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Ryerson (Ted Rogers) Thread Master
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Re: John would have reduced the time it took him to drive from his home to [#permalink]
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RSOHAL wrote:
John would have reduced the time it took him to drive from his home to a certain store by 1/3 if he had increased his average speed by 15 miles per hour. What was John's actual average speed, in miles per hour, when he drove from his home to the store?

(A) 25
(B) 30
(C) 40
(D) 45
(E) 50

Source: GMAT Focus


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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Re: John would have reduced the time it took him to drive from his home to [#permalink]
Engr2012 wrote:
RSOHAL wrote:
John would have reduced the time it took him to drive from his home to a certain store by 1/3 if he had increased his average speed by 15 miles per hour. What was John's actual average speed, in miles per hour, when he drove from his home to the store?

(A) 25
(B) 30
(C) 40
(D) 45
(E) 50

Source: GMAT Focus


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.


I like the way you approached the question.
I tried to solve the question in a similar way, but without success :P

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?
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Re: John would have reduced the time it took him to drive from his home to [#permalink]
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pepo wrote:
Engr2012 wrote:
RSOHAL wrote:
John would have reduced the time it took him to drive from his home to a certain store by 1/3 if he had increased his average speed by 15 miles per hour. What was John's actual average speed, in miles per hour, when he drove from his home to the store?

(A) 25
(B) 30
(C) 40
(D) 45
(E) 50

Source: GMAT Focus


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.


I like the way you approached the question.
I tried to solve the question in a similar way, but without success :P

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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Re: John would have reduced the time it took him to drive from his home to [#permalink]
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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.

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John would have reduced the time it took him to drive from his home to [#permalink]
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
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Re: John would have reduced the time it took him to drive from his home to [#permalink]
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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.
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Re: John would have reduced the time it took him to drive from his home to [#permalink]
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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Re: John would have reduced the time it took him to drive from his home to [#permalink]
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