Combined transformation of reflection, rotation, translation
and enlargement.
Composition of reflection and rotation
Let P(x, y) be a point on the plane. Then the image of the point P(x, y) under the
reflection R0 in the line x-axis is
P(x, y) x-axis P'(x, – y) i.e. R0(x, y) = (x, –y).
Again, the image of the point P(x, y) under the rotation R1 about (0, 0) through +90°
is
P(x, y) R[(0, 0), + 90°] P'(–y, x) i.e. R1(x, y) = (–y, x).
Now the reflection in x-axis followed by rotation through +90° is
R1oR0 (x, y) = R1[R0(x, y)]
= R1(x, –y)
= (y, x)
Hence, the reflection in the line x-axis followed by the rotation through +90° about
(0, 0) is equivalent to the reflection in the line y = x.
Composition of reflection and translation
Let P(x, y) be a point on the plane. The image of the point P(x, y) under the reflection
R in the line y = –x is
P(x, y) y = –x P'(–y, –x) i.e. R(x, y) = (–y, –x)
Again, the image of the point P(x, y) under translation T with the translation vector
a T= a P'(x + a, y + b) i.e. T(x, y) = (x + a, y + b).
b b
is P(x, y)
Now, the reflection in the line y = –x followed by the translation with vector a is
b
ToR(x, y) = T[R(x, y)]
= T(–y, –x)
= (–y + a, –x + b)
Composition of reflection and enlargement
Let P(x, y) be a point on the plane. The image of the point P(x, y) under the
reflection R in the line y-axis is
P(x, y) y-axis P'(–x, y) i.e. R(x, y) = (–x, y).
Infinity Optional Mathematics Book - 10 301
Again, the image of the point P(x, y) under the enlargement E with centre
(0, 0) and scale factor k is
P(x, y) E[(0, 0), k] P'(kx, ky) i.e. E(x, y) = (kx, ky)
Now, the enlargement with centre (0, 0) and scale factor k followed by the reflection
in the line y-axis is
RoE(x, y) = R[E(x, y)]
= R(kx, ky)
= (–kx, ky)
Composition of rotation and Translation
Let P(x, y) be a point on the plane. The image of the point P(x, y) under the rotation
R about (0, 0) through 180° is
P(x, y) R[(0, 0), 180°] P'(–x, –y) i.e. R(x, y) = (–x, –y)
Again, the image of the point P(x, y) under the translation T with translation vector
a T= a P'(x + a, y + b) i.e. T(x, y) = (x + a, y + b)
b is P(x, y) b
Now, the rotation about (0, 0) through 180° followed by the translation with the
translation vector a is ToR(x, y) = T[R(x, y)]
b
= T(–x, –y)
= (–x + a, –y + b)
Composition of rotation and enlargement
Let P(x, y) be a point on the plane. The image of the point P(x, y) under the rotation
R about origin through 90° clockwise is P(x, y) R[(0, 0), –90°] P'(y, –x)
i.e. R(x, y) = (y, –x).
Again, the image of the point P(x, y) under the enlargement E with centre
(0, 0) and scale factor k is
P(x, y) E[(0, 0), k] P'(kx, ky) i.e. E(x, y) = (kx, ky)
Now, the rotation about origin through –90° followed by the enlargement with
centre (0, 0) and scale factor k is
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EoR(x, y) = E[R(x, y)]
= E(y, –x)
= (k.y, k.(–x)) = (ky, –kx)
Composition of translation and enlargement
Let P(x, y) be a point on the plane. The image of the point P(x, y) under the translation
a
T with translation vector b is
T= a P'(x + a, y + b) i.e. T(x, y) = (x + a, y + b)
P(x, y) b
Again, the image of the point P(x, y) under the enlargement E with the centre (0, 0)
and scale factor k is P(x, y) E[(0, 0), k] P'(kx, ky) i.e. E(x, y) = (kx, ky)
Now, the enlargement with centre (0, 0) and scale factor k followed by the translation
a
with the translation vector b is
ToE(x, y) = T[E(x, y)]
= T(kx, ky) = (kx + a, ky + b)
WORKED OUT EXAMPLES
1. Point (–6, 11) is rotated about origin through –90° and image so formed
is reflected in the line y = –x. Find the co-ordinates of the images so
formed.
Solution: Under the rotation about origin through –90°,
P(x, y) R[(0, 0), –90°] P'(y, –x)
A(–6, 11) A'(11, 6)
Under the reflection in the line y = –x.
P(x, y) y = –x P'(–y, –x)
A'(11, 6) A''(–6, –11)
Hence, the co-ordinates of the images are A'(11, 6) and A''(–6, –11)
2. T= 2 and F denote the translation and reflection on the line y-axis
3
respectively. Find the image of the point (5, 6) under the combined
transformation FoT.
Solution: Under the translation with T = 2
3
Infinity Optional Mathematics Book - 10 303
T= 2 P'(x + 2, y + 3) i.e. T(x, y) = (x + 2, y + 3).
P(x,y) 3
Under the reflection on the line y-axis.
P(x, y) y-axis P'(–x, y) i.e. F(x, y) = (–x, y).
Now, FoT (x, y) = F[T(x, y)]
= F(x + 2, y + 3)
= (–(x + 2), y + 3) = (–x – 2, y + 3)
So, FoT (5, 6) = (–5 – 2, 6 + 3) = (–7, 9)
Hence, the image of the point (5, 6) under the combined transformation FoT is
(–7, 9).
3. What point would have the image (–3, 21) under the translation with
1
vector T = 2 followed by the enlargement with centre (0, 0) and scale
factor 3.
Solution: Let A(a, b) be a point. Then A''(–3, 21) be its final image.
Under the translation with vector T = 1 .
2
1
P(x, y) T= 2 P'(x + 1,y + 2)
A(a, b) A'(a + 1, b + 2)
Under the enlargement with center (0, 0) and scale factor 3.
P(x, y) E[(0,0), k] P'(kx, ky)
A'(a + 1, b + 2) E[(0,0), 3] A''{(a + 1).3, (b + 2).3} = A''(3a + 3, 3b + 6)
Now, A''(3a + 3, 3b + 6) = A''(–3, 21)
By comparing the corresponding elements, we get,
3a + 3 = – 3 3b + 6 = 21
or, 3a = – 6 or, 3b = 15
or, a = –2 or, b = 5
∴ The co-ordinates of the point is A(–2, 5)
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3. Translate the ∆ABC with vertices A(2, 4), B(–1, 2) and C(5, –1) by the
translation vector T = –2 and then reflect the image so formed in the
3
line x + y = 0. Write the co-ordinates of the vertices of the image and
draw the object and the images on the same graph paper.
Solution: The given vertices of ∆ABC are A(2, 4), B(–1, 2) and C(5, –1).
Under the translation with the translation vector T = –2 ,
3
–2
T= 3 P'(x + (–2), y + 3) = P'(x – 2, y + 3)
P(x, y)
A(2, 4) A'(2 – 2, 4 + 3) = A'(0, 7)
B(–1, 2) B'(–1 – 2, 2 + 3) = B'(–3, 5)
C(5, –1) C'(5 – 2, –1 + 3) = C'(3, 2)
Under the reflection in the line x + y = 0 i.e. x = –y
P(x, y) x = –y P'(–y, –x)
A'(0, 7) A''(–7, 0)
B'(–3, 5) B''(–5, 3)
C'(3, 2) C''(–2, –3)
∆ABC and its images ∆A'B'C' and ∆A''B''C'' are shown in the following graph.
B' Y
B'' A'
A
B
C'
X' A'' O X
C
C''
Y'
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Exercise 7.5
Section 'A'
1. (a) Which transformation is equivalent to the reflection on y-axis followed
by a rotation about origin through +90°?
(b) Which transformation is equivalent to the rotation about origin through
–90° followed by a reflection on the line y = x?
Section 'B'
2. (a) The image formed by reflecting the point (3, 4) on the y-axis is rotated
about (0, 0) through +90°, find the co-ordinates of this image.
(b) Find the co-ordinates of image of a point A(5, –3) when it is first reflected
on the line x = 2 and then translated by the translation vector –2 .
–1
(c) If R is the reflection in the line y = 2 and E is the enlargement with
centre (0, 0) and scale factor 3 i.e. E[(0, 0), 3], find the image of a point
A(4, 5) under the combined transformation EoR.
3. (a) If R1 is the reflection in the line x-axis and R2 is the rotation through –90°
about the origin, find the image of a point A(3, –6) under the combined
transformation of R1oR2.
(b) Find the co-ordinates of the image of a point (–4, 3) when it is first rotated
about origin O through +90° and then translated by 2 .
5
(c) Find the co-ordinates of the image of a point B(–3, –5) under the rotation
about (0, 0) through 180° followed by the enlargement about (–2, 0) and
scale factor 1 i.e, E[(–2, 0), 1].
4. (a) What is the image of a point F(4, 5) when it is first translated by –2
3
and then reflected on the y-axis?
(b) Under the enlargement with centre origin and scale factor 2, the image
of the point A(2, –4) is A'. Again A' is translated by vector 6 to A''. Find
8
the co-ordinates of A' and A''.
(c) Let T be the translation with vector –5 and R be the rotation about
2
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origin through quarter turn anticlockwise direction. Find the image of a
point M(–3, 1) under the combined transformation RoT.
5. (a) T = –1 denotes the translation and F the rotation about the origin
–2
through +90°. If the image of the point A(p, q) is A'(8, 9) under the
combined transformation ToF, find the values of p and q.
(b) What point would have the image (–6, 8) under the enlargement with
centre (0, 0) and scale factor 2 i.e. E[(0, 0), 2] followed by the rotation
about (0, 0) through –270° i.e. R[(0, 0), –270°].
(c) T denotes a translation vector 2 and R denotes the reflection in the
3
line y = 3. If RoT (a, 3) = (5, b), find the values of a and b.
Section 'C'
6. (a) Translate the ∆ABC with vertices A(–2, –4), B(–4, –2) and C(–6, –4) by
translation vector T = –3 and then reflect the image so obtained in the
–2
line y = x. Write down the co-ordinates of the vertices of the images and
draw the images on the same graph paper.
(b) Draw ∆PQR having the vertices P(2, 0), Q(3, 1) and R(1, 1) on a graph
paper. It is rotated about the origin O through –90° and present ∆P'Q'R'
on the same graph paper. Then the ∆P'Q'R' is reflected in the line
x + y = 0 and plot ∆P''Q''R'' on the same graph paper. Write the co-
ordinates of vertices of the ∆P''Q''R''.
