Opt Math 6 - 2078 (Final)

a. Y b. Y
X' A(2, 4) A(2, 4)

B(4, 2)

B(– 4, 1)

O X X' O X
C(1, – 2)
C(– 2, – 2)

Y' Y'

4. Find the image of DPQR in each of the following figures after the reflection in X-axis. Also
find the co-ordinates of the vertices of the image.

a. Y b. Y

P(2, 4) P(– 2, 4)

Q(4, 2)

R(– 2, 1) Q(– 4, 1)
X' O
X X' OX
R(3, – 1)

Y' Y' Transformation
5. a. If the vertices of DABC are A(2, 5), B(6, 4) and C(4, 0), find the image of DABC under

the reflection on Y-axis and the co-ordinates of the vertices of image.
b. Find the image of DPQR where vertices are P(– 1, 5), Q(4, 3) and R(5, 7) under the

reflection on X-axis and the co-ordinates of the vertices of image.
c. Find the image of quadrilateral PQRS where vertices are P(1, 5), Q(4, 6), R(5, – 2) and

S(2, – 3) under the reflection on Y-axis and the co-ordinates of the vertices of image.
d. Find the image of quadrilateral ABCD where vertices are A(– 3, 1), B(– 4, 5), C(3, 4)

and D(2, 2) under the reflection on Y-axis and the co-ordinates of the vertices of image.
6. a. If A'B' is the image of AB under the reflection and the co-ordinates of A, B, A', B'

are A(5, 7), B(6, 3), A'(– 5, 7), B'(– 6, 3), find the axis of reflection by drawing.
b. If P'Q' is the image of PQ under the reflection and the co-ordinates of P, Q, P' and Q'

are P(2, 6), Q(– 3, 3), P'(2, – 6), Q'(– 3, – 3), find the axis of reflection by drawing.

Reflection 95

7.3 Rotation

Introduction

In a rotation, the given object is rotated about a fixed point through a certain angle. The fixed point
is known as the centre of rotation and the angle through which it is rotated is known as the angle
of rotation. The hands of a clock rotate around the centre of the clock. The position of the minute
hand from 3:00 to 3:15 (i.e. within 15 minutes) changes its tip from 12 to 3. The minute hand turns
through 90° in the clockwise direction about the centre.

11 12 1 – 90°
10 2
93 A
84
90°
765
O

The hands of a clock rotates The cartoon rotates through The shape of F is rotated

about its centre. 90°about centre O. through – 90° about centre A.

P' A'
B'

60° B 90° P – 90°

O Q AD

It describes the rotation It describes the rotation It describes the rotation
of point B through 60° in of point P through 90° in of point A through 90° in
anticlockwise direction about anticlockwise direction about clockwise direction about
point O. B' is the image of B point Q. P' is the image of P point D. A' is the image of A
under this rotation. under this positive rotation under this negative rotation
(i.e. 90°). (i.e – 90°).

To rotate any figure, we need the following information.
a. The centre of rotation
b. The angle of rotation
c. The direction of rotation

96 Perfect Optional Mathematics Grade 6

Example 1: B

Find the image of line AB under the rotation
through – 90° about M.

Solution: A
M
Draw an arc taking MA as radius and M as centre and A
we have to rotate AB through 90° in clockwise direction B
about M. So join AM and BM by a dotted line.
A'
Make 90° in clockwise direction with AM at M as shown
and the intersection of the line making 90° and the arc A' M
is the image of A. Similarly, find the image B' of point B
and join A'B'. C

Example 2: B'

Rotate DABC given in the figure through 90° in B
positive direction about N.
B
N

Solution: B' A Transformation
C' A'
To have the image of DABC, find the images C
of points A, B and C under the given rotation.
Join AN, BN and CN and draw the arc of radius A
NA and centre N in positive direction as shown.
Draw a line NA' making 90° with NA which
intersects the arc at A' so A' is the image of A
under the given rotation.

N

Similarly, find the images of B and C under the
same rotation and join A'B', B'C' and C'A' to get
image DA'B'C' of DABC.

Rotation 97

Rotation Using Co-ordinates

Let P(– 4, 3) be a point and O(0, 0) be the centre of P(– 4, 3) Y C'(2, 5)
rotation. When P(– 4, 3) is rotated through 90° in anti- A'(3, 2)
clockwise direction (i.e. 90°) about O, we will have
P'(– 3, – 4). Taking two more points A(2, – 3) and
C(5, – 2) and their images under the same rotation, we
have the following result.

Object Image X' X
P(– 4, 3) P'(– 3, – 4) C(5, – 2)

A(2, – 3) A'(3, 2) A(2, – 3)
C( 5, – 2) C'(2,  5)
P'(–3, – 4)

Y'

We see the image in each case is obtained by interchanging the x and y co-ordinates and also
the sign of y co-ordinates. i.e. We can conclude it as the image of P(x, y) when rotated through
90° about origin is P'(– y, x).

