Problem Solving

Two cars painted yellow and red are racing against each other on a track of length 4000 meters. The race ends in a tie as both cars finish the race in exactly the same time. However at first when the race starts the yellow car moves 50% faster than the red car. The red car then quickened its pace and for the remaining distance moved 50% faster than the yellow car. When the red car quickened its pace what distance had it already covered?

(A) 1200meters
(B) 1600 meters
(C) 2400 meters
(D) 2800meters

[spoiler]OA= B[/spoiler]

I solved this question and I am getting the answer C.

Need expert help!

No, its [spoiler]{B}[/spoiler] only

AIM TO CRACK GMAT wrote:Two cars painted yellow and red are racing against each other on a track of length 4000 meters. The race ends in a tie as both cars finish the race in exactly the same time. However at first when the race starts the yellow car moves 50% faster than the red car. The red car then quickened its pace and for the remaining distance moved 50% faster than the yellow car. When the red car quickened its pace what distance had it already covered?

(A) 1200meters
(B) 1600 meters
(C) 2400 meters
(D) 2800meters

[spoiler]OA= B[/spoiler]

Initial speed of the red car = R
Initial speed of the yellow car = Y = 1.5R

After covering a distance of d, new speed of the red car = 1.5Y = 2.25R

Total time taken by the yellow car to cover 4000 meters = distance/speed = 4000/Y = 4000/(1.5R)

Total time taken by the red car to cover 4000 meters = d/R + (4000 - d)/(2.25R)

Since both the cars reach the finish line at the same time,
4000/(1.5R) = d/R + (4000 - d)/(2.25R)

Multiply both sides by 2.25R:
4000 * 1.5 = d * 2.25 + (4000 - d)
6000 = 2.25d + 4000 - d
2000 = 1.25d
d = 1600

Choose B

Cheers

The posted problem is virtually the same as the one below:

Two cars painted Black and Blue are racing against each other on a track of length 4000 meters. The race ends in a tie as both cars finish the race in exactly the same time. However at first when the race starts the Black car moves 50% faster than the Blue car. The Blue car then quickened its pace and for the remaining distance moved 50% faster than the Black car. When the Blue car quickened its pace what distance had it already covered?

(A) 1200 meters
(B) 1600 meters
(C) 2400 meters
(D) 2800 meters
(E) 3000 meters

Before the blue car speeds up:
Here, the black car travels 3/2 as fast as the blue car.
Thus, the black car travels 3/2 the distance of the blue car.

After the blue car speeds up:
Here, the blue car travels 3/2 as fast as the black car.
Thus, the blue car travels 3/2 the distance of the black car.

We can plug in the answers, which represent the distance traveled by the blue car before it speeds up.

Answer choice C: 2400m
Before the blue car speeds up:
Distance traveled by the blue car = 2400m.
Distance traveled by the black car = (3/2) * 2400 = 3600m.

After the blue car speeds up:
Remaining distance traveled by the black car = 4000-3600 = 400m.
Distance traveled by the blue car = (3/2) * 400 = 600m.

Total distance traveled by the blue car = 2400+600 = 3000m.
Doesn't work: the total distance traveled by the blue car must be 4000m.

To INCREASE the blue car's total distance to 4000m, the blue car must speed up EARLIER.
Thus, the distance traveled by the blue car before it speeds up must DECREASE.
Eliminate C, D and E.

Answer choice B: 1600m
Before the blue car speeds up:
Distance traveled by the blue car = 1600m.
Distance traveled by the black car = (3/2) * 1600 = 2400m.

After the blue car speeds up:
Remaining distance traveled by the black car = 4000-2400 = 1600m.
Distance traveled by the blue car = (3/2) * 1600 = 2400m.

Total distance traveled by the blue car = 1600+2400 = 4000m.
Success!

The correct answer is B.

Hi Friends, kindly help me on this

initial speed of Red car -- S(R1)
initial speed of Yellow car -- S(Y1) = 1.5 S(R1)
Latter Speed of Red car = S(R2) = 1.5 S(Y1) = 2.25 S(R1)

Hence, S(R1) / S(R2) = 1 / 2.25
Since Speed directly proportinal to distance,
can I equate S (R1) / S (R2) = D 1 / D2 , where D1 ,D2 distances covered by Red care initially and later. We are asked to find D1

Hence, D1 / D2 = 1 / 2.25 . since we know total distance = 4000
cant we get D1 = 1 / (1 + 2.25) * 4000.
I am not getting OA as my answer, kindly tell me where am I wrong in my approach!!

riz_gmat wrote: Hence, S(R1) / S(R2) = 1 / 2.25
Since Speed directly proportinal to distance,
can I equate S (R1) / S (R2) = D 1 / D2 , where D1 ,D2 distances covered by Red care initially and later.

riz_gmat, I believe the problem with the approach you used is the proportionality constant.
Yes, speed is proportional to Distance.. but the constant used is "Time"

If you write: S (R1) / S (R2) = D 1 / D2 , .. then you are assuming that Time was same for both the trips; that's not mentioned in the question.