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speed/distance

by BTGmoderatorRO » Sun Oct 15, 2017 9:43 am
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
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by DavidG@VeritasPrep » Sun Oct 15, 2017 10:37 am
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.
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by Brent@GMATPrepNow » Sun Oct 15, 2017 12:45 pm
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,
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by Jeff@TargetTestPrep » Fri Jan 05, 2018 7:16 am
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 = 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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by Brent@GMATPrepNow » Tue Feb 27, 2018 1:23 pm
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
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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by Scott@TargetTestPrep » Thu Mar 01, 2018 5:42 pm
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

Answer: B

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