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

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Bunuel
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On a long drive, a truck covered P percent of the total distance on lo [#permalink] New post   22 Dec 2019, 23:33
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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) \(50 - \frac{P}{5}\)

(C) \(\frac{(2P+1000)}{40}\)

(D) \(\frac{10000}{(P+200)}\)

(E) \(\frac{15000}{(2P+300)}\)
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E
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Re: On a long drive, a truck covered P percent of the total distance on lo [#permalink] New post   23 Dec 2019, 01:11
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Let total distance be x
Avg speed= x/((px/3000)+(x-px)/5000) = 1/p/1000(1/3-1/5)= 15000(2P+300)

OA:E
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Re: On a long drive, a truck covered P percent of the total distance on lo [#permalink] New post   23 Dec 2019, 03:01
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1) dist - p
Speed - 30
Time - p/30

2) dist - 100-p(because in percentage)
Speed - 50
Time - (100-p)/50

Av vel. = Total dist/total time
Av vel. = (P + 100 - p)/{(p/30) +
(100-p)/30}
Av vel. = 15000/(2p+300)
IMO- E
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Re: On a long drive, a truck covered P percent of the total distance on lo [#permalink] New post   23 Dec 2019, 10:03
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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−P5

(C) \(\frac{(2P+1000)}{40}\)

(D) \(\frac{10000}{(P+200)}\)

(E) \(\frac{15000}{(2P+300)}\)

Let Distance be = 1 unit
\(t1 = \frac{0.P}{30}\)
\(t2 = \frac{1 - 0.P}{50}\)
Avg. vel. = \(\frac{D}{t}\)
= \(\frac{1}{(0.P/30 + (1 - 0.P)/50)}\)
= \(\frac{150}{0.2P+3}\)

Answer E.
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Re: On a long drive, a truck covered P percent of the total distance on lo [#permalink] New post   23 Dec 2019, 11:21
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let total distance = 100
and P = 30% ; so P distance = 30
time = 30/30 ; 1 hr
and 70/50 ; 1.4 hrs
total time ; 2.4 hrs
total avg speed= 100/2.4 ; 41.6
IMO E 15000/(2P+300)

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 lo [#permalink] New post   23 Dec 2019, 12:26
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Quote:
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
(C) (2P+1000)/40
(D) 10000/(P+200)
(E) 15000/(2P+300)


METHOD 1
average rate = d / (t1+t2)
t1=(pd/100)/30=pd/3000
t2=(1-(p/100))d/50=d(100-p)/5000
rate=d/[pd/3000+d(100-p)/5000]
rate=1/[p/3000+100-p/5000]
rate=1/[(5p+300-3p)/15*1000]
rate=1/[(2p+300)/15000]
rate=15000/(2p+300)

Ans (E)

METHOD 2
if p=50 then,
avg rate=2*r_1*r_2/(r_1+r_2)
avg rate=2*30*50/(30+50)
avg rate=2*1500/80
avg rate=150/4=37.5
substitute p=50 in answer choices and find 37.5

Ans (E)
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Re: On a long drive, a truck covered P percent of the total distance on lo [#permalink] New post   23 Dec 2019, 19:53
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Let x be the total distance of the road.
—> p% —( p/100)*x

\(Ave.speed= Distance/(t_1+t_2)\)

—> x/(px/3000+ (x—px/100)/50))=

x/ (px/3000 + (100x—px)/5000))=

x/ (5000px + 300000x—3000px)/15000000 =

15000000x/(2000px +300000x )=

15000/ (2p+ 300)

The answer is E .

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Re: On a long drive, a truck covered P percent of the total distance on lo [#permalink] New post   26 Dec 2019, 21:17
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Bunuel wrote:

Competition Mode Question



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) \(50 - \frac{P}{5}\)

(C) \(\frac{(2P+1000)}{40}\)

(D) \(\frac{10000}{(P+200)}\)

(E) \(\frac{15000}{(2P+300)}\)



We can let the total distance = 100 miles. Since average rate = (total distance)/(total time), we have:

100 / (P/30 + (100 - P)/50)

100 / (5P/150 + (300 - 3P)/150)

100 / ((300 + 2P)/150)

15000 / (300 + 2P)

Answer: E
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Re: On a long drive, a truck covered P percent of the total distance on lo [#permalink] New post   04 May 2023, 20:56
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Re: On a long drive, a truck covered P percent of the total distance on lo [#permalink]
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