what is the solution of this sum about speed,distance,and time?

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.

Hope I Helped! If so, thumbs up! :)

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

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.

Speed=Distance/Time