(c) D(2, 1), E(5, 3) and F(7, –1) are the vertices of ∆DEF. Find the
co-ordinates of the images of ∆DEF under the reflection on the line y =
0 followed by the enlargement E[(0, 0), 2]. Present ∆DEF and its images
on the same graph.
7. (a) The vertices of ∆ABC are A(1, –2), B(3, 4) and C(–4, 0). Find the image
of ∆ABC under the rotation about origin through negative quarter turn
followed by the translation by vector –1 . Represent the object and
3
images on the same graph paper.
(b) M(–1, 3), N(4, 1) and O(2, 5) are the vertices of ∆MNO. Find the
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coordinates of the vertices of the image of ∆MNO under the rotation of
positive 90° about origin followed by E[(0, 0), 2]. Represent the object
and images on the same graph paper.
(c) The vertices of ∆PQR are P(1, –2), Q(3, 4) and R(–4, 0). If the centre of
enlargement is (2, 2) and scale factor 2, then enlarge ∆PQR and translate
by the translation vector 4 . Present ∆PQR and its images on the same
5
graph paper.
7.3 Inversion Transformation and Inversion Circle
Discuss the following questions.
(a) What is the equation of circle with centre at origin?
(b) What is the equation of circle with centre at (h, k)?
(c) Find the centre and radius of the following equations of circle.
(i) x2 + y2 = 25 (ii) (x – 3)2 + (y + 4)2 = 9
(iii) x2 + y2 + 8x – 6y + 2 = 0 (iv) (x – 2)2 + y2 = 36
(d) Define similar triangles. Under what condition the triangles are similar?
(e) Define co-linear points. Under what conditions the points are co-linear?
Inversion Point
AA A
rr r
O P = P' OP P' O P' P
BC BC BC
Fig.(i) Fig.(ii) Fig.(iii)
In the above figure, O be the centre of a circle ABC and r be its radius. Let P be
a point and P' be its image. In the figure (i), P lies in the circumference of a circle
and its image P' also lies in the circumference of the circle such that OP × OP' = r2.
In the figure (ii), the point P lies inside the circle on the line through O and P' lies
outside the circle on the line OP satisfying OP × OP' = r2. In the figure (iii), P lies
outside the circle on the line through O and its image P' lies inside the circle on the
line OP such that OP × OP' = r2. The image P' is called inversion point of P with
respect to the circle ABC. The circle ABC is called inversion circle and its centre O
is called inversion centre.
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Let O be the centre of a circle ABC with radius r. Let P be a point (except
centre) on the line through O. Then there is a point P' on the line OP such that
OP ×OP'= r2. The point P' is called as inversion point of P with respect to the circle ABC.
Inversion Transformation
The inversion transformation is a rule of changing the position of a point P with
respect to a circle to the other point P' (called inverse of the point P) such that
OP × OP' = r2. The point P' is called image of the point P under the inversion
transformation.
Properties of inversion transformation
1. If P lies on the circumference of a inversion circle then P'(inverse of P) also
lies on the circumference of the inversion circle i.e. P and P' coincide. Here P'
is said to be the invariant point.
2. If P lies outside (or inside) the inversion circle then P' lies inside (or outside)
the inversion circle. Here P' is the inversion of P and vice versa i.e. (P')' = P.
1
3. If the radius of inversion circle is 1 then OP × OP' = 1. So, OP = OP' or OP' =
1 distance of the inversion point
OP . i.e. the from the centre is reciprocal (or
inverse) of the distance of the given point from the centre of inversion.
Inversion point of a given point with respect to the circle with centre not
at origin.
Let C(h, k) be the centre of the circle and r be its radius. Let P(x, y) be a point and
P'(x', y') be its image. C, P and P' lie in the same straight line.
∴ CP × CP' = r2
Draw CD⊥OC, PE⊥OX and P'F⊥OX. Y
Again, draw CB⊥PE and CA⊥P'F.
Then, P'(x',y')
CB = DE = OE – OD = x – h
PB = PE – BE = y – k
CA = DF = OF – OD = x' – h P(x,y) A
C(h,k) B
P'A = P'F – AF = y' – k
In right angled triangle PBC,
PC2 = PB2 + BC2
∴ PC2 = (y – k)2 + (x – h)2
Since ∆PBC ∼ ∆P'AC, the ratio of the
corresponding sides are equal.
CA = P'A = P'C O DE F X
CB PB PC
or, CA = P'A = P'C × PC
CB PB PC PC
CA P'A r2
or, CB = PB = PC2 [ PC × P'C = r2]
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or, xx' –– h = y' – k = (x – r2 – k)2
h y–k h)2 + (y
Taking first and second ratios,
x' – h = r2
x– h (x – h)2 + (y – k)2
or, x' – h = (x – r2(x – h) k)2
h)2 + (y –
∴ x' = (x – r2(x – h) k)2 + h
h)2 + (y –
Taking second and third ratios,
y' – k = (x – r2 – k)2
y–k h)2 + (y
or, y' – k = (x – r2(y – k) k)2
h)2 + (y –
∴ y' = (x – r2(y – k) k)2 + k
h)2 + (y –
Hence, the inversion point of a given point with respect to the circle with center not
at origin is
(x', y') = (x – r2(x – h) k)2 + h, (x – r2(y – k) k)2 +k
h)2 + (y – h)2 + (y –
If the centre of a circle is origin i.e. (h, k) = (0, 0), then
x' = r2x and y' = r2y
x2 + y2 x2 + y2
Hence, the inversion point of the given point with respect to the circle with the
centre at origin is
(x', y') = r2x , r2y
x2 + y2 x2 + y2
WORKED OUT EXAMPLES
1. From the adjoining figure, locate the inversion 4 10
points of A and B with respect to the given O2
circle. A B
Solution:
Hence, radius of inversion circle (r) = 4, OA = 2 and
310 Infinity Optional Mathematics Book - 10
OB = 10.
Let A' and B' be the inversion points of A and B respectively.
Now, OA × OA' = r2
or, 2 × OA' = 42
or, OA' = 16 = 8
2
Again, OB × OB' = r2 4 B' 10
O2
or, 10 × OB' = 42 A B
or, OB' = 16
10
∴ OB' = 1.6 A'
A' and B' are shown in the figure alongside.
2. Find the inversion point of A(3, 5) with respect to the circle of equation
x2 + y2 = 25.
Solution:
Here, the equation of circle is
x2 + y2 = 25
or, (x – 0)2 + (y – 0)2 = 52
∴ Centre (h, k) = (0, 0) and radius (r) = 5 unit.
Here, (x, y) = A(3, 5), (x', y') = ?
We know that,
x' = r2x = 52 × 3 = 25 × 3 = 75
x2 + y2 32 + 52 9 + 25 34
y' = r2y = 52 × 5 = 25 × 5 = 125
x2 + y2 32 + 52 9 + 25 34
∴ (x', y') = 75 , 125 .
34 34
Hence, the inversion point of A(3, 5) is A' 75 , 125 .
34 34
3. Find the inverse point of (2, 2) with respect to the circle of equation
(x – 2)2 + (y – 1)2 = 9.
Solution:
Here, the given equation of a circle is (x – 2)2 + (y – 1)2 = 9
or, (x – 2)2 + (y – 1)2 = 32
Comparing above equation with (x – h)2 + (y – k)2 = r2
∴ Centre (h, k) = (2, 1) and radius (r) = 3.
Here, (x, y) = (2, 2), (x', y') = ?
We know that,
Infinity Optional Mathematics Book - 10 311
x' = (x – r2(x – h) k)2 + h = (2 32 (2 – 2) 1)2 + 2 = 2
h)2 + (y – – 2)2 + (2 –
Again, y' = (x – r2(y – k) k)2 + k = (2 32 (2 – 1) 1)2 + 1 = 10
h)2 + (y – – 2)2 + (2 –
∴ (x', y') = (2, 10)
Hence, the inverse point of (2, 2) is (2, 10)
Exercise 7.6
Section 'A'
1. From the figure alongside, find out the following. P
(a) inversion circle
(b) inversion radius
(c) inversion centre A
(d) inversion point of A
O A'
(e) inversion point of A' QR
2. From the figure given below, locate the inversion points of P, Q and
R with respect to the circle γ.
γ
3 P
O
Q
R
Section 'B'
3. (a) Let P' be the inverse point of P with respect to the circle with centre O.
If OP' = 12 units and radius of the circle (r) = 3 units then find OP.
(b) Let A' be the inversion point of A with respect to the circle having centre
at C. If CA = 9 and CA' = 4 then find the radius of the circle.
(c) From the figure alongside, find OB'.
A'
8A B
2
O?
5B'
4. (a) Find the inversion points of A(0, 3) and B(–2, 5) with respect to the circle
312 Infinity Optional Mathematics Book - 10
of equation x2 + y2 = 9.
(b) Find the inverse points of (6, 8) and (3,4) with respect to the circle of
equation x2 + y2 = 50.
Section 'C'
5. (a) Find the inversion points of P(3, 4) and Q(–3, –1) with respect to the
circle of equation (x – 2)2 + (y – 3)2 = 24.
(b) Find the inverse points of (0, 4) and (–2, 7) with respect to the circle of
equation (x – 1)2 + (y – 3)2 = 25.
6. (a) Find the inversion points of M(5, 0) and N(–3, –1) with respect to the
inversion circle x2 + y2 – 2x – 4y = 20.
(b) Find the inverse points of (1, 3) and (–1, 4) with respect to the inversion
circle of equation x2 + y2 – 8x – 6y – 2 = 0.
7.4 Transformation Using Matrix
The transformation of a point or an object using matrix can be done by using the
matrices of order 2 × 1 and 2 × 2.
Transformation using 2 × 1 matrix
Let P(x, y) be a point. P'(x + a, y + b) be the image of the point P(x, y) under
a
a T = b
inb . That is P(x, y)
translation by translation vector T = the matrix form as P'(x + a, y + b).
