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To solve this question, it is a simple formula substitution.

s = d/t

300 miles = 60% of d i.e., full distance

Therefore, d = 500 miles.

We only need to look at the last 200 miles, as for the first 300 miles there is no change in speed or time.

The bus rests for 1/2 hour, but still arrives on time. Speed and time are inversely proportional, meaning if S insreaces t will decrease.

S has increased by 20 miles, and t has decreased by 30 1/2 h.

Original Speed => S = 200/t

New Speed => S+20 = 200/t-1/2

As we have to find the original speed S, we will take t1 - t2 = 1/2, original time - new time taken to complete 200 miles.

t1 = 200/s
t2 = 200/s+20

=> 200/s - 200/s+20 = 1/2

when we solve this we get ---> 8000 = s(s+20)

Hence s = 80 miles.
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kabhere
how to solve this w/o quadratic eqn?

how to use options efficiently ?
Compare the time before delay and after the delay for the last 200 km and substitute the options in it.

Let i be the initial speed and n be the new speed.

We know, n = i + 2

Substitute the options in below equation as i

200/i - 200/n

The option giving 0.5 as output is the correct option as 0.5 is the delay given.

Checking for option E

200/80 - 200/(80+20)

=0.5

Answer: Option E

By this method you can utilize the options for finding the answer.

Hope this helps...
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Once you get the equation:
300/x+1/2+200/x+20 = 500/x

Use values and substitute you will get 80 that gets you LHS=RHS
kabhere
how to solve this w/o quadratic eqn?

how to use options efficiently ?
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