SOLUTION: Hi A man walks at 4/5 of his speed to his office and is late by 5 min. What is the time it takes to reach his office if he walks at 3/4 of his speed. Thanks

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Question 1156576: Hi
A man walks at 4/5 of his speed to his office and is late by 5 min. What is the time it takes to reach his office if he walks at 3/4 of his speed.
Thanks
Answer by ikleyn(52767) About Me  (Show Source):
You can put this solution on YOUR website!
.

The original formulation in the post is  FAR  from to be  PERFECT.

The  PERFECT  and  PRECISELY  CORRECT  formulation is  THIS: 


    A man walks at 4/5 of his highlight%28REGULAR%29speed to his office and is late by 5 min. 
    What is the time it takes to reach his office if he walks at 3/4 of his highlight%28REGULAR%29 speed.


I will solve the problem with this formulation. 


Let "d" be the distance to the office, and let "u" be the regular speed.


Then the condition produces this time equation

    d%2F%28%284%2F5%29%2Au%29 - d%2Fu = 5 minutes.


Simplify

    %285d%29%2F%284u%29 - d%2Fu = 5 minutes,   or

    d%2F%284u%29 = 5 minutes,   or

     d%2Fu = 4*5 = 20 minutes.


So, his REGULAR time to get the office is 20 minutes.


Now, if he moves at  3%2F4  of his refular speed, it will take  20%2A%284%2F3%29 = 80%2F3 = 262%2F3 minutes = 26 minutes and 40 seconds

to get the office.     ANSWER

Solved.


Same idea can be presented in other form. 

Let T be the regular time to get the office (moving at regular speed).


Then with the speed of 4%2F5  of the regular, the time will be  %285%2F4%29T.

Then the condition says us that

    %285%2F4%29T - T = 5 minutes,  or

    %281%2F4%29T = 5 minutes,

which means  T = 4*5 = 20 minutes (regular time).


Then with the speed of  3%2F4  of the regular, the time will be  20%2A%284%2F3%29 = 80%2F3 = 26 minutes and 40 seconds (the same answer).