Let us make Ceasar Cipher table first. Here, we have to set to + 4 table. For that, we have to start letter e to set as A, f as B … likewise d as Z. Now, the + 4 Ceasar Cipher table looks like

The given plain text is
fvieo mr gshiw ger fi xvmgoc
To crack this secret code, follow the steps given below.
Step 1: Using Ceasar Cipher table, let us first match the most repeated letters. This will help us to progress faster.
fvieo mr gshiw ger fi xvmgoc

Step 2: Then, let us find remaining letters to complete the code.

Thus, the secret sentence is, BREAK IN CODES CAN BE TRICKY

Given, the train runs constantly at a speed of 80 km/hr.
i.e For 1 hour = 80 km
2 hours = 2 × 80 = 160 km
3 hours = 3 × 80 = 240 km

We can tabulate as above, Take a suitable scale
1) Mark the number of hours on the x-axis.
2) Mark the distance inKm on the y-axis.
3) Plot the points (1,80) (2,160) (3,240) (4,320) and (5,400).
4) Join the points and get a straight line3.
5) From the graph.
(i) Time taken to cover 240 km is 3 hrs.
(ii) Th e distance covered in 5 ½ hrs 440 km.

1. Take a suitable scale on both the axes Here, we take on the x axis
1cm = 5 litres on the, y axis
1cm = 100 rupees.
2. Mark number of litres of milk along the x -axis.
3. Mark the cost of milk along the y-axis.
4. Plot the points (5,150) (10,300) (15,450) (20,600).
5. Join the points.
This graph can help us to estimate few more things also. Suppose we like to find the cost of 25 litres of milk. Mark 25 on the x-axis, follow the line parallel to y-axis through 25 till we meet the drawn line at P. From P we take a horizontal line to meet the y-axis. This meeting point of y-axis is the required answer.
Thus, the cost of 25 litres of milk is Rs.750. This is the graph of linear equation in two quantities, and hence they are in direct variation.

The given equation y = 5 x means that for any value of x , y takes five
times of x value.
Plot the point (−3,−15) (−1,−5) (0,0) (2,10) (3,15)

Given:
BE = 7 cm, BA = 7.5 cm and ㄥBEA = 80⁰


Steps:
1. Draw a line segment BE = 7 cm.
2. Make an angle ㄥBEX =80° at E on \(\overset { \_ \_ }{ BE } \).
3. With B as centre, draw an arc of radius 7.5 cm cutting \(\overset { \_ \_ }{ EX } \)  at A and Join BA.
4. With B as centre, draw an arc of radius equal to the length of \(\overset { \_ \_ }{ AE } \).
5. With A as centre, draw an arc of radius 7 cm. Let both arcs cut at R.
6. Join BR and AR.
7. BEAR is the required parallelogram.
Calculation of area:
Area of the parallelogram BEAR = bh sq. units
= 7 x 4.1 = 28.7 sq. cm

Given:
CA = 7 cm, CF = 6 cm and AF = 10 cm


Steps:
1. Draw a line segment CA = 7 cm.
2. With C and A as centres, draw arcs of radii 7 cm and 6 cm respectively. Let them cut at F.
3. Join CF and AF.
4. With A and F as centres, draw arcs of radii 6 cm and 7 cm respectively. Let them cut at L.
5. Join AL and FL.
6. CALF is the required parallelogram.
Calculation of area:
Area of the parallelogram CALF = bh sq. units
= 7 x 5.9 = 41.3 sq. cm

Given:
BI = 6.5 cm, IR = 5 cm and ㄥBIR = 70⁰


Steps:
1. Draw a line segment BI = 6.5 cm.
2. Make an angle ㄥBIX = 70⁰ at I on \(\overset { \_ \_ }{ BI } \).
3. With I as centre, draw an arc of radius 5 cm cutting IX at R.
4. With B and R as centres, draw arcs of radii 5 cm and 6.5 cm respectively. Let them cut at D.
5. Join BD and RD.
6. BIRD is the required parallelogram.
Calculation of area:
Area of the parallelogram BIRD = bh sq. units
= 6.5 × 4.7 = 30.55 sq. cm

2x = 9 − 5
2x = 4
x = \(\frac { 4 }{ 2 } \)
x = 2

x – 7 = 6
x – 7 + 7 = 6 + 7
x = 13

Let the number be n.
When 7 is added to this number we get n + 7.
This result is to give 19.
Therefore, the equation is n + 7 = 19.

