Three runners A, B and C run a race, with runner A finishing 12 meters

Can someone better explain this?

|---------------------------L-------------------|L=Length of the track
|-----------------------------C------B----------|A
|------------------------------------|<--12---->| A finishes 12 m ahead of B
|-----------------------------|<------18------->| A finishes 18 m ahead of C
|-----------------------------|<-6->|---------->| So, B was 6 meters ahead of C when A finished the race ------------1

|----------------------------------------C------|B
|----------------------------------------|<-8-->| B finishes 8 meter ahead of C--------------2


From 1 and 2, we can deduce that
B ran a distance of "L-12" meters taking a lead of 6 meters on C.

L-12->6
1 meter-> 6/(L-12) {:This is the per meter lead by B on C}

For the entire race of L meters, the lead would be
L meters->L*6/(L-12)

And we know the final lead was 8 meters
So,

6L/(L-12)=8
3L/(L-12)=4
3L=4L-48
L=48m

Ans: "B"

Karishma,

can you pls xplain this part



Im nt able to logically deduce how we got the length of the race?

c ----b -----A(finishes)

Think of the logic this way:

There are 2 people P and Q.
They both start running simultaneously at different but uniform speeds.
In the time P has covered 2 m, Q has covered only 1 m. What will be the distance between them by the time P covers a total of 4 m? P and Q are running at uniform speeds. Since P created a gap of 1 m in every 2 m he ran, the gap between them must be 2 m now.

Now imagine that P is standing ahead of Q by 6 m. They both start running. What happens by the time P covers 2 m? What is the gap between them? It must be 7 m now since another 1 m gap would have got created.

The logic used here is exactly the same.

B and C had a distance of 6 m between them at one point in time. B covered another 12 m and now the distance between them is 8 m. So basically a gap of 2 m was created by B when he ran 12 m. Overall, at the end of the race, the gap between B and C is 8 m. How much must B have run to create a gap of 8 m ( = 4*2)? He must have run 4*12 = 48 m

Let d=distance of race.
Between A's finish and B's finish, B runs 12 meters while C runs 10 meters.
B's rate to C's rate=6/5.
When B finishes, he is 8 meters ahead of C.
6/5=d/(d-8)
d=48 meters

Although solution by karishma is more logical and faster approach
i providing one conceptual way for solving...

Let speed of A,B,C be A,B &C respectively
& total distance be D

then for the first case when A finishes,, the time run by all three will be same
D/A=(D-12)/B &&
D/A=(D-18)/C
or (D-12)/B=(D-18)/C----------(i)

again for second case when B finishes,, time taken by B & C will be same
thus D/B=(D-8)/C
C={(D-8)*B}/D-------(ii)

putting value of C from (ii) to (i) and solving for D we get
D=48

Ans B

So, when A finished the race, B is 12 m away from finish line and C is 18 m away from finish line. Hence, effectively gap between B and C is 6 m. ----lets call this scenario 1
But when B finished the race, C is 8 m away from B (and hence from the finish line). --- say this is scenario 2
This means, when B covered a distance of 12m to finish the line, it increased the gap between B and C by 2m (from 6 m in first scenario to 8 m in second scenario).
So, every 12 meters distance traveled by B increase the gap between B and C by 2 meters. and as from scenario 2, at the end of the race, there is a gap of 8 meters between B and C.

So, in order to create a gap of 8 meters which is 4 times of 2 meters, B has to run a race which is 4 times of 12 meters = 48 meters

Let the distance be d, speeds (A,B,C) be v1, v2, v3 (resp.) and time taken to cover the distance d be t1, t2, t3 (resp.)
d=v1*t1 .....(1)
d-12=v2*t1 .....(2)
d-18=v3*t1 .....(3)
on dividing (2) by (3) we get--

(d-12)/(d-18) = v2/v3....... (4)

d=v2*t2 .....(5)
d-8=v3*t2 .....(6)
on dividing (3) by (4) we get--

d/(d-8) = v2/v3..................(7)

On equating (4) & (7), we get--

d/(d-8) = (d-12)/(d-18) (This eq. can be deduced logically too like earlier posts)

On solving this we get d=48

From the question, time taken for A to complete the race = D/ Va = d-12/Vb = d-18/Vc

Also given that B finishes the race 8m ahead of C.=> d/Vb = d-8/Vc

Solving for Vb & Vc, we get D = 48.

