Walking at 3/4 of his normal speed, Mike is 16 minutes late in reachin

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
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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

Let's start with a word equation

Distance traveled at 3/4 speed = Distance traveled at regular speed

Let v = regular walking speed (in miles/minute)
So, 3v/4 = REDUCED walking speed (in miles/minute)

Let t = regular travel time (in minutes)
So, t + 16 = travel time (in minutes) when walking 3/4 speed

Distance = (speed)(time)

So, we get: (3v/4)(t + 16) = vt
Expand: 3vt/4 + 12v = vt
Multiply both sides by 4 to get: 3vt + 48v = 4vt
Subtract 3vt from both sides: 48v = vt
Rewrite as: vt - 48v = 0
Factor: v(t - 48) = 0
So, EITHER v = 0 PR t = 48
Since the speed (v) cannot be zero, it must be the case that t = 48

Answer: B

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Would you explain how do you get t=48?

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

Lets assume his normal speed S=12; then T =T
3/4 of his speed S=9; then T = T-16
12T = 9(T-16)
T = 48

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


Don't forget some kudos if this helped you

3v/4*(t+16)=vt
t/4=12
t=48
Option B

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Solution:

As Mike walks at 3/4th of his speed, he has reduced his original speed by 1-3/4 =1/4

=> Increase in time = 1/(4-1) = 1/3

1/3 = 16 min

=>Originally he took 3*16 =48 min to reach (option b)

Devmitra Sen
GMAT SME

When the distance is constant, the speed traveled at is Inversely Proportional to the Time taken

If one of the quantities DECREASES by:

— (N) / (D + N)

The OTHER quantity will INCREASE by:

+ (N) / (D)

We are told he drives at 3/4th of his speed

This is equivalent to a Decrease in speed by: — (1 / 4)

since the distance is unchanging, the TIME taken will INCREASE by: + (1/3)

and this +1/3rd increase in Time corresponds to the + 16 minutes that he was late by

Let T = normal time

(1/3) T = 16 minutes

T = 48 minutes

B

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