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11 Sep 2013, 02:23
Kudos
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B
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Difficulty:
45%
(medium)
Question Stats:
71% (02:51) correct
29%
(02:19)
wrong
based on 246
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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
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?
\(\frac{D}{40} - \frac{3}{60} = \frac{D}{60} + \frac{3}{60}\)
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?
11 Sep 2013, 02:33
Kudos
Bookmarks
fozzzy
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?
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).
11 Sep 2013, 02:45
Kudos
Bookmarks
fozzzy
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
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?
\(\frac{D}{40} - \frac{3}{60} = \frac{D}{60} + \frac{3}{60}\)
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
