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Bunuel
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Mr. Smitherly leaves Cedar Rapids at 8 a.m. and drives north on the hi 26 Jan 2016, 02:46
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C
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Difficulty:

25% (medium)

Question Stats:

77% (01:55) correct 23% (02:35) wrong based on 208 sessions

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Mr. Smitherly leaves Cedar Rapids at 8 a.m. and drives north on the highway at an average speed of 50 miles per hour. Mr. Dinkle leaves Cedar Rapids at 8:30 a.m. and drives north on the same highway at an average speed of 60 miles per hour. Mr. Dinkle will

A. overtake Mr. Smitherly at 9:30 a.m.
B. overtake Mr. Smitherly at 10:30 a.m.
C. overtake Mr. Smitherly at 11:00 a.m.
D. be 30 miles behind at 8:35 a.m.
E. never overtake Mr. Smitherly
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C
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Mr. Smitherly leaves Cedar Rapids at 8 a.m. and drives north on the hi 26 Jan 2016, 04:12
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Bunuel
Mr. Smitherly leaves Cedar Rapids at 8 a.m. and drives north on the highway at an average speed of 50 miles per hour. Mr. Dinkle leaves Cedar Rapids at 8:30 a.m. and drives north on the same highway at an average speed of 60 miles per hour. Mr. Dinkle will

A. overtake Mr. Smitherly at 9:30 a.m.
B. overtake Mr. Smitherly at 10:30 a.m.
C. overtake Mr. Smitherly at 11:00 a.m.
D. be 30 miles behind at 8:35 a.m.
E. never overtake Mr. Smitherly
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Since they are travelling in the same direction, Relative Speed= (60-50)mph=10 mph
Mr. Smith travelled 30 min early so the distance covered by him=50*1/2=25miles
Now,
Time=Distance between them/Relative Speed=25/10=2.5 hour
Since. Mr.Dinkle started at 8:30 am he will overtake Mr.Smith in 2.5 hour i.e. at 11:00 am
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Mr. Smitherly leaves Cedar Rapids at 8 a.m. and drives north on the hi 27 Jan 2016, 03:00
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Another approach if stuck with above solution with relative speed. Although I know it, I suddenly forgot the relative speed, maybe was sleepy :lol: :lol:

Scanning answers quickly so can do it manually

Time 8:30 9:30 10:30 11:00
D 0 60 120 150 (Not that D covers 60 mile in 1 hour, so he covers 30 in 1/2 hour)
S 25 75 125 150 (Not that D covers 50 mile in 1 hour, so he covers 25 in 1/2 hour)

Answer: C
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Mr. Smitherly leaves Cedar Rapids at 8 a.m. and drives north on the hi 27 Jan 2016, 11:55
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Bunuel
Mr. Smitherly leaves Cedar Rapids at 8 a.m. and drives north on the highway at an average speed of 50 miles per hour. Mr. Dinkle leaves Cedar Rapids at 8:30 a.m. and drives north on the same highway at an average speed of 60 miles per hour. Mr. Dinkle will

A. overtake Mr. Smitherly at 9:30 a.m.
B. overtake Mr. Smitherly at 10:30 a.m.
C. overtake Mr. Smitherly at 11:00 a.m.
D. be 30 miles behind at 8:35 a.m.
E. never overtake Mr. Smitherly
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My Take:C
According to concept of Relative :-
Distance traveled by Mr S in 30 mnts(the difference in both) = 25 Miles.
As both were travelling in same direction, Relative speed of Mr. D -> 60-50 =10 Miles/ Hr
Hence, Time needed to cover 25 Miles with a relative Speed of 10 Miles/ Hr. by Mr. D = 25/10 = 2.5 hrs.
8.30 AM + 2.5 = 11 AM
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Mr. Smitherly leaves Cedar Rapids at 8 a.m. and drives north on the hi 19 Apr 2017, 16:13
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Bunuel
Mr. Smitherly leaves Cedar Rapids at 8 a.m. and drives north on the highway at an average speed of 50 miles per hour. Mr. Dinkle leaves Cedar Rapids at 8:30 a.m. and drives north on the same highway at an average speed of 60 miles per hour. Mr. Dinkle will

A. overtake Mr. Smitherly at 9:30 a.m.
B. overtake Mr. Smitherly at 10:30 a.m.
C. overtake Mr. Smitherly at 11:00 a.m.
D. be 30 miles behind at 8:35 a.m.
E. never overtake Mr. Smitherly
Show more

Let’s set this up as a catch-up problem in which distance of Mr. Smitherly = distance of Mr. Dinkle. Since Mr. Smitherly leaves Cedar Rapids at 8 a.m. and Mr. Dinkle leaves Cedar Rapids at 8:30 a.m., we can let t = the time of Mr. Dinkle, and thus t + 1/2 = the time of Mr. Smitherly. Thus, the distance of Mr Dinkle is 60t and the distance of Mr. Smitherly is 50(t + 1/2) = 50t + 25. Thus:

60t = 50t + 25

10t = 25

t = 25/10 = 2.5 hours

We see that Mr. Dinkle will overtake Mr. Smitherly at 8.5 + 2.5 = 11:00 a.m.

Answer: C
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Mr. Smitherly leaves Cedar Rapids at 8 a.m. and drives north on the hi 02 May 2023, 21:51
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