Hi All,

This question is based on a math 'truism' that many Test Takers don't know: Increasing your speed by 50% decreases the amount of travel time by 1/3. Here's why:

Distance = (Rate)(Time)

D = (R)(T)

Since the distance remains the same (you're just changing the rate and time), any increase in rate or time is met with a change in the other term. Increasing the rate by 50% would give us...

D = (3/2)(R)(?)

The change in time has to 'offset' the (3/2) that is now part of the calculation. The fraction (2/3) creates that 'offset', which gives us....

D = (3/2)(R)(2/3)(T)

D = (6/6)(R)(T)

In this prompt, we're told that the time is decreased by 1/3 after the 15 miles/hour increase in speed. We're asked for the ORIGINAL speed, which means that the 15 miles/hour represents the 50% increase. Thus, the original speed must be 30 miles/hour.

Final Answer:

GMAT assassins aren't born, they're made,
Rich
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Let's start with a word equation:
(John's travel time at faster speed) = 2/3(John's regular travel time)

Let r = John's regular speed
So, r+15 = John's faster speed
Let d = the distance traveled at regular speed
So d = the distance traveled at regular speed

time = distance/speed
So, we get: d/(r + 15) = (2/3)(d/r)
Rewrite as: d/(r + 15)= 2d/3r
Cross multiply: (d)(3r) = 2d(r + 15)
Expand: 3rd = 2rd + 30d
Rearrange to get: rd = 30d
Divide both sides by d to get: r = 30

Answer: B

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Let v,t be the original speed and time taken by John

As the total distance (=speed*time) must be the same,

v*t = (2t/3)*(v+15) ---> v=30 miles per hour.

B is the correct answer.
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I like the way you approached the question.
I tried to solve the question in a similar way, but without success

the new time given is equal to x-1/3, while the distance is the same and the John's rate changed, which is equal to Disatnce/time - 1/3.
In this way I got this equation: x-1/3=D/T-1/3... after that I was stucked.

Where did I fail?

The mistake you are doing is that when you are told that the time is reduced by 1/3 it's not x-1/3 but x-x/3 = 2x/3
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Hi pepo,

In this question, the changes in rate and time are based on 'ratios' (NOT on an absolute number).

For example, if you're traveling 60 miles/hour and you reduce THAT speed by 1/3, the calculation is NOT 60 - 1/3.... it's 60 - (1/3)(60) = 40.

Since most of your other math 'skills' seem fine, this issue is ultimately about your organization and how you take your notes. Instead of just writing down "- 1/3", you should think about what that difference represents (it represents a 1/3 decrease in speed) and add a bit more detail to your work.

GMAT assassins aren't born, they're made,
Rich
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There a re different approaches for the question. Engr2012 & Rich demonstrated great algebraic and logic way.


Another way that we can mix and match between both.

distance is constant. Any decrease in time means increased speed (or vice versa)

So time decreased by 1/3..........>remaining time is 2/3.........> speed is 3/2= V2/V1= (V1+15)/V1 (where V2: new speed & V1:original speed)

TESTing The Answers, it is easily to find that

(30+15)/15=45/30=3/2

Answer: B

suppose r=his rate, t his time, rt = distance traveled
new time 2t/3, new rate r+15, distance is the same = rt.

now, (r+15)* 2t/3 = rt
2rt/3 + 10t = rt - multiply by 3 to get rid of the fractions:
2rt +30t = 3rt
30t = rt | divide by t
30=r.

his rate was 30.

This is my approach.
Time (t)= distance(d)/speed(s)
This means t is inversely proportional to speed
In this case the distance is the constant since it’s the same and can be ignored since it will cancel out in the equations.
So we can say that t=1/s ........(a)
Where t is the original time taken by John to travel home at a speed of s

We are told that he can reduce his speed by 1/3 had he increased his average speed by 15mph.

So new time would now be 2t/3 and will correspond to a speed of s+15
Hence 2t/3=1/(s+15)
So t=3/(2(s+15)) ........(b)

equating (a) to (b)

1/s=3/(2s+30)
3s=2s+30
Hence s=30

Answer is therefore B

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