Question

Walking at 5 km/hr a student reaches his school from

his house 15 minutes early and walking at 3 km/hr he

is late by 9 minutes. What is the distance between his

school and his house ?
​

Answer 1

here

suppose ,

• t is time of school
• t1 is time taken by student with speed 5 km/hr =5/60 km/min
• t2 is time taken by student with speed 3 km/hr = 3/60 km/min

Now

relationship between t and t1,t2

t = t1+15

t1=t-15

t = t2-9

t2=t+9

now the distance between school and home d = time × speed

t1 × s1 = t2 × s2

(t-15)(5/60)=(t+9)(3/60)

5t-75=3t+27

2t=102

t=51min

now for t1

t= t1+15

51= t1+15

t1=36 min

d=t1×s1

d=36×5/60

d=180/60 km

d=3 km

sorry for my mistake,

i think now it is correct...if any further mistake you found , please feel Free to correct me.

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