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

Question

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−P
(C) (2P+1000)/40
(D) 10000/(P+200)
(E) 15000/(2P+300)

(A)  P-10

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

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

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

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

yatharthdas · Accepted Answer

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
   Posted from my mobile device

madgmat2019 · Answer

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

unraveled · Answer

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)

(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.

Archit3110 · Answer

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
(C) (2P+1000)/40
(D) 10000/(P+200)
(E) 15000/(2P+300)

(A) P−10

(B) 50−P/5

(C) (2P+1000)/40
(D) 10000/(P+200)

(E) 15000/(2P+300)

exc4libur · Answer

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/ 
rate=1/ 
rate=1/ 
rate=1/ 
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)

lacktutor · Answer

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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ScottTargetTestPrep · Answer

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

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