Let s = his normal speed
t = his normal time

Then
D = (3/4)s * (t+16)
Since the distance is the same we can equate this to his regular day which is D = s*t
s*t = (3/4)s * (t+16)
t=48

Answer: B

Look.
Vt=S (normal speed, normal time)
3/4V(t+16)=S (slow speed, late)
We can, for example divide second equation on first equation, thus we eliminate 2 variables.
3/4V(t+16) / Vt = S / S = 1
3/4(t+16)/t=1 ----> 3/4t+12=t ----> t = 48 mins

We can let r = Mike’s normal speed and t = the time he normally takes to reach his office from home. Thus, the distance between his home and his office is rt.

Since we are given that he is 16 minutes late in reaching his office when he walks at 3/4 of his normal speed, we can say that his new speed = (3/4)r and his new time = t + 16. Thus, the distance between his home and his office, in terms of the new speed and new time, is (3/4)r(t + 16).

Since the distance between his home and his office doesn’t change in relation to speed and time, we have:

rt = (3/4)r(t + 16)

Divide both sides by r, we have:

t = (3/4)t + 12

(1/4)t = 12

t = 48

Answer: B
_________________

Let the distance between home & office be 'D' and his normal speed be 'x'.

Mike usually takes (D/X) minutes or hours to reach office from home. This is what we need to find.

His new rate is 3x/4 so the time it takes is D/(3x/4) = 4D/3x

It says travelling at his new rate, he is 16 minutes late so
--> 4D/3x - D/x = 16 minutes
--> 4D-3D/3x = 16 minutes
--> D/3x = 16 minutes
--> D/x = 48 minutes


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