1.Table

Time ( in hours ) ( x)	4	8	12	24
Amount in Rs ( y )	60	120	180	360

2.Variation :
Direct Variation
3. Equation
y = 5x 
\(k=\frac{y}{x}=\frac{60}{4}=\frac{120}{8}=\ldots \ldots \ldots=15\) 
4.Points :
(4,60),(8,120),(12,180),(24,320)
5.Solution :
(i). If the parking time is 6 hours , then the parking charge = Rs. 90
(ii). If the amount Rs.150 is paid , then the Parking time = 10 hours

1. Table

x	24	12	8	6	4	3	2	1
y	1	2	3	4	6	8	12	24

2. Variation:
Indirect Variation
3. Equation
xy = k
xy = 24 x 1 = 12 x 2 = 8 x 3 =... = 12
xy = 24
4. Points
(24,1),(12,2),(8,3),(6,4)
(4,6),(3,8),(2,12),(1,24)
5. Solution
From the graph
(i) If x = 3, then, y = 8
(ii) If y = 6 then, x = 4

Construction
Step 1 : Draw a line segment BC = 8 cm.
Step 2 : At B, draw BE such that \(\angle\)CBE = 60^o.
Step 3 : At B, draw BF such that \(\angle\)EBF = 90^o .
Step 4 : Draw the perpendicular bisector to BC, which intersects BF at O and BC at G.
Step 5 : With O as centre and OB as radius draw a circle.
Step 6 : From B, mark an arc of 6 cm on BC at D.
Step 7 : The perpendicular bisector intersects the circle at I. Joint ID.
Step 8 : ID produced meets the circle at A. Now join AB and AC. Then \(\triangle\)ABC is the required triangle.

Given a triangle ABC, we are required to construct another triangle whose sides are \(\frac { 6 }{ 5 } \) of the corresponding sides of the ABC

Steps of construction:
1. Constructed a  ABC with any measurement
2. Drawn a ray BX making an acute angle with BC on the side opposite to the vertex A.
3. Joined B₅ to C and drawn a line through B₆ parallel to B₅C intersecting the extended line segment BC at C.
4. Drawn a line through C' parallel to CA intersecting the extended BA at A'.
5. Then A'BC' is the required triangle each of whose sides is six fifths of the corresponding sides of ABC.

Mean \(\bar { x } \)  = \(\frac { 16+13+17+21+18 }{ 5 } =\frac { 85 }{ 5 } \)  = 17

x	d = x - 17	d2
16	-1	1
13	-4	16
17	0	0
21	4	16
18	1	1
	Σd = 0	Σd2 = 34

σ =\(\sqrt { \frac { \Sigma d^{ 2 } }{ n } } =\sqrt { \frac { 34 }{ 5 } } =\sqrt { 638 } \)
σ = 2.61
Co-efficient of variation
CV = \(\frac { \sigma }{ \bar { x } } \) x 100 = \(\frac { 2.61 }{ 17 } \) x 100
= 15.35%

