Question

B is 50% faster than a. A starts at 10 a.M. And b starts at 12 p.M. a travels at a speed of 60 km/hr. If a and b are 420 kms apart, the time when they meet when they travel in opposite direction is?

Answer 1

Answer:

14.66  

Step-by-step explanation:

Given B is 50% faster than a. A starts at 10 a.M. And b starts at 12 p.M. a travels at a speed of 60 km/hr. If a and b are 420 kms apart, the time when they meet when they travel in opposite direction is?

B is 50% of A = 25 kms/hr

So B travels at 50 + 50% of 50 = 50 + 25 = 75 km / hr

Now A starts at 10 AM and B starts at 12 PM

A starts 60 kms in 1 hr

So when B starts they are 420 - 60 = 360 kms apart.

Together speed = 60 + 75 = 135 km/h

Now time = distance / speed

                = 360 / 135

        time = 2.66 hours to meet

So 12 + 2.66 = 14 .66 or they meet at 3.06 PM

We can do in a simple type also

 60(x + 1) + 75x = 420

 135 x = 420 - 60

    x = 360 / 136

   x = 2.66 hrs

Answer 2

Answer:

Step-by-step explanation:

B is 50% faster than A => B travels at a speed of 60 + (60 x 0.5) = 90 km/h.

Distance between A and B is 420 kms.

A travels at 60 kms/h, so at 12 pm, when B starts, A has travelled 60 kms x 2 = 120 kms.

Hence, at 12 pm they are 420 – 120 = 300 kms apart.

Together speed = 60 + 90 = 150 km/h

Now time = distance / speed

                = 300 / 150  

Now time = 2 hours to meet

So 12 + 2 = 14.

Hence they meet at 14 pm.