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10 Aug 2011, 19:38
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
85%
(hard)
Question Stats:
20% (05:14) correct
80%
(01:24)
wrong
based on 5
sessions
History
Date
Time
Result
Not Attempted Yet
In a race of lrngth " L " metres, Johnson beats Lewis by "M" metres and Greene by "N" metres, By how many metres does Lewis beat Greene in the same race ? (M<N)
A) L(L-N) / L-M
B) L(N-M) / L-M
C) L-N
D) M-N
E) M+L
A) L(L-N) / L-M
B) L(N-M) / L-M
C) L-N
D) M-N
E) M+L
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10 Aug 2011, 20:45
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the ans is quite simple N-M but since the ans is not given, the best way to 'find' the correct option is to plug in the numbers. For my work, i took the following numbers:
L = 100
M = 10
N =20.
So basically, L beat G by 10 meters. Now whatever options comes close to the figure 10 is the right ans..
looking the the opion C to E, one can make out that they are incorrect, leaving us with just two options..A and B.
Out of which, i choose B.
L = 100
M = 10
N =20.
So basically, L beat G by 10 meters. Now whatever options comes close to the figure 10 is the right ans..
looking the the opion C to E, one can make out that they are incorrect, leaving us with just two options..A and B.
Out of which, i choose B.
10 Aug 2011, 22:37
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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-M)
and Sj/Sg = L/(L-N)
=> Sg/Sl = (L-N)/(L-M)
Therefore the speeds of Lewis and Greene are in the ratio (L-M)/(L-N)
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-N)/(L-M) * L
=> Lewis beats Greene by L - L*(L-N)/(L-M) = L [ 1 - (L-N)/(L-M)] = L (N-M) / (L-M)
Option (B) is therefore correct.
If Sj,Sl,Sg are the speeds of Johnson, Lewis, and Greene respectively, then
Sj/Sl = L/(L-M)
and Sj/Sg = L/(L-N)
=> Sg/Sl = (L-N)/(L-M)
Therefore the speeds of Lewis and Greene are in the ratio (L-M)/(L-N)
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-N)/(L-M) * L
=> Lewis beats Greene by L - L*(L-N)/(L-M) = L [ 1 - (L-N)/(L-M)] = L (N-M) / (L-M)
Option (B) is therefore correct.
