In a race of length L metres, Johnson beats Lewis by X metres and Greene by Y metres, By how many ...
In a race of length L metres, Johnson beats Lewis by X metres and Greene by Y metres, By how many metres does Lewis beat Greene in the same race ? (X<Y)
Answer/Solution
L(Y-X) / L-X
Steps/Work
As the time is constant, the ratio of distances will be the same as the ratio of speeds.
If Sj,Sl,Sg are the speeds of Johnson, Lewis, and Greene respectively, then
Sj/Sl = L/(L-X)
and Sj/Sg = L/(L-Y)
=> Sg/Sl = (L-Y)/(L-X)
Therefore the speeds of Lewis and Greene are in the ratio (L-X)/(L-Y)
When Lewis finishes the race, the time run by him and Greene are same
=> The ratio of the speeds of Lewis and Greene will be the same as the ratio of distances run by them.
=> Distance run by Greene when Lewis finishes the race = (L-Y)/(L-X) * L
=> Lewis beats Greene by L - L*(L-Y)/(L-X) = L [ 1 - (L-Y)/(L-X)] = L (Y-X) / (L-X)
Option (B) is therefore correct.
If Sj,Sl,Sg are the speeds of Johnson, Lewis, and Greene respectively, then
Sj/Sl = L/(L-X)
and Sj/Sg = L/(L-Y)
=> Sg/Sl = (L-Y)/(L-X)
Therefore the speeds of Lewis and Greene are in the ratio (L-X)/(L-Y)
When Lewis finishes the race, the time run by him and Greene are same
=> The ratio of the speeds of Lewis and Greene will be the same as the ratio of distances run by them.
=> Distance run by Greene when Lewis finishes the race = (L-Y)/(L-X) * L
=> Lewis beats Greene by L - L*(L-Y)/(L-X) = L [ 1 - (L-Y)/(L-X)] = L (Y-X) / (L-X)
Option (B) is therefore correct.