Question

While going from
place A to B. Astha
travels 40% of the
oisiance at 12 km/hr,
then travels 50% of
the remaining
Distance at 18 km/hr
to reach C. If the
time taken to travel
from A to C is 15
hours, what is the
distance between A
and Brinkm)?​

Answer 1

Answer : 300 km

Explanation :

Distance of A and K = d/100 * 40

Distance of A and K = d/100 * 40Distance of K and C = d/100 * 30

Path A K

ath A K speed =distance /time

ath A K speed =distance /time

ath A K speed =distance /time 12km/h = (d/100 *40) / x

ath A K speed =distance /time 12km/h = (d/100 *40) / xd=30x Eq. (1)

ath A K speed =distance /time 12km/h = (d/100 *40) / xd=30x Eq. (1)Now for path K C

ath A K speed =distance /time 12km/h = (d/100 *40) / xd=30x Eq. (1)Now for path K C18km/h = (d/100*30) / (15 - x)

ath A K speed =distance /time 12km/h = (d/100 *40) / xd=30x Eq. (1)Now for path K C18km/h = (d/100*30) / (15 - x) Solving this we get

ath A K speed =distance /time 12km/h = (d/100 *40) / xd=30x Eq. (1)Now for path K C18km/h = (d/100*30) / (15 - x) Solving this we get

ath A K speed =distance /time 12km/h = (d/100 *40) / xd=30x Eq. (1)Now for path K C18km/h = (d/100*30) / (15 - x) Solving this we get 900 - 60x = d Eq. (2)

ath A K speed =distance /time 12km/h = (d/100 *40) / xd=30x Eq. (1)Now for path K C18km/h = (d/100*30) / (15 - x) Solving this we get 900 - 60x = d Eq. (2)Putting together both equations (1) and (2)

ath A K speed =distance /time 12km/h = (d/100 *40) / xd=30x Eq. (1)Now for path K C18km/h = (d/100*30) / (15 - x) Solving this we get 900 - 60x = d Eq. (2)Putting together both equations (1) and (2)We get

ath A K speed =distance /time 12km/h = (d/100 *40) / xd=30x Eq. (1)Now for path K C18km/h = (d/100*30) / (15 - x) Solving this we get 900 - 60x = d Eq. (2)Putting together both equations (1) and (2)We get 900 - 60x = 30x

ath A K speed =distance /time 12km/h = (d/100 *40) / xd=30x Eq. (1)Now for path K C18km/h = (d/100*30) / (15 - x) Solving this we get 900 - 60x = d Eq. (2)Putting together both equations (1) and (2)We get 900 - 60x = 30x900 = 60x + 30x

ath A K speed =distance /time 12km/h = (d/100 *40) / xd=30x Eq. (1)Now for path K C18km/h = (d/100*30) / (15 - x) Solving this we get 900 - 60x = d Eq. (2)Putting together both equations (1) and (2)We get 900 - 60x = 30x900 = 60x + 30xx = 10

ath A K speed =distance /time 12km/h = (d/100 *40) / xd=30x Eq. (1)Now for path K C18km/h = (d/100*30) / (15 - x) Solving this we get 900 - 60x = d Eq. (2)Putting together both equations (1) and (2)We get 900 - 60x = 30x900 = 60x + 30xx = 10

ath A K speed =distance /time 12km/h = (d/100 *40) / xd=30x Eq. (1)Now for path K C18km/h = (d/100*30) / (15 - x) Solving this we get 900 - 60x = d Eq. (2)Putting together both equations (1) and (2)We get 900 - 60x = 30x900 = 60x + 30xx = 10 putting value of x in Eq. (1)

ath A K speed =distance /time 12km/h = (d/100 *40) / xd=30x Eq. (1)Now for path K C18km/h = (d/100*30) / (15 - x) Solving this we get 900 - 60x = d Eq. (2)Putting together both equations (1) and (2)We get 900 - 60x = 30x900 = 60x + 30xx = 10 putting value of x in Eq. (1)d=30(x)

ath A K speed =distance /time 12km/h = (d/100 *40) / xd=30x Eq. (1)Now for path K C18km/h = (d/100*30) / (15 - x) Solving this we get 900 - 60x = d Eq. (2)Putting together both equations (1) and (2)We get 900 - 60x = 30x900 = 60x + 30xx = 10 putting value of x in Eq. (1)d=30(x)d=30(10)

ath A K speed =distance /time 12km/h = (d/100 *40) / xd=30x Eq. (1)Now for path K C18km/h = (d/100*30) / (15 - x) Solving this we get 900 - 60x = d Eq. (2)Putting together both equations (1) and (2)We get 900 - 60x = 30x900 = 60x + 30xx = 10 putting value of x in Eq. (1)d=30(x)d=30(10)d = 300 km