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Mr Ben leaves his house for work at exactly 8:00 AM every mo

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fozzzy
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Mr Ben leaves his house for work at exactly 8:00 AM every mo [#permalink] New post   11 Sep 2013, 02:23
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Mr Ben leaves his house for work at exactly 8:00 AM every morning. When he averages 40 miles per hour, he arrives at his workplace three minutes late. When he averages 60 miles per hour, he arrives three minutes early. At what average speed, in miles per hour, should Mr. Ben drive to arrive at his workplace precisely on time?

A) 45
B) 48
C) 50
D) 55
E) 58

Show Spoiler
Here is what I did

\(\frac{D}{40} - \frac{3}{60} = \frac{D}{60} + \frac{3}{60}\)

Solve this equation to get

\(D = 12\)

\(\frac{D}{40} - \frac{3}{60}\) ( will this equation give me the time?)

we get \(\frac{1}{4}\)

R * T = D

\(\frac{1}{4} * X = 12\)

X = 48

Is this correct?
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Bunuel
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Re: Mr Ben leaves his house for work at exactly 8:00 AM every mo [#permalink] New post   11 Sep 2013, 02:33
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fozzzy wrote:
Mr Ben leaves his house for work at exactly 8:00 AM every morning. When he averages 40 miles per hour, he arrives at his workplace three minutes late. When he averages 60 miles per hour, he arrives three minutes early. At what average speed, in miles per hour, should Mr. Ben drive to arrive at his workplace precisely on time?

A) 45
B) 48
C) 50
D) 55
E) 58

Here is what I did

\(\frac{D}{40} - \frac{3}{60} = \frac{D}{60} + \frac{3}{60}\)

Solve this equation to get

\(D = 12\)

\(\frac{D}{40} - \frac{3}{60}\) ( will this equation give me the time?)

we get \(\frac{1}{4}\)

R * T = D

\(\frac{1}{4} * X = 12\)

X = 48

Is this correct?


Yes, that's correct. From D=12 miles, you can get the time T in which he should get to the work (T=D/40-3/60) and then the required rate (R=D/T=12/(1/4)=48 miles).
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Re: Mr Ben leaves his house for work at exactly 8:00 AM every mo [#permalink] New post   11 Sep 2013, 02:45
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fozzzy wrote:
Mr Ben leaves his house for work at exactly 8:00 AM every morning. When he averages 40 miles per hour, he arrives at his workplace three minutes late. When he averages 60 miles per hour, he arrives three minutes early. At what average speed, in miles per hour, should Mr. Ben drive to arrive at his workplace precisely on time?

A) 45
B) 48
C) 50
D) 55
E) 58

Show Spoiler
Here is what I did

\(\frac{D}{40} - \frac{3}{60} = \frac{D}{60} + \frac{3}{60}\)

Solve this equation to get

\(D = 12\)

\(\frac{D}{40} - \frac{3}{60}\) ( will this equation give me the time?)

we get \(\frac{1}{4}\)

R * T = D

\(\frac{1}{4} * X = 12\)

X = 48

Is this correct?


I too followed the same approach. Here 8AM is an extra information.

Let the distance from house to work be x
& the right time required to reach office on time be t.

At 40 miles/hr, he is 3 minutes late from time
therefore,\(\frac{x}{40}\) = t + \(\frac{3}{60}\)

At 60 miles/hr, he is 3 minutes early
i.e. \(\frac{x}{60}\) = t - \(\frac{3}{60}\)

Now, either you equate t or subtract one equation from another,
you will get x=12

Now find t, putting x in any equation
\(\frac{12}{60}\)=t - \(\frac{3}{60}\)

t will come out to be, t = \(\frac{1}{4}\)

So Average speed to reach office precisely on time = \(\frac{x}{t}\) =\(\frac{12}{(1/4)}\) = 12*4 = 48

