krishna kumar wrote:Hi,

Could anyone solve this ?


The speed of a bicycle varies directly as the square of the weight of the rider and inversely as the amount of air in the tyre. If the volume of air in the tyre is doubled, then what would be the percentage change in the weight of the bicycle-rider to maintain same speed?

A. 20 B. 41.4 C. 140 D. 120 E . 60

OA B

thanks

If w is the weight and v is the volume of air in tyre, then the speed s of the bicycle-rider could be given by

s = k w^2/v, or w = √ (v s/k), where k is some constant of proportion.

Now, when v is 2 v, let's say w has to be w' in order to maintain same s,

Or, w' = √ (2 v s/k)

So, the new weight is √2 times the old one, that is an increase of (√2 - 1) × 100 percent, [spoiler]41.4 is the best choice.

B[/spoiler]

All u need is to form the equation with the description given.

consider weight as 'w' and air volume as 'a'

so since speed is direct proportional to the weight and inversely proportional to the air volume

s= w^2/a (assume the constant term = 1 for simplicity)

next the question asks to keep the ratio same when a = 2a

then s= w^2/2a now to keep the ratio same as w^2/a

we will keep have to multiple the ratio with 2

therefore s= 2w^2/2a

now we need to look for change in value for w

so we take the 2 within the square sign of w

=> s= [root(2)w]^2/2a

so this implies the new value of w is root(2)w

which is 1.414w

obviously the difference is 1.414w - w = 0.414w

and percentage change will be 41.4%

IMO B

Hope this helps...