Question

Two cars travel the
same distance
starting
at
10:00 AM
and
11:00 AM,
respectively on the
same day. They
reach their common
destination at the
same time. If the first
car travelled for at
least 6 hrs, then the
highest possible
value
the
percentage by which
the speed of second
car
ar could exceed
that of the first car is:
(Assume both cars
travel with uniform
speed)
of​

Answer 1

Answer:

20%

Explanation:

given,

departure time of

A= 10.00 am and B= 11.00 am

since, car A travels for 6 hrs. and both the cars reach at the same time to the destination,

therefore,

arrival time of

A = B = 10.00 am + 6hrs = 4 pm

giving, travel time of A is 6hrs and B is 5 hrs.

Let the distance from departure to destination be x.(km)

then speed of A = x/6 & B = x/5 (km/h)

so difference in speed = x/5 -x/6 =x/30

exceeding percentage = [{x/30} \ {x/6}]*100 %

= x/30 * 6/x * 100

=1/5*100

=20%

Ans

Answer 2

Answer:

your answer are

20%

Explanation:

I hope it's helpful to you dear