This transformation can be expressed
Object (O) = x
y
2 × 1 matrix (M) = a
b
Image matrix (I) = x+a = a + x
y+b b y
∴ Image matrix = 2 × 1 matrix + Object matrix i.e. I = M + O
Transformation using 2 × 2 matrix
Let P(x, y) be a point. P'(x', y') be the image of the point P(x, y) under the transformation
ab
matrix cd . This transformation can be expressed in the matrix form as:
matrix x
Object (O) = y
2 × 2 matrix (M) = a b
c d
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Image matrix (I) = x'
y'
Now, x' = a b . x = ax + by
y' c d y cx + dy
∴ Image matrix = 2 × 2 matrix × object matrix i.e. I = M × O
Reflection using 2 × 2 matrices
(i) Reflection on x-axis
Let P'(x', y') be the image of a point P(x, y) under the reflection on x-axis
i.e. y = 0 line. So, P(x, y) P'(x', y') = P'(x, –y)
Thus,
x' = x = 1.x + 0.y
y' = –y = 0.x + (–1).y
Writing the above equations in matrix form as:
x' 1 0 x
y' 0 –1 y
= .
1 0
Hence, 0 –1 is the required matrix for reflection on x-axis.
Similarly, we can find the matrices for the reflection on other lines.
(ii) –1 0 is the matrix for reflection on y-axis.
0 1
iii) 0 1 is the matrix for reflection on y = x line.
1 0
(iv) 0 –1 is the matrix for reflection on y = -x line.
–1 0
Rotation using 2 × 2 matrices
(i) Rotation through –90° or + 270° about origin.
Let P'(x', y') be the image of a point P(x, y) under the rotation through –90° or
+270° about origin. So, P(x, y) P'(x', y') = P'(y, –x)
Thus,
x' = y = 0.x + 1.y
y' = –x = (–1).x + 0.y
Writing above equations in the matrix form as:
x' 01 x
y' = –1 0 . y
Hence, 01 is the required matrix for R[(0, 0), –90°].
–1 0
314 Infinity Optional Mathematics Book - 10
Similarly,
(i) 0 –1 is the matrix for R[(0, 0), + 90° or – 270°]
10
(ii) –1 0 is the matrix for R[(0, 0), 180°]
0 –1
Enlargement using 2 × 2 matrix
Let P'(x', y') be the image of the point P(x, y) under the enlargement about
centre (0, 0) and scale factor k i.e. E[(0, 0), k].
So, P(x, y) P'(x', y') = P'(kx, ky)
Thus,
x' = kx = k.x + 0.y
y' = ky = 0.x + k.y
Writing above equations in the matrix from as:
x' = k0 . x
y' 0k y
Hence, k0 is the required matrix for the enlargement with centre (0, 0)
0k
and scale factor k.
Unit Square Y
C (0, 1) B(1, 1)
In the figure alongside OABC is an unit square having
O (0, 0) A(1, 0)
vertices O(0, 0), A(1, 0), B(1, 1) and C(0, 1). So, the
unit square can be expressed in the matrix form as:
0 1 1 0 . X
0 0 1 1
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WORKED OUT EXAMPLES
1. Find the image of a point A(2, 6) under the translation by vector
–1
2 by using matrix method.
Solution: Here, 2
6
Object matrix (O) =
2 × 1 matrix (M) = –1
2
Image matrix (I) = ?
Now, I = M + O
= –1 + 2 = –1+2 = 1
2 6 2+6 8
∴ The image of A(2, 6) is A'(1, 8).
2. Find the image of a point (–5, 4) under the rotation through +90° about
origin by using matrix method.
Solution:
Here, object matrix (O) = –5
4
2 × 2 matrix which represents rotation through +90° about origin
(M) = 0 –1
10
Image matrix (I) = ?
Now, I = M × O
= 0 –1 × –5
10 4
= 0 × (–5) + (–1) × 4
1 × (–5) + 0 × 4
= 0–4 = –4
–5 + 0 –5
∴ The image of (–5, 4) is (–4, –5).
316 Infinity Optional Mathematics Book - 10
3. Find the transformation represented by the matrix 0 –1 . Find the
image of a point A(6, –2) using the matrix. –1 0
Solution:
Here, 2 × 2 matrix (M) = 0 –1
–1 0
Let P(x, y) be a point and P'(x', y') be its image.
Object matrix (O) = x
y
Image matrix (I) = x'
y'
Now, I = M × O
x' = 0 –1 × x = 0.x + (–1).y = –y
y' –1 0 y (–1).x + 0.y –x
∴ Image of P(x, y) is P'(x', y') = P'(–y, –x)
i.e. P(x, y) P'(–y, –x)
Hence, the required transformation is the reflection on the line y = –x.
Again, 2 × 2 matrix (M) = 0 –1
–1 0
Object matrix (O) = 6
–2
Image matrix (I) = ?
Now, I = M × O
= 0 –1 . 6 = 0 × 6 + (–1) × (–2) = 2
–1 0 –2 (–1) × 6 + 0 × (– 2) –6
∴ Image of (6, –2) is (2, –6).
4. Find the image of ∆PQR with vertices P(–1, 3), Q(2, 0) and R(–4, –1)
under the rotation through +90° about origin using matrix method.
Solution: Here,
The matrix formed from the vertices of ∆PQR = –1 2 –4
3 0 –1
i.e. object matrix (O) = –1 2 –4
3 0 –1
Now, P(x, y) R[(0,0), +90°] P'(–y, x) = P'(x', y')
x' = –y = 0.x + (–1).y
y' = x = 1.x + 0.y
Infinity Optional Mathematics Book - 10 317
Writing above equations in matrix form as:
x' = 0 –1 . x
y' 10 y
∴ 2 × 2 matrix (M) = 0 –1
10
Image matrix (I) = ?
Now, I = M × O
= 0 –1 × –1 2 –4
10 3 0 –1
= 0.(–1) + (–1).3 0.2 + (–1).0 0.(–4) + (–1).(–1) = –3 01
1.(–1) + 0.3 1.2 + 0.0 1. (–4) + 0. (–1) –1 2 –4
∴ The image of vertices of ∆PQR are P'(–3, –1), Q'(0, 2), R'(1, –4)
5. Find 2 × 2 transformation matrix which transforms a unit square to
0 3 4 1
the parallelogram 0 1 3 2 .
Solution: Here,
Object matrix (O) = unit square = 0110
0011
Image of matrix (I) = 0341
0132
2 × 2 matrix (M) = ?
Let M = a b . Now, I = M × O
c d
or, 00 3 4 1 = ab 0110
1 3 2 cd 0011
or, 00 3 4 1 = a.0 + b.0 a.1 + b.0 a.1 + b.1 a.0 + b.1
1 3 2 c.0 + d.0 c.1 + d.0 c.1 + d.1 c.0 + d.1
or, 00 3 4 1 = 0 a a+b b
1 3 2 0 c c+d d
Comparing the corresponding elements, we get, a = 3, b = 1, c = 1, d = 2
∴ Required transformation matrix is ab = 3 1 .
cd 1 2
6. Find a 2 × 2 matrix which transforms ∆PQR having vertices P(4, 3),
Q(6, 4) and R(8, 1) with the image ∆P'Q'R' having vertices P'(–3, –4),
Q'(–4, –6) and R'(–1, –8)
Solution: Here, 468
341
The matrix formed from the vertices of ∆PQR =
318 Infinity Optional Mathematics Book - 10
i.e Object matrix (O) = 468 –3 –4 –1
341 –4 –6 –8
The matrix formed from the vertices of ∆P'Q'R' =
–3 –4 –1
i.e. Image matrix (I) = –4 –6 –8
2 × 2 transformation matrix (M) = ?
ab
Let M = cd
Now, I = M × O
or, ––43 –4 –1 a b 468
–6 –8 = c d × 341
or, ––43 –4 –1 = 4a + 3b 6a + 4b 8a + b
–6 –8 4c + 3d 6c + 4d 8c + d
By comparing the corresponding elements, we get,
4a + 3b = – 3 ............ (i) 6a + 4b = – 4 ............ (ii)
4c + 3d = – 4 ............. (iii) 6c + 4d = – 6 ............. (iv)
Solving equations (i) and (ii), we get, a = 0 and b = –1
Solving equations (iii) and (iv), we get, c = –1 and d = 0
a b 0 –1
Hence, the required 2 × 2 matrix = c d = –1 0 .
Exercise 7.7
Section 'A'
1. (a) Write a matrix to represent reflection on the line y-axis.
(b) Which matrix does represent rotation about origin through 90° in
clockwise direction.
(c) Find the matrix which is equivalent to E[(0, 0), 3]
2. (a) What does –1 0 represent?
0 1
–1 0
(b) Write the transformation which is equivalent to 0 –1 .
(c) Name the transformation which represents 2 0 .
0 2
Section 'B'
3. Transform the points A(3, 2) and B(–5, 1) using the following 2 × 1
matrices. –4
–2
(a) 0 (b) 1 (c) –3 (d)
3 2 5
4. (a) Find a 2 × 1 matrix which transform a point (p, q) into the point
Infinity Optional Mathematics Book - 10 319
(p + 2, q – 3). Using the same 2 × 1 matrix, transform the point (3, 1).
(b) Find a 2 × 1 matrix which transform a point (a, b) into the point
(a + 3, b – 4). Using the same 2 × 1 matrix, find the image A(4, 5).
5. (a) If a point (a, b) is transformed into (b, –a) by a 2 × 2 transformation
matrix, find the matrix.
(b) Find the 2 × 2 matrix which transforms A(–3, 1) to A'(–1, –3).
6. (a ) Find the transformation matrix from the equations x' = 3x and y' = 3y.
Using the matrix to find the image of P(2, 5).
(b) Find the transformation matrix from the equations x' = x – y and
y' = 2x + 3y. Use the matrix to find the image of a point (–7, –1).
7. (a) Find the transformation matrix which represents the reflection on the
line y = –x. Using this matrix, find the image of A(3, –5).
(b) Find the transformation matrix which represents the rotation through
–90° about (0, 0). Use this matrix to find the image of a point (–8, –9).
8. (a) Find the transformation represented by the matrix 1 0 . Also find
the image of A(2, –5) using the matrix. 0 –1
(b) Find the transformation represented by the matrix –2 0 . Find the
image of a point (–3, 9) using the matrix. 0 –2
9. (a) If A(a, b) is transformed by 0 2 to A'(–10, –8), find the values of a
and b. 2 0
(b) If a point (a, 5) is transformed by a matrix –2 1 , the image of the
point (7, b + 3). Find the values of a and b. 1 –2
10. Prove that the following using matrix method.
(a) The reflection on y = –x line followed by the rotation through +90° about
origin is equivalent to the reflection on x-axis.
