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100 points for $49 worth of Veritas practice GMATs FREE VERITAS PRACTICE GMAT EXAMS Earn 10 Points Per Post Earn 10 Points Per Thanks Earn 10 Points Per Upvote speed/distance tagged by: Brent@GMATPrepNow This topic has 5 expert replies and 0 member replies speed/distance 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 oa is b Can any experts help me with this? How to come up with the correct answer? Thanks GMAT/MBA Expert Legendary Member Joined 14 Jan 2015 Posted: 2667 messages Followed by: 122 members Upvotes: 1153 GMAT Score: 770 Roland2rule 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 oa is b Can any experts help me with this? How to come up with the correct answer? Thanks Rate and Time have a reciprocal relationship, so if someone walks at 3/4 their normal speed, they'll take 4/3 as much time as they typically do. If Mike took 16 minutes longer than he typically does, and he typically takes T minutes, then we know that T + 16 = (4/3)T, and T = 48. The answer is B. _________________ Veritas Prep | GMAT Instructor Veritas Prep Reviews Save$100 off any live Veritas Prep GMAT Course

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GMAT/MBA Expert

GMAT Instructor
Joined
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Posted:
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GMAT Score:
770
Roland2rule 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
oa is b
Can any experts help me with this? How to come up with the correct answer?
Thanks
We can also solve this question algebraically.
Let d = distance to office
Let x = Mike's REGULAR walking speed
Time = distance/rate
So, Mike's REGULAR travel time to office = d/x

If Mike walks at 3/4 of his normal speed, then his NEW speed = (3/4)x = 3x/4
So, Mike's NEW travel time to office = d/(3x/4) = 4d/3x = (4/3)(d/x)
ASIDE: We can see that this matches what David stated above. That is, the NEW Mike's NEW travel time is 4/3 that of his regular travel time

Mike is 16 minutes late in reaching his office
We can say (Mike's NEW travel time) = (Mike's REGULAR travel time) + 16
Or we can write: (4/3)(d/x) = d/x + 16
Or.....(4/3)(d/x) = (1)(d/x) + 16
Subtract (d/x) from both sides of the equation to get: (1/3)(d/x) = 16
Multiply both sides by 3 to get: d/x = 48
Since d/x represents Mike's REGULAR travel time to office, the correct answer is B

Cheers,
Brent

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Posted:
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Roland2rule 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 = Mikeâ€™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 doesnâ€™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

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Jeffrey Miller
jeff@targettestprep.com

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GMAT/MBA Expert

GMAT Instructor
Joined
08 Dec 2008
Posted:
12985 messages
Followed by:
1249 members
5254
GMAT Score:
770
Roland2rule 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

_________________
Brent Hanneson â€“ Creator of GMATPrepNow.com
Use my video course along with

And check out all of these free resources

GMAT Prep Now's comprehensive video course can be used in conjunction with Beat The GMATâ€™s FREE 60-Day Study Guide and reach your target score in 2 months!

GMAT/MBA Expert

GMAT Instructor
Joined
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Posted:
2801 messages
Followed by:
18 members
43
Roland2rule 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 the usual speed = r and the usual time = t and create the equation:

rt = (3r/4)(t + 16)

rt = 3rt/4 + 12r

t = 3t/4 + 12

Multiplying by 4 we have:

4t = 3t + 48

t = 48

_________________

Scott Woodbury-Stewart
Founder and CEO
scott@targettestprep.com

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