Hence, P(x, y) 90°, (0, 0) P'(– y, x)
Rotation

Example 3:

The vertices of a triangle ABC are A(– 1, 2), B(– 2, 5) and C(– 3, 3). Rotate DABC negative

quarter turn (i.e. 90° clockwise direction) about origin and find the image of DABC. Also

draw DABC and the image DA'B'C' on the same graph.
Solution:

The vertices of a triangle ABC are A(– 1, 2), B(– 2, 5) BY
and C(– 3, 3). Draw DABC on the graph and find the
image of points A, B and C as in the above example C C'
as follows. A A'

Object Image B'
X
A(– 1, 2) A'(2, 1) X'
B(– 2, 5) B'(5, 2)

C(– 3, 3) C'(3, 3)

Joining A'B', B'C' and C'A', we get the image of DABC. Y'
And we can conclude the the image of P(x, y) when
rotated through – 90° about origin is P'(y, – x).

Hence, P(x, y) – 90°, (0, 0) P'(y, – x)
Rotation

98 Perfect Optional Mathematics Grade 6

Exercise 7.3

1. Rotate the following line segments about the point O by the mentioned angle of rotation.
a. O b. A

P
B

Q O
Negative quarter turn (– 90°) Positive quarter turn (+ 90°)

2. Rotate the following figures about the point M by the mentioned angle of rotation.

a. M b. a. PM
P

QS

R

Q R
Negative quarter turn (– 90°) Positive quarter turn (+ 90°)

3. Find the images of the following points under the rotation through – 90° about the origin.

a. A(3, 4) b. B(– 4, 3) c. P(– 5, – 4) d. Q(6, – 4)

4. Find the images of the following points under the rotation through – 90° about the origin.

a. Y b. Y Transformation
P(– 2, 4)
A(– 2, 4)
C(0, 3)

B(– 4, 1) O Q(– 4, 1) R(3, – 1) X
X' X X' O

Y' Y'
Rotation 99

5. Find the images of the following points under the rotation through 90° about the origin.

a. Y b. Y

P(2, 4)

Q(4, 3)

R(0, 1) B(2, 0)

X' O X X' O X

C(4, – 2)

A(1, – 3)

Y' Y'

6. Find the images of the following points under the rotation through given angles about the
origin.

a. Y b. Y

D(– 2, 4) A(1, 4) P(2, 4)
Q(3, 3)
S(– 2, 3)

C(– 3, 1) B(0, 1) R(0, 1)
O
X' O X X' X

Y' Y'
90° – 90°
7. DABC has the vertices A(2, 4), B(6, 7) and C(4, 1). Rotate DABC about the origin through:
a. 90° in the anticlockwise direction
b. 90° in the clockwise direction
8. Quadrilateral ABCD has vertices A(5, 2), B(3, 5), C(– 2, 4) and D(– 4, 7). Rotate quadrilateral
ABCD about the origin through:
a. 90° in the anticlockwise direction
b. 90° in the clockwise direction

100 Perfect Optional Mathematics Grade 6

8 STATISTICS

Learning Objectives

classify the given data into a frequency distribution table.
find Arithmetic mean of an individual series.
find Arithmetic mean of an discrete series.
find Median of an individual series.
find Median of an discrete series.
find Mode of an individual series.
find Mode of an discrete series.

Introduction

Statistics is a branch of science which deals with various methods of data collection, tabulation
or classification, analysis of the collected data and drawing conclusion from the analysis.

Classification of Data

Suppose that 26 students got the following marks in a class test.

3, 5, 7, 8, 4, 5, 7, 5, 4, 6,
6, 4, 5, 5, 4, 5, 3, 6, 3, 5,
7, 5, 6, 4, 3, 5.
Frequency distribution table is prepared by using tally bars as follows.

3, 5, 7, 8, 4, 5, 7, 5, 4, 6,
6, 4, 5, 5, 4, 5, 3, 6, 3, 5,
7, 5, 6, 4, 3, 5.

Marks Tally bars Frequency Statistics
3 //// 4
4 //// 5
5 9
6 //// //// 4
7 //// 3
8 /// 1
/ 26
Total

 101

8.1 Arithmetic Mean

Measure of Central Tendency

When we have the group of numerical values like marks, heights, ages of students of any class,
the entire values can be represented by a single value called “average.” The measure of such a
single value is known as the measure of central tendency. In this class, we introduce the following
three measures of central tendencies.

a. Arithmetic mean b. Median c. Mode

Arithmetic Mean of an Individual Series

The value obtained by dividing the sum of the given values by the number of values is known as
arithmetic mean and is denoted by x (x bar). If the heights of 3 students of class 6 are 120cm,
130cm and 140cm, the average height or the arithmetic mean height of those students can be
obtained by dividing the sum of their heights by the number of students.

i.e. Arithmetic mean or Average = 120 + 130 + 140 = 390 = 130cm.
3 3
If x1, x2, ... ... ... ..., xn be n values of variable, then their arithmetic mean or average is denoted

by x (x bar) and defined by

x = x1+ x2 + ... ... + xn = ∑x
n n
Example 1:

Find the arithmetic mean of 20, 22, 25, 28, 30.
Solution:

The given numbers are 20, 22, 25, 28, 30

Number of values (n) = 5 x1+ x2 + x3 + x4 + x5
5
Hence, the mean value is x =

=20 + 22 + 25 + 28 + 30 = 125 = 25
Example 2: 5 5

Find the arithmetic mean of all factors of 12.
Solution:

Here, the factors of 12 are 1, 2, 3, 4, 6, 12.

