Ramesh says, “Driving at an average speed of 60 kmph, I reach office 10 minutes early. However, if I drive at a speed 10 kmph lesser than the earlier, I get late by half an hour”. Find the distance between Ramesh’s office and home.

Let distance be D
With speed 50km/hr (10 kmph less than the earlier 60 kmph), he is 30 minutes late
With speed 60 km/hr he is 10 minutes early
Difference between two times = 30+10 = 40min = [40 / 60]hours.

Note: Difference between two given times can also be easily measured or checked by looking at a watch or imagining a watch.

D = S x T
==> T = D / S (or) S = D / T
Also, time = T = D / S
==> [ (D / 50) – (D / 60) ] = [ 40 / 60 ]
==> [ 60D – 50D / 3000] (L.C.M) = [ 40 / 60 ]
==> [10D / 3000] = [ 40 / 60 ]
==> D = 200 km.

No, but now, it is modified and here is the solution:

Let distance be D
With speed 50km/hr (10 kmph less than the earlier 60 kmph), he is 30 minutes late
With speed 60 km/hr he is 10 minutes early
Difference between two times = 30+10 = 40min = [40 / 60]hours.

Note: Difference between two given times can also be easily measured or checked by looking at a watch or imagining a watch.

D = S x T
==> T = D / S (or) S = D / T
Also, time = T = D / S
==> [ (D / 50) – (D / 60) ] = [ 40 / 60 ]
==> [ 60D – 50D / 3000] (L.C.M) = [ 40 / 60 ]
==> [10D / 3000] = [ 40 / 60 ]
==> D = 200 km.

komal says

is this option is right

Saran Harika says

No, but now, it is modified and here is the solution:

Let distance be D

With speed 50km/hr (10 kmph less than the earlier 60 kmph), he is 30 minutes late

With speed 60 km/hr he is 10 minutes early

Difference between two times = 30+10 = 40min = [40 / 60]hours.

Note: Difference between two given times can also be easily measured or checked by looking at a watch or imagining a watch.

D = S x T

==> T = D / S (or) S = D / T

Also, time = T = D / S

==> [ (D / 50) – (D / 60) ] = [ 40 / 60 ]

==> [ 60D – 50D / 3000] (L.C.M) = [ 40 / 60 ]

==> [10D / 3000] = [ 40 / 60 ]

==> D = 200 km.