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Adam began driving from home on a trip averaging 30 miles per hour. Ho

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Adam began driving from home on a trip averaging 30 miles per hour. Ho  [#permalink]

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New post 26 Dec 2016, 12:01
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Adam began driving from home on a trip averaging 30 miles per hour. How many miles per hour must jimmy drive on average to catch up to him in exactly when Adam drives for 3.5 hours, if jimmy leaves 30 minutes after Adam?

A. 39
B. 55
C. 35
D. 40
E. 60
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Re: Adam began driving from home on a trip averaging 30 miles per hour. Ho  [#permalink]

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New post 26 Dec 2016, 12:26
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Distance traveled by Adam = 30 * 3.5 = 105

Jimmy must complete 105 miles in 3 hours --> Speed = 105/3 = 35

Answer: C
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Re: Adam began driving from home on a trip averaging 30 miles per hour. Ho  [#permalink]

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New post 23 Aug 2017, 22:33
Somehow I am not arriving at the right answering using Shrinking/Expanding Distance approach. Any help?
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Re: Adam began driving from home on a trip averaging 30 miles per hour. Ho  [#permalink]

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New post 23 Aug 2017, 23:19
1
Jimmy starts after half an hour. Adam will have traveled 15 miles by now.
So Jimmy will have to cover 15 miles in 3 hours. (Because jimmy starts half an hour after Adam)
So the relative difference between there speeds when travelling in same direction \(s_1\)-\(s_2\)= \(\frac{15}{3}\) = 5
Adam's speed is 30
so Jimmy's speed is 30+5 =35
Option C
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Adam began driving from home on a trip averaging 30 miles per hour. Ho  [#permalink]

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New post 24 Aug 2017, 10:36
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rohit8865 wrote:
Adam began driving from home on a trip averaging 30 miles per hour. How many miles per hour must jimmy drive on average to catch up to him in exactly when Adam drives for 3.5 hours, if jimmy leaves 30 minutes after Adam?

A. 39
B. 55
C. 35
D. 40
E. 60

1. Gap = Distance
When Adam begins driving before Jimmy, Adam creates a distance "gap" between them:

\(D = rt\), so \(D_{gap} = 30 mph *
\frac{1}{2}\) hour = 15 miles

2. Relative speeds / rates

Jimmy must be driving faster than Adam to catch Adam.

Use relative speeds where, if two travelers move in same direction ("chase"), subtract slower rate from faster one.

Relative speed = \(Rate_{Jimmy} - Rate_{Adam}\)

Call them \(R_2\) and \(R_1\)

3. Equation and answer

(\(R_2 - R_1) * t = D\) -->

(relative) \(r * t = D\), so

relative \(r = \frac{D}{t}\)

Time is 3 hours (to close the gap when both are driving)

Jimmy's rate is \(R_2\)
Adam's rate is 30

relative \(r = \frac{D}{t}\), where RHS = \(\frac{15}{3}\)

\(R_2\) - 30 = \(\frac{15}{3}\)

\(R_2\) - 30 = 5

\(R_2\) = 35

Answer C
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Re: Adam began driving from home on a trip averaging 30 miles per hour. Ho  [#permalink]

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New post 24 Aug 2017, 13:01
genxer123 wrote:
rohit8865 wrote:
Adam began driving from home on a trip averaging 30 miles per hour. How many miles per hour must jimmy drive on average to catch up to him in exactly when Adam drives for 3.5 hours, if jimmy leaves 30 minutes after Adam?

A. 39
B. 55
C. 35
D. 40
E. 60

1. Gap = Distance
When Adam begins driving before Jimmy, Adam creates a distance "gap" between them:

\(D = rt\), so \(D_{gap} = 30 mph *
\frac{1}{2}\) hour = 15 miles

2. Relative speeds / rates
.
.
.
.

Time is 3 hours (to close the gap when both are driving)




Thanks, Genxer but why is the time 3 hours and not 3.5? Is it because we have already accounted for 30 mins when Jimmy will be catching up with Adam?
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Adam began driving from home on a trip averaging 30 miles per hour. Ho  [#permalink]

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New post 24 Aug 2017, 13:47
Blackbox wrote:
genxer123 wrote:
rohit8865 wrote:
Adam began driving from home on a trip averaging 30 miles per hour. How many miles per hour must jimmy drive on average to catch up to him in exactly when Adam drives for 3.5 hours, if jimmy leaves 30 minutes after Adam?

A. 39
B. 55
C. 35
D. 40
E. 60

1. Gap = Distance
When Adam begins driving before Jimmy, Adam creates a distance "gap" between them:

\(D = rt\), so \(D_{gap} = 30 mph *
\frac{1}{2}\) hour = 15 miles

2. Relative speeds / rates
.
Time is 3 hours (to close the gap when both are driving)


Thanks, Genxer but why is the time 3 hours and not 3.5? Is it because we have already accounted for 30 mins when Jimmy will be catching up with Adam?

Blackbox - Hmm. I'm not sure exactly what you mean when you say "we have already accounted for [the first 30 minutes]."

If you mean "we ignore that extra 30 minutes because in this and similar gap problems it can't be used to find the time for catch-up or the rate of catch-up [and we've used it to find the distance of catch-up]," then yes, that is correct. We have used that 30-minute time block to find the distance of the gap between the drivers; it has nothing to do with the chase while the chase is on.

If you mean, "we ignore that extra 30 minutes because it is used only to find the distance/gap (and hence is accounted for)," then the answer is -- not all the time.

Here? Yes. That's not true in all problems. If the question asked you to find the ratio of miles driven by Jimmy to miles driven by Adam, you would have to go back and use that 30-minute time block again (to find Adam's total distance.) I'm just being cautious, because these problems can get confusing.

Generally, in gap problems, when you are asked about anything that pertains only to the time when both travelers are traveling, yes, you use the time the second traveler started. Generally.

Finally (wow am I trying to be careful!): In "chase" problems, if you are using relative speed and trying to find the rate of one person during the chase, or the time it takes to catch up, although you are using the distance created by the person who started first, you may only use the time that both are traveling at the same time -- which is, as you note, when the second person starts traveling.

It helped me to think about the time issue this way: When "solving for the gap" . . . . that gap only begins to close when the second driver starts (and gradually catches up to the slower one). So if the question has to do with the chase, you use the second traveler's beginning time. Does that make sense? :-)
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Re: Adam began driving from home on a trip averaging 30 miles per hour. Ho  [#permalink]

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New post 29 Aug 2017, 16:14
rohit8865 wrote:
Adam began driving from home on a trip averaging 30 miles per hour. How many miles per hour must jimmy drive on average to catch up to him in exactly when Adam drives for 3.5 hours, if jimmy leaves 30 minutes after Adam?

A. 39
B. 55
C. 35
D. 40
E. 60


We can let Jimmy’s rate = r, Jimmy’s time = 3, and Adam’s time = 3.5; thus:

3r = 30(3.5)

3r = 105

r = 35

Answer: C
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Adam began driving from home on a trip averaging 30 miles per hour. Ho  [#permalink]

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New post 14 Sep 2018, 08:21
OA:C

Speed of Adam = 30 miles/hr
Time duration of Adam's travel = 3.5 hours
Speed of Jimmy = x miles/hr
Time duration of Jimmy's travel = 3.5 hours - 0.5 hours = 3 hours
Distance by both Adam and Jimmy is same.

\(30*3.5=x*3\)

\(x=\frac{30*3.5}{3}=35\) miles/hr
Adam began driving from home on a trip averaging 30 miles per hour. Ho &nbs [#permalink] 14 Sep 2018, 08:21
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