The sum of the factors (Sx) = 1 + 2 + 3 + 4 + 6 + 12

\ Sx = 28

Number of factors (n) = 6 ∑x 28 14
n 6 3
Now, Arithmetic mean (x) = = = = 4 .67

\ x = 4 .67

102 Perfect Optional Mathematics Grade 6

Example 3:

If the arithmetic mean of 6, 10, p, 12 and 16 is 11, find the value of p.
Solution:

The given numbers are 6, 10, p, 12 and 16.

Arithmetic mean (x) = 11

Sum of the numbers (Sx) = 6 + 10 + p + 12 + 16 = 44 + p

Number of terms (n) = 5

Now we have, ∑x
n
Arithmetic mean (x) =

or, 11 = 44 + p
n
or, 44 + p = 55

or, p = 55 – 44

\ p = 11

Hence, the value of p is 11.

Arithemtic Mean of Descrete Series

If the frequencies of n variable values x1, x2, ... ... ..., xn are f1, f2, ... ... ..., fn respectively.

The arthmetic mean of these data is x = f1x1 + f2x2 + ... ... ... + fn×xn = ∑fx
f1 + f2 + ... ... ... + fn N

where N = sum of frequencies.

Example 4:

The following are the heights of 72 plants in a garden.

Heights (cm) 58 60 62 64 66 68

No. of Plants 12 14 20 13 8 5

Find the mean height of the plants.
Solution:

Calculation of Arithmetic Mean

Heights (x) No. of plants (f) fx

58 12 696 Statistics
60 14 840
62 20 1240
64 13 832
66 8 528
68 5 340

∑f = 72 ∑fx = 4476

From formula, ∑fx 4476
N 72
Arithmetic mean (x) = = = 62.17

Arithmetic Mean 103

Exercise 8.1

1. Find the arithmetic mean of the following data.

a. 2, 4, 6, 8, 10 b. 10, 14, 16, 20, 22

c. 3, 8, 10, 15, 18 d. 20, 50, 30, 40, 60, 30

e. 25, 35, 55, 65, 75, 95 f. 45, 25, 35, 10, 15, 5

2. Find the arithmetic mean of the following numbers.

a. The first 8 natural numbers b. The first 10 odd numbers

c. The first 5 prime numbers d. The first 6 even numers

e. The first 7 multiples of 5 f. The first 13 counting numbers

3. Find the value of x in the following cases.

a. The arithmetic mean of 6, 10, x and 12 is 18.

b. The arithmetic mean of 6, 8, 9, x and 13 is 10.

c. The arithmetic mean of 20, 22, x, 28 and 30 is 25.
d. The arithmetic mean of 12, 3, 4, x and 12 is 723.
e. The arithmetic mean of x, x + 3, x + 6, x + 9, x + 12 and x + 15 is 18.

f. The arithmetic mean of x, x + 2, x + 4, x + 6 and x + 8 is 13.

4. Find the arithmetic mean of the following data.

a. x 8 10 15 20

f 5884

b. x 10 12 20 25 35
f
3 10 15 7 5

c. x 5 15 25 35 45

f 7 18 20 10 5

d. x 10 30 50 70 80
f 7 8 10 15 10

e. Weight (kg) 60 63 66 69 72

No. of workers 43221

f. Values 56789

Frequencies 4 8 14 10 4

104 Perfect Optional Mathematics Grade 6

8.2 Median and Mode

Median is that value of observation which divides the whole observations into two equal parts

after arranging them into ascending or descending order. Median is denoted by Md.

Median of an Individual Series

We apply the following methods to find the median of individual series.

1. Arrange the given data in the ascending or descending order.

2. Count the data and find n. n + 1 th
2
3. Find the median by using median = observation.

Example 1:

A student obtained the following marks in 7 subjects in an examination : 50, 53, 61, 49,

45, 63, 48. Find the median mark.
Solution:

Here, arranging the given marks in the ascending order,

45, 48, 49, 50, 53, 61, 63

Number of given values (n) = 7 and n + 1 = 7 + 1 = 4
2 2

\ Median = value of 4th term = 50

Example 2:

Find the median of the following data : 10, 75, 3 , 81, 18, 27, 4, 48, 12, 47, 9, 15
Solution:

Arranging the given data in the ascending order,

3, 4, 9, 10, 12, 15, 18, 27, 47, 48, 75, 81

\ Number of observations (n) = 12 and n + 1 = 12 + 1 = 6.5
2 2

In this case the median lies between the 6th and 7th terms.

\ Median = Average of 6th and 7th term.

= Average of 15 and 18 = 15 + 18 = 16.5
Example 3: 2

The median of the following data arranging in the ascending order is 18. Find the value Statistics

of x. 11, 13, 14, x + 2, 20, 22, 25, 30
Solution:

The given data are 11, 13, 14, x + 2, 20, 22, 25, 30

\ Number of observations (n) = 8 and n + 1 = 8 + 1 = 4.5
2 2

Median and Mode 105

In this case the median lies between the 4th and 5th terms.

