Oxford Entrance Exam Question

$$ Let $\color{#D61F06}{s_1}$ be the distance covered by the red car and $\color{#3D99F6}{s_2}$ be the distance covered by the blue car, then $$ $\begin{aligned} \color{#D61F06}{s_1}&=\color{#3D99F6}{s_2}\\ v_1t_1&=v_2t_2\\ 40\left(t_2+\frac{3}{60}\right)&=60t_2\\ 20t_2&=2\\ t_2&=\frac{1}{10}\text{ h}. \end{aligned}$ $$ Thus, the distance that two cars covered when the cars are level is $s_2=v_2t_2=60\cdot\dfrac{1}{10}=\boxed{6\text{ km}}$ .

It has been proven without any reasonable doubt. Now, I'm going to Oxford! V(^_^)V #LOL $$

$\text{\# }\mathbb{Q.E.D.}\text{ \#}$

It is very simple. Red car travels 2 km in 3minutes----given. Blue car travels 1 km in 1 minutes----given --- 3 km in 3 minutes. Thus, when blue car starts travelling, red has already covered 2 km.` As we have to find out the shortest distance where the cars meet, we have to find out the LCM (Least Common Multiple) of 2 and 3, which is 6. Therefore, answer is: the cars meet after travelling 6 km.

Both cars will cover same distance .

let time taken by car b be x.

Therefore time taken by car a = x + 0.05 (0.05 because 3 minutes is equal to 0.05 hours)

We have distance = speed * time

distance traveled by car a = 40(x+0.05) ---> equation 1

distance traveled by car b = 60*x----> equation 2

As both cars will cover same distance

we will get equation as

speed of car a * time by it = speed of car * time taken by it

i.e 60* x = 40*(x+0.05)

20*x=2

therefore x= 1/10

now we have to find distance

now put x value in place of time in equation 1 or equation 2.

i.e 40 * (1/10 + 0.05)=4*1.5

distance = 6 km

First I converted the speeds to kilometres per minute, being 2/3 km/m and 1km/m. From there I used simple algebra.

2/3x + 2 = x

multiply all by three

2x + 6 = 3x

subtract 2x from both sides

6 = x

My solution is not a a reliable solution. But it answers the question.

Thus from the given example, 3 minutes is equal to 1/20 of 60 minutes. So every 3 minutes the red car covers 40/20 kilometers which is equal to 2 kilometers. And every 3 minutes in bile car covers 60/20 kilometers which is equal to 3 kilometers. So I came up with a list like this: Time Red Car Blue Car 3 minutes 2 km 0 km 6 minutes 4 km 3 km 9 minutes 6 km 6 km

So therefore I conclude that after 9 minutes they will be on the same level of kilometers, which is 6 kilometers.

I also conclude that some solutions can be made even though your not very good on that field, you just need to use your logical senses!

red car starts 3 min early at 40 km/hr therefore by converting 3 min to hr you get 0.05 hr multiplying it to 40km/hr you get 2 km after another 3 min the blue running at 60km/hr would reach 3km in this case the red car have traveled a total of 4 km in 6 mins while the blue have traveled 3 km in 3 mins so basically 3 mins = 2 km for red and 3 mins = 3 km for blue since red started earlier 3 mins. adding another 3 mins would give them both a 6km distance

sorry for my solution i'm new to this :)

I set up an easy equation

$40 \times 3 + 40 x = 60 x$

Red car's 3 minutes + it's speed times the time we do not know = the blue car's speed times the time x.

If we simplify

$x=6$ mins or $\frac{1}{10}$ of an hour

$60 \text{ km/h} \times \frac{1}{10} = 6 \text{ km}$

red car has a speed of 40kph means it can travel 2 km in 3 mints. after 3 mints blue also starts and it can travel 3 km in 3 mints. now , after 6 mints red car will be at 4 km and blue is 3 km, and in next three mints , i.e, after 9 mints red car is at 6 km and blue is also at 6 km

s1 = s2 <=> 40t + 40x0.05 = 60t <=> 40t + 2 = 60t => t=0.1

After 0.1 hour two cars will be level (s1=s2) S1=S2= 0.1 X 60 = 6km

Since the red car has already traveled for 3 minutes before the blue one started: s0=(3/60) 40=2kms Now, we can use simple algebra to calculate the time that both will meet 60t=40t+2kms t=0.1hr => distance=60 (0.1)=6kms

