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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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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
[Reveal] Spoiler: OA

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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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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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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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New post 13 Aug 2017, 10:16
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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Walking at 3/4 of his normal speed, Mike is 16 minutes late in reachin [#permalink]

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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

Kudos [?]: 25 [1], given: 157

Walking at 3/4 of his normal speed, Mike is 16 minutes late in reachin   [#permalink] 13 Aug 2017, 14:59
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