At a movie set a horse gallops away with the heroine at a speed of 60 km/hr for sometime before the hero jumps on its back from a tree and brings the horse to a comfortable trot of 12 km/hr.If in total 26 minutes,the horse covered a distance of 10 km ,calculate for how much distance and time the horse galloped and trotted
Joao
Let's say you have 2 variables, both of time, in hours, that the horse moved:
x - time it galloped
y - time it trotted
km/h x h = km right? And with the data you have:
60x + 12y = 10
x + y = 26/60 (26 minutes is 26/60 of an hour)
x + y = 26/60 <=> x = 26/60 - y
60x + 12y = 10 <=> 60(26/60 - y) + 12y = 10 <=> 26 - 60y + 12y = 10 <=> 26 - 48y = 10 <=> -48y = -16 <=> y = 16/48 = 1/3
x + y = 26/60 <=> x + 1/3 = 26/60 <=> x = 26/60 - 1/3 = 1/10
x = 1/10 hours = 1/10 x 60 minutes = 6 minutes galloping
y = 1/3 hours = 1/3 x 60 minutes = 20 minutes trotting
Since km/h x h = km,
Galloping speed is 60 km/h then 60 x 1/10 = 6 km galloping;
Trotting speed is 12 km/h then 12 x 1/3 = 4 km trotting.
A: The horse galloped 6 minutes and a distance of 6 km and trotted 20 minutes and a distance of 4km.
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ranjankar
Let the no of minutes the horse galloped be x minutes
THEN , the no of minutes the horse trotted will be ( 26-x) minutes
Distance galloped = Speed X time = 60 X x/60 = x km
Distance trotted = 12 X (26-x) /60 = (312 - 12x)/60 km
x + (312 -12x) /60 = 10
60x + 312 -12x = 600
48x = 288
x = 6
ANSWER 6km ( gallop) and time = 6 minutes
4 km ( trotted) and time = 20 minutes
Iggy Rocko
Let g = time galloped in minutes. This makes the time trotted equal to 26 - g.
60 km/hr = 1 km/min
12 km/hr = 1/5 km/min
g(1) + (26 - g)1/5 = 10
g + 26/5 - g/5 = 10
4g/5 + 26/5 = 50/5
4g/5 = 24/5
g = 6 min
26 - 6 = 20 min
6(1) = 6 km
20(1/5) = 4 km
The horse galloped 6 km for 6 min and trotted 4 km for 20 minutes.
Vaneet
Speed=Distance/Time