"THE RATIO OF THE SPEEDS OF TWO MOVING OBJECTS IS EQUAL TO THE DISTANCES THEY COVER IN ANY GIVEN TIME"
Let the length of the race be 'x' meters and the speeds of B and C be Vb and Vc respectively. Then:
(a) Vb/Vc = (d-12)/(d-18)
(b) Vb/Vc = d/(d-8)
(d-8)(d-12) = d(d-18)....> d = 48. ANS: B

Let X be total distance, therefore
X - (How much is A ahead of B)/X - (How Much is A ahead of C) = How much is A ahead of B/ How much is A ahead of C - How much is B ahead of C.

(X - 12)/(X - 18) = 12/(18 - 8)

On solving we get X = 48. This is simplest method. I hope this helps. Thank you.

Posted from my mobile device: iPhone

Posted from my mobile device

How did B gain 4 metres if the speed is constant?

But how did B gain 4 metres in the latter half if the speed is constant?

Part 1: A finishing 12m ahead of runner B and 18m ahead of runner C.
At this point:
Since B is 12 meters behind A, B must travel 12 more meters to finish the race.
Since C is 18 meters behind A, C must travel 18 more meters to finish the race.

Part 2: B finishes 8m ahead of runner C
At this point:
Since B finishes the race, B just traveled the remaining 12 meters.
Whereas the remaining distance for C in Part 1 was 18 meters, the remaining distance for C in Part 2 is 8 meters.
Implication:
C just traveled 10 meters.
Since B travels 12 meters in the time it takes C to travel 10 meters, the rate ratio for B to C \(= \frac{12}{10} = \frac{6}{5}\).

We can PLUG IN THE ANSWERS, which represent the total distance.
When the correct answer is plugged in, Part 1 will yield for B and C a rate ratio of \(\frac{6}{5}\).
In Part 1, B finishes 12 meters behind A, while C finishes 18 meters behind A.

D: 72
Part 1 --> A=72 meters, B=72-12=60 meters, C=72-18=54 meters
Rate ratio for \(\frac{B}{C} = \frac{60}{54} = \frac{10}{9}\)
Eliminate D.

B: 48
Part 1 --> A=48 meters, B=48-12=36 meters, C=48-18=30 meters
Rate ratio for \(\frac{B}{C }= \frac{36}{30} = \frac{6}{5}\)
Success!

Speed is Constant not Same: There is difference between Same and Constant. Speed being constant means they are running 2 meters in 1 minute, 4 in 2 minutes, 6 in 3 minutes and so on, for example. If speed would have not been constant, they would increase speed in 2nd minute and then slow in 3rd minute, that is not the case.

I hope this helps.

Posted from my mobile device

Another way to use the answer choices:

Part 1: runner A finishing 12m ahead of runner B and 18m ahead of runner C
Part 2: runner B finishes 8m ahead of runner C
When the correct answer is plugged in, the distance ratio for B to C will be the same in each part.

D: 72
Part 1: A=72, B=72-12=60, C=72-18=54 --> B:C = 60:54 = 10:9
Part 2: B=72, C=72-8=64 --> B:C = 72:64 = 9:8
Since the ratios are not the same, eliminate D.

B: 48
Part 1: A=48, B=48-12=36, C=48-18=30 --> B:C = 36:30 = 6:5
Part 2: B=48, C=48-8=40 --> B:C = 48:40 = 6:5
Success!

we can get the answer by using options.
use the first and last piece of information and calculate A:B:C, further check if it satisfies the section condition.