Two cars painted Black and Blue are racing against each

Thanks ... both the question and solution are good!

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.

Answer: B

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

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.

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.
Answer:
y=1600

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

Apologies a typo error possibly - mistyped 400 which should be 4000 in both plug in answer explanations

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

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.

speed of blue car be x
speed of black car 1.5x
total distance = 4000 mtrs

distance covered by black car 1.5x
distance covered by blue car x
pending distance to be covered by blue car ;
(4000-1.5x)
total distance covered by blue car = 4000
x+(4000-1.5x)*1.5 = 4000
solve for x
1600
option B

Speed=3:2
So, at first black will Cover 1/3 more than blue
To make up this Blue has to cover (2:3) i.e,1/2 distance
So Distance covered by
Black:Blue=1/3:1/2=2:3
5=4000
2=1600

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Rather than setting up algebra, I'd just go with Plugging In The Answers (PITA). I like trying B and D.

B: Blue starts out at 2m/sec. Black starts out at 3m/sec. Blue goes 1600m; that takes 800 seconds. In 800 seconds, Black goes 2400m. Black has 1600m to go and the new speed will be 2m/sec. It will take 800 seconds. Blue has 2400m to go and the new speed will be 3m/sec. It will take 800 seconds. Yay!

Answer choice B.


ThatDudeKnowsPITA