Ans B
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Re: Mr Ben leaves his house for work at exactly 8:00 AM every mo [#permalink] New post   11 Sep 2013, 02:56
A good thing about this question is that even if you miss the 3 minutes late/early part i.e. if by mistake solve for 3 hours late/early, you will get the same Answer, I suppose.
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Re: Mr Ben leaves his house for work at exactly 8:00 AM every mo [#permalink] New post   24 Sep 2013, 23:29
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Home...........................................Workplace
40 m/h t+3
60 m/h t-3
x m/h t

We know that distance is constant hence

40(t+3) = 60 (t-3) = Distance

t=15
Distance =720

X = Distance/t

hence x=48
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Re: Mr Ben leaves his house for work at exactly 8:00 AM every mo [#permalink] New post   26 Sep 2014, 04:54
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fozzzy wrote:
Mr Ben leaves his house for work at exactly 8:00 AM every morning. When he averages 40 miles per hour, he arrives at his workplace three minutes late. When he averages 60 miles per hour, he arrives three minutes early. At what average speed, in miles per hour, should Mr. Ben drive to arrive at his workplace precisely on time?

A) 45
B) 48
C) 50
D) 55
E) 58

Show Spoiler
Here is what I did

\(\frac{D}{40} - \frac{3}{60} = \frac{D}{60} + \frac{3}{60}\)

Solve this equation to get

\(D = 12\)

\(\frac{D}{40} - \frac{3}{60}\) ( will this equation give me the time?)

we get \(\frac{1}{4}\)

R * T = D

\(\frac{1}{4} * X = 12\)

X = 48

Is this correct?


Can someone tell me is there is any problem if I try and solve it like this:

Let usual time taken = t
Distance = d

Time taken if he's 3 minutes late = t + 3
Time taken if he's 3 minutes early = t - 3

Distance, d = 40 x (t + 3) [when he's late]
Distance, d = 60 x (t - 3) [when he's early]

Hence, 40 x (t + 3) = 60 x (t - 3)
t = 15

if we plug the value of t in any of the equations:
d = 40 x (15 + 3) = 40 x 18 = 720

Average speed = Distance / Time = 720 / 15 = 48
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Re: Mr Ben leaves his house for work at exactly 8:00 AM every mo [#permalink] New post   26 Sep 2014, 07:09
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iaratul wrote:
fozzzy wrote:
Mr Ben leaves his house for work at exactly 8:00 AM every morning. When he averages 40 miles per hour, he arrives at his workplace three minutes late. When he averages 60 miles per hour, he arrives three minutes early. At what average speed, in miles per hour, should Mr. Ben drive to arrive at his workplace precisely on time?

A) 45
B) 48
C) 50
D) 55
E) 58

Show Spoiler
Here is what I did

\(\frac{D}{40} - \frac{3}{60} = \frac{D}{60} + \frac{3}{60}\)

Solve this equation to get

\(D = 12\)

\(\frac{D}{40} - \frac{3}{60}\) ( will this equation give me the time?)

we get \(\frac{1}{4}\)

R * T = D

\(\frac{1}{4} * X = 12\)

X = 48

Is this correct?


Can someone tell me is there is any problem if I try and solve it like this:

Let usual time taken = t
Distance = d

Time taken if he's 3 minutes late = t + 3
Time taken if he's 3 minutes early = t - 3

Distance, d = 40 x (t + 3) [when he's late]
Distance, d = 60 x (t - 3) [when he's early]

Hence, 40 x (t + 3) = 60 x (t - 3)
t = 15

if we plug the value of t in any of the equations:
d = 40 x (15 + 3) = 40 x 18 = 720

Average speed = Distance / Time = 720 / 15 = 48


Yes, your approach is correct.
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Re: Mr Ben leaves his house for work at exactly 8:00 AM every mo [#permalink] New post   03 Dec 2017, 10:45
Bunuel wrote:
iaratul wrote:

Time taken if he's 3 minutes late = t + 3
Time taken if he's 3 minutes early = t - 3
Yes, your approach is correct.

Wouldn't t+3 be t+3/60?
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Re: Mr Ben leaves his house for work at exactly 8:00 AM every mo [#permalink] New post   22 Apr 2022, 09:11
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Re: Mr Ben leaves his house for work at exactly 8:00 AM every mo [#permalink]
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