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A truck departed from Newton at 11:53 a.m. and arrived in Far City, 2

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Bunuel
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A truck departed from Newton at 11:53 a.m. and arrived in Far City, 2 [#permalink] New post   13 Feb 2018, 05:46
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A truck departed from Newton at 11:53 a.m. and arrived in Far City, 240 miles away, at 4:41 p.m. on the same day. What was the approximate average speed of the
truck on this trip?

(A) 20 mph

(B) 34 mph

(C) 50 mph

(D) 54 mph

(E) 72 mph
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C
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Re: A truck departed from Newton at 11:53 a.m. and arrived in Far City, 2 [#permalink] New post   13 Feb 2018, 07:42
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Answer is E.

Simple way to do it is to just add 7 minutes to both start time and arrival time, making it 12:00 and 4:48 pm, making it easily calculate it in hours form -> 4 hours 48 minutes -> 4.8 hours

240 miles/4.8 hours = 50 mph.
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Re: A truck departed from Newton at 11:53 a.m. and arrived in Far City, 2 [#permalink] New post   13 Feb 2018, 08:21
Bunuel wrote:
A truck departed from Newton at 11:53 a.m. and arrived in Far City, 240 miles away, at 4:41 p.m. on the same day. What was the approximate average speed of the
truck on this trip?

(A) 20 mph

(B) 34 mph

(C) 50 mph

(D) 54 mph

(E) 72 mph


11:53 to 12:53 = 1 Hour
12:53 to 1:53 = 1 Hour
1:53 to 2:53 = 1 Hour
2:53 to 3:53 = 1 Hour
3:53 to 4:11 = 18 min

Total we have 258/60 Hr = 4.30 Hours

Final speed is 240 /4.30 = 55.81 Miles/Hour, IMHO (D)
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Re: A truck departed from Newton at 11:53 a.m. and arrived in Far City, 2 [#permalink] New post   13 Feb 2018, 12:03
Abhishek009 wrote:
Bunuel wrote:
A truck departed from Newton at 11:53 a.m. and arrived in Far City, 240 miles away, at 4:41 p.m. on the same day. What was the approximate average speed of the
truck on this trip?

(A) 20 mph

(B) 34 mph

(C) 50 mph

(D) 54 mph

(E) 72 mph


11:53 to 12:53 = 1 Hour
12:53 to 1:53 = 1 Hour
1:53 to 2:53 = 1 Hour
2:53 to 3:53 = 1 Hour
3:53 to 4:11 = 18 min

Total we have 258/60 Hr = 4.30 Hours

Final speed is 240 /4.30 = 55.81 Miles/Hour, IMHO (D)

The arrived time is 4:41. So the answer is C: 50 mph
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A truck departed from Newton at 11:53 a.m. and arrived in Far City, 2 [#permalink] New post   13 Feb 2018, 17:57
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Bunuel wrote:
A truck departed from Newton at 11:53 a.m. and arrived in Far City, 240 miles away, at 4:41 p.m. on the same day. What was the approximate average speed of the
truck on this trip?

(A) 20 mph

(B) 34 mph

(C) 50 mph

(D) 54 mph

(E) 72 mph

11:53 a.m. to 12:00 = 7 minutes
12:00 to 4:00 p.m. = 4 hours
4:00 to 4:41 = 41 minutes
41 + 7 = 48 minutes total
4 hours total

48 minutes = \(\frac{48}{60} = \frac{4}{5} = 0.8\) hours
4 hours + 0.8 hours = 4.8 hours = time taken

\(\frac{D}{T}= R\)

\(\frac{240miles}{4.8hrs}= 50\) miles per hour

Answer C

To make the arithmetic faster, use \(4\frac{4}{5}\) hours = \(\frac{24}{5}\) hours
\(\frac{240}{\frac{24}{5}} = 240 * \frac{5}{24} = 50\) mph
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Re: A truck departed from Newton at 11:53 a.m. and arrived in Far City, 2 [#permalink] New post   16 Feb 2018, 10:29
Expert Reply
Bunuel wrote:
A truck departed from Newton at 11:53 a.m. and arrived in Far City, 240 miles away, at 4:41 p.m. on the same day. What was the approximate average speed of the
truck on this trip?

(A) 20 mph

(B) 34 mph

(C) 50 mph

(D) 54 mph

(E) 72 mph


The trip took 4 hours and 48 minutes = 4 hours and 48/60 hours = 4 hours and 4/5 hours = 24/5 hours.

Thus, the average speed is 240/(24/5) = 1200/24 = 50 mph.

Answer: C
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Re: A truck departed from Newton at 11:53 a.m. and arrived in Far City, 2 [#permalink] New post   21 Oct 2018, 03:51
Time taken by Newton to travel = 11:53 am - 4:41 pm= 4 hours 48 mins= 24/5 hrs
Distance= 280 miles

Speed= distance/time= 280*5/24= 50 mph

Ans(C)
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Re: A truck departed from Newton at 11:53 a.m. and arrived in Far City, 2 [#permalink] New post   03 May 2023, 08:43
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Re: A truck departed from Newton at 11:53 a.m. and arrived in Far City, 2 [#permalink]
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