

55°

65°

Q. No.4

A rhombus is circumscribed by a circle. The angle made by sides of rhombus on the centre of circle measures

A. 60 , 120 , 60 , 120B. 30 , 60 , 120 , 150C. 75 , 105 , 75 , 105D. 90 , 90 , 90 , 90

Q. No.5

Statement-I: The angles subtended by a chord AB at any two points C and D of a circle are equal in measure.Statement-II: Two chords of lengths 20 cm and 18 cm can be at the distances 10 cm and 9 cm, respectively from the centre.

A. Statement-I is trueB. Statement-II is trueC. Both the statements are trueD. Both the statements are false

Q. No.6

If any line segment AB joining two points A and B, subtends equal angles at two other points C and D, lying on the same side of theline containing the line segment, then the four points A, B, C and D are _____.

Q. No.7

In a circle of diameter AB and radius _____ units, a chord CD of length 24 units can be drawn parallel to the diameter at the distanceof 9 units from the centre.

Q. No.8

The total number of distinct circles that can pass from the vertices of a square, taking three vertices at a time, is _____.

Q. No.9

The diameter of the given circle is 10 cm. The measure of the BOC would be _____, given that the length of the chord is 5 cm.

° ° ° °

° ° ° °

° ° ° °

° ° ° °

∠

Q. No.10

In the given figure, the measure of ∠ABD is _____.

Circles_SolutionsSolution 1

Since AB, CD and EF are equal chords, therefore, they will subtend equal angle at the centre.

The angle subtended by the arc AD at point O = We know that the angle subtended by an arc at the centre is double the angle subtended by it at any other point on the remaining part ofthe circle.


Solution 2

If the sum of a pair of opposite angles of a quadrilateral is 180º, the quadrilateral is cyclic.Moreover, in case of a rectangle and a square, the sum of a pair of opposite angles is also 180º.Hence, the correct answer is option D.

Solution 3

We have ∠AOB = ∠COD = 50° (Equal chords subtend equal angles at the centre)Now, ∠AOB + ∠BOC + ∠COD = 180° (AD is a straight line)⇒ 50° + ∠BOC + 50° = 180°⇒ ∠BOC = 80°

Now, ∠CBO = ∠BCO (Angles opposite to the equal sides)Using, angle sum property in the triangle OBC, we get ∠CBO + ∠BCO + ∠BOC = 180°∠BCO + ∠BCO + 80° = 180°⇒ ∠BCO = 50°Hence, the correct answer is option A.

Solution 4

A rhombus circumscribed by a circle is actually a square as shown in the figure.And the angles made by the sides of the square on the centre of circle measures each.Hence, the correct answer is option D.

Solution 5

Statement-I: If two points lie in the same segment only, then the angles will be equal otherwise they are not equal.Statement-II: The larger chord is basically closer to the centre of the circle as compared to the shorter chord.Both statements are false.Hence, the correct answer is option D.

Solution 6

If any line segment AB joining two points A and B, subtends equal angles at two other points C and D, lying on the same side of the linecontaining the line segment, then the four points A, B, C and D are concyclic.

Solution 7

∴ ∠AOB = ∠COD = ∠EOF = 30°∠AOD = ∠AOB+ ∠BOC + ∠COD = 30° + 10° + 30° = 70°

∴ ∠AGD = = = 35°∠AOD

2

70°

2

90°

It is clear that MD = 12 units and OM = 9 units.Using the Pythagoras theorem, we get OD = 15 units, which is the radius of the circle.

Solution 8

Since a square has all the angles as right angles with the sum of pair of opposite angles equal to .Therefore, this quadrilateral will act as a cyclic quadrilateral when circumscribed by a circle.Thus, the required number of circles is one.

Solution 9

The diameter of the given circle is 10 cm.Therefore, the radius, OA = OB = 5 cm = AB

OAB is equilateral triangle.

Solution 10

We know that the angle in a semi-circle is a right angle.∴ ∠AEC = 90°Applying angle sum property of triangles in ΔAEC, we get∠AEC + ∠EAC + ∠ECA = 180°⇒ 90° + 68° + ∠ECA = 180°⇒ ∠ECA = 22°Therefore, ∠ACD = ∠ECA + ∠ECD = 22° + 46° = 68°It is also known that the angles in the same segment are equal.∴ ∠ABD = ∠ACD = 68°.