(b) The rotation through 180° about origin followed by the reflection in
y = x line is equivalent to the reflection in x + y = 0 line.
(c) The reflection in the line x – y = 0 followed by the reflection in the line
x + y = 0 is equivalent to the rotation through 180° about origin.
Section 'C'
11. (a) Find the co-ordinates of the vertices of image which is transformed
by the transformation of square PQRS having vertices P(2, 1),
Q(4, 1), R(4, 3) and S(2, 3) under the transformation by a matrix
1 2 .
2 3
320 Infinity Optional Mathematics Book - 10
(b) The vertices of ∆ABC are A(1, –1), B(3, 2) and C(0, 2). If the ∆ABC is
transformed under the matrix transformation through enlargement
with centre (0, 0) and scale factor –3, find the vertices of image ∆A'B'C'
so formed.
12. (a) Find the co-ordinates of the vertices of the image of the unit square
transformed by the matrix 4 2 .
1 2
(b) Transform a parallelogram 0 3 5 2 under the rotation through
0 –1 –2 –1
quarter turn in anticlockwise direction about origin by using matrix
method.
13. (a) Find the 2 × 2 transformation matrix which transforms a unit square
0 1 1 0 into the parallelogram 0 6 8 2 .
0 0 1 1 0 2 6 4
(b) The matrix ab make a unit square into the parallelogram
cd
0 2 2 0 a b
0 0 2 2 . Find the matrix c d .
14. (a) Find 2 × 2 matrix which transform 0 2 2 0 onto the rectangle
0 0 1 1
0 2 2 0
0 0 –1 –1 . What is the transformation of this mapping.
(b) Find 2 × 2 matrix which transform 0 2 3 1 into
0 3 5 2
0 3 5 2
0 2 3 1 . Also state the transformation of this mapping.
15. (a) Find 2 × 2 transformation matrix which transforms a triangle ABC
with vertices A(1, 2), B(5, 2) and C(1, 5) into a triangle A'B'C' with
vertices A'(1, –2), B'(5, –2) and C'(1, –5)
(b) A square ABCD having vertices A(2, 3), B(4, –3), C(4, 5) and D(–2, 5) is
mapped into the square A'B'C'D' with vertices A'(–3, 2), B'(3, 4), C'(–5, 4)
and D'(–5, –2) by a 2 × 2 transformation matrix. Find the 2 × 2 matrix.
Infinity Optional Mathematics Book - 10 321
UNIT
8 STATISTICS
Review
The marks obtained by two students A and B of class 9 in eight subjects are given
below.
S.N. Subjects Marks Obtained
AB
1. Nepali 60 35
2. English 70 60
3. C. Mathematics 50 80
4. Science 55 65
5. Social Studies 62 40
6. Population 71 85
7. Optional Mathematics 65 90
8. Computer Science 75 50
Representing the given information in a graph.
Students B
A
0 5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100
Marks
Study the above information with graph and discuss on the following questions.
(i) What are the total marks obtained by A and B? Also, write their average
marks.
(ii) Whose obtained marks is more scattered?
322 Infinity Optional Mathematics Book - 10
(iii) Out of two students whose achievement is higher? why?
(iv) What are the methods to measure the consistency and variability of the
statistical data?
(v) How to compare the marks obtained by students A and B?
The term dispersion is commonly used to mean scatter, variation, deviation,
fluctuation or variability. However, the common use of the term is to mean
scatter or deviation. The measures of central tendency bring out only one of the
characteristics of the whole distribution. They give us an idea of the concentration
of the observations about the central part of the distribution. If the average value
alone is known, but it cannot form a complete idea about the distribution.
The measure of dispersion shows the scatterings of the data. It tells the variation
of the data from one another and gives a clear idea about the distribution of the
data. The measure of dispersion shows the homogeneity or the heterogeneity of the
distribution of the observations. The main idea about the measure of dispersion is
to get to know how the data are spread. It shows how much the data vary from their
average value.
Therefore, variations or the scatterness of the items from the central value are
known as 'Dispersion'. Measures of such scatterness are known as the measures
of dispersion.
8.1 Classification of measures of Dispersion
The measures of dispersion is categorized as
(i) An absolute measure of dispersion.
(ii) A relative measure of dispersion.
(i) An absolute measure of dispersion : Absolute measures defined in such
a way that they have units such as meters, grams etc. same as those of the
original measurements. Absolute measures cannot be used to compare the
variation of two or more sets of data. Most common absolute measures of
variability are:
(a) Range (b) Quartile deviation
(c) Mean deviation (d) Standard deviation
(e) Variance
(ii) Relative measures:
The relative measures have no units as these are ratios, coefficients, or
percentages. Relative measures are independent of units of measurements
and are useful for comparing data of different natures. The relative measures
Infinity Optional Mathematics Book - 10 323
of variability are
(a) Coefficient of range.
(b) Coefficient of quartile deviation.
(c) Coefficient of mean deviation.
(d) Coefficient of standard deviation
(e) Coefficient of variation
Quartile deviation (Semi-interquartile range)
Quartiles divide the whole observation into four equal parts. So, there are three
quartiles. The first quartile or lower quartile (Q1) is the middle number between the
smallest number and the median of the data. The second quartile (Q2) is the median
(md). The third quartile or upper quartile (Q3) is the middle number between
median and the largest number.
The difference between the upper quartile (Q3) and the lower quartile (Q1) is known
as the interquartile range. Half of the interquartile range is known as the semi-
interquartile range or quartile deviation. For determining semi-interquartile
range, we need first and third quartile.
Here, we have to find Q1 and Q3 for continuous (grouped) data as
The position of Q1 = N th N th
Now, 4 4
item and the position of Q3 = 3 item.
N – c.f 3N – c.f
First quartile (Q1) = l + 4 f × i Third quartile (Q3)= l + 4 f
× i
Where, l = lower limit of quartile class
c.f. = cumulative frequency of class just before quartile class.
f = frequency of quartile class
i = width of class interval
∴ Semi-interquartile range or Quartile deviation = Q3 – Q1
2
Quartile deviation is also called average deviation of first quartile (Q1) and
third quartile (Q3) from median (Q2)
∴ Q.D. = (Q3 – Q1) + (Q2 – Q1) = Q3 – Q1
2 2
324 Infinity Optional Mathematics Book - 10
To compare the variability we need coefficient of quartile deviation. It is
relative measure. Thus, coefficient of quartile deviation is defined as : Q3 – Q1
Q3 + Q1
Thus, coefficient of Q. D. = Q3 – Q1
Q3 + Q1
Less the coefficient of quartile deviation less the variability and more the
coefficient of quartile deviation more the variability.
Quartile Scale:
0% 25% 50% 75% 100%
Q2 Q3
Q1
Merits of Quartile deviation
Some advantages of quartile deviations are as follows.
i) It is easy to calculate and easy to understand.
ii) It is not affected by extreme observations.
iii) It can be calculated even for open end classes.
Demerits of Quartile deviation
Some disadvantages of quartile deviation are as follows.
1. It is not based on all observations.
2. It is affected by fluctuation of sampling.
3. Arrangement of data are necessary.
WORKED OUT EXAMPLES
1. In a continuous series, quartile deviation is 10 and coefficient of
1
quartile deviation is 2 then find Q1 and Q3.
Solution: Here,
Quartile deviation (Q.D) = 10
and Coefficient of Q.D. = 1
2
By formula,
Q.D. = Q3 – Q1
2
Infinity Optional Mathematics Book - 10 325
10 = Q3 – Q1
2
Q3 – Q1 = 20
Q3 = 20 + Q1 ............ (i)
and coefficient of Q.D. = Q3 – Q1
Q3 + Q1
or, 21 = 20 + 20 + Q1 [ From (i)]
Q1
or, 2Q1 + 20 = 40
or, 2Q1 = 20
Q1 = 10
Substituting Q1 = 10 in equation (i) we get
Q3 = 20 + 10 = 30
∴ Q1 = 10 and Q3 = 30
2. Obtain the semi-interquartile range (Q.D.) and its coefficient.
Height (in Inches) 60 – 62 62 – 64 64 – 66 66 – 68 68 – 70 70 – 72
No. of students 4 6 8 12 7 2
Solution: Here,
Tabulating the given data in ascending order for the calculation of Q.D. and
its coefficient.
Height (in inches) No. of students (f) Cumulative frequency (c.f)
60 – 62 4 4
62 – 64 6 10
64 – 66 8 18
66 – 68 12 30
68 – 70 7 37
70 – 72 2 39
The position of first quartile (Q1) = N th item. = 39 th item = 9.75th
item 4 4
In c.f. column, 10 is just greater than 9.75 so its corresponding class is 62 – 64.
∴ l = 62, f = 6, c.f. = 4, i = 2
N – c.f
Now, Q1 = l + 4 f × i
326 Infinity Optional Mathematics Book - 10
= 62 + 9.75 – 4 × 2
6
= 62 + 1.92 = 63.92 inches
Similarly, the position of third quartile (Q3) = 3 N th
4
item
= 3(9.75)th item
= 29.25th item
In C. f. column 30 is just greater than 29.25 so its corresponding class is
66 – 68.
∴ l = 66, f = 12, c.f. = 18 and i = 2
N – c.f
Now, Q3 = l + 4 f × i
= 66 + 29.25 – 18 × 2
12
= 66 + 1.875 = 67.875 inches
Again, Q.D. = Q3 – Q1 = 67.875 – 63.92 = 1.97
2 2
and coefficient of Q.D. = QQ33 – Q1 = 67.875 – 63.92 = 3.955 = 0.03
+ Q1 67.875 + 63.92 131.795
3. Find the quartile deviation and its coefficient of the data given below.
Ages (in yrs) Below 25 25 – 30 30 – 35 35 – 40 40 – 45 Above 45
No. of students 4 11 21 24 16 8
Solution : Here,
Tabulating the given data in ascending order.
Ages (in yrs) No. of students (f) Cumulative frequency (c.f)
Below 25 4 4
25 – 30 11 15
30 – 35 21 36
35 – 40 24 60
40 – 45 16 76
Above 45 8 84
Σf = N = 84
The position of first quartile (Q1) = N th 84 th item = 21th item
4 4
item =
Infinity Optional Mathematics Book - 10 327
In c.f. column, 36 is just greater than 21 so its corresponding class is 30 – 35.
