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

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Bunuel
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Walking at 3/4 of his normal speed, Mike is 16 minutes late in reachin [#permalink] New post   Updated on: 15 Mar 2019, 02:34
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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
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B

Originally posted by Bunuel on 20 Mar 2016, 09:46.
Last edited by Bunuel on 15 Mar 2019, 02:34, edited 1 time in total.
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Re: Walking at 3/4 of his normal speed, Mike is 16 minutes late in reachin [#permalink] New post   04 Oct 2017, 16:45
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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


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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Re: Walking at 3/4 of his normal speed, Mike is 16 minutes late in reachin [#permalink] New post   20 Mar 2016, 10:31
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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
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Walking at 3/4 of his normal speed, Mike is 16 minutes late in reachin [#permalink] New post   Updated on: 20 Apr 2021, 06:36
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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


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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Originally posted by BrentGMATPrepNow on 27 Feb 2018, 14:23.
Last edited by BrentGMATPrepNow on 20 Apr 2021, 06:36, edited 1 time in total.
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Re: Walking at 3/4 of his normal speed, Mike is 16 minutes late in reachin [#permalink] New post   13 Aug 2017, 10:16
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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

Answer: B


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] New post   13 Aug 2017, 14:59
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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

Answer: B


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] New post   01 Oct 2017, 18:01
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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] New post   27 Jul 2018, 11:22
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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


Don't forget some kudos if this helped you :)
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Walking at 3/4 of his normal speed, Mike is 16 minutes late in reachin [#permalink] New post   21 Aug 2019, 00:48
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] New post   21 Apr 2021, 05:11
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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] New post   11 Sep 2022, 21:47
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] New post   03 Dec 2023, 18:25
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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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