jlgdr wrote:

guerrero25 wrote:

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

is there a way to solve it without equation ?

Any experts opinions on how to solve this one faster?

Maybe Karishma can provide some insight on as to how to solve this with ratios @Karishma

Cheers!

J

Yes, you can do it with the use of ratios and weighted average formula. In race questions, make a diagram.

Attachment:

Ques3.jpg [ 8.99 KiB | Viewed 1170 times ]
For some distance, ratio of speeds is Black: Blue = 3:2 (Speed of black car is 3x and that of Blue is 2x)

For rest of the distance, ratio is Black:Blue = 3:4.5 (Speed of black is still 3x, that of blue is 4.5x)

Note that they tie so the average speed of blue car is 3x. (the constant speed with which it should have run to cover the same distance in same time)

Ratio of time taken in first part and second part = t1/t2 = (4.5 - 3)/(3 - 2) = 3:2 (the weights in case of average speed is always time, not distance)

Ratio of distance covered in first part and second part will be inverse i.e. 2:3 (think from the perspective of the black car which has constant speed)

So distance covered in first part of journey = (2/5)*4000 = 1600

Answer (B)

P.S. - If you are looking for directions from me, pm me the link to ensure that I see the question.

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