∴ l = 30, f = 21, c.f. = 15, i = 5
Now, by formula,
N – c.f
Q1 = l + 4 f × i
= 30 + 21 – 15 × 2
21
= 30 + 1.42 = 31.42
Again, the position of third quartile (Q3) = 3 N th item = (3 × 21)th item
4
= 63th item
In c.f. column, 76 is just greater than 63 so its corresponding class is 40 – 45.
∴ l = 40, f = 16, c.f. = 60, i = 5
3N – c.f
4
Q3 =l+ ×i
f
= 40 + 63 – 60 × 5 = 40 + 0.93 = 40.93
16
Now, Q.D. = Q3 – Q1 = 40.93 – 31.42 = 4.75
2 2
and coefficient of Q. D. = Q3 – Q1 = 40.93 – 31.42 = 4.75 = 0.06
Q3 + Q1 40.93 + 31.42 72.35
Note: If the data is open end data (lower and upper limit respectively of lowest
class and highest class unknown) range can not be calculated and in such
case quartile deviation is useful.
Exercise 8.1
Section 'A'
1. (a) What is dispersion?
(b) What are the methods to measure the dispersion? Write them.
2. (a) Define quartile deviation. Write the formula to find quartile deviation.
(b) What do you mean by coefficient of quartile deviation?
3. (a) Write any two merits and demerits of quartile deviation.
(b) How we can interpret the results of coefficient of quartile deviation?
328 Infinity Optional Mathematics Book - 10
Section 'B'
4. (a) In a continuous series, the first quartile and third quartiles are 35 and
75 respectively. Find the quartile deviation and its coefficient.
(b) In a continuous series, the first quartile and third quartiles are 25 and
50 respectively. Find the quartile deviation and its coefficient.
5. (a) In a continuous data, quartile deviation and the first quartile are 5 and
40 respectively. Find the third quartile and the coefficient of quartile
deviation.
(b) In a continuous data, quartile deviation and the third quartile are 7 and
56 respectively. Find the first quartile and the coefficient of quartile
deviation.
6. (a) In a continuous series, the quartile deviation and its coefficient are 15
(b)
7. (a) and 3 respectively. Find the first quartile and third quartile.
7
In a continuous data, the quartile deviation and its coefficient are 14
and 7 respectively. Find the first quartile and its coefficient.
22
The third quartile and the coefficient of quartile deviation of continuous
data are 45 and 1 respectively. Find the first quartile and the interquartile
4
range of the data.
(b) The first quartile and the coefficient of quartile deviation of continous
data are 20 and 7 respectively, find the third quartile and the
22
interquartile range of the data.
Section 'C'
8. Find the quartile deviation and the coefficient of quartile deviation
from the following distribution.
(a) Marks obtained 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70
No. of students 12 10 8 10 7 5
(b) Age (in yrs.) 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60
No. of workers 2 6 22 13 7
(c) Weight (in kg.) 12 – 14 14 – 16 16 – 18 18 – 20 20 – 22 22 – 24
No. of students 41 53 67 72 47 32
Infinity Optional Mathematics Book - 10 329
(d) Wages (Rs.) 0-250 250-500 500-750 750-1000 1000-1250 1250-1500 1500-1750 1750-2000
No. of Workers 1 3 7 12 25 39 11 2
9. (a) Marks obtained by 27 students of grade X are given below.
14, 16, 12, 18, 13, 15, 13, 14, 13, 15, 12, 15, 16, 14, 16, 12, 14, 12, 13, 18,
15, 16, 15, 13, 14, 15, 12. Construct a frequency distribution table by
taking 12 – 14 as first class interval and find the quartile deviation.
(b) Marks obtained by 28 students of grade 'X' in an internal examination
are given below. Construct a frequency distribution table by taking 10
– 20 as class first interval and find the quartile deviation.
66, 75, 34, 64, 53, 20, 47, 13, 74, 38, 30, 22, 70, 28, 62, 14, 40, 56, 29, 34,
50, 48, 60, 21, 45, 57, 15, 41.
10. Find the quartile deviation and its coefficient from the data given
below.
(a) Ages (in yrs) 0<x< 10 10<x<20 20<x<30 30<x<40 40<x<50 50<x<60
No. of Persons 8 26 30 40 20 16
(b) Marks less than less than less than less than less than less than
obtained 10 20 30 40 50 60
No. of students 8 34 64 104 124 140
(c) Age (in yrs) Below 10 10 – 20 20 – 30 30 – 40 40 –50 50 – 60 Above 60
No. of people 4 7 14 11 6 5 3
11. The data are given below. By using coefficient of quartile deviation
state which is more variable.
Class interval 10 – 20 20 – 30 30 – 40 40 –50 50 – 60 60 – 70
For A (Frequency) 10 18 32 40 22 18
For B (Frequency) 18 22 40 32 29 10
330 Infinity Optional Mathematics Book - 10
Mean deviation:
The marks obtained by 38 boys and 38 girls of a school of class X in an internal
examination are given below.
Marks 40 – 50 50 – 60 60 – 70 70 –80 80 – 90 90 – 100
No. of boys 4 6 12 8 7 3
No. of girls 5 4 10 3 12 6
Study the data given above and discuss on the following questions.
(i) What are the average marks of boys and girls?
(ii) Whose (boys or girls) marks obtained is more consistency? why?
(iii) How we can compare the marks obtained by boys and and girls?
(iv) What are the median marks obtained by boys and girls?
(v) Write the model class of marks obtained by boys and girls.
By using mean deviation, we can find the uniformity and variability of the above
data. As we know that range depends on the largest and smallest value of the
distribution and quartile deviation depends on 50% of the total observation. They
are not based on all the observations and they do not measure the scatterdness of
the items from the average value. Thus, they are not consider as good measure of
dispersion. But mean deviation measures the variation of each observation of the
total distribution from the average.
Mean deviation is the arithmetic average of deviations of all items from the mean,
median or mode. All deviations considered positive. It is also known as average
deviation. When deviations are taken from mean, it is known as the mean deviation
from mean and if deviation are taken from median, it is known as the mean deviation
from median. Mean deviation is denoted by M.D.
Calculation of mean deviation in continuous series
Let, m1, m2, m3, ......, mn be the mid values of the corresponding classes, x1, x2, x3,
......., xn and f1, f2, f3, ....... fn be their respective frequencies.
Σf |m – x| Σf|D|
(i) Mean deviation from mean = N = N
Where, m = mid value
|D| = |m – x|, x = Actual mean
Σf |m – Md| Σf|D|
(ii) Mean deviation from median = N = N
Where, m = mid value, Md = median and |D| = |m – Md|
Infinity Optional Mathematics Book - 10 331
Coefficient of mean deviation
Mean deviation is an absolute measure. So, to compare two or more series having
different units, the relative measure corresponding to mean deviation is used,
which is called coefficient of mean deviation.
(i) Coefficient of M.D. from mean = M.D from mean = M.D
Mean X
(ii) Coefficient of M.D. from median = M.D from median = M.D
Median Md
WORKED OUT EXAMPLES
1. In a continuous series, Σf = 20 , Σf|m – X| = 110 and Σfm = 140, find
mean deviation and its coefficient.
Solution: Here,
N = Σf = 20, Σfm = 140, Σf|m – X|= 110
By formula, Σf |m – X|
N
Mean deviation (M.D.) = = 110 = 5.5
Again, X = Σfm 140 20
N = 20 = 7
and coefficient of M.D. = M.D = 5.5 = 0.78
X 7
2. Find the mean deviation from mean and its coefficient from the
following data.
Age (in yrs) 10 – 20 20 – 30 60 – 70 40 – 50 50 – 60 30 – 40
No. of people 6 8 3 14 8 11
Solution : Here,
Tabulating the given data in ascending order for the calculation of M.D. and
its coefficient.
Age No. of people Mid value fm |m–X|=|D| f|D|
(in yrs) (f) (m)
10 – 20 6 15 90 23.8 142.8
20 – 30 8 25 200 13.8 110.4
30 – 40 11 35 385 3.8 41.8
40 – 50 14 45 630 6.2 86.8
50 – 60 8 55 440 16.2 129.6
60 – 70 3 65 195 26.2 79.8
Σf = N = 50 Σfm=1940 Σf|D|= 591.2
Mean X = Σfm = 1940 = 38.8
N 50
332 Infinity Optional Mathematics Book - 10
Again, Mean deviation from mean = Σf|D| = 591.2 = 11.824
N 50
and coefficient of M. D. = M.D = 11.824 = 0.304
X 38.8
3. Calculate the mean deviation from median and its coefficient from
the following data.
Marks 0-10 10-20 20-30 30-40 40-50 50-60 60-70
No. of students 4 6 10 20 10 6 4
Solution: Here, Tabulating the given data in ascending order for the calculation
of M.D. and its coefficient.
Marks No. of students Mid value C.f |m – Md|=|D| f|D|
(f) (m)
0 – 10 4 54 30 120
10 – 20 6 15 10 20 120
20 – 30 10 25 20 10 100
30 – 40 20 35 40 0 0
40 – 50 10 45 50 10 100
50 – 60 6 55 56 20 120
60 – 70 4 65 60 30 120
Σf = N = 60 Σf|D|= 680
N th 60 th
2 2
The position of median = item = item = 30th item
In c.f. column, 40 is just greater than 30 so its corresponding class is 30 – 40.
∴ l = 30, f = 20, c.f = 20, i = 10
Now, median (Md) = l + N – c.f × i
2 f
= 30 + 30 – 20 × 10 = 30 + 5 = 5
20
∴ Median (Md) = 35
Again, M.D from median = Σf|D| = 680 = 11.33
N 60
and coefficient of M.D. = M.D = 11.33 = 0.32
Md 35
Infinity Optional Mathematics Book - 10 333
Merits and Demerits of mean deviation
(a) Merits
(i) It is based on all observation.
(ii) Deviations are taken from, the central value.
(iii) It is easy to understand and simple to calculate.
(iv) It is least affected by extreme values.
(b) Demerits
(i) Ignore of sign is deviation is the major defect.
(ii) It cannot be calculated for open - end classes.
(iii) It is affected by fluctuation of sampling.
Exercise 8.2
Section 'A'
1. (a) Define mean deviation.
(b) What is coefficient of mean deviation? Describe its uses with example.
