Question

A car starts 10 minutes late in a race but
reaches the finsh line 5 minutes earlier. It was
found that the race driver had increased the
speed by R %. The options below give
different values of possible time and
corresponding R. Choose the correct one.
(1) 55 minutes, 37.5
(II) 45 minutes, 50
(III) 45 minutes, 100
(IV) 65 minutes, 30​

Answer 1

Answer:

55min, 37.5

Step-by-step explanation:

Let track length be equal to T.

Time taken to meet for the first time = Trelativespeed = T6−b or Tb−6

Time taken for a lap for A = T6

Time taken for a lap for B = Tb

So, time taken to meet for the first time at the starting point = LCM (T6,Tb) = THCF(6,b)

Number of meeting points on the track = Time taken to meet at starting point/Time taken for first meeting = Relative speed / HCF (6,b).

So, in essence we have to find values for b such that 6−bHCF(6,b) = 2 or b−6HCF(6,b) = 2

Answer 2

Step-by-step explanation:

Let the track length be d and with speed s, he takes t minutes. Then one equation is

t=d/s

The 2nd equation is

t-15=d/(s+s.R/100)

t-15=d/{s(1+R/100)}

t-15=t/(1+R/100)

There are 4 unknowns and you have two equations, you can not solve this equation