Bacteria at the end of 3 hours
a = p\(\left( 1+\frac { a }{ 100 } \right) \left( 1-\frac { b }{ 100 } \right) \left( 1+\frac { c }{ 100 } \right) \)('-' because ‘decrease’)
= 10000\(\left( 1+\frac { 5 }{ 100 } \right) \left( 1-\frac { 8 }{ 100 } \right) \left( 1+\frac { 10 }{ 100 } \right) \)
= 10000 x \(\frac { 105 }{ 100 } \times \frac { 92 }{ 100 } \times \frac { 110 }{ 100 } \)
A = Rs.10626

Let the population in 2016 be ‘P’.
Then, A = P\({ \left( 1+\frac { r }{ 100 } \right) }^{ n }\)
⇒ 238765 = p\({ \left( 1+\frac { 6 }{ 100 } \right) }^{ 2 }=P{ \left( \frac { 53 }{ 50 } \right) }^{ 2 }\)
⇒ P = 238765 x \(\frac { 50 }{ 53 } \times \frac { 50 }{ 53 } \)
∴ P = 212500
Let the population in 2020 be ‘A’
Then, A= P\({ \left( 1+\frac { r }{ 100 } \right) }^{ n }\)
∴ A = 238765\({ \left( 1+\frac { 6 }{ 100 } \right) }^{ 2 }\)
= 238765 x \(\frac { 53 }{ 50 } \times \frac { 53 }{ 50 } \)
= 95.506 x 53 x 53
A = 268276
∴ The population in the year 2016 is 212500 and that in the year 2020 is 268276.

Now, by altitude-on-hypotenuse theorem,
AB² = AD x AC gives,
10² = x × 26
\(\Rightarrow x=\frac { 100 }{ 26 } =\frac { 50 }{ 13 } units\quad and\)
BC² = CD × AC gives,
24² = y × 26
\(\Rightarrow y=\frac { 576 }{ 26 } =\frac { 288 }{ 13 } units\quad and\)
In Δ ABC, AB² + BC² = 10² + 24² = 676 = 26² = AC² Therefore, Δ ABC is a right angled triangle.

Here, the hypotenuse is 29 m. One side of the right triangle is 20 m. let the other side be ‘l’ m Therefore, by Pythagoras theorem,
l² = 29² − 20² = 841 − 400 = 441 = 21²
∴ l = 21m
Therefore, the area rectangular plot of land = l × b square units. = 20 × 21 = 210 m².

From Δ LMO, by Pythagoras theorem,
LM² = OL² − OM²
⇒ LM² = 13² −12² = 169 −144 = 25 = 5²
∴ LM = 5 units
From Δ NMO, by Pythagoras theorem,
MN² = ON² − OM²
= 15² −12² = 225 −144 = 81 = 9²
∴ MN = 9 units
Hence, LN = LM + MN = 5 + 9 = 14 units
Area of Δ LON = \(\frac { 1 }{ 2 } \) × base × height
= \(\frac { 1 }{ 2 } \) × LN × OM
= \(\frac { 1 }{ 2 } \) × 14 × 12
= 84 square units.

Take a = 5, b = 12 and c = 13
Now, a² + b² = 5² +12² = 25 +144 = 169 = 13² = c²
By the converse of Pythagoras theorem, the triangle with given measures is a right angled triangle.

Let the maximum marks be x.
Now, 80% of x = 576
\(\frac { 80 }{ 100 } \)\(\times\) x = 576
⇒ x = 576 x \(\frac { 100 }{ 10 } \)
x = 720 marks
Therefore, the total marks in the examination = 720.

Let his income be rs x.
Income after 10% increase is
\(100+100 \times \frac{10}{100}=Rs. 110\)
Now, income after 10% decrease is
\(110-110 \times \frac{10}{100}=110-11=Rs. 99\)
Net change in his income = 100 – 99 = 1
Percentage change \(=\frac{1}{100} \times 100 \%=1 \%\)
That is, income of the person is reduced by 1%.
Aliter
Let his income be Rs. 100
Income after 10% increase is
100 + 100 × \(\frac{10}{100}\) = Rs.110
Now, income after 10% decrease is,
110 – 110 × \(\frac{10}{100}\) = 110 – 11 = Rs. 99
∴ Net change in his income = 100 – 99 = 1
Percentage change = × = \(\frac{1 }{100}\) × 100% = 1% 
That is, income of the person is reduced by 1%.

Let the original price of Orid dhall be Rs x.
New price aft er price of 20% increase = x +\(\frac { 20 }{ 100 } x=\frac { 120x }{ 100 } \)
Given that, 96 = \(\frac { 120x }{ 100 } \)
∴ x = \(\frac { 96\times 100 }{ 200 } \)
∴ Original price of Orid dhall per kg, x = Rs. 80

x% of 600 = 450
\(\frac { x }{ 100 } \)× 600 = 450
\(x=\frac { 450 }{ 6 } \)
x = 75