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

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Walking at 3/4 of his normal speed, Mike is 16 minutes late in reachin [#permalink]
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Bunuel wrote:
Walking at 3/4 of his normal speed, Mike is 16 minutes late in reaching his office. The usual time taken by him to cover the distance between his home and his office is

A. 42 minutes
B. 48 minutes
C. 60 minutes
D. 62 minutes
E. 66 minutes

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

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Re: Walking at 3/4 of his normal speed, Mike is 16 minutes late in reachin [#permalink]
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jn30 wrote:
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

Would you explain how do you get t=48?
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Re: Walking at 3/4 of his normal speed, Mike is 16 minutes late in reachin [#permalink]
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pclawong wrote:
jn30 wrote:
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

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
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Re: Walking at 3/4 of his normal speed, Mike is 16 minutes late in reachin [#permalink]
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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
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Re: Walking at 3/4 of his normal speed, Mike is 16 minutes late in reachin [#permalink]
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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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Walking at 3/4 of his normal speed, Mike is 16 minutes late in reachin [#permalink]
3v/4*(t+16)=vt
t/4=12
t=48
Option B

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Re: Walking at 3/4 of his normal speed, Mike is 16 minutes late in reachin [#permalink]
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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)

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Walking at 3/4 of his normal speed, Mike is 16 minutes late in reachin [#permalink]
Bunuel wrote:
Walking at 3/4 of his normal speed, Mike is 16 minutes late in reaching his office. The usual time taken by him to cover the distance between his home and his office is

A. 42 minutes
B. 48 minutes
C. 60 minutes
D. 62 minutes
E. 66 minutes

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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Re: Walking at 3/4 of his normal speed, Mike is 16 minutes late in reachin [#permalink]
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Re: Walking at 3/4 of his normal speed, Mike is 16 minutes late in reachin [#permalink]
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