HOT Competition 2 Sep/8AM: Three athletes A, B and C run a 1000 m race

Official Solution:

Three athletes A, B and C run a 1000 m race. When A is at a finish line, B is 50 m left to run and C is 88 m left to run. If all three athletes run at a constant speed, how many meters will C have to run to finish the race when B is at the finish line?

A. \(19\)
B. \(36\)
C. \(38\)
D. \(39\)
E. \(40\)


By the time A covered 1000 m, B covered 950 m, and C covered 912 m.

The ratio of the distances covered by B and C in the same time period is \(\frac{950}{912}\).

By the time B covered 1000 m, B covered additional 50 m, and C covered additional \(x\) m.

The ratio of the distances covered by B and C in the same time period is \(\frac{50}{x}\).

Since B and C run at constant speeds, then the ratio of the distances covered in the same period of time must be equal:

\(\frac{950}{912}=\frac{50}{x}\);

\(x=48\).

So, when B is at the finish line, C is left to run \(88-48=40\) m.


Answer: E

Three athletes A, B and C run a 1000 m race. When A is at a finish line, B is 50 m left to run and C is 88 m left to run. If all three athletes run at a constant speed, how many meters will C have to run to finish the race when B is at the finish line?

A. 19
B. 36
C. 38
D. 39
E. 40

In t seconds, A is at 1000m, B is at 950m, & C is at 912m.

For B to run another 50m, (50×t)/(950)= t/19
In t/19 s, C will run (912/t)×(t/19)=48

So C will have another 88-48 = 40m left to run to finish the race.
IMO E.

Posted from my mobile device

Distance travelled by B when A reached finish line - 950m -- (1)

Distance travelled by C when A reached finish line - 912m -- (2)

Time taken for traveling both (1) & (2) are equal
Let speed of A and B be S1 and S2 respectively.

950/S1 = 912/S2
S2/S1 = 912/950 = 456/475 -- (3)

Then when B reaches finish line, let x be the distance travelled by C

50/S1 = x/S2
S2/S1 = x/50

Substituting (3) in above equation,

x/50 = 456/475
x = 456*50/475
= 48

Since C has travelled 48m out of 88m, the remaining distance that C needs to travel is 40m

Option E

Posted from my mobile device

When A finished; ___________________________C____B______A
--> C=912m, B=950m, A=1000m
--> Assume t=10s
--> \(Speed_C\)=\(\frac{912}{10}\)=91.2m/s
--> In 950m, B use 10s, so 50m left, B will use \(\frac{10}{19}\)s

When B finished; __________________________________C____B
--> B use \(\frac{10}{19}\)s to finish last 50m.
--> In \(\frac{10}{19}\)s, C can run \(91.2*\frac{10}{19}=48\)m more.
--> Therefore, C have to run 1000-(912+48)=40m to finish the race.

ANS E.

P.S. Super speed three athletes.

Posted from my mobile device

Total distance = 1000 m
When A is at a finish line, B is 50 m left to run and C is 88 m left to run.
So Sa = 1000 m
Sb = 950 m
Sc = 912 m

To solve this question, we need to find Vb and Vc.
Assuming that Ta = 100 sec, so Va = \(\frac{1000}{100}\).
Vb = \(\frac{950}{100}\)
Vc = \(\frac{912}{100}\)

The question is how many meters will C have to run to finish the race when B is at the finish line?
1st; We need to know the distance when B is at the finish line.
Tb = Sb/Vb = \(\frac{ 50*100}{950}\) (Tb in 50 m)
2nd; We need to know the remaining distance of C when B is at the finish line.
Sc2 = VC*Tb = \(\frac{912*50*100}{(100*950)}\)
Sc2 = 48 m
So Sc+Sc2 = 912 +48 = 960
The remaining distance of C when B is at the finish line is 40 m

So I choose E.

Posted from my mobile device

First leg:
In the 1000-meter race, B has 50 meters left, so the distance traveled by B in the first leg = 1000-50 = 950 meters
C has 88 meters left, so the distance traveled by C in the first leg = 1000-88 = 912 meters
Thus:
\(\frac{B's-distance}{C's-distance} = \frac{950}{912}\)

Second leg:
Here, B runs the last 50 meters, so the distance traveled by B in the second leg = 50 meters
Let C's distance in the second leg = C
Thus:
\(\frac{B's-distance}{C's-distance} = \frac{50}{C}\)

Since B and C run at constant speeds, the blue ratios above must be EQUAL:
\(\frac{950}{912} = \frac{50}{C}\)
\(950C = 50*912\)
\(C= 48\)

Of the 88 meters remaining for C after the first leg, C runs 48 meters.
Thus, the remaining distance for C = 88-48 = 40

it's game of approximation i will walk you through why the hell is A given is still a mystery for me now only B and C is required threfore when 950 is covered by B , C covers 912 from question then when B covers 50 C covers approximately 47+< 48 meters propotionality then C covers 40 meters 88-48 = 40 imo E

After covering a distance of 950m (1000 - 50) B takes a lead of 38m (88 - 50) over C.
After covering a distance of 1000m, B will take a lead of (38/950)*1000 = 40m over C.

ANS: E

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can you please tell how did you arrive at this conclusion-"So, by the time B covered 1000m, C would have covered 960m, thereby having 40m to reach the finish line

The difference between B and C when C is at 950 is 38 m.

38/950 = (9,5*4)/(9,5*100)

C is 4 % slower.

4 % of 1000 = 40.

Posted from my mobile device

Concept of constant speed and relative speed: for a given set distance, since B is faster than C, B will put himself ahead of C by a constant amount of miles.

Since the speeds B and C do not change and are constant, the longer that B and C run the MORE of a lead that B will gain relative to C.


For the first 950 miles:

B was able to put himself ahead of C by 38 miles


For a certain proportional decrease in miles traveled by B——- you can apply the SAME proportional decrease in the miles that B is able to put himself ahead of C

What I mean is the following:

950 meters traveled by B ————-> B is able to put himself ahead of C by 38 meters across this distance

If for instance, B traveled (1/2) this distance after ——-> B will only be able to put himself ahead of C by (1/2) of (38) or 19 meters

1st: fractional decrease in meters traveled by B and C

(950 - 50) / 950 = 900 / 950 = -18/19 decrease ———-> = Multiplier of (1/19)


2nd: apply the same fractional decrease to the amount of meters by which B is able to put himself ahead of C

(38 meters) * (1/19) = 2 meters

So for the last 50 meters that B and C both run, B is able to increase the distance between them by another +2 meters in addition to the 38 meters he is already of C by.

38 + 2 = 40 meters that B will create as a lead when he runs the entire 1,000 meters (finishes the race)

Answer E. 40

Also, you can eliminate answers A, B, and C right off the bat. Since the speeds are constant, the longer that B and C travel, the more distance that B will put between himself and C

Therefore, if he travels another 50 meters, B will end up putting more than 38 meters between himself and C (since B has already created the lead of 38 meters in the first 950 meters)

Thus A, B, and C can NOT be the answers

Posted from my mobile device

Hi Bunuel - I never thought this day would come but I think at the very end you meant, please correct me otherwise -

When B is at the finish line, C is left to run = 1000 (Total) - 912 (Already ran) - 48 (Amount ran when B was finishing his remaining 50m) = 40 m (ANS E)

__________________
Fixed the typo. Thank you!

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