180°

⇒△

∴ ∠AOB = 60°

⇒ ∠BOC = 180° − 60° = 120°

Heron's Formula_Questions

Q. No.1

What is the area of an isosceles triangle whose base is a unit and equal sides are b units each?

A.

B.

C.

D.

Q. No.2

The area of an equilateral triangle ABC with each side a = , and that of an isosceles triangle PQR with base a and equal sides

b is .Then ar(ABC) = ar(PQR) if

A.B.C.D.

Q. No.3

What will be the area of the isosceles triangle, if the perimeter is 18 cm and the base is 8 cm?

A. 12 cm2

B. cm2

C. cm2

D. cm2

Q. No.4

The area of a triangular field is 150 m2 and its sides are in the ratio 3 : 4 : 5. Then the perimeter of this triangle is

A. 50 cmB. 60 cmC. 62.5 cmD. 67.5 cm

Q. No.5

√4b2 − a2 square unitsa

4

√2b2 − a2 square unitsa

4

√b2 − a2 square unitsa

4

√a2 + b2 square unitsa

4

a2√3

4

√4b2 − a2a

4

a = b

a = −b

a = ±b

2a = b

12√2

12√3

12√6

Anmol decided to find the area of triangle ABC. He measured the sides of the triangle and calculated a few things. The observationsmade by Anmol are given as:

, where are the sides of the triangle and s is the semi-perimeter.The perimeter of this particular triangle is

A. 1 unitB. 2 unitsC. 3 unitsD. 6 units

Q. No.6

The area of a right-angled triangle whose base length and hypotenuse respectively are 3 cm and 5 cm is _____.

Q. No.7

The area of the given right angled triangle is 60 cm2. The length of the side corresponding to the biggest angle is _____.

Q. No.8

In a triangle, the sides are given as 5 cm, 6 cm and 7 cm. The length of the altitude is _____ cm corresponding to the side havinglength 6 cm.

Q. No.9

If each side of a triangle is halved, then the area of the given triangle is_____ times the area of the new triangle thus formed.

s − a = s − b = s − c = 1 a, b, c

Heron's Formula_SolutionsSolution 1

Using Pythagoras theorem in ADC, we get

Therefore, the area of ABC = Hence, the required answer is option A.

Solution 2

Given, ar(ABC) = ar(PQR)

=


Solution 3

Let the two equal sides be x cmPerimeter = 18 cm

x + x + 8 = 18 x = 5 cm

Thus, the sides of triangle are 5 cm, 5 cm and 8 cm.

The area is given by


Solution 4

Let the sides be 3x, 4x, 5x

The area is given by

△

h = √b2 − ( )2

= √ =a

2

4b2−a2

4

√4b2−a2

2

△ × BC × AD = × a × = √4b2 − a212

12

√4b2−a2

2a

4

⇒ a2√3

4√4b2 − a2a

4

⇒ √3a = √4b2 − a2

⇒ 3a2 = 4b2 − a2

⇒ 4a2 = 4b2

⇒ a2 = b2

⇒ a = b (a, b can not be negative)

⇒⇒

∴ s = = 95+5+8

2

√9 (9 − 5) (9 − 5) (9 − 8)

= √9 × 4 × 4 × 1

= 12 cm2

∴ s = = 6x3x+4x+5x

2

√6x (6x − 3x) (6x − 4x) (6x − 5x)

= √6x (3x) (2x) (1x)

= 6x2

Now, Therefore, the three sides are 15 cm, 20 cm and 25 cmAnd the perimeter is 60 cm.Hence, the correct answer is option B.

Solution 5

The observation made by Anmol are:, where are the sides of the triangle and s is the semi-perimeter.

Adding the three equations, we get

And therefore, perimeter = 2s = 6 units.


Solution 6

Let AV = 5 cm and VC = 3 cm.Using Pythagoras theorem. we can get AC = 4 cm.So, the area of

Solution 7

We have, area of the triangle = 60 cm2

Using Pythagoras theorem, we obtain AC = 17 cmNow, the length of the side corresponding to the biggest angle, which is angle B, is 17 cm.

