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On a long drive, a truck covered P percent of the total distance on

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On a long drive, a truck covered P percent of the total distance on  [#permalink]

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New post 22 Jul 2014, 08:00
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  35% (medium)

Question Stats:

79% (02:28) correct 21% (02:36) wrong based on 234 sessions

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On a long drive, a truck covered P percent of the total distance on local road, where it drove only 30 mph. The rest of the trip, it traveled at 50 mph. In terms of P, which of the following represents the average velocity of the entire trip?

(A) P-10
(B) 50 - p/5
(C) (2P+1000)/40
(D) 10000/(p+200)
(E) 15000/(2P+300)
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Re: On a long drive, a truck covered P percent of the total distance on  [#permalink]

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New post 22 Jul 2014, 19:36
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Refer diagram below:

Time taken to travel distance p \(= \frac{p}{30}\)

Time taken to travel remaining distance \(= \frac{100-p}{50}\)

Total time consumed \(= \frac{p}{30} + \frac{100-p}{50}\)

\(= \frac{300+2p}{150}\)

Total Distance Travelled = 100

Average Velocity\(= \frac{100}{\frac{300+2p}{150}}\)

\(= \frac{15000}{300+2p}\)

Answer = E
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Re: On a long drive, a truck covered P percent of the total distance on  [#permalink]

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New post 22 Jul 2014, 11:46
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if total distance is D then time taken in first part of journey=PD/(100*30)
and time taken in rest of journey= (100-P)D/(100*50)
now avg velocity= total distance /total time
=15000/(2P+300)
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Re: On a long drive, a truck covered P percent of the total distance on  [#permalink]

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New post 22 Jul 2014, 12:41
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In my view, one quicker approach would be to apply smart numbers in the problem statement to come up with the desired value (average velocity of entire trip) and then cross-check that value with answer choices. For e.g.,

I'd take total distance of the trip as 150 miles (LCM of 30 and 50). Now for P, first value that comes to my mind is 20% - that makes first part of my calculation very easy - time taken to cover P% of total distance, i.e., time taken to cover 30 miles (20% of 150 miles) = 1 hour (=30 miles / 30 mph). Time taken to cover rest of the distance (120 miles) = 120/50 = 2.4 hr. So, average velocity of the entire trip = Total distance / total time taken = 150 / (1 + 2.4) = 1500/34 mph.

Scanning the answer choices, E comes out as clear winner. Although in real test, there should not be any need to seriously evaluate other answer choices, but still, for ease of understanding, checking other choices as well:

A. P - 10 = 20 - 10 = 10 --> Not Correct
B. 50 - P/5 = 50 - 20/5 = 46 ---> Not Correct
C. (2P + 1000) / 40 = 1040/40 = 26 --> Not Correct
D. 10000 / (P + 200) = 10000 / 220 --> Wouldn't even bother to solve --> Not Correct
E. 15000 / (2P + 300) = 15000 / 2*20 + 300 = 15000/340 = 1500/34 --> Correct.

Correct answer choice - E.

Overall, this approach might look bit longer to you but if you are good in numbers, I don't think this would take more than 60-70 secs on real test.
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Re: On a long drive, a truck covered P percent of the total distance on  [#permalink]

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New post 13 Nov 2016, 12:37
Q: Ask you to find the total average rate in terms of a percentage ex: P%

Step1: define P% and build your RTD chart

RTD
30(t1) = some portion of d (?)
50(t2) = some portion of d
(?) (tt) = total distance (d)

p% = the percentage of the total distance travelled at 30mph.
So I need (?)s from above. The total avg rate and the % distance at 30mph.

Step2: assign an easy split (50/50) for total distance (d). P = 50%

Step3: Break/solve the two pieces for t1 and t2 and add to get the total time (tt)
rtd=> 30(t1)=50 = 5/3 and rtd => 50(t2) = 50 = 1. now => 5/3 + 3/3 = 8/3.

Step4: find the average total rate: r (8/3) = 100 => r = 300/8

Step5: remember to solve p = 50% to get a result of r = 300/8 when you try it in options A through E.
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Re: On a long drive, a truck covered P percent of the total distance on  [#permalink]

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New post 14 Nov 2016, 00:18
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Maksym wrote:
On a long drive, a truck covered P percent of the total distance on local road, where it drove only 30 mph. The rest of the trip, it traveled at 50 mph. In terms of P, which of the following represents the average velocity of the entire trip?

(A) P-10
(B) 50 - p/5
(C) (2P+1000)/40
(D) 10000/(p+200)
(E) 15000/(2P+300)

Would be grateful for any quick approach


Let total distance be 100

Travelled @ 30mph

Distance = \(p\)
Speed = \(30\)
Time = \(\frac{p}{30}\)

Travelled @ 50mph

Distance = \(100 - p\)
Speed = \(50\)
Time = \(\frac{100 - p}{50}\)

Total time taken = \(\frac{p}{30}\) + \(\frac{100 - p}{50}\)

Or, Total time taken = \(\frac{p}{30}\) + \(\frac{100 - p}{50}\) = \(\frac{2p + 300}{150}\)


Hence, average speed = \(100\)/\(\frac{2p + 300}{150}\) = \(\frac{15000}{(2P+300)}\)

Hence, Correct answer must be (E)...


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Re: On a long drive, a truck covered P percent of the total distance on  [#permalink]

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Re: On a long drive, a truck covered P percent of the total distance on &nbs [#permalink] 17 Jan 2018, 16:52
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