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In a 10 km race. A, B, and C, each running at uniform speed, get the

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Math Expert
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In a 10 km race. A, B, and C, each running at uniform speed, get the  [#permalink]

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New post 02 Apr 2020, 10:29
00:00
A
B
C
D
E

Difficulty:

  55% (hard)

Question Stats:

61% (02:12) correct 39% (01:25) wrong based on 41 sessions

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Director
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Re: In a 10 km race. A, B, and C, each running at uniform speed, get the  [#permalink]

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New post 02 Apr 2020, 10:40
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2
The distance between B and C increases at the rate 1km for every 10km

So when B runs 9km, the distance between B and C must be 9/10=0.9km=900m

Distance AC = AB+BC = 1000+900 = 1900m

Answer is C

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Re: In a 10 km race. A, B, and C, each running at uniform speed, get the  [#permalink]

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New post 02 Apr 2020, 10:57
1
when a reaches finish line b reaches 9km (i)
when b reaches finish line c reaches 9km (ii)


from (i) same equating time

S denotes speed

10/S(a)=9/S(b)

=> 10/9S(a) =1/S(b)

similarly from ii

10/s(b) = 9/s(c)

1/s(b)= 9/10s(c)

equating for 1/s(b), we get relation between s(a) and s(c)

10/9S(a) =9/10s(c) => s(c)=81S(a)/100


running at uniform speed c could only cover 81/100 of distance covered by A

so while covering 10000 metres c could only cover 81/100*10000=8100


so answer is c 1900m
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Re: In a 10 km race. A, B, and C, each running at uniform speed, get the  [#permalink]

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New post 02 Apr 2020, 11:27
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Speed of A, B, and C are \(S_A\), \(S_B\) and \(S_C\) respectively.

\(S_A : S_B = 10:9\)

\(S_B : S_C = 10:9\)

\(S_A : S_B : S_C = 10*10 : 9*10 : 9*9 = 100 : 90 : 81\)


A beat C by 1.9 (10-8.1) Km or 1900m.



Bunuel wrote:
In a 10 km race. A, B, and C, each running at uniform speed, get the gold, silver, and bronze medals, respectively. If A beats B by 1 km and B beats C by 1 km, then by how many metres does A beat C?

A. 1750
B. 1800
C. 1900
D. 1950
E. 2000
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Re: In a 10 km race. A, B, and C, each running at uniform speed, get the  [#permalink]

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New post 02 Apr 2020, 12:23
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nick1816 wrote:
Speed of A, B, and C are \(S_A\), \(S_B\) and \(S_C\) respectively.

\(S_A : S_B = 10:9\)

\(S_B : S_C = 10:9\)

\(S_A : S_B : S_C = 10*10 : 9*10 : 9*9 = 100 : 90 : 81\)


A beat C by 1.9 (10-8.1) Km or 1900m.



Bunuel wrote:
In a 10 km race. A, B, and C, each running at uniform speed, get the gold, silver, and bronze medals, respectively. If A beats B by 1 km and B beats C by 1 km, then by how many metres does A beat C?

A. 1750
B. 1800
C. 1900
D. 1950
E. 2000



An excellent way to solve it!!!!
Took me close to 4 min by the traditional (R*T=D) method
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Re: In a 10 km race. A, B, and C, each running at uniform speed, get the  [#permalink]

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New post 04 Apr 2020, 13:40
Bunuel wrote:
In a 10 km race. A, B, and C, each running at uniform speed, get the gold, silver, and bronze medals, respectively. If A beats B by 1 km and B beats C by 1 km, then by how many metres does A beat C?

A. 1750
B. 1800
C. 1900
D. 1950
E. 2000


Let’s let the time A runs the 10 km (or 10,000-meter) race be 1000 seconds. Thus, A’s speed is 10,000/1000 = 10 m/s. Since B is 1 km behind A when A finishes the race, B runs 9 km, or 9000 meters, in 1000 seconds. Thus, B’s speed is 9000/1000 = 9 m/s. In other words, B runs the 10 km race in 10,000/9 seconds. Since C is 1 km behind B when B finishes the race, C runs 9 km, or 9000 meters, in 10,000/9 seconds. So C’s speed is 9000/(10,000/9) = 9/(10/9) = 81/10 = 8.1 m/s.

Now, let’s compare A and C when A finishes the race. A finishes the 10 km race in 1000 seconds (see above). During this time, C runs 8.1 x 1000 = 8100 meters. Therefore, A beats C by 10,000 - 8100 = 1900 meters.

Answer: C

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Re: In a 10 km race. A, B, and C, each running at uniform speed, get the   [#permalink] 04 Apr 2020, 13:40

In a 10 km race. A, B, and C, each running at uniform speed, get the

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