Solution 8

The area of the triangle is given by

Also, the area of triangle =

Solution 9

For the sake of simplicity, let's consider a right-angled triangle with sides 6 cm, 8 cm and hypotenuse 10 cm.Its area = .

6x2 = 150 ⇒ x = 5

s − a = 1, s − b = 1, s − c = 1 a, b, c

3s − (a + b + c) = 3

⇒ 3s − 2s = 3 [∵ s = ]

⇒ s = 3

a+b+c

2

△ AVC = × AC × VC = × 4 × 3 = 6 cm21

2

1

2

⇒ × AB × BC = 60 cm2

⇒ × 8 × BC = 60

⇒ BC = = 15 cm

121

2604

s = = 95+6+7

2

√9 (9 − 5) (9 − 6) (9 − 7)

= √9 × 4 × 3 × 2

= 6√6 cm2

× height × 612

⇒ × height × 6 = 6√6

⇒ height = 2√6 cm

12

× 6 × 8 = 24 cm212

Now, let's take another triangle whose sides are half of the sides of given triangle i.e. 3 cm, 4 cm and 5 cmIts area = Therefore, the area of the given triangle is four times the area of the new triangle thus formed.

× 3 × 4 = 6 cm212

Surface Areas and Volumes_Questions

Q. No.1

Out of a cone and a cylinder, the volume of which figure will become thrice the original volume if the radius is tripled and heightbecomes one third.

A. Cone onlyB. Cylinder onlyC. Both cone and cylinderD. Neither cone nor cylinder

Q. No.2

The radii of a sphere and a hemisphere are in the ratio of 1 : 1. The ratio of their total surface area is

A. 1 : 1B. 2 : 1C. 3 : 2D. 4 : 3

Q. No.3

If a hemisphere is inscribed in a cube, then the ratio of the volume of the hemisphere to the volume of the cube is

A.B.C.D.

Q. No.4

A cone, a cylinder and a sphere have same bases, and their heights are equal to their radius. The ratio of their volumes is

A. 1 : 2 : 3B. 1 : 2 : 4C. 1 : 3 : 4D. 1 : 4 : 3

Q. No.5

A sphere and a right circular cone of the same radius have equal volumes. Which of the following relations is true?

A.B.C.

π : 12π : 6π : 3π : 2

h = 4r

h = 2r

h = r

D.

Q. No.6

The total surface area of two hemispheres of radius 6 cm is equal to the total surface area of a sphere whose radius is _____ cm.

Q. No.7

If 2412 cuboidal boxes of dimensions cm3 can be placed inside a room with dimensions of floor cm2, thenthe required height of the room is _____.

Q. No.8

The radius of the cone difference of whose curved surface area and total surface area is 22 cm is _____.

Q. No.9

Of all the three dimensional figures, _____ is the only figure whose curved surface area is same as the total surface area.

Q. No.10

The length of the longest rod that can be placed inside a cuboid of dimensions is _____.

r = 4h

(20 × 12 × 1) (144 × 60)

(3 × 1 × 2) cm3

Surface Areas and Volumes_SolutionsSolution 1

For cone:The original volume is

Now if , we have the new volume = , which is three times the original volume.

For cylinder:The original volume is

Now if , we have the new volume = , which is three times the original volume.


Solution 2

The radius of the sphere and the hemisphere is same = r(say).The total surface area of sphere = The total surface area of hemisphere = The required ratio is 4 : 3Hence, the correct answer is option D.

Solution 3

The volume of the hemisphere = and that of the cube =

The required ratio = Hence, the correct answer is option A.

Solution 4

Given that for a cone, a cylinder and a sphere, we have h = r.The volume of the cone =

The volume of the cylinder = The volume of the sphere = The required ratio is 1 : 3 : 4Hence, the correct answer is option C.

