Question

a man travels from a to b at 5 kmph and reaches b 20 minutes late .if his speed had been 7.5kmph he would have reached his office 12 minutes early find the distance between a to b

Answer 1

b=office

Answer 2

increase in 2.5 mph = reduction of 32 minutes in driving time... does this help?

Answer 3

how?

Answer 4

hint - convert all to one unit (so miles per minute, or convert 32 minutes into a decimal of how many hours that is)

Answer 5

do we have two equations

d = 5*(t+20)
d = 7.5*(t-12)

Answer 6

\(\large \color{black}{\begin{align} d=5(t+\dfrac{20}{60})=7.5(t-\dfrac{12}{60})\hspace{.33em}\\~\\
\end{align}}\)

Answer 7

key formula:
\(\huge \frac {miles~ travelled}{speed} = time~of ~travel\)

journey 1
\(\large x~ miles \times \frac {1~ hour}{5~ miles} = y~ hours\)
journey 2
\(\large x ~miles \times \frac {1~ hour}{7.5 ~miles} = (y-0.2)~ hours\)

rearrange equation 1 in terms of x, and substitute into equation 2
journey 1
\(\large x~ miles \times \frac {1~ hour}{5~ miles} = y~ hours\)
\(\large x = 5y \)

sub x = 5y into journey 2 equation
journey 2
\(\large x~ miles \times \frac {1 ~hour}{7.5 ~miles} = (y-0.2) hours\)
\(\large 5y~ \times \frac {1 }{7.5} = (y-0.2)~ hours\)
you now have only one variable in this equation, can you solve this and find y (which was the time taken for the original trip).
Then use this to solve the distance.

Answer 8

bugger... the 0.2 is incorrect, that's 12 minutes in hours... it should be 32 minutes difference, sooooo 8/15 hours, not 2/10 hours... sorry

Answer 9

\(\large 5y~ \times \frac {1 }{7.5} = (y-(\frac 8{15})~ hours\)