The following is the information provided, put into the format of a table:

From the information presented to us in the question, we can derive that Blue car has driven x amount of time, while Red car has driven x+0.05 amount of time (0.05 hours=3 minutes). We also know the the speed of the Blue car is 60 km/h, while the speed of the Red Car is 40 km/h. From this data, we can say that the Blue car has driven a distance of 60x kilometers, while the Red car has driven a distance of 40(x+0.05). Since the question provides us with the knowledge that the cars are to meet during the time of the questioned event, an thus have traveled the same distance, we can formulate the equation: 40(x+0.05)=60x We can insert the information, that we have extracted from the question, into our table:

Now all we have to do is solve our latest equation to find the amount to time that the car's traveled:

40(x+0.05)=60x

40x+2=60x

2=60x-40x

2=20x

x=0.1

So that the Blue car had traveled for 0.1 hours before they met, while the Red car had traveled for 0.15 hours (It got a 3 minute head start, remember? Well, 3 minutes is 0.05 hours). But that's not what we needed to know. We needed to know how many kilometers the Red car had traveled up to the point at which the Blue car caught up with it. Well, we already know the speed at what the Red car was travelling: 40 km/h We also know how many hours the Red car spent travelling:0.15 So all we have to do is multiply 40 by 0.15, which gives us 6!

Thus, we can conclude that the Red car had driven 6 kilometers before the Blue car caught up with it.

first find how far the red car travels in three minutes:

3 / 60 X 40 = 2

therefore the red car got a 2km head start

let x rep the time spent travelling until level

40x + 2 = 60x

40x + 2 = 60x

2 = 20x

2/20 = x

0.1 = x

then plug the values in and check if equal

40(0.1)+2 = 60(0.1)

4+2 = 6

6 = 6

therefore the cars traveled 6 km before they were level with each other

Use a graph of D vs. T where for the red car, d= 2/3 x (t+3), and the blue car d=t. The crossover shows the distance travelled when side by side.

60km/h = (1) km/min

40km/h = (2/3) km/min

In 3 minutes red travels 2km. Blue car 0km.

2km+(2/3)(km/min)(t)=0km+(1)(km/min)(t) -> t=6min

0km+6(1)(km/min) =6km.

First convert km/h to km/min:

40km/h ÷ 60 = 0.6km/min 0.6km * 3min = 2km

60km/h ÷ 60 = 1km/min

After the 3 minute head start, the 40km/h car is 2km ahead, gaining 0.6km per minute after that. Double that and the car is 4km ahead after 6 minutes.. Keep in mind that the car got a head start so the next is 6 kilometres in 9 minutes, but you have subtract the initial 3 minute head start so its actually 6 minutes. As for the the second car, it'll make it to the other car, 6 kilometers later, in 6 minutes.

speed of A is 2000/3 m/min. speed of B is 1000 m/min A starts initially and at 3 min it travels 2000m, simultaneously B starts from 0. there for at t=6 min A travels 4000m thus total 6000m and B also travels 6000m = 6 km therefore the distance at which both ,meet is 6 km

let time taken by red car be x+3 and by blue car be x . speed of red one = 40km/h & speed of blue one = 60km/h . distance travelled by red = distance travelled by blue . distance = speed * time . 40(x+3) = 60x . x=6km

The head start gives the red car a 2km head start (workings-> $\frac { 40km }{ h } \quad \rightarrow \quad \frac { \frac { 4 }{ 6 } km }{ minute } \rightarrow 3min\cdot \frac { \frac { 4 }{ 6 } km }{ min } \rightarrow 2km$ ) then we can express it like this $2+\frac { 40km }{ h } t=\frac { 60km }{ h } t$ them solve it anyliticly/via a graphing calculator to get $0.1hours$ then take $0.1\cdot \frac { 60km }{ hour }$ or $(0.1\cdot \frac { 40km }{ h } )+2$ to get the answer which is 6

Let time taken by red car be x ... then for blue car it will be (x-3) Now convert velocity into km/min ... Now when the cars meet S¹ = S²....calculate x and then find S=vt...

First, convert three minutes into hours.

3 min * 1/60 min = 1/20 h

Then find out how many kilometers are in 1/20 min.

40/1 = x/1/20

40/20 = x

2 = x

Then make a proportion.

40t + 2 = 60t, t being the amount of time.

2 = 20t

1/10 = t

Then plug it back in.