Solution 5

Given that a sphere and a right circular cone of the same radius (r) have equal volumes.Let the height of the cone be h.Therefore,


Solution 6

Let the required radius be r.Therefore, we have

Solution 7

πr2h13

r = 3r, h =h

3π(3r)

2( ) = πr

2h

1

3h3

πr2h

r = 3r, h = h

3π(3r)2 ( ) = 3πr2h

h3

4πr2

3πr2

πr32

3(2r)3 = 8r

3

πr3 : 8r3 = π : 1223

πr2h = πr313

13

πr2h = πr3

πr343

πr3 = πr2h43

13

⇒ 4r = h

2 × 3π(6)2 = 4πr2

⇒ 2 × 3π(6)2 = 4πr2

⇒ 216π = 4πr2

⇒ r2 = 54

⇒ r = √54 = 3√6 cm

Let the required height be h cm.According to the question,

Solution 8

We know that for a cone, Total surface area = Curved surface area + Area of base Total surface area Curved surface area = Area of base 22 = Area of base

Solution 9

We know that the curved surface area of the sphere = the total surface area of the sphere = .Hence, the required figure is a Sphere.

Solution 10

We know that the longest rod that can be placed inside a cuboid is along its diagonal.The diagonal of the cuboid is , which is the required answer.

= 2412

⇒ = 2412

⇒ h = = 67 cm

Volume of the room

Volume of cuboidal box

144×60×h

20×12×1

2412×20×12

144×60

⇒ −⇒⇒ πr2 = 22

⇒ r2= 22

⇒ r2 = 7

⇒ r = √7 cm

227

4πr2

√l2 + b2 + h2 = √32 + 12 + 22 = √14 cm

Statistics_Questions

Q. No.1

A data consists of 7 observations. If each observation of the data is increased by 1, then which of the following option is true?

A. Mean will be increased by 1B. Median will be increased by 1C. Both mean and median will remain sameD. Both mean and median are increased by 1

Q. No.2

If 2n + 2 observations are given, where n is an odd number, then the median of this data is given by the

A.

B.

C. the mean of

D. the mean of

Q. No.3

5 people reviewed Brand 1 and gave it an average rating of 4.6. Also, 8 people reviewed Brand 2 with an average rating of 4.8. Themean of the two rating is

A. 4.7B. 4.72C. 4.8D. 4.88

Q. No.4

Which of the following measure of central tendencies always take up the value from the given collection of observations only?

A. MeanB. MedianC. ModeD. All of the above

Q. No.5

Let m be the mid-point and l be the lower-class limit of a class in a continuous frequency distribution. The upper-class limit of the classis

A. m + l

(n + 1)th term

(n + 2)th term

( )th

term and ( + 1)th

term2n+1

2

2n+1

2

(n + 1)th term and (n + 2)th term

B.C.D.

Q. No.6

5 students of grade 8 got an average score of 93. Also, 8 students of grade 9 got an average score of 97. The mean of the two scoresis _____.

Q. No.7

A given data consists of 5 observations i.e. 1, 2, 3, 4 and 5. If each observation of the data is multiplied with 2, then the new mean willbe multiplied by _____.

Q. No.8

In a frequency distribution, the upper limit of the class is 4 and the width of the class is 10. The mid-value of a class is _____.

Q. No.9

The marks obtained by 9 students in a mathematics test (out of 100) are given below:94, 100, 82, 70, 46, 52, 75.Then, the range of the data is _____ (<,=,>) the median of the data.

Q. No.10

If the mean of the five observations is 5, then the value of x is _____.

l − m

2m − l

2 (m − l)

x − 3, x, , , x + 52x+7

3

3x

2

Statistics_SolutionsSolution 1

Let the n odd observations be .

Let their mean be and median be

If 1 is added to each observation, then the observations become

Their mean will be And the median will be . Both mean and median are increased by 1.Hence, the correct answer is option D.

Solution 2

If n is an odd number, then 2n + 2 will be an even number.And we know that the median of the data with even number of observations is given by the mean of

i.e. Hence, the correct answer is option D.

Solution 3

5 people reviewed Brand 1 and gave it an average rating of 4.6Therefore, the total rating point of Brand 1 will be .

Also, 8 people reviewed Brand 2 with an average rating of 4.8Therefore, the total rating point of Brand 2 will be .