A coin is tossed thrice by the sample space
S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
n(S) = 8
Let A be the event of getting exactly two heads
A = { HHT, HTH, THH}
n(A) = 3
\(P(A)=\frac{n(A)}{n(S)}=\frac{3}{8}\)
Let B be the event of getting atleast one tail.
B = {HHT, HTH, HTT, THH, THT, TTH, TTT}
n(B) = 7
\(\mathrm{P}(\mathrm{B})=\frac{n(\mathrm{~B})}{n(\mathrm{~S})}=\frac{7}{8}\)
Let C be the event of getting consecutively two heads
C = {HHT, THH, HHH}
n(C) = 3
\(\mathrm{P}(\mathrm{C})=\frac{n(\mathrm{C})}{n(\mathrm{~S})}=\frac{3}{8}\)
A n B  = {HHT, HTH, THH}
n(A n B) = 3
\(\mathrm{P}(A \cap B) =\frac{n(\mathrm{~A} \cap \mathrm{B})}{n(\mathrm{~S})}=\frac{3}{8} \)
\(B \cap C =\{\mathrm{HHT}, \mathrm{THH}\} \)
\(n(\mathrm{~A} \cap \mathrm{C}) =2 \)
\(P(\mathrm{~B} \cap \mathrm{C}) =\frac{n(\mathrm{~B} \cap \mathrm{C})}{n(\mathrm{~S})}=\frac{2}{8} \)
\(A \cap C =\{\mathrm{HHT}, \mathrm{THH}\}\)
\(n(\mathrm{~A} \cap \mathrm{C}) =2 \)
\(P(B \cap C) =\frac{n(\mathrm{~B} \cap \mathrm{C})}{n(\mathrm{~S})}=\frac{2}{8} \)
\(A \cap C =\{\mathrm{HHT}, \mathrm{THH}\} \)
\(n(\mathrm{~A} \cap \mathrm{C}) =2 \)
\(P(\mathrm{~A} \cap \mathrm{C}) =\frac{n(\mathrm{~A} \cap \mathrm{C})}{n(\mathrm{~S})}=\frac{2}{8} \)
\(A \cap B \cap C =\{\mathrm{HHT}, \mathrm{THH}\} \)
\(n(\mathrm{~A} \cap \mathrm{B} \cap \mathrm{C})=2 \)
\(P(\mathrm{~A} \cap \mathrm{B} \cap \mathrm{C})=\frac{n(\mathrm{~A} \cap \mathrm{B} \cap \mathrm{C})}{n(\mathrm{~S})}=\frac{2}{8} \)
\(P(\mathrm{~A} \cup \mathrm{B} \cup \mathrm{C})=P(\mathrm{~A})+\mathrm{P}(\mathrm{B})+\mathrm{P}(\mathrm{C}) \) \(\ -P(A \cap B)-P(B \cap C) \)
\(P(A \cup B \cup C)= \frac{3}{8}+\frac{7}{8}+\frac{3}{8}-\frac{3}{8}-\frac{2}{8}-\frac{2}{8}+\frac{2}{8} \)
\(= \frac{3+7+3-3-2-2+2}{8} \)
\(= \frac{8}{8}=1 \)
The required probability is 1.

When two dice are rolled together, there will be 6 x 6 = 36 outcomes. Let S be the sample space. Then n(S) = 36.
Let A be the event of getting a doublet and B be the event of getting face sum 4.
Then A = {(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)}
B = {(1,3),(2,2),(3,1)
Therefore, A∩B = {(2,2)}
Then, n(A) = 6, n(B) = 3, n(A∩B) = 1
P(A) = \(\frac { n(A) }{ n(S) } =\frac { 6 }{ 36 } \)
P(B) = \(\frac { n(B) }{ n(S) } =\frac { 3 }{ 36 } \)
P(A∩B) = \(\frac { n(A\cap B) }{ n(S) } =\frac { 1 }{ 36 } \)
Therefore, P (getting a doublet or a total of 4) = P(AUB)
P(AUB) = P(A) + P(B) - P(A∩B)
=\(\frac { 6 }{ 36 } +\frac { 3 }{ 36 } -\frac { 1 }{ 36 } =\frac { 8 }{ 36 } =\frac { 2 }{ 9 } \)
Hence, the required probability is \(\frac { 2 }{ 9 } \).

When two unbiased dice are rolled, the Sample Space
s = {(1, 1) (1, 2) (r,3) (1,4) (1,5) (1,6)
(2, 1) (2,2) (2, 3) (2, 4) (2,5) (6, 6)
(3, 1) (3,2) (3, 3) (3, 4) (3, 5) (3, 6)
(4, 1) (4,2) (4,3) (4,4) (4, 5) (4,6)
(5, 1) (5,2) (5,3) (5,4) (5,5) (6,6)
(6, 1) (6,2) (6, 3) (6,4) (6, 5) (6, 6)}
n(S) = 36
(i) Let A be the event of getting a doublet
A = {( 1, 1) (2,2) (3,3) (4, 4) (5, 5) (6, 6)}
n(A) = 6
\(\mathrm{P}(\mathrm{A})=\frac{n(A)}{n(S)}=\frac{6}{36}=\frac{1}{6}\)
(ii) Let B be the event of getting the product as a prime number.
B = {(1,2) (1,3) (1, 5) (2,1) (3, 1) (5, 1)}
n(B) = 6
\(P(B)=\frac{6}{36}=\frac{1}{6}\)
(iii) Let C be the event of getting the sum as a prime number.
c = {(1, 1) ( 1, 2) (1, 4) ( 1, 6) (2, 1) (2, 3) (2, 5) (3,2) (3, 4) (4, 1) (4,3) (5,2) (5,6) (6, 1) (6,5)}
n(C) = 15
\(\mathrm{P}(\mathrm{C})=\frac{n(C)}{n(S)}=\frac{15}{36}=\frac{5}{12}\)
(iv) Let D be the event of getting the sum as 1. Since it is an impossible event.
n(D) = 0 and P(D) = g

