walking at 3/4 of his usual speed,
a man is late by 2.5 hrs. What is he usual time?
3/4 speed takes 4/3 the usual time.
So,
4/3 x = x + 2.5
1/3 x = 2.5
x = 7.5
usual speed --- x km/h
time at usual speed to go 1 km = 1/x hrs
time at 3/4 his normal speed = 1/(3x/4) = 4/(3x)
4/(3x) - 1/x = 2.5
times 3x
4 - 3 = 7.5x
7.5x = 1
x = 1/7.5 = 2/15 km/h
Steve's answer and mine seem to contradict each other, but we defined x differently.
Steve defined x as his usual time, so the answer of 7.5 is correct for his usual time
I defined x to be the usual speed, and got x = 2/15
so to go 1 km would take
1/(2/15) = 15/2 = 7.5 , the same as Steve.
To find the man's usual time, we can set up an equation based on the information given. Let's start by assigning some variables:
Let's say the man's usual speed is represented by "S" (in units per hour).
Let's say the man's usual time is represented by "T" (in hours).
According to the given information, the man is walking at 3/4 of his usual speed. So his current speed can be represented by (3/4)S.
We also know that he is late by 2.5 hours. This means the difference between his usual time and his delayed time is 2.5 hours. Mathematically, we can represent this as:
T - (T + 2.5) = 2.5
Let's simplify the equation:
T - T - 2.5 = 2.5
-2.5 = 2.5
Since -2.5 is not equal to 2.5, this equation has no solution. It seems there is an inconsistency in the problem statement. Please double-check the given information or provide any additional details you may have so we can help you further.