Total rating points = Total people who gave review =

So, the mean of the rating of two brands = .Hence, the correct answer is option B.

Solution 4

The mean of data may or may not belong to the given collection of observations. For ex., the mean of 1 and 2 is 1.5, which is neither 1 nor2.Similarly, the median of 1 and 2 is 1.5, which is neither 1 nor 2.But, the value of mode always lies in the given collection.Hence, the correct answer is option C.

Solution 5

It is given that m be the mid-point and l be the lower-class limit.

We know that,


Solution 6

5 students of grade 8 got an average score of 93.Therefore, the total score of grade 8 will be .

Also, 8 students of grade 9 got an average score of 97.Therefore, the total score of grade 9 will be .

Total scores = Total number of students =

So, the mean of the scores is = .

Solution 7

a1, a2, a3, . . . , a7

x =a1+a2+a3+...+a7

7( ) th term i. e. 4 th term i. e. a4

7+1

2

a1 + 1, a2 + 1, a3 + 1, . . . , a7 + 1

= = + = x + 1(a1+1)+(a2+1)+(a3+1)+...+(a7+1)

7

(a1+a2+a3+...+a7)+7

7

(a1+a2+a3+...+a7)

7

7

74 th term i. e. a4 + 1

( )th

term and ( + 1) i. e. ( )th

term2n+2

2

2n+2

2

2n+4

2

(n + 1)th

term and (n + 2)th

term

4. 6 × 5 = 23

4. 8 × 8 = 38. 4

23 + 38. 4 = 61. 45 + 8 = 13

= 4. 7261.4

13

mid − point =lower limit+upper limit

2

⇒ m =

⇒ upper limit = 2m − l

l+upper limit

2

93 × 5 = 465

97 × 8 = 776

465 + 776 = 12415 + 8 = 13

= 95. 461241

13

Let the 5 odd observations be 1, 2, 3, 4 and 5.Let their mean be .If 2 is multiplied with each observation, then the observations become

Their mean will be = i.e. mean will be multiplied with 2 as well.Hence, if each observation of the data is multiplied with 2, then the mean will be multiplied by 2.

Solution 8

Since the upper limit of the class is 4 and the width of the class is 10, therefore, the given class interval is - 4.And its mid-point is .

Solution 9

The given data can be arranged as 30, 33, 46, 52, 70, 75, 82, 94, 100.The range is And the median is 70.Therefore, the range of the data is equal the median of the data.

Solution 10

Given that the mean of the five observations is 5

i.e.

x = = 31+2+3+4+5

5

2, 4, 6, 8, 10

= 62+4+6+8+10

5

−6

= −1−6+4

2

100 − 30 = 70

x − 3, x, , , x + 52x+7

3

3x

2

= 5x−3+x+ + +x+5

2x+7

3

3x

2

5

⇒ = 5

⇒ = 5

⇒ = 5

⇒ 31x + 26 = 150

⇒ x = 4

x−3+x+ + +x+52x+7

3

3x

2

5

3x+ + +22x+7

3

3x

2

5

18x+4x+14+9x+12

30

Probability_Questions

Q. No.1

The numerical value of e truncated to 50 decimal places is 2.71828182845904523536028747135266249775724709369995. Anumber is chosen randomly from the decimal places. The probability of getting number 2 is

A.

B.

C.

D. None of the above

Q. No.2

The experimental probability of an event E is given by

A. P(E) =

B. P(E) =

C. P(E) =

D. P(E) =

Q. No.3

The probability of getting all heads if one coin is tossed is a and when two coins are tossed is b. Then

A. 2a = bB. a > bC. a = bD. a = 2b

Q. No.4

Two dice are thrown together and the sum of two numbers appearing on their tops is noted. What is the probability that the sum is aprime number?

A.

B.

C.

D.

Q. No.5

4

25

7

50

8

51

Number of trials

Total number of trialsNumber of trials in which the event has happened

Number of trials in which the event has not happened

Total number of trials

Number of trials in which the event has happened



5

12

1

215

18

17

36

Which of the following sample spaces is incorrect?

