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Bunuel
Car A travels at three times the average speed of car B. Car A started to travel at 12:00 o'clock, car B started to travel at 16:00 o'clock. What is the speed of car B (in Km/h) if the total distance that both cars traveled until 18:00 was 1000 Km?

A. 10.
B. 25.
C. 30.
D. 38.
E. 50.

Total distance= Distance travelled by A +Distance travelled by car B

Distance= speed*time
Distance of A= 3x * 6

Distance of B= x*2

(3x * 6) + (x*2) = 1000
x= 50 KM/hr

E is the answer
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let Sa and Sb be the speed of car A and B
and Da and Db be the distance travelled by car A and car B
as per the question
Sa=3Sb
Da=3Sb*6=18Sb
Db=Sb*2=2Sb
18Sb+2Sb=1000
Sb=1000/20=50Km/h
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\(3x:1x\) ratio of the rates, where \(x\) is the rate of car b
\(6:2\) ratio of the times
\(18x:2x\) multiply them together to get ratio of the distances
Therefore, since the sum of the two distances is the unknown multiplier and equals the distance of 1000, \(20x = 1000\), \(x = 50\)

Option E.
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Let x be the average speed of B. That gives average speed of A as 3x

Total distance = 3x*(18-12) + x*(18-16) = 1000

Solving this, x = 50

Ans: E
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Bunuel
Car A travels at three times the average speed of car B. Car A started to travel at 12:00 o'clock, car B started to travel at 16:00 o'clock. What is the speed of car B (in Km/h) if the total distance that both cars traveled until 18:00 was 1000 Km?

A. 10.
B. 25.
C. 30.
D. 38.
E. 50.

We can let x = the speed of car B and thus 3x = the speed of car A. We see that car A has traveled for 6 hours and car B has travel for 2 hours. Thus we have:

6(3x) + 2x = 1000

20x = 1000

x = 50

Answer: E
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Bunuel
Car A travels at three times the average speed of car B. Car A started to travel at 12:00 o'clock, car B started to travel at 16:00 o'clock. What is the speed of car B (in Km/h) if the total distance that both cars traveled until 18:00 was 1000 Km?

A. 10.
B. 25.
C. 30.
D. 38.
E. 50.

Let average speed of car B = X, hence speed of car A = 3X

Total distance traveled = 1000 km

Car A is driven for 6 hrs & Car B is driven for 2 hrs.

We have 6*(3X) + 2*X = 1000

hence X = 50 km per hr


Answer E.


Thanks,
GyM
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Given: Car A travels at three times the average speed of car B. Car A started to travel at 12:00 o'clock, car B started to travel at 16:00 o'clock.
Asked: What is the speed of car B (in Km/h) if the total distance that both cars traveled until 18:00 was 1000 Km?

Let the speed of car B be x kmh
Speed of car A = 3x kmh

Time taken by car A until 18:00 = 18 - 12 = 6 hours
Time taken by car B until 18:00 = 18 - 16 = 2 hours
Total distance travelled = 6*3x + 2x = = 20x = 1000 km
x = 1000/20 = 50 kmh

IMO E
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