Let us draw a figure as per given question.
Let AB = 5 meter and CD = 2 meter be the heights of a lamp post and the men respectively.
At any time t BD = x meter and shadow of man ED = y meter.
Then, dx/dt = 6 m/min
Now, right triangles ABE and CDE are similar, then
AB/CD = BE/DE
? 5/2 = (x + y)/y
? 5y = 2x + 2y
? 5y - 2y = 2x
? 3y = 2x
? 3 dy/dt = 2 dx/dt
? 3 dy/dt = 2 x 6
? dy/dt = 2 x 6 / 3
? dy/dt = 2 x 2 = 4
Hence, length of his shadow increase at the rate of 4 m/min.
We have the important relation, More work, More time (days)
? A piece of work can be done in 6 days.
? Three times of work of same type can be done in 6 x 3
= 18 days
? = 750.0003 ÷ 19.999
? ? ? 750 ÷ 20
? ? ? 375 ? 38
Subtract 20, 25, 30, 35, 40, 45 from successive numbers. So 0 is wrong.
NA
Each previous number is multiplied by 2.
? 8 m shadow means original height = 12 m
? 1 m shadow means original height = 12/8 m
? 100 m shadow means original height = (12/8) x 100 m
= (6/4) x 100 = 6 x 25 = 150 m
Let 8% of 96 = y of 1/25
? (8 x 96)/100 = y/25
? y = (8 x 96 x 25)/100 = 192
Since the principal is not given, so data is inadequate.
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