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06 Apr 2020, 02:15
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In a race of three horses, the first horse beat the second horse by 11 metres and the third horse by 90 metres. If the second horse beat the third horse by 80 metres, what was the length, in metres, of the race?
A. 800
B. 860
C. 870
D. 880
E. 900
Are You Up For the Challenge: 700 Level Questions
ArunSharma12
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06 Apr 2020, 03:10
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Quote:
In a race of three horses, the first horse beat the second horse by 11 metres and the third horse by 90 metres. If the second horse beat the third horse by 80 metres, what was the length, in metres, of the race?
A. 800
B. 860
C. 870
D. 880
E. 900
A. 800
B. 860
C. 870
D. 880
E. 900
let the race be x meters long.
when A finished the race, B was (x-11) meters behind and C was (x-90) meters behind.
when B finished the race, C was (x-80) meters behind.
so, when B ran 11 meters, C ran 10 meters.
when B ran (x-11) meters, C ran = \(\frac{10*(x-11)}{11} = x-90\)
\(\frac{10x}{11} -10 = x-90\)
\(\frac{x}{11} = 80\)
x =880 meters
Ans D
06 Apr 2020, 02:28
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Length of track = L
\(V_F : V_S : V_T = L : (L-11) : (L-90)\)
Also, \(V_S : V_T = L : (L-80)\)
\(\frac{(L-11)}{(L-90)} = \frac{L}{(L-80)}\)
(L-11)(L-80) = (L)(L-90)
\(L^2 -91L +880 = L^2-90L\)
L= 880
\(V_F : V_S : V_T = L : (L-11) : (L-90)\)
Also, \(V_S : V_T = L : (L-80)\)
\(\frac{(L-11)}{(L-90)} = \frac{L}{(L-80)}\)
(L-11)(L-80) = (L)(L-90)
\(L^2 -91L +880 = L^2-90L\)
L= 880
