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Walking at 3/4 of his normal speed, Mike is 16 minutes late in reachin
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20 Mar 2016, 08:46
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Question Stats:
87% (01:50) correct 13% (02:13) wrong based on 304 sessions
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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
Walking at 3/4 of his normal speed, Mike is 16 minutes late in reachin
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13 Aug 2017, 13:59
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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
Re: Walking at 3/4 of his normal speed, Mike is 16 minutes late in reachin
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04 Oct 2017, 15:45
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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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Walking at 3/4 of his normal speed, Mike is 16 minutes late in reachin
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27 Feb 2018, 13:23
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Top Contributor
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
Walking at 3/4 of his normal speed, Mike is 16 minutes late in reachin
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27 Jul 2018, 10:22
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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
gmatclubot
Walking at 3/4 of his normal speed, Mike is 16 minutes late in reachin &nbs
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27 Jul 2018, 10:22
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87% (01:50) correct 
