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# Two cars painted Black and Blue are racing against each

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Two cars painted Black and Blue are racing against each  [#permalink]

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02 Oct 2013, 16:53
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Difficulty:

75% (hard)

Question Stats:

67% (03:22) correct 33% (03:21) wrong based on 224 sessions

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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 ?
Math Expert
Joined: 02 Sep 2009
Posts: 49384
Re: Two cars painted Black and Blue are racing against each  [#permalink]

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27 Nov 2013, 06:04
5
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 ?

At first when the race starts the Black car moves 50% (3/2) faster than the Blue car:
The Black car covers 3/2 the distance of the Blue car.

After the Blue car quickens and moves 50% (3/2) faster than the Black car:
The Blue car covers 3/2 the distance of the Black car.

Plug-in the options:
(C) if the Blue covered 2400 meters before it quickened, then the Black car covered 2400*3/2=3600 meters. The remaining distance covered by the Black car is then 400-3600=400, thus the Blue car covered 400*3/2=600. Total distance covered by the Blue car 2400+600=3000, which is less than the actual distance of 4000 meters.

This means that the Blue car must have quickened earlier then 2400 meters.

Check (B):
If the Blue covered 1600 meters before it quickened, then the Black car covered 1600*3/2=2400 meters. The remaining distance covered by the Black car is then 400-2400=1600, thus the Blue car covered 1600*3/2=2400. Total distance covered by the Blue car 1600+2400=4000. Bingo!

Hope it's clear.
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Re: Two cars painted Black and Blue are racing against each  [#permalink]

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02 Oct 2013, 20:24
8
I think the answer is 1600 meters.

I have attached the solution I wrote down, so that you can understand the steps :- Hope it helps!

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Re: Two cars painted Black and Blue are racing against each  [#permalink]

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02 Oct 2013, 21:03
1
Thanks ... both the question and solution are good!
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Two cars painted Black and Blue are racing against each  [#permalink]

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Updated on: 14 Jul 2014, 06:36
3
Much easier way to solve this problem:

Let's say the blue car traveled $$x$$ meters at the first speed; then the Black car traveled $$1.5x$$, meaning the black car needs to travel $$4000 - 1.5x$$, during which the blue car travels $$(4000 - 1.5x)(1.5).$$

Therefore, the total distance traveled by the blue car = $$x + (4000 - 1.5x)(1.5) = 4000$$

$$x+ 6000 - 9x/4 = 4000$$

$$2000 = 5x/4$$

$$x =1600$$ meters.

Originally posted by eaze on 01 Apr 2014, 11:27.
Last edited by eaze on 14 Jul 2014, 06:36, edited 1 time in total.
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Re: Two cars painted Black and Blue are racing against each  [#permalink]

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01 Apr 2014, 21:39
1
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 2646 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

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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Re: Two cars painted Black and Blue are racing against each  [#permalink]

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03 Apr 2014, 11:23
1
First: We know that the Black Car starts 50% faster than the Blue car. Lets traduce this into variables:
Rate Blue Car: x
Rate Black Car: (3/2)*x

Second: The Blue car quickened its pace and for the remaining distance moved 50% faster than the Black car.
NEW Rate Blue Car: (3/2)*(3/2)x = (9/4)*x
Rate Black Car: (3/2)*x

Note tha Rate of the Black Car remains constant over the entire race.

Third: We know that the race end in a tie so both cars took the same time to complete the 4000 meters. Knowing this we can equal the Time that each car took.
Time=Distance/Rate

Time Black Car: (4000/(3/2)*x)
Time Blue Car: (y/x + (4000-y))/ (9/4)*x
Time Black Car= Time Blue Car
Note you can eliminate the x in the denominator.
y=1600
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Re: Two cars painted Black and Blue are racing against each  [#permalink]

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28 Jun 2014, 07:12
Another way to solve the problem:

x = distance covered by blue car before it picked up pace (this is what is asked for)
Total distance = 4000

For distance x, let Blue car's speed = 40
So, Black car's speed = 60 (1.5 * 40 - 50% more than blue car's speed)
For distance 4000 - x, Blue car's speed = 90 (1.5 * 60 - 50% more than black car's speed)

Since both blue and black car reach at the same time, time taken by blue car = time taken by black car
x/40 + (4000-x)/90 = 4000/60

Solving for x, x = 1600
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Two cars painted Black and Blue are racing against each  [#permalink]

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17 Jul 2016, 13:57
Bunuel 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 ?

At first when the race starts the Black car moves 50% (3/2) faster than the Blue car:
The Black car covers 3/2 the distance of the Blue car.

After the Blue car quickens and moves 50% (3/2) faster than the Black car:
The Blue car covers 3/2 the distance of the Black car.

Plug-in the options:
(C) if the Blue covered 2400 meters before it quickened, then the Black car covered 2400*3/2=3600 meters. The remaining distance covered by the Black car is then 400-3600=400, thus the Blue car covered 400*3/2=600. Total distance covered by the Blue car 2400+600=3000, which is less than the actual distance of 4000 meters.

This means that the Blue car must have quickened earlier then 2400 meters.

Check (B):
If the Blue covered 1600 meters before it quickened, then the Black car covered 1600*3/2=2400 meters. The remaining distance covered by the Black car is then 400-2400=1600, thus the Blue car covered 1600*3/2=2400. Total distance covered by the Blue car 1600+2400=4000. Bingo!

Hope it's clear.

Apologies a typo error possibly - mistyped 400 which should be 4000 in both plug in answer explanations
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Re: Two cars painted Black and Blue are racing against each  [#permalink]

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17 Jul 2016, 16:21
let r=initial rate of blue car
let d=distance blue car covers before speeding up
total race time for black car=4000/1.5r
total race time for blue car= d/r+(4000-d)/2.25r
we know that total race times for each car are equal
d=1600 m
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Re: Two cars painted Black and Blue are racing against each  [#permalink]

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04 Dec 2017, 00:32
Let d=distance covered by the Blue car (B1) before it speeds up. The distance covered by the Black car (B2) at this point is therefore 1.5d. The distances still to be covered by B1 and B2 are thus (4000-d) and (4000-1.5d) respectively.Now, since the race is a tie, they cover these remaining distances in the same time. Now, we know that the ratio of the distances covered by two vehicles IN ANY GIVEN (SAME) TIME is equal to the ratio of their speeds. In the second leg of the race, (after B1 quickens its pace), the ratio of speeds of B1 and B2 is 3:2 and the distances covered are (4000-d) and (4000-1.5d) meters respectively. Therefore: (4000-d)/(4000-1.5d)=3/2; d=1600m.
Re: Two cars painted Black and Blue are racing against each &nbs [#permalink] 04 Dec 2017, 00:32
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# Two cars painted Black and Blue are racing against each

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