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30 Nov 2021, 02:25
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E
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Difficulty:
55%
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Question Stats:
70% (03:18) correct
30%
(03:07)
wrong
based on 20
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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. \(40\)
A. \(19\)
B. \(36\)
C. \(38\)
D. \(39\)
E. \(40\)
30 Nov 2021, 02:25
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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. \(40\)
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
15 Mar 2025, 02:59
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This is a high-quality question, and that's for sure, Bunuel , I made it bit elaborate to understand where the ratio comes in first place,
Does this make sense, let me know your thought.
If we count the standard formula for rate problems.
s = v1 *t
(B) 950 = v1 * t
(C) 912 = v2 * t
t = 950/v1
t = 912/v2
=> 950/v1 = 912/v2
v1/v2 = 950/912
We don't know the value of constant speed, right?
For B to complete the remaining 50m
t1 = 50/v1
t1 = x / v2
50/v1 = x/v2
v1/v2 = 50/x
So,
950/912 = 50/x
x = 48
Eventually, the remaining meters to finish for C is 40!
Does this make sense, let me know your thought.
If we count the standard formula for rate problems.
s = v1 *t
(B) 950 = v1 * t
(C) 912 = v2 * t
t = 950/v1
t = 912/v2
=> 950/v1 = 912/v2
v1/v2 = 950/912
We don't know the value of constant speed, right?
For B to complete the remaining 50m
t1 = 50/v1
t1 = x / v2
50/v1 = x/v2
v1/v2 = 50/x
So,
950/912 = 50/x
x = 48
Eventually, the remaining meters to finish for C is 40!
