Exercise 13.2

1. Draw the graphs for the following tables of values, with suitable scales on the axes.

(a) Cost of apples

Number of apples	Cost (in Rs.)
1	5
2	10
3	15
4	20
5	25

Instructions

(b) Distance travelled by a car

Time (in hours)	Distances (in km)
6 a.m.	40
7 a.m.	80
8 a.m.	120
9 a.m.	160

(i) How much distance did the car cover during the period 7.30 a.m. to 8 a.m?

(ii) What was the time when the car had covered a distance of 100 km since it’s start?

Instructions

(i) How much distance did the car cover during the period 7.30 a.m. to 8 a.m? km

(ii) What was the time when the car had covered a distance of 100 km since it’s start? a.m.

(c) Interest on deposits for a year.

Deposit (in Rs.)	Simple Interest (in Rs.)
1000	80
2000	160
3000	240
4000	320
5000	400

(i) Does the graph pass through the origin?

(ii) Use the graph to find the interest on Rs. 2500 for a year.

(iii) To get an interest of Rs. 280 per year, how much money should be deposited?

Instructions

(i) Does the graph pass through the origin?

(ii) Use the graph to find the interest on Rs. 2500 for a year. Interest on Rs. 2500 is Rs. for a year.

(iii) To get an interest of Rs. 280 per year, how much money should be deposited? Rs. should be deposited for the interest of Rs. 280.

Solution:

(a) From the graph, we can observe that the line will pass through the origin. We can also approach this question logically by asking- if we deposit Rs.0 (no money at all), what will be the simple interest on it? The answer will be also zero. Thus, the point (0,0) satisfies the relation.

(b) Using the point (2000,160) and (3000, 240) we can interpolate for the value x = 2500. Let the simple interest on Rs.2500 be a. We get:

3000−2000240−160 = 2500−2000a−160

100080 = 500a−160

a - 160 = 40

a = Rs. 200

(c) Similarly, like in (b), we now have the value y = 280 and we need to find the corresponding x value. Let the amount to be deposited be b. Using interpolation between the points (3000,240) and (4000,320):

4000−3000320−240 = 4000−b320−280

100080 = 4000−b40

4000 - b = 500

b = Rs. 3500

2. Draw a graph for the following.

(a) Side of square and corresponding perimeter:

Side of square(cm)	Perimeter (cm)
2	8
3	12
3.5	14
5	20
6	24

Instructions

Is it linear?

(b) Side of square and corresponding area:

Side of square(cm)	Area (cm²)
2	4
3	9
4	16
5	25
6	36

Instructions

Is it linear?

Chapter 13: Introduction to Graphs

Glossary

Exercise 13.2

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Chapter 13: Introduction to Graphs

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Glossary

Exercise 13.2