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Bunuel
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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Bunuel
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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Bunuel
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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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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