The mean of marks is 35.9 which is not an integer. Hence we take assumed mean, A=35,n=10.

xi	di = xi - A di = x - 35	di2
25	-10	100
29	-6	36
30	-5	25
33	-2	4
35	0	0
37	2	4
38	3	9
40	5	25
44	9	81
48	13	169
	\({ \Sigma d }_{ i }\) = 9	\(\frac { { \Sigma d }_{ i }^{ 2 } }{ n } \) = 453

Standard deviation
σ =\(\sqrt { \frac { { \Sigma d }_{ i }^{ 2 } }{ n } -\left( \frac { { \Sigma d }_{ i } }{ n } \right) ^{ 2 } } \)
= \(\sqrt { \frac { 453 }{ 10 } -\left( \frac { 9 }{ 10 } \right) ^{ 2 } } \)
= \(\sqrt { 45.3-0.81 } \)
= \(\sqrt { 44.49 } \)
σ ≃ 6.67
(ii) Step deviation method
Let x₁, x₂, x_n,... be the given data. Let A be the assumed mean.
Let c be the common divisor of x_i-A
Let d_i = \(\frac { x_{ i }-A }{ c } \)
Then x_i = d_ic + A             ....(1)
Σx_i = Σ(d_ic + A) = cΣd_i + A x n
\(\frac { \Sigma { x }_{ i } }{ n } =c\frac { \Sigma d_{ i } }{ n } +A\)
\(\bar { x } \) = c \(\bar { d } \)+A            ..(2)
\({ x }_{ i }-\bar { x } =c{ d }_{ i }+A-c\bar { d } -A=c(\bar { d_{ i } } -\bar { d } )\)(using (1) and (2))
σ = \(\sqrt { \frac { \Sigma ({ x }_{ i }-\bar { x } )^{ 2 } }{ n } } =\sqrt { \frac { \Sigma (c({ d }_{ i }-\bar { d } )^{ 2 } }{ n } } =\sqrt { \frac { { c }^{ 2 }\Sigma ({ d }_{ i }-\bar { d } )^{ 2 } }{ n } } \)
σ = \(c\times \sqrt { \frac { { \Sigma d }_{ i }^{ 2 } }{ n } -\left( \frac { { \Sigma d }_{ i } }{ n } \right) ^{ 2 } } \\ \)

P (will rain) + P (will not rain) = 1
P (will not rain) = 1 - P (will rain)
= 1 - 0.85 = 0.15
Probability that will not rain tomorrow =.0.15.

\( \mathrm{P}(A \cup B) =\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(A \cap B) \)
\(\frac{1}{3} =\frac{2}{3}+\frac{2}{5}-\mathrm{P}(A \cap B) \)
\(\mathrm{P}(A \cap B) =\frac{2}{3}+\frac{2}{5}-\frac{1}{3} \)
\(\mathrm{P}(A \cap B) =\frac{10+6-5}{15}=\frac{16-5}{15}=\frac{11}{15} \)

When a coin is tossed thrice, the outcome will be

The sample space s = {(HHH), (THH), (HTH), (HHT), (HTT), (THT), (TTH), (TTT)}
n(S) = 3
Let A be the event of getting two consecutive tails
4 = {HTT, TTH, TTT}
n(A) = 3
\(\Rightarrow P=\frac { n\left\{ F \right\} }{ n\{ O\} } =\frac { 3 }{ 8 } \)
Probability of getting two consecutive tails = \(\frac{3}{8}\)

When we toss three coins the outcome will be

The sample space = { HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
Total number of outcomes = 8

Standard deviation \(\sigma=6.5\)
Mean \(\bar{x}=12.5\)
Coefficient of variation C.V  \(=\frac{\sigma}{x} \times 100 \% \)
\(=\frac{6.5}{12.5} \times 100 \% \)
\(=\frac{65}{125} \times 100 \% \)
\(=\frac{13}{25} \times 100 \% \)
= 52 %
Co-efficient of variation is 52%

Standard deviation of first n natural numbers
\(=\sqrt{\frac{n^{2}-1}{12}}\)
SD of first 21 natural numbers
\( =\sqrt{\frac{21^{2}-1}{12}} \)
\(=\sqrt{\frac{441-1}{12}}=\sqrt{\frac{440}{12}} \)
\(=\sqrt{36.6666}=6.05\)
Standard deviation of first 21 natural numbers = 6.05