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Emdad
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Running at 5/4 of his usual speed, an athlete improves his t 02 Oct 2021, 11:30
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

25% (medium)

Question Stats:

76% (01:42) correct 24% (01:45) wrong based on 92 sessions

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Running at 5/4 of his usual speed, an athlete improves his timing by 5 minutes. The time he usually takes to run the same distance is?
A. 30 min
B. 28 min
C. 25 min
D. 30 min
E. None of these

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C
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Running at 5/4 of his usual speed, an athlete improves his t 02 Oct 2021, 12:19
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Emdad
Running at 5/4 of his usual speed, an athlete improves his timing by 5 minutes. The time he usually takes to run the same distance is?
A. 30 min
B. 28 min
C. 25 min
D. 30 min
E. None of these

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\(\frac{Distance}{Speed}\) = Time

Let D is the usual distance and T is the usual time

\(\frac{D}{S}\) -\(\frac{ 4D}{5S}\) = 5

=> \(\frac{D}{5S }\) =5
=>\(\frac{D}{S} \)=25

Hence C
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Running at 5/4 of his usual speed, an athlete improves his t 02 Oct 2021, 21:43
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We know, \(Speed ∝\frac{1}{time}\)

Let, his usual speed be x. If his speed is x, time taken is 1/x, if his speed is 5x/4 then the time taken is 4x/5.

Here, late time~usual time = 5 minute
So, \(\frac{4x}{5}\)~\( x=5\)
Therefore, x=25 minute.

Answer is C

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Running at 5/4 of his usual speed, an athlete improves his t 27 Oct 2021, 20:04
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Running at 5/4 of his usual speed, an athlete improves his timing by 5 minutes. The time he usually takes to run the same distance is?
A. 30 min
B. 28 min
C. 25 min
D. 30 min
E. None of these

Solution:

Speed: Usual-4 Present-5
Time: Usual-5 Present-4
Hence, difference in time is 1 i.e, (5-4) which equals 5 minutes according to question

So, 1 equals 5 minutes
5 equals 5*5= 25 minutes(Ans)

Tusher Karmoker
Love from Bangladesh🇧🇩

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Running at 5/4 of his usual speed, an athlete improves his t 27 Oct 2021, 22:35
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Running at 5/4 of his usual speed, an athlete improves his timing by 5 minutes. The time he usually takes to run the same distance is?
A. 30 min
B. 28 min
C. 25 min
D. 30 min
E. None of these

#Approach 1:

In both cases , the distance traveled is constant and we Distance = Speed * Time .

Let the usual speed be S and time taken is T hours.
The new speed =\(\frac{ 5}{4}\) S and new time taken = T - \(\frac{5}{60}\) as he reduces the time by 5 min.( 5 min = 5/60 hours)

Since distance covered is same in both cases, we can equate both.
Distance = Speed * Time .

S*T =\(\frac{ 5}{4}S*(T-\frac{5}{60})\)

T =\(\frac{5}{4}\)*(T-\(\frac{5}{60}\))

4T = 5T - 5*\(\frac{5}{60}\)

T = \(\frac{25}{60}\) hours = 25 mins

Option C is the answer.

#Approach 2:

When the Distance covered is constant, the speed and time are inversely proportional.
For Example: If you double your speed, the time taken to cover the same distance would be halved.

We can apply the same property of Speed and Time here as well as the distance covered is the same.

Here the speed of the person is increased from S to 5/4 S , that means his time taken should decrease from T to 4/5T.
If the normal time taken by him is T mins, then the new time taken = \(\frac{4}{5}\)T mins

We know that he improves his timing by 5 mins,i.e T -\(\frac{ 4}{5}\)T = 5 mins

Solving the equation,
T/5 = 5
T= 25 mins.

Option C is the answer.

Thanks,
Clifin J Francis,
GMAT SME
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