40 (1/10) + 2 = 60 (1/10)

4 + 2 = 6

6 = 6

6 kilometers.

Both of the cars will meet when traveling with the same distance ( $S_{red-car}$ = $S_{blue-car}$ ). Let see: $t$ is the time of a red car for following blue car had been moving 3 minutes earlier, then the blue car when will have met the red car will have been moving during $t + 3$ , with the assumption of constant velocity, then:

$S_{red-car}$ = $S_{blue-car}$ = S; and: $distance (S) = velocity (v) \times time (t)$

$40 \times (t + 3) = 60 \times t$

$40t + 120 = 60t$

$20t = 120$ $\Rightarrow t = 6$ $second$

$S = 40 \times (t + 3) = 60 \times t = 60 \times \frac{6}{60} = 6$ $km$

Well I dunno if this was how anyone else did this but: the red car is traveling at 2/3km per min. That puts red at 2 km out. Blue speed relative to red is 1-2/3=1/3km per min. 2 km at 1/3km per min takes 6 mins. 6 min at 1 km/min is 6 km

60 km/h = 1 km/min

40 km/h = 2/3 km/min

x = time traveled in min

1x = 2/3x + 3

1/3x = 3

x = 9

Red car travels 2/3 km/min (9 min) = 6 km

-As Red travel 2 km in 3 minutes. -And Blue travel 3 km in 3 minutes.

Time 0 = Red = 0 and Blue = 0

Time 3 = Red = 2 and Blue = 0

Time 6 = Red = 4 and Blue = 3

Time 9 = Red = 6 and Blue = 6

In 3 minutes, the red car has travelled 2km. Therefore we need to find the time it takes for the blue car to overtake the red car by 2 km. In 1 hour, red car has travelled 40 km and blue car has travelled 60 km. Therefore in 1 hour the difference in their distance is 20 km. Or the difference in the distance between blue car and red car is 20km per hour. 20km divided by 10 = 2 km. 20km/h divided by 10 = 2 km/ 6 minutes. It takes 6 minutes for the blue car to surpass red car by 2 km, so 6 minutes is the time needed for blue car to catch up to the red car who has had a headstart of 3 minutes or 2 km.

What if the trip were only 3 kilometers? But if the trip is 6km or longer... Easiest to change the units to km/min then use s1 = s2 where v1 is 2/3 km/min, t1 is x+3 and v2 is 1 , t2 is x.

Red distance is 40t Blue distance is 60(t-0.05) because t is different by 0.05hrs or 3mins So when they meet: 40t=60(t-0.05) 40t=60t-3 t=3/20 hrs = 0.15 hrs before they meet. Because time = speed X distance Red has travelled: 40km/h x 0.15hrs = 6km

Simply find how far the red car will travel in 3 minutes, which is 2 km.

Then set the equation 40x + 2 = 60x Solve for x to find the time before they meet. Plug that into 60x to find the distance!!! You get 6 km.

Can i get into Oxford? I did it in my head!!!

Speed = (distance) /(time) Distance =speed x time Distance at which two cars meet means the distance is same for red and blue car. Hence, 40(x+3)=60x
20x=120 X=120/20 =6km

The red car travels 2/3 of a kilometer per minute. After 3 minutes, it would have traveled 2 kilometers, thus giving it a 2 kilometer head start.

The blue car travels 1 kilometer per minute and needs to make up a 2 kilometer difference.

The blue car travels 1/3 of a kilometer per minute faster than the red car, so it will take 6 minutes to catch up on that 2 kilometer head start.

So 6 minutes after the blue car takes off, they are neck and neck or even,

but the red car started off 3 minutes before that 6 minutes for the blue car, so the red car traveled 9 minutes at 2/3 of a kilometer per minute....6 kilometers

In 3 minutes, the red car travels 40*(3/60)=2 km. Therefore, the blue car needs to catch up 2 km.

The speed difference between the blue car and red car is 60-40=20 km/h, which is 20/60=1/3 of the speed of the blue car. Therefore, the distance between the red car and blue car changes by 1/3 of the distance blue car travels. Hence the blue car travels 2/(1/3)=6 km when the cars are level, which is the total distance blue car travels.

i've guessed it in just 6 seconds.. there are 3 tries.. i answer 1st 2km, then 3 km, and then the right answer which is 6km.. i've noticed that the red car is ahead of only 3 minutes.. and the answer would likely a short distance and likely divisible by 2 or 3..haha..i have no pen here and i'm too lazy to think more..

