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Bunuel
In a race of 500 m, L beats M by 40 seconds and beats N by 125 m. If M and N run a 500 m race, M beats N by 40 seconds. What is the time taken (in seconds) by M to run the race?

(A) 160
(B) 240
(C) 280
(D) 320
(E) 330


Are You Up For the Challenge: 700 Level Questions

Let L, M & N be the speeds of L, M and N respectively

\(\frac{500}{M}-\frac{500}{L}=40\) -------------- (1)
and \(\frac{500}{N}-\frac{500}{M}=40\) ----------- (2)

We also have \(\frac{L}{N}=\frac{500}{500-125}=\frac{4}{3}\)

Or \(L=\frac{4}{3}N\) ------------ (3)

Put (3) in (1)

\(\frac{500}{M}-\frac{500}{(\frac{4}{3})N}=40\)
Or, \(\frac{500}{M}-\frac{375}{N}=40\) ------------ (4)

Solving (2) and (4) we get,

\(\frac{500}{N}-40=\frac{375}{N}+40\), That is \(\frac{125}{N}=80\)

\(N=\frac{125}{80}=\frac{25}{16}\)

Time taken by N to run 500m = \(\frac{500}{\frac{25}{16}}= 320\) seconds

Since M takes 40 seconds less than N, M runs the race in \(280\) seconds

Answer is (C)
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In a race of 500 m, L beats M by 40 seconds and beats N by 125 m. If M and N run a 500 m race, M beats N by 40 seconds. What is the time taken (in seconds) by M to run the race?

Distance = 500 m
x----------------- 500 m ---------------------x
\(t_M = x \) sec, time taken by M to complete the 500 mtr race
\(x = t_N-40 \) sec, because M beats N by 40 sec, meaning M took 40 sec less than the time taken by M
\(v_N = \frac{500}{t_N} = \frac{500}{(x+40)} \)
Now, when L finished the race (500 mtr in x-40 sec), N completed only 375 mts.
\(\frac{500}{(x+40)}*(x-40) = \) distance covered by N in x-40 sec = 375
solving it, we get x = 280
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IMO C

A bit lengthy approach

Let the speed of L,M,N be l, m, n respectively

time taken by L = 500/l
time taken by M = 500/m
time taken by N = 500/n

Given::
L beats M by 40 seconds
\(\frac{500}{m}\) -\(\frac{500}{l}\) = 40
=> \(\frac{1}{m}\) -\(\frac{1}{l}\) = \(\frac{2}{25}\) ------eq 1


L beats N by 125 m.
time taken by L. * Speed of N = 500-125 (i.e. distance covered by N when L finished the race)
\(\frac{500}{l}\) * n = 375

=> \(\frac{n}{l}\) = 3/4 --- eq 2

M beats N by 40 seconds

\(\frac{500}{n}\) -\(\frac{500}{m}\) = 40
=> \(\frac{1}{n}\) -\(\frac{1}{m}\) = \(\frac{2}{25}\) ------eq 3

Add eq 1 & eq 3

\(\frac{1}{n}\) - \(\frac{1}{l}\) = 4/25

from eq 2 n = \(\frac{3l}{4}\)

\(\frac{4}{3l}\) - \(\frac{1}{l}\) = 4/25

l = \(\frac{25}{12}\)

from eq 1

\(\frac{1}{m}\) -\(\frac{12}{25}\) = \(\frac{2}{25}\)
m = \(\frac{25}{14}\)

Time taken by M = \(\frac{500}{m}\) = 500*14/25 = 280 sec
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Bunuel
In a race of 500 m, L beats M by 40 seconds and beats N by 125 m. If M and N run a 500 m race, M beats N by 40 seconds. What is the time taken (in seconds) by M to run the race?

