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sjuniv32
Can anyone show a detailed solution to the problem above?


Let's take:
Distance travelled be: D
Initial Speed: S
Time taken by travelling at Speed S as: T (T=D/S)


It's given that speed increased by 25%;
So, new Speed will be: 125*S/100

Let's calculate new time taken:
New time= D/(125S/100) = 4D/5S


% change in time: [(D/S) - (4D/5S)]/(D/S) * 100 = 20%

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sjuniv32
Can anyone show a detailed solution to the problem above?

D = S * T
Since, D is the same in both the cases :

S1 -> initial Speed
T1 -> initial time
S2 -> final Speed
T2 -> final time

S1 * T1 = S2 * T2

S2 = S1 + (\(\frac{25}{100}\)) * S1 => (\(\frac{125}{100}\))S1

We need to find T2

There fore,

S1 * T1 = (\(\frac{125}{100}\)) S1 * T2 => (\(\frac{5}{4}\)) S1 * T2

T2 = (\(\frac{4}{5}\)) T1

Now, multiplying numerator and denominator by 20 :

T2 = ( \(\frac{4* 20}{5*20}\) )* T1

T2 = (\(\frac{80}{100}\)) T1 => (\(\frac{100-20}{100}\)) *T1

Therefore, T2 is 20% less than T1

Answer (B)

Hope this helps!
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This is how I get the answer

Let john drive at 100 mph for 100 miles ===> take 1 hour
if John drive at 125 mph(25% increase) ===> 100/125 => take 4/5 hour
Therefore, time reduction = 1-(4/5)/ 1 = 1/5 => 20%
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BrentGMATPrepNow, expecting a solution from you for the problem above. Thanks for your attempt in advance!
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sjuniv32
What would be the percent reduction in John’s commute time if he were to drive along the same route at an average speed that is 25% higher?

(A) 15
(B) 20
(C) 25
(D) 30
(E) 35

Let's assign some nice values...

Let 500 km be the distance of John's commute
And let 100 kmh be John's regular speed

Time = distance/rate
So, John's regular driving time = 500/100 = 5 hours

If John's regular driving speed (100 kmh) is increased by 25%, we get a new speed of 125 kmh
So John's new driving time = 500/125 = 4 hours

So John's driving time decreases from 5 hours to 4 hours, which represents a 20% decrease in driving time

Answer: B
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