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Example 5: l varies directly as m and l is equal to 5, when m = . Find l when m = 3 .
3
Solution: Given, l varies directly as m. So, l = km, where k is a constant.
2
When l = 5, then m = (Given)
3
So, 5 = k × 2 Think and Answer
3
3 15 1. Mention any three situations where
⇒ k = 5 × =
2 2 two quantities are in direct variation.
15 2. The gravitational force of attraction (F)
Therefore, l = × m
2 between an object and the Earth is directly
16 proportional to the mass (m) of the object.
Since, m = If the force of attraction is 147 N when the
3 object’s mass is 15 kg, find the constant of
15 16
Then, l = × = 40 proportionality.
2 3
Thus, the required value of l is 40.
Example 6: An electric pole 18 m high casts a shadow of 14 m. If the height of a tree is 31.5 m, find
the length of its shadow under similar conditions.
Solution: Let the length of the tree’s shadow be x m. We will form a table as follows:
Height of the object (in m) 18 31.5
Length of the shadow (in m) 14 x
Observe that more the height of an object, more would be the length of its shadow. Hence, this
is a case of direct proportion.
.
x x 18 31 5
That is, 1 = 2 ⇒ =
y 1 y 2 14 x
9 315
⇒ = ⇒ 90x = 315 × 7
7 10x
315 × 7 35 × 7 245
⇒ x = = ⇒ x = = 24.5
90 10 10
Therefore, the length of the shadow of tree of height 31.5 m is 24.5 m.
Example 7: A train is moving at a uniform speed of 100 km/hr.
(a) Find the distance covered by it in 32 minutes.
(b) In how much time will it cover 250 km?
Solution: Let the distance covered by the train in 32 minutes be x km and let it cover 250 km in
y minutes. Then we form a table as follows:
Distance covered (in km) 100 x 250
Time taken (in minutes) 60 32 y
Since, the speed is uniform, more distance will be covered in more time.
So, this is a case of direct variation.
100 x 250
Therefore, = =
60 32 y
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