A. Sample space of tossing a coin = {Head, Tail}B. Sample space of tossing two coins = {(Head, Head); (Head, Tail); (Tail, Tail)}C. Sample space of rolling a die = {1, 2, 3, 4, 5, 6}D. Sample space of drawing a marble from a bag containing 1 red marble, 1 grey marble and 1 yellow marble = {Red, Grey, Yellow}

Q. No.6

In a piggy bank, there are ten coins of Rs. 1, fourteen coins of Rs 2, five coins of Rs 5 and one coin of Rs 10. A coin is taken out fromthe piggy bank at random. The probability of getting a coin being a factor of 10 is _____.

Q. No.7

In the past 75 cricket matches between two teams A and B, Team A had won 41 matches and lost 27 matches and the remainingmatches were drawn. The experimental probability that the next cricket match between the two teams will result in a draw is _____.

Q. No.8

A survey was conducted among randomly selected 100 youths from a locality to know their hobby. The result of the survey is shown inthe given table.

Hobby Number of youthsReading magazines 25Watching television 24

Surfing internet 30Playing 16Others 5

The experimental probability that the hobby of a randomly selected youth from the locality is reading magazines is _____.

Q. No.9

The given table shows the number of times different numbers appear on dice in 1000 throws.Number on dice 1 2 3 4 5 6

Number of times it appears 315 124 117 144 210 90The probability that an even number appeared on the dice is _____.

Q. No.10

In an office comprising 69 employees, 15 are found to be vegetarian. The experimental probability that an employee selected atrandom in the office is non-vegetarian is _____.

Probability_SolutionsSolution 1

The numerical value of e truncated to 50 decimal places is 2.71828182845904523536028747135266249775724709369995.

Therefore, the probability of getting 2 in the decimal place is = Hence, the correct answer is option B.

Solution 2

P(E) = Hence, the correct answer is option D.

Solution 3

The probability of getting all heads if one coin (Possible outcomes = H, T) is tossed is a.

The probability of getting all heads if two coins (Possible outcomes = HH, HT, TH, TT) are tossed is b


Solution 4

The prime numbers and their possible appearances are given below:2: (1, 1)3: (1, 2), (2, 1)5: (1, 4), (2, 3), (3, 2), (4, 1)7: (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)11: (5, 6), (6, 5)The total possible outcomes are 36.Therefore, the required probability is Hence, the correct answer is option A.

Solution 5

Consider the experiment of tossing two coins.There are four possible outcomes, which are:Face on 1st coin Face on 2nd coin

Head HeadHead TailTail HeadTail Tail

∴ Sample space = {(Head, Head); (Head, Tail); (Tail, Head); (Tail, Tail)}Thus, the sample space of the experiment given in alternative B is incorrect.Hence, the correct answer is option B.

Solution 6

Since, all the denominations i.e. Rs 1, Rs 2, Rs 5 and Rs 10 are the factors of 10.Therefore, the event of getting a coin being a factor of 10 is a sure event.And the probability of a sure event is 1.

Solution 7

Total number of cricket matches the two teams had played = 75Number of matches won by team A = 41Number of matches lost by team A = 27Therefore, number of matches which were drawn = 75 − (41 + 27) = 7Thus, the experimental probability that the next cricket match between the two teams will result in a draw =

Solution 8

=Number of times 2 appears in decimal place

Total number of decimals

7

50



⇒ = a

⇒ 2a = 1

1

2

⇒ = b

⇒ 4b = 1

1

4

∴ 2a = 4b i. e. a = 2b

=15

36

5

12

7

75

Total number of youths = 100Number of youths whose hobby is reading magazines = 25

Thus, experimental probability that the hobby of a randomly selected youth from the locality is reading magazines =

Solution 9

Number of throws of dice = 1000Number of times even numbers appeared = 124 + 144 + 90 = 358∴ Required probability

Solution 10

Total number of employees working in the office = 69Number of employees found to be vegetarian = 15∴ Number of employees found to be non-vegetarian = 69 − 15 = 54Thus, the experimental probability that an employee selected at random in the office is non-vegetarian is .

= = 0. 358358

1000

=54

69

18

23

grade ix mathematics...number systems_solutions solution 1 we know that, by definition, the decimal...

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