Let distance travelled by the cars before meeting=x Km, Time taken by Red car to cover x Km= x/40 - 3/60.......(1) Time taken by Blue car to cover x Km= x/60 ...............(2) From eqn (1) & (2), x/40 - 3/60= x/60 Thus, x=6

we need to find equation that the car equals another car

the red car is ahead 3 minutes in front of the blue car and to figure out how many kilometers its gone from 40 kmh, we need to what 40 kmh means, 40 km/1 hour * 1 hour/60 minutes 40/60 = 2/3 kmm (kilometers per minute) the red car has traveled 3 minutes, therefore 2/3 kmm * 3 minutes/1 = 6/3 = 2 so the red car is already 2 kilometers ahead of the blue car the blue car is traveling at 60 kmh, which is 60 km/1 hour * 1 hour/60 minutes = 60/60 = 1 so the blue car is at 1 kmm (kilometers m=per minute now we need to know when they equal x represents kilometers blue car = red car 1x = 2/3x+2 -2/3x 1/3x = 2 (3)1/3x = 2(3) x=6 meaning for both cars to be equal, they would have traveled exactly 6 kilometers when they are level

3 minute head start equals 3/60 of an hour or .05 hours. This times 40km/hr equals a 2km head start.

So, $\frac{x}{60} \times 40 + 2 = \frac{x}{60} \times 60$

Simplify: $\frac{x}{1.5} + 2 = x$

Further: $x + 3 = 1.5x$

Then: $3 = .5x$

Finally: 6 = x

the first we determine the red car distance with 3 minutes and 40 km/h, and we get 2 km. from 3 minutes also we determine the blue car distance with 60 km/h, and we have 3 km. from 2 km and 3 km we find the distance when red car and blue car same level with Least Common Multiple (LCM) between 2 and 3, and finally we get 6.

HAhaha.....LOL

i have done it manually though its not a good practice my technique is

Red car has a head start of 3 min = 0.05hr

So it has a head distance of 0.05hr*40km/hr=2km

Relative velocity of both cars is

60km/h-40km/h = 20 km/h

It's takes

2km/(20km/h) = 0.1hr

time for the blue to reach red

0.1hr*60km/hr=6km

• $Distance = rate \times time$
• $D_R = \text{Distance traveled by red car}$
• $D_B = \text{Distance traveled by blue car}$
• $t = \text{time}$

$D_R=40\dfrac{km}{hr}\times(\dfrac{3}{60}\text{hr} + t) \\D_B=60\dfrac{km}{hr}\times{(t)} \\\text{ } \text{When the red car leveled with the blue car } D_R = D_B \\\text{ } \\\text{Thus, } \\\text{ } \\40\dfrac{km}{hr}\times(\dfrac{3}{60}\text{hr} + t) = 60\dfrac{km}{hr}\times{(t)} \\\text{ } \\t=6\text{ minutes} \\\text{ } \\\text{Solving for the distance traveled by red car} \\\text{ } \\D_R=40\dfrac{km}{hr}\times(\dfrac{3}{60}\text{hr} + \dfrac{6}{60}\text{hr}) \\\text{ } \\D_R = 6 \text{km}$

The red car and the d car must cover the same distance if they're going to be level with each other, so they both travel a distance of 'd': Using distance=speed x time you get these 2 eqns....

blue car: 60t = d

red car: 40 (t + 0.05) = d --> 40t +2km =d --> 40t = d-2

1) Substitute d=60t into the red car's equation to give: 40t= 60t - 2 --> 2 = 20t --> 1/10 = t

2) Substitute t= 1/10 into the blue car's equation: 60(1/10) = d --> 60/10 = d --> 6 = d

Therefore, d = 6km

My solution is: If the red car (R) starts 3 minutes before the blue (B) one,then it's 2km before the blue one too ( based on 40km/h => 2/3 km/min).So hypothetically, R and B are not start at the same point, R starts 2km nearer than B. Then they both start the same time.It takes 6 minutes for R to drive 4 km and B to drive 6km (60km/h) when they are level (plus 2km nearer for R so it's 6km) . So the answer is 6km for B.