\ Median = Average of 4th and 5th term.

or, 18 = Average of (x + 2) and 20

or, 18 = x + 2+ 20 = x + 22
2 2
or, 36 = x + 22

\ x = 36 – 22 = 14

Median of Discrete Series

Process of finding the median from the discrete series.

1. Arrange the given data in the ascending order with corresponding frequencies.

2. Find the less than cumulative frequency table.
th
n + 1
3. Find the median using, Median = 2 item.

4. The variable corresponding to the cumulative frequency equal to or just greater than

n + 1 th
2
is the median.

Example 4:

Find the median from the following data.

Size of shoes 4.0 4.5 5.0 5.5 6.0 6.5 7.0 7.5 8.0

Frequency 10 18 22 25 40 15 10 8 7

Solution:

Size of shoes (x) Frequency (f) Cumulative frequency (cf)

4.0 10 10
4.5 18 28
5.0 22 50
5.5 25 75
6.0 40 115
6.5 15 130
7.0 10 140
7.5 8 148
8.0 7 155

N = 155

Here, N = 155 th th

We have, Median = N + 1 item = 155 + 1 item = 78th term
2 2

Cumulative frequency just greater than 78 is 115. Hence, the median is the corresponding value
of cumulative frequency 115.
i.e. Median = 6

106 Perfect Optional Mathematics Grade 6

Mode

Mode is the value of the variable that occurs most frequently i.e. mode is the variable with
maximum frequency. Study the following examples:

Example 5:

Find the mode of the following data.
10, 13, 12 , 10, 15, 13, 14, 13, 12, 15, 12, 13
Solution:

Arranging the given data in ascending order,
10, 10, 12, 12, 12, 13, 13, 13, 13, 14, 15, 15
In above data 13 is repeated most (four times)

Mode = 13.

Example 6:

Find the mode of the following data.

Value (x) 5678

Frequecny (f) 15 25 20 10

Solution:

The maximum frequency in the above data is 25.
The mode is the corresponding value of variable = 6.

Exercise 8.2

1. Find the median of the following data. b. 4, 8, 12, 16, 20, 23, 28, 32
a. 2, 3, 5, 7, 9 d. 13, 22, 25, 8, 11, 19, 17, 31, 16, 10
c. 60, 33, 63, 61, 44, 45, 51 f. Prime numbers between 51 and 80
e. The first 10 prime numbers

2. a. The median of the following data arranging in the ascending order is 24. Find the value of x.

11, 12, 14, 18, x + 2, 30, 32, 35, 41

b. The median of the following data arranging in the ascending order is 15. Find the value of a.

10, 11, 12, 13, a + 2, a + 4, 17, 19, 19, 23.

3. a. The following table shows the income of 73 families of a town. Find the median of the Statistics

data.

Income (Rs) 100 120 130 150 180 200

No. of families 10 22 20 13 6 2

Median and Mode 107

b. The following are the marks obtained by 38 students. Find the median mark.

Marks 20 90 50 70 60 95

No. of students 4 5 8 10 6 5

c. Find the median wage of the following data.

Wages (Rs) 20 21 22 23 24 25 26 27 28

No. of workers 8 10 11 16 20 25 15 9 6

4. a. Find the arithmetic mean of the first 10 natural numbers and show that it is equal to the
their median.

b. Find the arithmetic mean of the first 11 odd numbers and show that it is equal to the their
median.

5. Find the mode of the following data.

a. 8, 5, 6, 8, 3, 4, 6, 10, 8, 2

b. 1, 2, 3, 3, 3, 5, 6, 8, 8, 8, 8 9, 8

c. 3, 5, 6, 6, 5, 3, 5, 3, 5, 3, 6, 5, 3, 5, 7, 6, 5, 7, 5

d. 3, 4, 7, 11, 4, 3, 4, 5, 6, 4, 1, 4, 2, 4, 4

6. Find the mode of the following data.

a. 10 12 15 20 25 35 45 50 60
Marks

No. of students 4 6 10 14 20 19 10 6 3

b. 250 300 350 400 450 500 550 600 650
Wages (Rs.) 2 3 10 8 6 5 2 1 1
No. of workers

7. a. For what value of x, the mode of the following data is 11?
5, 11, 6, 11, 12, 9, 15, x, 12, 15, 14.
b. For what value of y, the mode of the following data is 22?
13, 21, 26, 22, 22, 21, 20, y + 12, 19, 18, 23.
8. Find the mean, median and mode of the following marks obtained by 16 students in a class test.
10, 10, 20, 20, 30, 30, 30, 40, 50, 50, 50, 50, 60, 60, 70, 80
9. A boy student secured the following marks in various examinations of 20 full marks.
15, 17, 16, 7, 10, 12, 14, 16, 19, 12, 16
a. What is his mode mark?
b. What is his mean mark?
c. What is his median mark?

