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by [email protected] » Thu Nov 28, 2013 3:44 am
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

Not clear as to what is being asked and how to convert it into an equation
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Source: — Problem Solving |

by GMATGuruNY » Thu Nov 28, 2013 4:00 am
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.
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by theCodeToGMAT » Thu Nov 28, 2013 7:32 am
Initially
BK = 1.5 BL

Later
BL = 1.5BK = 1.5 * 1.5 * BL

To find: D1

Before Speed change
D1/BL = D2/BK ==> D1/BL = D2/1.5BL ==> D1/D2 = 2/3 - (1)

After Speed Change
(4000-D1)/(2.25 BL) = (4000-D2)/(1.5BL)
4000 - D1 = 1.5 (4000 - D2)

Using (1)
4000 - D1 = 3/2 (4000 - 3/2D1)
16000 - 4D1 = 24000 - 9D1
5D1 = 8000
D1 = 1600

[spoiler]{B} [/spoiler]
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by Mathsbuddy » Thu Nov 28, 2013 7:53 am
Black speed: 1.5B constantly
Blue speed: B then 2.25B

Time = Distance/Speed

Black:
T = 4000/1.5B

Blue times:
t1 + t2 = T

d1/B + d2/2.25B = 4000/1.5B

d1 + d2/2.25 = 4000/1.5

But d2 = 4000 - d1

d1 +(4000-d1)/2.25 = 4000/1.5

d1(2.25-1) = 2.25*8000/3 - 4000

d1 = 1600
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by Mathsbuddy » Thu Nov 28, 2013 8:00 am
theCodeToGMAT wrote:Initially
BK = 1.5 BL

Later
BL = 1.5BK = 1.5 * 1.5 * BL

To find: D1

Before Speed change
D1/BL = D2/BK ==> D1/BL = D2/1.5BL ==> D1/D2 = 2/3 - (1)

After Speed Change
(4000-D1)/(2.25 BL) = (4000-D2)/(1.5BL)
4000 - D1 = 1.5 (4000 - D2)

Using (1)
4000 - D1 = 3/2 (4000 - 3/2D1)
16000 - 4D1 = 24000 - 9D1
5D1 = 8000
D1 = 1600

[spoiler]{B} [/spoiler]
Hi there,

I got the same answer as yours, but with a different method.
I like the simplicity of yours, but don't see where your equation D1/BL = D2/BK came from.
It implies that time t1 = t2, but the question does not directly state that.
Could you please let me know. Maybe I'm just missing the obvious!

Thanks.
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by theCodeToGMAT » Thu Nov 28, 2013 8:35 am
Mathsbuddy wrote: Hi there,

I got the same answer as yours, but with a different method.
I like the simplicity of yours, but don't see where your equation D1/BL = D2/BK came from.
It implies that time t1 = t2, but the question does not directly state that.
Could you please let me know. Maybe I'm just missing the obvious!

Thanks.
Hello Mate,

As the question stated that both cars moved and then one car changed it speed, so we can deduce that both cars travelled for equal time before one car changed its speed; both speed & distance must have varied.

So, I equated the time of both cars

D1/BL = D2/BK

where,
BL = Speed of Blue
BK = Speed of Black
D1 = Distance traveled by Blue Car before change of speed
D2 = Distance traveled by Black Car before change of speed
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by Mathsbuddy » Thu Nov 28, 2013 8:39 am
theCodeToGMAT wrote:
Mathsbuddy wrote: Hi there,

I got the same answer as yours, but with a different method.
I like the simplicity of yours, but don't see where your equation D1/BL = D2/BK came from.
It implies that time t1 = t2, but the question does not directly state that.
Could you please let me know. Maybe I'm just missing the obvious!

Thanks.
Hello Mate,

As the question stated that both cars moved and then one car changed it speed, so we can deduce that both cars travelled for equal time before one car changed its speed; both speed & distance must have varied.

So, I equated the time of both cars

D1/BL = D2/BK

where,
BL = Speed of Blue
BK = Speed of Black
D1 = Distance traveled by Blue Car before change of speed
D2 = Distance traveled by Black Car before change of speed
Ahh, I see. I stupidly thought D2 was the 2nd distance travelled! (Probably because my method uses suffixes 1 and 2 to denote first and second legs of the race.)

Thanks.
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by theCodeToGMAT » Thu Nov 28, 2013 8:51 am
Mathsbuddy wrote: Ahh, I see. I stupidly thought D2 was the 2nd distance travelled! (Probably because my method uses suffixes 1 and 2 to denote first and second legs of the race.)

Thanks.
Variables Ambiguity ;)
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