Also going Oxford . MairalaaAmare

I wish Oxford will accept me if I can solve this....they probably won't because almost everyone can solve it.... x=2+40t=60t x=6

Another method We can easily convert 40km/h to 0.66667km/min and convert 60km/h to 1 km/min.... . In 3mins, red car travelling at 0.6667km/min would have traveled 2km, and the blue car would not have moved. In 6mins, the red car would have traveled 4km and the blue car would have traveled 3km. In 9mins, the red car would habe traveled 6km and so would have the blue car. So by analysis, the answer is 6km.

I solved this using the concept of relative velocity which is 20 in this case

If the red car is traveling at a rate of 40 Km / h: 40 Km / 60 m 2/3 Km / m

If the blue car is traveling at a rate of 60 Km / h: 60 Km / 60 m 1 Km / m

assuming the 3 minute head start, the red car has traveled 2 Km, while the blue car is at 0. Applying the above rates of motion, the following must be true:

minute 0: Red 2 Km, Blue 0 Km minute 1: Red 2 + 2/3 Km, Blue 1 Km minute 2: Red 3 + 1/3 Km, Blue 2 Km minute 3: Red 4 Km, Blue 3 Km minute 4: Red 4 + 2/3 Km, Blue 4 Km minute 5: Red 5 + 1/3 Km, Blue 5 Km minute 6: Red 6 Km, Blue 6 Km

The cars are even when they are 6 Km from their starting point.

Equaling the final positions s1 = s2 find a time when common 360 s or 0.1 h. Now just multiply that time by 60 km / h or 16.666 m / s proven the answer.

A = 40 km / h , 40 / 60 = 2 km / 3 minute

B = 60 km / h, 60 / 60 = 1 km / 1 minute

let's multiply B by three (just to be easier to calculate) = 3 km / 3 minute

A = 2

         +   2      +    2       =    6

B = 0

         +   3      +    3       =    6

After 3 mins, the red car travels 40/20 miles (2 miles). Then, you can write the equation x=2/3x+2 because in every minute, the blue car travels 1 km and the red car travels 2/3 km. You simplify into 2=1/3x which is x=6

Should have gone to Oxford if this is how easy the questions are for the entrance exam... don't even need a calculator or scrap paper! Since the first car goes 2 km in 3 min which is the head start it is safe to assume the other car will catch up in multiples of 3 min so 2 iterations is all it takes to get how much time/distance is needed to get to 9 minutes and 6 km.

The "correct" answer is incorrect. They will have both traveled 9 km. The red car would have traveled 3 before the puzzle started and 6 by the time the blue car catches up; the blue car will travel 9 from the time the puzzle starts.

Everyone else got it in some crazy math way. I'm not smart so I just used the Distance = Rate/Time (+2km for red car) equations for each car. Then put the equation to equal each other (since we are finding the point where ranges are equal). Then I solved for time and inputted that back into the original equation for red car to find the distance at that time.

Red=40/T+2

Blue=60/T

60/T=40/T+2

60=40+2T

T=10

Red=40/10+2=6

Blue=60/10=6

Glad I had a second chance, for some reason I thought it said how long, in time...

At the instant blue car starts, red car travelled (40km per hour,so 40km per 60min ,so 2km per 3min) 2km..the time taken by blue car to meet red car can be found out using the formula time = distance / relative speed, so time =2/20,which is 1/10hr,which is 6 min.distance travelled by blue car in 6 min is (60km per 60min) is 6kms...Answer:6kms

First convert the calculation in minutes for easy calculation. hence red car Travels at 40000/60 = 666.66667 meters per minute. Hence in 3 minutes the car would have traveled 666.66667 x 3 = 2000 meters. the blue car travels at 60000/60 = 1000 meters per minute. The cars would meet each other when the blue car covers the extra 2000 meters traveled by the red car while going on the road to catch up. As red car travels at 40000 meters per hour therefore the blue car must travel at 40000 meters per hour to travel the same distance as the red car. Now the Blue car needs to cover the extra 2000 meters at the extra speed of 20000 meters per hour = 333.33333 meters per minute. so it will take 2000/333.3333 = 6 min for the blue car to reach the red car which will then be traveling for 9 min. hence it will take 6 x 1000 = 6000 meters = 6 kilometers or 9 x 666.666667 = 6000 meters = 6 km.

Given that both the cars travelled the same distance, so let it be x. An the time period , when both would be at same level will be same for both the cars. Hence , Time taken - blue car= time taken - red car Now, using formula- (Time =distance/speed) calculate the value of x Note: Don't forget to add 3 sec. In case of red car. Thank you !