2. (a) Write any two merits and demerits of mean deviation.
(b) Why mean deviation is better than range and quartile deviation?
3. (a) Write the formula to find the mean deviation from mean in continuous
series.
(b) In a continuous series, write the formula to find the mean deviation from
median.
(c) Write the formula to find the coefficient of mean deviation from mean
and median in a continuous series.
Section 'B'
4. (a) In a continuous series, Σfm = 750, Σf = 25 and Σf|m – X| = 140. Find the
mean deviation from mean and its coefficient.
(b) In a continuous series, Σf = 20, Σf|m – Md| = 250 and Md = 50. Find the
mean deviation from median and its coefficient.
5. In a continuous series, Σfm = 210, mean deviation from mean and median
(M.D.) = 4.5, Σf = 30 and mean (X) = median (Md) then find the coefficient of
M.D from mean and median.
Section 'C'
6. Calculate the mean deviation from mean and its coefficient from the
following data.
(a) Marks 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70 70–80
No. of students 4 7 9 18 12 7 3
334 Infinity Optional Mathematics Book - 10
(b) Weight (in kg) 20 – 25 25 – 30 30 – 35 35 – 40 40 – 45 45 – 50
No. of people 7 3 6 4 8 2
(c) Wages (in Rs.) 0-400 400-800 800-1200 1200-1600 1600-2000 2000-2400
No. of students 7 7 10 15 7 6
(d) Marks 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50
No. of students 3 6 9 5 2
7. Calculate the mean deviation from median and its coefficient from
the following data.
(a) Marks obtained 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50
No. of students 4 6 10 18 6
(b) Distance cover (m) 95-105 105-115 115-125 125-135 135-145 145-155
No. of students 6 8 11 14 8 3
(c) Age (in yrs) 6 – 12 12 – 18 18 – 24 24 – 30 30 – 36
2 5 4 3
No. of students 3
(d) Age (in yrs.) 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70
7 8 6 4
No. of people 5
8. Calculate the mean deviation from mean and also calculate its
coefficient.
(a) Marks obtained 5< x<10 10<x<15 15<x<20 20<x <25 25<x<30
No. of students 7 4 5 6 3
(b) Weight (in kg) 0 – 10 0 – 20 0 – 30 0 – 40 0 – 50
No. of students 2 5 11 16 20
(c) Again (in yrs.) 20 – 30 20 – 40 20 – 50 20 – 60 20 – 70
No. of people 5 12 18 23 25
9. Find the mean (average) deviation and its coefficient from the
following data.
(a) Marks 0–9 10 – 19 20 – 29 30 – 39 40 – 49
No. of students 5 8 15 16 6
(b) Marks 11 – 19 21 – 29 31 – 39 41 – 49 51 – 59 61 – 69
No. of students 6 8 11 14 8 3
Infinity Optional Mathematics Book - 10 335
10. (a) Construct a frequency distribution table taking (0 – 10) as first class
interval and find the mean deviation from mean and its coefficient of the
given data.
5, 9, 16, 26, 14, 34, 19, 40, 15, 25, 43, 36, 24, 49, 33, 6, 11, 15, 25, 35, 38,
33, 18, 9, 46, 28, 36, 23, 18, 7
(b) Construct a frequency distribution table taking (20 – 30) as first class
interval and find the mean deviation from mean and median and the
coefficient of mean deviation of the given data.
64, 60, 52, 65, 50, 79, 39, 55, 69, 58, 49, 37, 42, 29, 21, 35, 44, 45, 71, 59
11. Find the mean deviation from median and also calculate its coefficient
from the data given below.
Marks less than less than less than less than less than less than
10 20 30 40 50 60
No. of students 3 7 15 20 26 30
Standard Deviation
Out of 100 full marks the marks obtained by two students A and B in six terminal
examination are given below.
Examination 1 2 3 4 5 6
A 50 60 70 65 62 57
B 40 71 82 90 60 80
Study the above table and answer the following questions.
(i) What is the average marks of A and B?
(ii) What are the dispersion of A and B?
(iii) Which method is more reliable for the calculation of deviation?
(iv) Who is better? (From mean)
(v) Who is intelligent? (From median)
(vi) If the consistency of performance is the criteria for awarding a price,
who should get the prize?
We can find the deviation from different method for the above distribution. But
standard deviation gives the uniform, correct and stable result. Standard deviation
is the most popular and useful measure of dispersion used in practice which the
drawback presents in other measure of dispersion are removed.
A standard deviation is defined as the positive square root of the arithmetic mean of
the square of the deviation taken from the arithmetic mean. It is denoted by Greek
letter σ (read as sigma). It is based on the mean, which gives a uniform result.
336 Infinity Optional Mathematics Book - 10
Standard deviation is the best measure of dispersion because
(i) Its value is based on all the observations.
(ii) The deviation of each item is taken from mean.
(iii) All algebraic signs are also considered.
A small standard deviation means a high degree of uniformity of the observation as
well as homogeneity of the standard deviation.
Standard deviation of continuous series
In a continuous (grouped) series we can find the standard deviation by four methods.
They are:
(i) Actual mean method
Let, m1, m2, m3, .... mn be the mid value of the corresponding classes x1, x2, x3,
.... xn and f1, f2, f3, ...... fn be their respective frequencies X is the actual mean.
Standard deviation(S.D. or σ) = Σf(m – X)2 = Σfd2
N N
where, m = mid value and d = m – X
(ii) Direct method
Standard deviation (S.D. or σ) = Σfm2 – Σfm 2
N N
where m = mid value
(iii) Assumed mean method (short cut method)
Standard deviation (S.D or σ) = Σfd2 – Σfd 2
N N
where d = m – A, A = Assumed mean.
(iv) Step-deviation method: This method is used to find the standard deviation
when the class interval is very large.
Standard deviation (S.D. or σ) = Σfd'2 – Σfd' 2
N N
×h
where, d' = m – A, A = Assumed mean
h
m = mid value, h = class size
Coefficient of standard deviation
The ratio of standard deviation to the arithmetic mean is known as the coefficient
of standard deviation. If X is the arithmetic mean and σ, the standard deviation,
then
Coefficient of standard deviation = σ
X
Infinity Optional Mathematics Book - 10 337
Variance
The square of the standard deviation is known as variance. By definition, σ2 is the
variance.
Coefficient of variation
The coefficient of dispersion based on standard deviation multiplied by 100 is known
as the coefficient of variance (C.V.) then
C.V. = σ × 100%
X
A distribution with lower value of C.V. is said to be more homogenous, more stable,
more uniform, more consistent, more equitable, less variable, less scattered and
vice versa.
Merits and demerits of standard deviation
Merits
(i) It is based on all observations.
(ii) Sign is not artificially made.
(iii) Deviations are taken from central value.
(iv) It is less affected by fluctuations of sampling.
Demerits
(i) It is difficult to calculate.
(ii) It gives greater weights to extreme value.
(iii) It can't be calculated for open end classes.
WORKED OUT EXAMPLES
1. In a continuous series, if Σfm2 = 720, Σfm = 72 and N = 10, find the
standard deviation.
Solution: Here,
By formula, Σfm2 – Σfm 2 720 – 72 2
N N 10 10
S.D (σ) = = = 72 – 51.84 = 4.48
∴ Standard deviation (σ) = 4.48.
2. Find the standard deviation by
(i) Actual mean method
(ii) Direct method
338 Infinity Optional Mathematics Book - 10
(iii) Assumed mean method
(iv) Step deviation method, for the following data.
Weights (in kg) 0 – 10 10 – 20 30 – 40 40 – 50 20 – 30
No. of students 5 461 4
Solution: Here,
Tabulating the given data in ascending order for the calculation of S.D.
(a) By actual mean method
Weights No. of Mid value fm d = m – X d2 fd2
(in kg) students (f) (m)
0 – 10 5 5 25 –17 289 1445
10 – 20 4 15 60 –7 49 196
20 – 30 4 25 100 3 9 36
30 – 40 6 35 210 13 169 1014
40 – 50 1 45 45 23 529 526
Σf = N = 20 Σfm= 440 Σfd2=3220
Now, Σfm 440
N 20
Mean (X) = = = 22
Again, Standard deviation (σ) = Σfd2 = 3220 = 12.68
N 20
(b) By direct method
Weight No. of students Mid value fm fm2
(in kg) (f) (m)
5
0 – 10 5 15 25 125
25 60 900
10 – 20 4 35 100 2500
45 210 7350
20 – 30 4 45 2025
Σfm = 440 Σfm2 = 12,900
30 – 40 6
40 – 50 1
Σf = N = 20
Standard deviation (σ) = Σfm2 – Σfm 2
N N
12,900 – 400 2
20 20
= = 645 – 484
= 161 = 12.68
Infinity Optional Mathematics Book - 10 339
(c) By assumed mean method :
Weights No. of Mid value m – A = d fd fd2 = fd × d
(in kg) students (f) (m)
–100 2000
0 – 10 5 5 –20 – 40 400
10 – 20 4 15 – 10 0 0
60 600
20 – 30 4 25 0 20 400
Σfd = –60 Σfd2=3400
30 – 40 6 35 10
40 – 50 1 45 20
Σf = N = 20
Let, assumed mean (A) = 25
By formula, standard deviation (σ) = Σfd2 – Σfd 2
N N
= 3400 – –60 2
20 20
= 170 – 9 = 161 = 12.68
(d) By step deviation method
Weight No. of mid value d = m–A d' = d fd' d'2 fd'2
(in kg) students (f) (m) h
0 – 10 5 5 –20 –2 –10 4 20
10 – 20 4 15 –10 –1 –4 1 4
20 – 30 4 25 0 0 0 0 0
30 – 40 6 35 10 1 6 16
40 – 50 1 45 20 2 2 44
Σf = N = 20 Σfd' = –6 Σfd'2 =
34
Let, assumed mean (A) = 25 and h = 10
By formula, standard deviation (σ) = Σfd'2 – Σfd' 2
N N
×h
= 34 – –6 2
20 20
× 10
= 1.7 – 0.09 × 10
= 12.68
340 Infinity Optional Mathematics Book - 10
3. From the following data, compute the following
(i) Standarad deviation (ii) Coefficient of S.D
(iii) Variance (iv) Coefficient of variance
Marks obtained 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60
No. of students 8 12 20 40 12 8
Solution: Here,
Tabulating the given data in ascending order, we get
Marks No. of Mid m–A=d fd fd2 = fd × d
obtained students (f) value (m)
–30 –240 7200
0 – 10 8 5 –20 –240 4800
10 – 20 12 15 –10 –200 2000
20 – 30 20 25
30 – 40 40 35 0 0 0
40 – 50 12 45 10 120 1200
8 55 20 160 3200
Σf = N = 100 Σfd = –400 Σfd2 = 18400
Let, assumed mean (A) = 35
(i) Standard deviation (σ) = Σfd2 – Σfd 2
N N
= 18400 – –400 2
100 100
= 184 – 16
=12.96
Again, Mean (X) = A + Σfd = 35 + –400 = 31
N 100
(ii) Coefficient of S.D. = S.D = 12.96 = 0.418
X 31
(iii) Variance = (σ)2 = (12.96)2 = 167.96 = 168
(iv) Coefficient of variation (C.V) = S.D × 100% = 0.418 × 100% = 41.8%
X
Infinity Optional Mathematics Book - 10 341
Exercise 8.3
Section 'A'
1. (a) Define the following terms
(i) Standard deviation (ii) Coefficient of S.D.