108 Perfect Optional Mathematics Grade 6

Exercise – 1.1

2. P = S, Q = D, R = A, B = C

3. a. x = 3, y = 4 3. b. x = 3, y = – 3, 3. c. x = 1, y = 8
3 3 1 2
3. d. x = 2, y = 4 3. e. x = 2 , y = 2 3. f. x = 2 , y = 3

4. a. A × B = {(1, 3), (1, 4), (2, 3), (2, 4)}, B × A = {(3, 1), (3, 2), (4, 1), (4, 2)}

4. b. P × Q = {(a, x), (a, y), (a, z), (b, x), (b, y), (b, z)}, Q × P = {(x, a), (x, b), (y, a), (y, b), (z, a), (z, b)}

5. a. A × B = {(2, 1), (2, 4), (5, 1), (5, 4)}, B × A = {(1, 2), (1, 5), (4, 2), (4, 5)}

5. b. M × N = {(– 2, 4), (– 2, – 1), (3, 4), (3, – 1)} , N × M = {(4, – 2), (4, 3), (– 1, – 2), (– 1, 3)}

6. a. A × B = {(a, m), (a, n), (b, m), (b, n)}, B × A = {m, a), (m, b), (n, a), (n, b)},

A × A = {(a, a), (a, b), (b, a), (b, b)}

6. b. P × Q = {(1, 4), (1, 5), (2, 4), (2, 5), (3, 4), (3, 5)}, Q × P = {(4, 1), (4, 2), (4, 3), (5, 1), (5, 2), (5, 3)},

Q × Q = {(4, 4), (4, 5), (5, 4), (5, 5)}

7 a. A × B = {(a, 1), (a, 2), (b, 1), (b, 2), (c, 1), (c, 2)}

7. b. A × B = {(4, 1), (4, 2), (4, 3), (5, 1), (5, 2), (5, 3)}

8. a. R1 = {(1, 1), (9, 3)} 8. b. R2 = {(1, 2), (1, 3)}
R2 = {(2, 3), (4, 1)}
8. c. R3 = {(1, 1)}
Domain = {1, 2, 3, 4}, Range = {4, 3, 2, 1}
9. a. R1 = {(2, 1), (4, 1), (4, 3)} 9. b.

10. a. R1 = {(– 1, 1), (– 1, 2), (– 1, 3), (– 1, 4), (2, 3), (2, 4)}

10. b. R2 = {(– 1, 4), (2, 1)}

11. a. Domain = {1, 2, 3}, Range = {2, 3, 4} 11. b

11. c. Domain = {a, c, e, x}, Range = {b, d, f, y} 11. d. Domain = {2, 3}, Range = {4, 5, 6}

12. a. Domain = {1, 2, 3}, Range = {3, 4, 5} 12. b. Domain = {2, 3}, Range = {3, 4}

13. a. Many-one 13. b. One-many 13. c. One-one 13. d. One-many

14. a. R – 1 = {(3, 2), (4, 3), (6, 5)} 14. b. R – 1 = {(x, a), (y, b), (z, c)}
1 2

14. c. R – 1 = {(6, 4), (5, 3), (3, 1), (3, 2)} 14. d. R – 1 = {(3, 2), (4, 3), (6, 5), (6, 3)}
3 4

15. a. Domain = {1, 2, 3, 4}, Co-domain = {2, 3, 4, 5, 6}, Range = {2, 3, 4, 5}

15. b. Domain = {– 2, 2, 3, 4}, Co-domain = {4, 9, 16, 17}, Range = {4, 9, 16}

15. c. Domain = {3, 0, 4, 2}, Co-domain = {4, 1, 5, 3}, Range = {4, 1, 5, 3}

15. d. Domain = {a, b, c, d}, Co-domain = {1, 2, 3}, Range = {1, 2, 3}

Exercise – 1.2

1. a, b, d, f 2. a. 3 2. b. 4 2. c. 5

2. d. 5 3. a. binomial 3. b. trinomial 3. c. monomial

3. d. polynomial 4. a. 2x2 + 3x + 8 or, 8 + 3x + 2x2 Answers

4. b. – 4xy4 + 2x2y2 + x2 + 8 or, 8 + x2 + 2x2y2 – 4xy4

4. c. 4x3 – 3x2 + 3x – 9 or, – 9 + 3x – 3x2 + 4x3

4. d. 4y6 – x2y2 + 3xy2 + 2xy or, 2xy + 3xy2 – x2y2 + 4y6

Answers 109

5. a. f(x) = g(x) 5. b. f(x) = g(x) 5. c. f(x) ≠ g(x) 5. d. f(x) ≠ g(x)

6. a. 6x4 + 6x3 + x2 + 10x + 2 and 4x4 – x2 + 4x – 2 4x4 + 2x3 – 14x2
6x3 – 8x2 + 9x – 12
6. b. 5x3 – 0.5x2 – 5x – 1 and x3 + 5.5x2 – 9x + 5

6. c. 8x3 + x2 + 8x + 4 and 5x2 + 2x – 4

6. d. 5x3 + 3x2 – 11x – 5 and 3x3 + 11x2 – 5x – 5

7. a. 3x2 – 19x + 28 7. b.

