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.

(Rearranging the like terms)
\(\frac { 4y }{ 3 } -\frac { 2 }{ 5 } =7\)
\(\frac { 20y-6y }{ 15 } =7\)
14y = 7 ×15
y = \(\frac { 7\times 15 }{ 2 } \)
y = \(\frac { 15 }{ 2 } \)

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

Let the distance be ‘ x ’ km. (Recall that time = \(\frac { Distance }{ Speed } \))
Time taken to cover ‘ x ’ km at 35 km/hr: T₁ = \(\frac { x }{ 35 } hr\)
Time taken to cover ‘ x ’ km at 50 km/hr: T₂ = \(\frac { x }{ 50 } hr\)
According to the problem, the difference between two timings
= 4–(–5)
= 4+5 = 9 minutes
= \(\frac { 9 }{ 60 } \)hour (changing minutes to hour)
Given, T₁ – T₂ =\(\frac { 9 }{ 60 } \)
\(\frac { x }{ 35 } \)-\(\frac { x }{ 50 } \) = \(\frac { 9 }{ 60 } \)
\(\frac { 10x-7x }{ 350 } =\frac { 9 }{ 60 } \)
\(\frac { 3x }{ 350 } =\frac { 9 }{ 60 } \)
x \(=\frac { 9 }{ 60 } \times \frac { 350 }{ 3 } \)
The distance to his office x = 17\(\frac { 1 }{ 2 } \) km.

Let the original fraction be \(\frac { x }{ y } \)
Given that y = x + 3. (Denominator = Numerator + 3).
Therefore, the fraction can be written as \(\frac { x }{ x+3 } \).As per the given condition, \(\\ \frac { x+2 }{ (x+3)+9 } =\frac { 5 }{ 6 } \)
By cross multiplication, 6( x +2) = 5 ( x +3+9)
6 x +12 = 5( x +12)
6 x +12 = 5 x +60
6 x − 5 x = 60 − 12
x = 60 − 12
x = 48.
Therefore, the original fraction is \(\frac { x }{ x+3 } =\frac { 48 }{ 48+3 } =\frac { 48 }{ 51 } \).

Given condition: After two years, Mother’s age = 4 times of Daughter's age
5 x +2 = 4 ( x +2)
5 x +2 = 4 x +8
5 x − 4x = 8 − 2
x = 6
Hence daughter’s present age = 6 years;
and mother’s present age = 5 x = 5 × 6 = 30 years 

Let the smaller number be x and the greater number be x + 8
Given: the sum of two numbers = 36
x + (x+8) = 36
2 x + 8 = 36
2 x = 36 − 8
2 x = 28
x = \(\frac { 28 }{ 2 } \)
x = 14
The smaller number, x = 14
The greater number, x + 8 = 14 + 8 = 22

(1) x = -7

(ii) y = 6

From the graph, the points (- 2, 4) and (- 3, 5) lie on the line, which is passing through (4, - 2) and (0, 2) in the second quadrant.

Murali’s age is 15 years old, Thenmozhi’s age is 20 years old

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