(iii) Coefficient of variation (iv) Variance
(b) Write the merits and demerits of standard deviation.
2. (a) Find the difference between mean deviation and standard deviation.
(b) What are the differences between coefficient of mean deviation and
coefficient of standard deviation?
Section 'B'
3. (a) In a continuous series, Σfm2 = 800, X = 12, N = 40, find the value of
standard deviation.
(b) In a continuous series, Σfm2 = 2230, Σfm = 470, X = 20 and N = 200, find
the standard deviation and is coefficient.
4. (a) In a continuous series, Σfd2 = 168, Σfd = –30, and N = 15 find the standard
deviation.
(b) In a continuous series, Σfd2 = 600, Σfd = 150, N = 50, find the standard
deviation and variance.
(c) In a continuous series, Σfd' = 43, Σfd'2 = 157, N = 89, h = 10 find the
standard deviation.
Section 'C'
5. Find the standard deviation and coefficient of standard deviation
from the following data.
(a) Marks obtained 20–30 30–40 40–50 50–60 60–70 70–80 80–90
No. of students 4 6 10 17 11 9 3
(b) Marks obtained 0 –10 10–20 20–30 30–40 40–50 50–60
No. of students 4 6 10 20 6 4
(c) Daily wages (in Rs.) 100–125 125–150 150–175 175–200 200–225
No. of workers 75 57 81 19 12
(d) Marks obtained 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60
No. of students 12 20 40 12 8 8
342 Infinity Optional Mathematics Book - 10
6. Find the standard deviation and coefficient of variation from the
following data.
(a) C.I. 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70
f 17 20 14 15 14 10
(b) Income (in Rs.) 300–400 400–500 500–600 600–700 700–800
No. of persons
8 12 20 6 4
(c) Profit (in Rs.) 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50
No. of shops 8
13 16 8 5
7. Find the standard deviation (S.D), coefficient of S.D and coefficient
of variation from the following data.
(a) Class 0<x<10 10<x<20 20<x<30 30 <x<40 40<x<50
f 12 33 30 15 10
(b) Marks less than less than less than less than less than less than
10 20 30 40 50 60
Frequency 4 10 20 40 46 50
8. (a) A sample of 60 cars of two marks P and Q is taken and their average
running life in years is recorded as follows.
Life (in years) 0–2 2–4 4–6 6–8 8 – 10
22 14 4
Mark P 8 12 19 12 5
Mark Q 10 14
(i) Find the mean life of each mark
(ii) Which mark shows greater consistency in performance and why?
(b) A purchasing agent obtained samples of incandescent lamps from two
suppliers. He had the samples tested in his own laboratory for length of
life with the following results.
Length of life (hrs) 700 –900 900 - 1100 1100 - 1300 1300 - 1500
8
Lamps Suppliers A 10 16 26 3
Suppliers B 3 42 12
(i) Which suppliers lamps gives a higher average life?
(ii) Which suppliers lamps are more uniform?
Infinity Optional Mathematics Book - 10 343
ANSWER SHEET
Exercise 1.1
1. (a), (b), (e), (f) are functions.
2. (a) one to one onto (b) one to one into (c) one to one into
(d) many to one onto (e) one to one into (f) many to one into
3. (b), (d)
5. (a) f-1 = {(1, 1), (4, 2), (9, 3), (16, 4) (b) g-1 = {(3, 1), (3, 2), (5, 3), (6, 4), (7, 5)}
(c) h-1 = {(3, a), (5, b), (7, c), (8, d), (9, e), (2, f)}
(d) f-1 = {(4, –4), (5, –5), (6, –6), (7, –7)}
(=x){(=–x25–, 53, (–45,
(e) g-1 –3), (– –4), –5)}
6. (a) f-1 3
(b) g-1 (x) = x + 5
x + 2 5 – x
(c) h-1(x) = 5 (d) t-1(x) = 7
(e) s-1(x) = 4x + 3 (f) f-1(x) = x
5 3
(g) g-1 (x) = 4x – 1 (h) h-1(x) = –4x – 7
3 5
(i) t-1(x) = 5 – 4x (j) f-1(x) = 5x – 3
x 2x
(k) f-1(x) = 2 6x (l) h-1(x) = 2x + 1
– 5x 3–x
(m) f-1(x) = x+2 (n) g-1 (x) = 3 – 4x
2x – 5 3x – 4
(o) h-1 (x) = –(xx–+11) (p) k-1(x) = {(x, y): y = 1 x ∈ R, y ∈R}
x
(q) t-1(x) = {(x, y): y = x, x ∈R} (r) f-1(x) = {x, 3x – 5): x ∈ R}
(s) h-1(x) = x, x + 3 :x∈R (t) g-1(x) = x, x + 1 :x∈R
5 3 – x
7
7. (a) f-1 (x) = x + 5 (b) f-1(2) = 3
3
(c) f-1 (–3) = 32 2x + 5
(d) f-1(2x) = 3
(e) f-1 3 = 13
2 6
8. (a) f-1(x) = 4x2–3 (b) f-1(3x) = 12x – 3
2
3 = –130 4x + 1
(c) f-1 5 (d) f-1(x + 1) = 2
(e) f-1(2x – 5) = 8x – 23
2
9. k = –1 10. x = 7
2
11. (a) f-1(x) = x + 1 (b) f-1(x) = x + 4
2 x – 1
344 Infinity Optional Mathematics Book - 10
(c) f-1(x) = 7x – 1x4 12. f(x) = x – 5 , f(3) = –2
3 – 3 3
Exercise 1.2
1. (a) {(1, 9), (2, 16), (3, 25)} (b) {(a, f), {b, e}, (c, d)}
(c) {(–2, 0), (–1, 2), (0, 2), (1, 0), (2, –1)}
(d) {(2,3), (2, 1), (3,3)}
2. (a) B = {1, 4, 9}, C = {2, 8, 18}, gof = {(1, 2), (2, 8), (3, 18)}
(b) B = {8, 10, 16}, C = {15, 19, 31}, {(4, 15), (5, 19), (6, 31)}
(c) B = {(1, 2, 3}, C = (3, 4, 5}
{(1, 3), (8, 4), (27, 5)}
(d) B = {2, –5, 0}, C = {5, 26, 1}
{(–1, 5), (–8, 26), (–3, 1)}
3. (a) {(1, 3), (3, 1), (4, 3)} (b) {(1, 4), (2, 5), (3, 6)}
(c) {(–2, 5), (–1, 2), (0, 1), (1, 2), (2, 5)}
4. (a) g = {(3, 5), (0, 2) (–3, –1)}
(b) gof (1) = 3, gof(3) = 5, fog (2) = 4, fog(3) = 2, fog(4) = 3
(c) goh(1) = 3, goh(8) = 4
5. (a) fog(x) = 21x, gof(x) = 21x (b) fog(x) = x – 2, gof(x) = x – 2
(c) gof (x) = 10x – 5, fog(x) = 10x – 2 (d) gof(x) = –7x – 3, fog(x) = 19 – 7x
(e) fog(x) = 15x – 16 , gof(x) = 5x – 4
3
(f) fog(x) = x, gof(x) = x (g) hog(x) = x, 5 + 4x , goh(x) = x
4
(h) fog(x) = 3x2 + 30x + 76, gof(x) = 3(x2 + 2)
(i) fog(x) = x2(x2 + 2), gof(x) = x4 – 2x2 + 2
(j) fog(x) = 2x + 1145, gof = 16x + 1
2x + 2x
6. (a) 6x – 1, 11, 6x + 3, – 15 (b) 6 – 4x, 18, 1 – 4x, – 5
(c) 53, 535 (d) 21, 4181 (e) 728, 49
7. (a) x + 9, 4 (b) 9x + 1, –44
3 7 7
(c) 4 7– x, 4 – x, 2, 53 (d) – x+8 , 10 – x, –511, 12
3 5 5 5
(e) x +5 2, x, x, 3 (f) x, x, 3, –3
8. (a) (i) 1 – 6x, (ii) 7 – 6x, (iii) 3 – 6x, (iv) 25 (b) (i) –6x – 9, (ii) –6x – 9, (iii) –8 – 6x, (iv) –10
9. (c) m = 2 (d) x = 9, 1 (e) x = 0, 5 3x + 2
(h) 1
(f) 6 (g) –171 (k) x + 3 (l)
(i) 5x3– 7,3x5+ 7 (j) 8 7–x7x,–1231
Infinity Optional Mathematics Book - 10 345
Exercise 1.4
1. (a) 3x4 + 4x3 – 9x2 – 17x – 14 (b) 10x4 – 25x3 – 6x2 + 29x – 35
(c) 7x7 + 21x6 + 14x5 – 2x3 – 6x2 – 4x (d) 10x6 + 15x5 – 6x3 + 5x2 + 3x – 27