(A) 160
(B) 240
(C) 280
(D) 320
(E) 330


Are You Up For the Challenge: 700 Level Questions

Since, M beats N by 40 seconds and L beats N by 125 m. Again L beats M BY 40 seconds. So, in 80 seconds, N can run 125 m. So to run 500 m, N needs 320 seconds and M needs 280 seconds.
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L beat M by 40sec................(1)
M beat N by 40sec.................(2)
L beat N by 125m..................(3)

so L beat N by 80 sec
so the time taken by N to run 125m is 80sec


for 500m M took 320sec

so from eq.. 1
total time taken by M is 320-40=280sec

therefore answer is (C)
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Bunuel
In a race of 500 m, L beats M by 40 seconds and beats N by 125 m. If M and N run a 500 m race, M beats N by 40 seconds. What is the time taken (in seconds) by M to run the race?

(A) 160
(B) 240
(C) 280
(D) 320
(E) 330



Solution:

We can let t, u and v be the time it takes L, M, and N to finish the 500-m race, respectively. We see that:

t = u - 40

and

u = v - 40

Thus, t = (v - 40) - 40 = v - 80 or v = t + 80.

Furthermore, since L beats N by 125 meters, for t seconds, N runs 500 - 125 = 375 meter (as opposed to L’s 500 meters). Since the rate of N is the constant during the race, we can create the equation:

375/t = 500/v

375/t = 500/(t + 80)

375(t + 80) = 500t

375t + 30,000 = 500t

30,000 = 125t

t = 30,000/125 = 240

Since t = u - 40, u = t + 40 = 240 + 40 = 280. Thus, it takes M 280 seconds to run the race.

Answer: C

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L beats M by 40 seconds and M beats N by 40 seconds, therefore L beats N by a total of 80 seconds, which is equal to 125 meters. 80 seconds = 125 meters, therefore 320 seconds = 500 meters. Hence M = 320 - 40 = 280
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By the time L covers 500 m, N covers 375 m
Distance Ratio = Speed Ratio => L/N = 500/ 375 = 4/3

Time taken by
L: t
M: t+40
N: t+80

Time Ratio: L/N = t/t+80

Time Ratio is Inversely Proportional to Speed Ratio
t/t+80 = 3/4
=> t= 240 sec

Time taken by M: t+40 = 240 +40 = 280 sec
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Let time taken by L,M and N to complete 500m are t, t+40, t+80 secs respectively.

we know L beats N by 125 M.

HENCE 500/L's speed = 375/N's speed

L's speed/N's speed= 4/3

Hence L's time / N's time = 3/4
t/t+ 80 = 3/4

solving for t = 240

time taken by M = t+ 40 = 240 + 40 = 280
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Distance ; Time
L ; 500 ; x-40
M ; 500 ; x
N ; 375 ; x-40
L ; 500 ; y-40
M ; 500 ; y

Since speed of L, M and N can be assumed to be same
Speed of N \(\frac{ 375}{(x-40)} = \frac{500}{y}\)
Speed of M \(\frac{ 500}{x }= \frac{500}{(y-40)} \)


Solve for x;
x= 280 option (c) is the right answer

Bunuel
In a race of 500 m, L beats M by 40 seconds and beats N by 125 m. If M and N run a 500 m race, M beats N by 40 seconds. What is the time taken (in seconds) by M to run the race?

(A) 160
(B) 240
(C) 280
(D) 320
(E) 330


Are You Up For the Challenge: 700 Level Questions
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Bunuel
In a race of 500 m, L beats M by 40 seconds and beats N by 125 m. If M and N run a 500 m race, M beats N by 40 seconds. What is the time taken (in seconds) by M to run the race?

(A) 160
(B) 240
(C) 280
(D) 320
(E) 330


Are You Up For the Challenge: 700 Level Questions
­
L beats N by 125 meters and also beats N by 80 seconds. 

125 = 1/4 of 500

Therefore 80 x 4 = 320 Seconds = Total Time Taken by N to Finish.

We know M finished 40 seconds before N so N - 40 = M.

320 - 40 = 280. 
Answer C­
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L-M = 40 seconds
L-N =125 Mts ......................(1)
M-N = 40 seconds ....................(2)

Hence, L-N= 80 seconds ...................(3)

So from (1) and (3), N takes 80 seconds to cover 125 mts.
Time taken to by N cover 125 mts = 80 seconds
Time taken by N to cover 500 mts = 320 seconds



From (2) M takes 40 seconds Less than N to cover 500 mts.

So M takes 280 seconds.

C
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