7. c. 2x3 + 3x2 + 6x + 9 7. d.

Exercise – 1.3

1. a. 2 1. b. 3 1. c. 5 1. d. 4
2. a. 2; 11, 13
3. a. 3n – 1, 29 2. b. 3; 18, 21 2. c. 4; 21, 25 2. d. 5; 27, 32
4 a. 2
5. a. 2; 32, 64 3. b. 2n + 3, 21 3. c. 4n – 3, 45 3. d. 5n – 2, 38
6. a. 4n – 1, 256
8. c. 1, 3, 5, 7 4. b. 2 4. c. 3 4. d. 4
9. c. 7, 17, 32
5. b. 4; 768, 3072 5. c. 5; 625, 3125 5. d. 5; 1250, 6250

6. b. 5 × 2n – 1, 160 8. a. 2, 3, 4, 5 8. b. 3, 6, 9, 12

8. d. 4, 6, 8, 10 9. a. 5, 7, 13 9. b. 16, 30, 51

9. d. 29, 45, 61

Exercise – 2

1. a. 50ml, 25ml, The glass will be empty

1. b. 0.675cm, 0.3375cm, The chocolate will be finished.

2. a. 25, 30 2. b. 23, 28 2. c. 0, – 2 2. d. 10, 0

3. a. 20, 25, 30; ∞ 3. b. 130, 150, 170; ∞

3. c. 7.99999, 7.999999, 7.9999999; 8 3. d. 7.00001, 7.000001, 7.0000001; 7

3. e. 2.300001, 2.3000001, 2.30000001; 2.3 3. f. 6.000001, 6.0000001, 6.00000001; 6

4. a. 5cm 4. b. 2.5cm 4. c. 40cm, 20cm, 10cm, 5cm, 2.5cm, ... ...

4. d. 0cm 5. a. 90cm, 45cm, 22.5cm 5. b. 11.25cm

5. c. 90cm, 45cm. 22.5cm, 11.25cm, ... ... 5. d. 0 cm

6. a. 16cm 6. b. 128cm, 64cm, 32cm, ... ... 6. c. 0 cm

7. a. 5, 7, 8, 11, 19 7. b. – 3, 1, 3, 9, 12

7. c. 4, 8, 13, 40, 61 7. d. 0, 8, 15, 48, 71

8. a. 9 8. b. 5

Exercise – 3.1

1. a. 2×2, 5, – 3 1. b. 2×2; 2, 4, 0, 6 1. c. 2×3, 3, 7, – 3
2. a. 2×4
3. a. square 2. b. 3×3 2. c. 4×1 2. d. 2×3

3. b. rectangular 3. c. rectangular 3. d. rectangular

4. a. 1 – 1 4. b. 5 8 4. c. 7 9 11 4. d. 3 2 1
4 2 7 10 12 14 16 7 6 5

5. a. Yes, because all corresponding elements of two matrices are same.

110 Perfect Optional Mathematics Grade 6

5. b. No, because the order of matrices are not same.

6. a. x = – 4 and y = 0 6. b. a = 3, b = 7

6. c. x = 5, y = 3 and z = 6 6. d. x = 0, y = 5, z = 6

7. a. x = 4, y = 1 7. b. x = 3, y = 1 7. c. x = 5, y = 2 7. d. x = 2, y = 1

Exercise – 3.2

1. a. 7 0 1. b. 5 2 1. c. – 11 17
6 1 5 10 14 27

2. a. – 6 – 9  , 10 15 2. b. – 2 18  , 4 – 36
12 15 –  20 –25 10 12 –  20 – 24

2. c. – 4 – 3  , 12 9 3. a. 3 0 3. b. – 6 –3
5 2 –  15 –6 – 4 – 13 10 1

3. c. – 3 – 14 4. a. 75 42  , 1 – 4
–   15 34 – 3 – 4
0 3  – 4 3
4. b. 4 2 5  , – 2 – 6 1 4. c. 2 13  , 8 – 1
9 5 – 1 3 – 5 9
6 0 – 6 8

4. d. 5 1 4  , – 1 1 2
5 14 – 9 9 2 1

Exercise – 4.1

1. A(3, 4), B(– 3, 2), C(– 3, – 4), D(4, – 2), E(5, 2)

3. a. 5 units 3. b. 10 units 3. c. 5 units 3. d. 10 units
4 a 2 units 3. h. (b – a) 2 units
3. e. 4 5 units 3. f. 13 units 3. g.
5 units
4. A(2, 5), B(4, 1), C(– 2, 4), D(– 4, – 2), E(– 1, – 4), F(3, – 1)

4. a. 2 5 units 4. b. 2 10 units 4. c.

