- A pedestrian left point A for a walk, going with
the speed of 5 km/h. When the pedestrian was at
a distance of 6 km from A, a cyclist followed him,
starting from A and cycling at a speed 9 km/h higher
than that of the pedestrian. When the cyclist overtook
the pedestrian, they turned back and returned to A
together, at the speed of 4 km/h. At what y will the
time spent by the pedestrian on his total journey from
A to A be the least?
(a) 5 km/h
(b) 6 km/h
(c) 6.1 kmph
(d) 5.5 km/h
Answers
Given :- A pedestrian left point A for a walk . When the pedestrian was at a distance of 6 km from A, a cyclist followed him, starting from A and cycling at a speed 9 km/h higher than that of the pedestrian. When the cyclist overtook the pedestrian, they turned back and returned to A together, at the speed of 4 km/h. At what speed will the time spent by the pedestrian on his total journey from A to A be the least ?
(a) 5 km/h
(b) 6 km/h
(c) 6.1 kmph
(d) 5.5 km/h
Answer :-
Let us assume that, the speed of pedestrian is x km/h .
so,
→ speed of cyclist = (x + 9) km/h .
given that, the cyclist starts from A after pedestrian covers 6 km. Let after y hours he cross the pedestrian .
so,
→ Distance travelled by pedestrian = (6 + xy) km..
→ Distance travelled by cyclist = (x + 9)y = (xy + 9y) km .
then,
→ 6 + xy = xy + 9y
→ 9y = 6
→ y = (2/3) hours.
now,
→ Total distance travelled by pedestrian from A = 6 + (2x/3) km.
then,
→ Time taken by pedestrian to cover this distance = D/S = [6 + (2x/3)]/x = {(6/x) + (2/3)} hours. -------- Eqn.(1)
given that, while returning there speed were 4km/h .
so,
→ Time taken by pedestrian to reach back to A = D/S = [6 + (2x/3)]/4 = {(6/4) + (2x/12)} = {(3/2) + (x/6)} hours -------- Eqn.(2)
we have to find value of x for which total time of pedestrian from A to A would be minimum .
so, using Eqn.(1) and Eqn.(2)
→ Total time taken by pedestrian from A to A = {(6/x) + (2/3)} + {(3/2) + (x/6)} = (6/x) + (x/6) + (2/3) + (3/2) = (6/x) + (x/6) + (4+9)/6 = {(6/x) + (x/6) + (13/6)} hours.
checking options for minimum value now, we get,
(a) 5km/h
→ (6/5) + (5/6) + (13/6)
→ (6/5) + (18/6)
→ 1(1/5) + 3
→ 4(1/5) hours = 4 hours + 12 minutes .
(b) 6 km/h
→ (6/6) + (6/6) + (13/6)
→ 1 + 1 + 2(1/6)
→ 4(1/6) hours = 4 hours + 10 minutes .
(c) 6.1 km/h
→ (6/6.1) + (6.1/6) + (13/6)
→ (6/6.1) + (19.1/6)
→ (60/61) + (191/60)
→ (3600 + 11651)/3660
→ (15251/3660)
→ 4(611/3660) ≈ 4 hours + greater than 10 minutes .
(d) 5.5 km/h
→ (6/5.5) + (5.5/6) + (13/6)
→ (60/55) + (18.5/6)
→ (12/11) + (37/12)
→ (144 + 407)/132
→ (551/132)
→ 4(23/132) = 4 hours + greater than 10 minutes .
therefore, we can conclude that, at 6km/h (B) the time spent by pedestrian will be minimum .
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Given : A pedestrian left point A for a walk, going with the speed of y km/h.
When the pedestrian was at a distance of 6 km from A, a cyclist followed him, starting from A and cycling at a speed 9 km/h higher than that of the pedestrian. When the cyclist overtook the pedestrian, they turned back and returned to A together, at the speed of 4 km/h.
To Find : At what y will the time spent by the pedestrian on his total journey from A to A be the least?
(a) 5 km/h
(b) 6 km/h
(c) 6.1 kmph
(d) 5.5 km/h
Solution:
pedestrian speed = y km/hr
cyclist speed = y + 9 km/hr
Cyclist relative speed = y + 9 - y = 9 km/hr
cyclist relative distance = 6 km/hr
Time taken to overtake = 6/9 = 2/3 hrs
Travel time of pedestrian from A to return point = time to cross 6 km + 2/3 hrs = 6/y + 2/3 hrs
Distance covered = 6 + (2/3)y km
Return speed = 4 km/hr
Return time = ( 6 + (2/3)y)/4 = 3/2 + y/6 hrs
Total travel time T of Pedestrian from A to A = 6/y + 2/3 + 3/2 + y/6
T = 6/y + y/6 + 13/6
dT/dy = -6/y² + 1/6
dT/dy = 0
=> -6/y² + 1/6
-6/y² + 1/6 = 0
=> y² = 36
=> y = 6
d²T/dy² = 12/y³ > 0 Hence Time is minimum at y = 6
At 6 km/hr the the time spent by the pedestrian on his total journey from
A to A be the least
Hence 6 km/hr is correct option.
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