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.

Using the formula for calculating the simple interest, the following table of values is prepared.

These are the points which are to be plotted in the graph sheet. Let us take the deposits along x-axis and
annual simple interest along y-axis.
We choose the scale as follows:
Then we plot the points and draw the straight line.
From the graph we find:
(i) Corresponding to Rs.300 on the x-axis, we get the interest as Rs.30 on the y-axis.
(ii) Corresponding to Rs.70 on the y-axis, we get the deposit as Rs.700 on the x-axis.

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 = \(\frac { 15 }{ 2 } \)

x – 7 = 6
x – 7 + 7 = 6 + 7
x = 13

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 breadth of the field be ‘ x ’ metres; then its length (x + 9) metres.
Perimeter of the P = 2(length + breadth) = 2(x + 9 + x) = 2(2x + 9)
Given that, 2(2x + 9) = 154.
4x + 18 = 154
4x = 154 − 18
4x = 136
x = 34
Th us, Breadth of the rectangular fi eld = 34m
Length of the rectangular fi eld = x + 9 = 34 + 9 = 43m

From the graph, the point S is (5, - 4).

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l = 8 cm, b = 24 cm

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