A loaded truck travels 14 km in 25 minutes. If the speed remains the same, how far can it travel in 5 hours?
Since the speed is constant,
∴ For longer distance, more time will be required.
So, it is a case of direct variation.
Let the required distance to be travelled in 5 hours (= 300 minutes) be x km.
We have:
|
Distance (in km) |
Time (in minutes) |
|
14 |
25 |
|
X |
300 |
Observe the following tables and find which pair of variables (here x and y) are in inverse proportion.
(i)
|
X |
50 |
40 |
30 |
20 |
|
y |
5 |
6 |
7 |
8 |
(ii)
|
X |
100 |
200 |
300 |
400 |
|
y |
60 |
30 |
20 |
15 |
(iii)
|
X |
90 |
60 |
45 |
30 |
20 |
5 |
|
y |
10 |
15 |
20 |
25 |
30 |
35 |
(i) ∵ X1 = 50 and Y1 = 5 
x1Y1 = 50 x 5 = 250
X2 = 40 and Y2 = 6 x2Y2 = 40 x 6 = 240
X3 = 30 and Y3 = 7 x3Y3 = 30 x 7 = 210
X4 = 20 and Y4 = 8 x4Y4 = 20 x 8 = 160
Question: Observe the following tables and find which pair of variables (here x and y) are in inverse proportion.
(i)
|
X |
50 |
40 |
30 |
20 |
|
y |
5 |
6 |
7 |
8 |
(ii)
|
X |
100 |
200 |
300 |
400 |
|
y |
60 |
30 |
20 |
15 |
(iii)
|
X |
90 |
60 |
45 |
30 |
20 |
5 |
|
y |
10 |
15 |
20 |
25 |
30 |
35 |
On a squared paper, draw five squares of different sides. Write the following information in a tabular form.
|
Square 1 |
Square 2 |
Square 3 |
Square 4 |
Square 5 |
|
|
Length of a side (L) |
|||||
|
Perimeter (P) |
|||||
|
L/P |
|||||
|
L/A |
Find whether the length of a side is in direct proportion to:
(a) The perimeter of the square.
(b) The area of the square.
We draw the following squares I, II, III, IV and V having the length of side as 1 cm, 2 cm, 3 cm, 4 cm and 5 cm respectively.
Now, we have
Obviously:
(a) The length of a side (L) and the perimeter of the square (P) are in the direct proportion.
(b) But the length of a square (L) and the area of the square (A) are not in the direct proportion.
A 5 m 60 cm high vertical pole casts a shadow 3 m 20 cm long. Find at the same time (i) the length of the shadow cast by another pole 10 m 50 cm high and (ii) the height of a pole which casts a shadow 5 m long.
(i) Let the required length of shadow be x cm.
|
Height of the pole |
Length of the shadow |
|
5 m 60 cm = 560 cm |
3 m 20 cm = 320 cm |
As the height of the pole increases the length of its shadow also increases in the same ratio. Therefore, it is a case of direct variation.
(ii) Let the required height of the pole be y cm.
∴ We have:
|
Height of the pole |
Length of the shadow |
|
560 cm y |
320 cm 5 m = 500 cm |
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