Exercise – 4.2

1. a. (2, – 1) 1. b. 21, 5 1. c. (1, 3) 1. d. (a, 2b)
2. a. A(0, – 5) 2. c. A(4, 6)
3. a. (5, 2) 2. b. B(2, 3) 3. c. (8, 9) 2. d. B(10, – 1)
4. a. (– 6, 13) 5. a. 13 units
3. b. (0, 5) 3. d. 12, 5
2
4. b. (– 5, 10)
5. b. 10 units

6. a. 17 units 6. b. (7, 2) 7. a. a = 8, b = 12 7. b. (4, – 7)

7. c. (– 7, 13) 8. a. 4:1 8. b. 1:3 9. a. 5:2

9. b. 3:2

Exercise – 5.1

1. a. 36000" 1. b. 66600" 1. c. 72630" 1. d. 19097" Answers
2. a. 900' 2. b. 618' 2. c. 310.5' 2. d. 530.67'
3. a. 45.5° 3. b. 15.5042° 3. c. 18.2625° 3. d. 10.2708°
4. a. 120000" 4. b. 181000" 4. c. 40816" 4. d. 487980"
5. a. 1500' 5. b. 4017' 5. c. 1316.24' 5. d. 2215.16'
6. a. 15.10g 6. b. 0.4518g 6. c. 10.4814g 6. d. 12.1624g

Answers 111

7. a. 63° 7. b. 18° 7. c. 31° 30' 7. d. 40° 57'
8. d. 11.4963g
8. a. 30g 8. b. 20.56g 8. c. 17.2269g 10. b. 135°

9. a. 50°, 60°, 70° 9. b. 30°, 60°, 90° 10. a. 120°

11. a. 66.67g 11. b. 50g

Exercise – 5.2

1. a. 81cm2 1. b. 289cm2 2. a. 36cm2 2. b. 17.64cm2, 4.2cm
3. a. 5cm 3. b. 6cm 3. c. 7.2cm 3. d. 29cm
3. e. 4.5cm 3. f. 15.6cm 4. a. 45m 4. b. 15m

Exercise – 5.3

1. a. h = AC, b = BC, p = AB. 1. b. h = PR, b = PQ, p = QR.

2. a. sin a = AACB, cos a = ABCC, cot a = ABCB 2. b. sin b = ABCB, cos b = ABCC, cot b = AC
AB
sin g = ABCC, cos g = ABCB, cot g = AABC,

3. a. sin2q 3. b. 1 3. c. sin2a 3. d. 1

4. a. 1 4. b. cos2b 4. c. tan2q 4. d. sec q

Exercise – 5.4

1. a. sin f = 1 , cos f = 1 , tan f = 1, cosec f = 2, sec f = 2, cot f = 1
2 2

b. sin a = 1 , cos a = 23, tan a = 13, cosec a = 2, sec a = 2, cot a = 3
2 3

2. a. cosec q = 5 , tan q = 4 2. b. cos a = 5 cot a = 5
4 3 13, 12

2. c. sin q = 1 , cot q = 1 2. d. cos f = 492, tan f = 7
2 42

2. e. sin b = 3 , sec b = 5 3. a. 2 3. b. 35
5 4 6

3. c. 1 3. d. 3 4. a. 2 2 4. b. 0

4. c. – 1 4. d. 29
4 6

Exercise – 5.5

1. a. C = 30°, BC = 2 3, AC = 4 1. b. P = 45°, PQ = 32, QR = 3
1. c. DF = 5 2, D = 45°, F = 45° 2

2. a. 10 3 m 2. b. 28.87m

2. c. 50m 3. a. 25m 3. b. 60m 3. c. 15m

Exercise – 6.1

1. a. u = 4 1. b. p = 4 1. c. d = – 4
2 – 2 – 3

2. a. 3 , x-component = 3, y-component = 6 2. b. – 5 , x-component = – 5, y-component = 2
6 2

112 Perfect Optional Mathematics Grade 6

2. c. – 2 , x-component = – 2, y-component = – 5 2. d. 6 , x-component = 6, y-component = – 3
– 5 – 3

2. e. 6 , x-component = 6, y-component = 0 2. f. 0 , x-component = 0, y-component = – 6
0 – 6

3. a. AB = 3 , BA = – 3 3. b. AB = – 5 , BA = 5
– 4 4 – 5 5

3. c. AB = 4 , BA = – 4 3. d. AB = – 5 , BA = 5
2 – 2 1 – 1

4. a. Q= 5 4. b. A = – 2
1 2

5. a.  PQ  = 5 units, PQ = 45, – 3 5. b.  AB  = 10 units, AB = 35, 4
5 5

5. c.  MN  = 65 units, MN = 675, 4 5. d.  CD  = 2 5 units, CD = – 51, 2
65 5

6. a. PQ = – 5 ,  PQ  = 34 units 6. b. PQ = 6 ,  PQ  = 10 units
– 3 8

c. PQ = – 4 ,  PQ  = 5 units 6. d. PQ = –  10 ,  PQ  =10 2 units
– 3 10

7. a. x = 3, y = 6 7. b. x = 5, y = 4 7. c. x = – 5, y = – 2 7. d. x = 7, y = 2

8. a. AB = 2 , CD = 2 8. b. AB = 6 , CD = 6
1 1 – 5 – 5

8. c. AB = – 2 , CD = – 2 8. d. AB = – 2 , CD = – 2
7 7 3 3

9. a. 5, a = 35, – 4 9. c. AB = – 4 , PQ = – 3
5 – 4 – 3

Exercise – 6.2

1. a. 4 , – 2 , 2 1. b. 1 , 3 , 4 1. c. – 4 , 2 , – 2
2 2 4 – 4 2 – 2 2 2 4

2. a. 2 , 2 , 0 2. b. 1 , 3 , – 2 2. c. – 4 , 2 , – 6
– 4 2 – 6 4 – 2 6 2 2 0

3. a. – 1 and 5 3. b. 3 and – 7 3. c. 5 and – 1
7 – 1 10 – 2 – 5 – 3

3. d. 2 and 6 4. a. – 1 , 5 , – 1 , – 5 4. b. – 3 , – 1 , – 3 , 1
– 1 –  11 10 0 10 0 3 – 9 3 9

