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An athlete decides to run the same distance in 1/4th less time that she usually took. By how much percent will she have to increase her average speed?
  • a)
    40 
  • b)
    44.4 
  • c)
    33.3 
  • d)
    22.2
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
An athlete decides to run the same distance in 1/4th less time that sh...
(3/4)t=1
t=4/3=1+1/3
She will have to increase her speed by 33.333% to run the same distance in 3/4s of the time..
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Most Upvoted Answer
An athlete decides to run the same distance in 1/4th less time that sh...
Given: An athlete decides to run the same distance in 1/4th less time that she usually took.

To Find: By how much percent will she have to increase her average speed?

Let us assume the original time taken by the athlete to cover the distance is 't' and the original speed is 's'. Hence, the distance can be calculated as:

Distance = Speed × Time = s × t

Now, as per the question, the athlete decides to run the same distance in 1/4th less time than she usually took. Therefore, the new time taken by the athlete can be calculated as:

New Time = t - (1/4)t = (3/4)t

Hence, the new speed of the athlete can be calculated as:

New Speed = Distance / New Time = (s × t) / (3/4)t = (4/3)s

We need to find the percentage increase in her speed:

% Increase = [(New Speed - Original Speed) / Original Speed] × 100

Substituting the values, we get:

% Increase = [(4/3)s - s / s] × 100 = (1/3) × 100 = 33.3%

Therefore, the athlete needs to increase her average speed by 33.3% to run the same distance in 1/4th less time. Hence, the correct answer is option 'C'.
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Community Answer
An athlete decides to run the same distance in 1/4th less time that sh...
(3/4)t=1
t=4/3=1+1/3
She will have to increase her speed by 33.333% to run the same distance in 3/4s of the time..
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An athlete decides to run the same distance in 1/4th less time that she usually took. By how much percent will she have to increase her average speed?a)40b)44.4c)33.3d)22.2Correct answer is option 'C'. Can you explain this answer?
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