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Verbal Problem:

 

A car leaves city A to city B,  the way's length is 210 km.

20 minutes after it, a courier on a bike leaves city A so he can pass a delivery to the car.His speed is 120 km/h.

Immmediately after he passes the delivery, he turns back to city A.The car keeps moving with the same speed, and at the moment it reaches city B, the courirer reaches city A.

Find car's speed.

SabiBoo  Apr 27, 2018
edited by SabiBoo  Apr 27, 2018
 #1
avatar
+1

Average speed = Distance travelled / Time

Let the distance of the meet-up = D

The distance from the meet-up to city B = 210 - D

Let speed of the car =S

Let the time taken by the car until they meet = T(hours)

The time taken by the biker to catch up to the car =T - 1/3(hours)

S*T = 120[T - 1/3], solve for S

S = 120 - 40/T - speed of the car

T*[120 -40/T] + 120[T - 1/3] = 2D,

[210 - D] / [120 - 40/T] =D/120, solve for D, T

D =120Km, and T =4/3 hours.

120/(4/3) =90 Km/h - The speed of the car.

 

 

 

 

Guest Apr 27, 2018
edited by Guest  Apr 27, 2018
edited by Guest  Apr 27, 2018
 #2
avatar+91046 
+1

Let R  be the car's rate

 

Call the time for the car to travel  (in hours)  =  210/R

 

Call the total time for the bike to travel  (in hours)  =   210/R - 1/3

 

Now.....the distance bike travels   =  its rate * its time  =  120 *  [ (210/R) - 1/3 ] 

 

And the distance the  bike travels to meet the  car  = 120* [ (210/R - 1/3] / 2  =    60[ (210/R - 1/3 ] = the same distance the bike travels back to "A"

 

So.....the distance the bike travels back to "A" / ts rate  =  some time   (1)

 

And the distance the car travels after meeting the bike / its rate  =  the same time   (2)

 

So.......equating  (1)  and (2) in algebraic terms, we have

 

( 210 - 60 [ 210/R - 1/3] ) / R  =    60[ 210/R-  1/3 ] / 120

 

(210  - 60 [ 210/R - 1/3 ] ) / R  =   [ 210/ R - 1/3 ] / 2        cross-multiply

 

2(210 - 60[ 210/R - 1/3 ] =  R [ 210/R - 1/3 ]

 

420  - 120 [ 210/R - 1/3 ]  =  210  -  R /3 

 

420  - 25200/ R  +  40  =  210 - R/3

 

460 - 25200/R  = 210 - R/3 

 

R/3 + 250 - 25200/R  = 0        multiply though by R

 

(1/3)R^2 + 250R - 25200  = 0       multiply through by 3

 

R^2  + 750R  - 75600  = 0          factoring, we get

 

(R - 90) (R + 840)  = 0

 

Only the first factor set to 0  will provide a positive solution for R  =  90 kmh   =   the car's rate

 

 

cool cool cool

CPhill  Apr 28, 2018

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