2. (a) Q(x) = 3x2 – 4x – 8 and R = – 21 (b) Q(x) = 5x4 + 10x3 + 17x2 + 34x + 73 and R = 150
(c) Q(x) = 2x2 – 27x + 14,R = – 1 (d) Q(x) = x5 + 3x4 + 9x3 + 27x2 + 81x + 243, R = 721
4
3. Q(x) = x2 – 5 x – 292, R = 40
3 9
Exercise 1.5
1. 11
3. (a) 6 (b) 21 (c) – 4
(d) 209 (e) 637 (f) 11(1 – 2)
4. (a) 2272 32
(d) 14 (c) –29
(b) 8 (f) 172
6. (a) No (c) No
(e) 4
(b) Yes
(d) Yes (e) No
7. (a) 1 (b) –75 (c) ± 7
(d) 73 (e) 53 (f) 2–47
(g) –6, 1
10. (a) 10, 5x2 + 8x + 3 (b) –6, 3x2 + x – 2 (c) –32, 2x2 – 6x + 13
(d) 0, x2 + 2x – 1 (e) 47, x3 + 3x2 + 6x + 18 (f) –5, x5 + x4 + x3 + x2 + x + 1
0, 2y2 – 2y + 1
(g) 145, 2x2 – 5 x + 5 (h)
2 4 (b) Q(x) = x2 – 2x – 1, R = – 7
(d) Q(x) = 2x2 – 3x + 3, R = –14
(i) –481, 3 p2 + 34P – 285
2
11. (a) Q(x) = 2x2 – 9x + 10, R = 0
(c) Q(x) = 32x2 + 43x – 187, R = – 18
12. 3
14. (a) –6, –21 (b) 7, 4 (c) 3, 1
(d) 3, – 3 (e) 2, –2 (f) 2, 5
(g) –5, 3
15. (a) (x + 1) (x + 2) (x + 3) (b) (x – 1) (x + 2) (2x + 1) (c) (x + 2) (2x + 3) (2x – 1)
(d) (z + 1) (z2 – z – 4) (e) (x + 1)2 (2x – 1) (f) (y + 2) (y + 3) (y – 5)
(g) (p + 2) (p + 6) (2p – 3)
16. (a) 1, – 2, – 2141 (b) –2, – 23, 12 (c) –1, –2, –3
(d) 1, – 12, – (e) 2, –1, –1
(f) 12, –1, – 2
3
346 Infinity Optional Mathematics Book - 10
(g) 4, –3, –1 (h) 1, 1, 2 (i) –2, –6, 3
2
(j) –3, – 21, 5
3
Exercise 1.6
2. (a) (i) d = 4 (ii) tn = 4n – 2 (iii) 14, 18, 22 (b) (i) d = –3 (ii) tn = –(3n + 2) (iii) –14, –17, – 20
(c) (i) d = –20 (ii) tn = 10(11 – 2n) (iii) 30, 10, –10 (iv) 131, 133, 5
(d) (i) d = –4 (ii) tn = 9 – 4n (iii) –7, –11, –15
3. (a) 7, 11, 15, 19 (b) –50, –40, –30, –20
(c) 125, 110, 95, 80
(d) 5,243, 246, 29
4
4. (a) 21, 61 (b) 13, 63 (c) 33, 103
(d) 68, –12 (e) 100, 150
5. (a) 0 (b) 312 6. (a) 4 (b) –10
7. (a) 8 (b) 20 (c) 11 (d) 8
8. (a) 12 (b) 44 9. (a) Yes (n = 34) (b) No
10. (a) a = 2, d = 2 (b) a = –3, d = –4 11. (a) 45 (b) –86
12. (a) 17 (b) No (c) 73 (d) No (e) 10
13. (a) 21
14. – 34, 41, 5
4
Exercise 1.7
1. (a) 10 (b) 10 (c) 4x – y (d) a2 + b2
2. (a) 10 (b) 40 (c) –1231 (d) 2507
3. (a) 7 (b) 22 (c) 8, 14 (d) 0, 10
(f) 2, 32
(e) 15 (g) 1, –6 (h) 7, 1
(i) –35 (b) 3, –1, –5, –9 7
4. (a) 7, 11, 15
(c) 1, 53, 37, 3, 11
3
5. (a) 5, 8, 11 (b) –2, 3, 8, 13 (c) 0, 5, 10
(b) a = 3, b = 24
(d) 233 , 25
3
6. (a) x = 2, y = 14 (c) p = –2, q = 33
7. (a) b = 35 and other means are 10, 15, 20, 25, 30
(b) k = –430, other means are 20 , 10 , 0, –310, –230, –30
3 3 6
8. (a) 5 (b) 5
9. (a) 28 (b) 8 (c) 8 (d) 48
Infinity Optional Mathematics Book - 10 347
Exercise 1.8
1. (a) 465 (b) 275 (c) 270 (d) 366
(e) –960
2. (a) 50 (b) 225 (c) 228
(c) 2950
3. (a) 425 (b) –500 5. (a) 2
4. (a) 3 (b) 10 (b) 147
6. (a) 4 (b) –4
7. (a) 12 (b) 5 (c) n = 30, d = 3 (d) n = 10, d = 3
(b) n = 10 or, 11, since t11 = 0
8. (a) n = 12, or 9, since t10 + t11 + t12 = 0
9. (a) 420 (b) 215
10. (a) 630 (b) 675 11. (a) 110 (b) 465
(c) 2, 8, 14 (d) 3, 4, 5
12. (a) 11, 18, 25 (b) 4, 7, 10
(e) 3, 5, 7 (f) 3, 7, 11
13. (a) 4392 (b) 24552 (c) 728 (d) 32850
(e) 1275 (f) 60702 (g) 2275
14. (a) 675 (b) 3800 (c) –1, 7, 15
(d) –8, –4 + 0 + 4 + 8 + 12 + ..... (e) 107 1
2
15. (a) 2 1 years (b) Rs. 200
2
Exercise 1.9
2. (a) r = 3, tn = 2 × 3n – 1, 54,162,486 7. (a) n = 9
(b) n = 8
(b) r = 3, tn = – 3n – 4; –1, – 3, – 9 (c) n = 9
(d) n = 11
(c) r = – 2, tn = – 1 (– 2)n–1, 2 2, – 4, 4 2 (e) n = 8
(d) r= 1 , tn = 5 × 26 – 2n, 5 , 5 , 5 8. (a) n = 11
4 4 16 64 (b) n = 7
(e) –3 (c) n = 9
r = – 2, tn = 2 (– 2)n – 1, 12, – 24, 48
9. (a) Yes
3. (a) 2, 6, 18, 54
(b) 2, 2, 2 2, 4
(c) 91, 31, –1, 3
(d) 16, 8, 4, 2
(e) 8 , 49, 23, 1 (b) Yes
27 (c) Yes
(d) No
4. (a) 21, – 4
10. (a) 64
(b) 94, 32
243
348 Infinity Optional Mathematics Book - 10
(c) 135, 3645 (b) 1
11. (a) 1024
5. (a) a = 2
(b) 81
(b) a = 1 (c) 128 2
(c) a = 27 12. (a) 8
(b) 9
6. (a) r = 2 (c) Yes (n = 2)
(b) r = 2 13. (a) 2 + 4 + 8 + 16 + 32 + 64 + 128
3 (b) 1
(c) r = – 2 729
(c) 59049
(d) r = 3
2
Exercise 1.10 2. (a) 8 3. (a) 4
(b) 1 (b) 7
1. (a) 8 6
(b) 30 (c) 6, 24
(d)
(c) 3 (c) 1 –13
16
(d) 2 (e) –3, 3
5 (d) 25 2
46 6. (a)
(e) 9 (b) 243, 1
4 5. (a) 1, 1, 3 3
(b) –32, –2 2, – 4, – 4 2 1,8
4. (a) 4, 8, 16 64
(c) 81, 27, 9, 3, 1
(b) –2, 2 2, –4, 4 2
(c) 2, 1, 3, 9, 27
3 24 8
(d) 1, 1, 4
4
7. (a) 12, 36, 108, 324, k = 972
(b) x = –16 & means = –2, – 2 2 , – 4, – 4 2 , – 8, – 8 2
8. (a) 5 9. (a) 4
(b) 5 (b) 7
(c) 8 (c) 11
Exercise 1.11 (b) 32767 (c) 340 (d) 25534
8 (g) 1023
1. (a) 29, 524
(f) 121 1
(e) 7 2 ( 2 + 1) 3
Infinity Optional Mathematics Book - 10 349
2. (a) 30 (b) 549 (c) 1– 1
(b) – 546 215
3. (a) 242
4. (a) – 3 (b) 1 (c) 7 (d) 1
2 (c) 2, – 3 (d) 4
5. (a) – 1
3 (b) 6 (d) 2
6. (a) 6 (b) 6 (c) 6
7. (a) 1 (b) 384
8. (a) 765
(b) 1023
9. (a) 254 32
10. (a) 2 (b) 1023
512
(b) 2
11. (a) 1, 2, 4 (b) 5, 10, 20 (c) 2, 8, 32
12. (a) r = 2, a = 1 (b) 2 + 6 + 18 + … (c) 2
4
(e) 49152, 98301
Exercise 1.12 (b) 16, 4 (c) 45, 5
1. (a) 64, 4 4. 2, 7, 12 or 26, 7, –12 5. 3, 6, 12 or 12, 6, 3
3. 3, 7, 11 or 12, 7, 2
6. 25 7. 9:1 or 1:9 8. 12, 108 9. 2, 1
10. 16, 4 2
Exercise 1.14 11. 32, 44, 56, 68
1. (a) Max. 12 and min. 0 (b) Max. 24 and Min. 9 (c) Max. 30 and min 17
(d) Max. 24 and min 12 (c) Max. 52, Min. 0
2. (a) Max. 26, Min. –4 (b) Max. 50, Min. 10
(d) Max. 14, Min 11. (e) Max. 6, Min. 2
3. (a) 3x + 5y ≤ 20, x – y ≤ 4, x ≥ 0, y ≥ 0, Max. 48, Min. 0.
(b) 2x – y ≤ 6, 2x + 3y ≤ 6, y ≤ 2, Max. 34, Min. 12.
Exercise 1.15
5. (a) –3, –1 (b) 5, –1 (c) –3, –4 (d) 3, 3
(e) –2, –2 (f) 3 , –1
6. (a) (–2, 4), (3,9) (c) (–2, 5), – 1, 13 (d) (–3, –5), (1 , –5)
7. (a) y = x2 2 44
(b) (0, 1), (3, 10)
(c) y = x2 + 4x + 4 (d) y = x2 + 6x – 8
(b) y = x2 – 2
350 Infinity Optional Mathematics Book - 10
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Shubharambha OPT Mathematics 10 Final for CTP 2077
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