1st in 3 min, red car goes @ 40km/hr =2km Relative speed of the both cars=(60-40)km/hr = 20km/hr Therefore, the two cars came to the same levels after ( 2/20)hr i.e.(distance/relative speed) =6 min Therefore ,in 6 min blue car travels @ 60 km/hr is =6 km(ans)

FORMULA (S1×S2)÷ (S1-S2) × (T1-T2) (60×40)÷(60-40)×(3÷60) =6

x/40 - x/60 = 3/60

solve for x.

Let the distance covered be x km. Red car:- Speed=40 km/hr. Time=x/40 hr. Blue car:- Speed=60 km/hr. Time=x/60 hr. Since red starts 3 mins earlier. Therefore, x/40 - x/60 = 3/60. x=6. Distance=60 km.

Speed of red car = 2/3 km per min

Speed of blue car = 1 km per min

If they meet after blue car has traveled t min, 2/3 (t+3) = t

Solving for t gives t=6 min.

If they meet after blue car has traveled 6 min, then distance traveled is 6 km

Three minutes pass for the red car. The distance it travels is

1. (40km/h)(1h/60minutes)(3minutes) = 2km -- I converted it to km/min (there are 60 minutes per hour) then multiplied by the 3 minutes. This means the red car is ahead by 2km before the blue even starts

Next we set up an equation in terms of total distance traveled by each car. We know the distance traveled is the speed multiplied by time -- (km/h)(h) = km.

We want to know WHEN the total distance for each car is equal.

1. 40t + 2km = 60t 40 and 60 represent the speeds of the two cars. We are finding at what time the cars meet. Each side of the equation represents the distance traveled. The red car already traveled an extra 2km so we must take that into account.

Solving we get

1. t = 1/10 (h).

Now substitute for either equation to get 6km

$f(x)=((4÷6)\times x)+2$ Where X is the kilometers covered by the blue car. Then we just need to observe or solve for: $f(6)=((4÷6)\times 6)+2=\boxed{6}$

distance traveled by red car(d)= its speed (KM/MIN)* (time+3),since red car starts 3 min before the blue car starts, same distance traveled by blue car(d)= its speed(KM/MIN) * (time), d red car=(40/60) (t+3), d blue car=(60/60) (t), equate both the equations & solve for time t we get ,t=6 min. put,t=6 in one of the above equation we get required distance,d=6 km.

We need to find the distance travelled from the starting point till the cars are at level. I'm going to do this by finding the distance travelled by the blue car.

Relative Distance = distance travelled by Red car in three minutes = 2 km

Relative speed of the cars = 60-40 = 20

Time = distance/speed Time = 2/20 = 1/10 hour or 6 minutes.

So distance travelled by blue car in 6 minutes at 60 km/hour = 6 km

Red car gets a head start of 3 min, during which it travels 2 km. Then Blue car sets off. The kilometers traveled once the cars are level can be found with the equation:

60 x = 2 +40 x

where x = time that Blue car travels or the time AFTER the head start that Red car travels.

Solve for x, and get x = 1/10 hrs. Plug this value back into one side of the equation, and you'll get 6 km.

6 kms. Red car travels 6kms in 9 mins where as blue car catches up in 6 mins covering the same distance

(Solution for people who TL:DR every equation ....like I do) So firstly the red car has a three minute head start. The answer needs to be in kilometres, so

$\frac{40}{60}=\frac{2}{3}$

Which gives a speed of 2/3 km per minute or 2 km every three minutes. Blue travels at 1 km per minute or 3 km every three minutes.

Minute 0: red car 0 km, blue car 0 km

Minute 3: red car 2 km, blue car 0 km

Minute 6: red car 4 km, blue car 3 km

Minute 9: red car 6 km, blue car 6 km.

so they are level at 6 km.

My method is rather specific, didn't manage to find a more general equation for this yet.

Let the time taken by the blue car from the starting pt. to the pt.where the two cars are at level be x. therefore, blue car travels 60x km, red car travels 40(x+3) km. so, 60x=40(x+3/60) or, 60x=40x+ 40.1/20 or, 20x=2 or, x=1/10 therfore,required distance would be 60x or 60.1/10 or 6 km .Thats the answer.

let the two car travels a distance xkm red car travels 40 km in 60m,then it can travels xkm in (60 x)/40 m similarly white car travels 60km in 60 m,then it can travels x km in x m by the question we can write, 60 x/40 =x+3 solving this equation we get x=6km