Question

please give me the answer fast...​

Answer 1

Answer:

Let the average speed be s, distance=d, time taken on average speed= t1 and

time taken by increasing the speed=t2

t2=t1-1 equation 1

t1=d/s=300/s equation 2

t2=d/(s+10)=300/(s+10) equation 3

put the values of t1 and t2 in equation 1

300/(s+10)=(300/s)-1

300{(1/(s+10)-1/s}=-1

300{-10/(s)(s+10)}=-1

s(s+10)=3000

s^2+10s-3000=0

you will get imaginary value because

b^2 is less than 4ac

10^2 is less than 4×1×3000

don't do this type of question with such a big number