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