A person goes 4/5 of his usual speed reaches 10 min late to
his destinaton, time taken?

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A person goes 4/5 of his usual speed reaches 10 min late to his destinaton, time taken? ..

50 min

 Is This Answer Correct ? 26 Yes 6 No

A person goes 4/5 of his usual speed reaches 10 min late to his destinaton, time taken? ..

explanation:
speed=distance/time
in both cases distance is same.
s1*t1=s2*t2
s*t=(4s/5)*(t+10)
t=40 min
but we want total time means 10 min late
40+10=50 min

 Is This Answer Correct ? 27 Yes 11 No

A person goes 4/5 of his usual speed reaches 10 min late to his destinaton, time taken? ..

ans is:50 min....4/5th of his usual speed leads 10 min
late....means..1/5th of his speed equals
10min...1/5*speed=10->50 min...

 Is This Answer Correct ? 14 Yes 6 No

A person goes 4/5 of his usual speed reaches 10 min late to his destinaton, time taken? ..

We know speed × time = distance, so ((4/5)×s)×(t+10)=D
On solving above equation we get
4/5×st+4/5s×10=D
4st+40s=5st. [ Since D=S×T]
(4t+40)= 5t
40=5t-4t
t=40 minutes.

 Is This Answer Correct ? 4 Yes 0 No

A person goes 4/5 of his usual speed reaches 10 min late to his destinaton, time taken? ..

speed is inversely proportional to time

ANS:50 min

 Is This Answer Correct ? 3 Yes 4 No

A person goes 4/5 of his usual speed reaches 10 min late to his destinaton, time taken? ..

let a person's usual speed be x km/min
let the distance be d km
given (d/(4 x/5))-(d/x)=10
when we solve this equation we will get d=40 x
t=?
t=d/(4 x/5)=(40 x/(4 x/5))=(5*40)/4=50

 Is This Answer Correct ? 0 Yes 1 No

A person goes 4/5 of his usual speed reaches 10 min late to his destinaton, time taken? ..

5 min

 Is This Answer Correct ? 2 Yes 8 No

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