5. a. 2 , 8 , 53 units, 65 units 5. b. 6 , – 2 , 10 units, 10  2 units
7 – 1 8 14

6. a. 4 , 6 , – 8 , 1 6. b. – 6 , 4 , 2 , – 2 7. a. 10
– 6 – 9 12 – 3 15 –   10 – 5 5 – 2
2 1 Answers
5 3 – 18
7. b. – 1 7. c. –  11 7. d. –   25

8. a. 4 , 10 , – 2 , 2 2 units 8. b. 2 , 1 , 290 units
– 6 –   10 – 2 1 –  17

Answers 113

Exercise – 7.1

3. a. P'(6, 1), Q'(10, – 1), R'(8, – 5) 3. b. P'(8, 9), Q'(12, 7), R'(10, 3)
3. c. P'(– 1, 0), Q'(3, – 2), R'(1, – 6) 3. d. P'(– 3, 7), Q'(1, 5), R'(– 1, 1)
4. A'(0, 8), B'(5, 9), C'(2, 4) 5. A'(– 4, 0), B'(– 3, – 3), C'(1, – 4), D'(2, 1)

Exercise – 7.2

2. a. A'(– 3, 4), B'(4, 2), C'(2, – 4) 2. b. A'(4, – 4), B'(– 4, – 3), C'(2, 4)

3. a. A'(– 2, 4), B'(– 4, 2), C'(– 1, – 2) 3. b. A'(– 2, 4), B'(4, 1), C'(2, – 2)

4. a. P'(2, – 4), Q'(4, – 2), R'(– 2, – 1) 4. b. P'(– 2, – 4), Q'(– 4, – 1), R'(3, 1)

5. a. A'(– 2, 5), B'(– 6, 4), C'(– 4, 0) 5. b. P'(– 1, – 5), Q'(4, – 3), R'(5, – 7)

5. c. P'(– 1, 5), Q'(– 4,6), R'(– 5, – 2), S'(– 2, – 3) 5. d. A'(3, 1), B'(4, 5), C'(– 3, 4), D'(– 2, 2)

6. a. Y-axis 6. b. X-axis

Exercise – 7.3

3. a. A'(4, – 3) 3. b. B'(3, 4) 3. c. P'(– 4, 5) 3. d. Q'(– 4, – 6)

4. a. A'(4, 2), B'(1, 4), C'(3, 0) 4. b. P'(4, 2), Q'(1, 4), R'(– 1, – 3)

5. a. P'(– 4, 2), Q'(– 3, 4), R'(– 1, 0) 5. b. A'(3, 1), B'(0, 2), C'(2, 4)

6. a. A'(– 4, 1), B'(– 1, 0), C'(– 1, – 3), D'(– 4, – 2) 6. b. P'(4, – 2), Q'(3, – 3), R'(1, 0), S'(3, 2)

7. a. A'(– 4, 2), B'(– 7, 6), C'(– 1, 4) 7. b. A'(4, – 2), B'(7, – 6), C'(1, – 4)

8. a. A'(– 2, 5), B'(– 5, 3), C'(– 4, – 2), D'(– 7, – 4) 8. b. A'(2, – 5), B'(5, – 3), C'(4, 2), D'(7, 4)

Exercise – 8.1

1. a. 6 1. b. 16.4 1. c. 10.8 1. d. 38.33
1. e. 58.33 1. f. 22.5 2. a. 4.5 2. b. 10
2. c. 5.6 2. d. 7 2. e. 20 2. f. 7
3. a. x = 44 3. b. x = 14 3. c. x = 25 3. d. x = 7.33
3. e. x = 10.5 3. f. x = 9 4. a. 12.8 4. b. 20
4. c. 23 4. d. 53.2 4. e. 64.25 kg 4. f. 7.05

Exercise – 8.2

1. a. 5 1. b. 18 1. c. 51 1. d. 16.5
1. e. 12 1. f. 67
3. a. Rs.130 3. b. 70 2. a. 22 2. b. 12
4. a. 5.5 4. b. 11
5. c. 5 5. d. 4 3. c. Rs.24
7. a. 11 7. b. 10
9. a. 16 9. b. 14 5. a. 8 5. b. 8

6. a. 25 6. b. Rs,350

8. Mean = 41.25, Median = 45, Mode = 50

9. c. 15

114 Perfect Optional Mathematics Grade 6