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Re: GRE Math Challenge #7-David drove to work at an average [#permalink]
Great Job!!!
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Re: GRE Math Challenge #7-David drove to work at an average [#permalink]
C : 360/7
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Re: GRE Math Challenge #7-David drove to work at an average [#permalink]
From home to work: 45 = t1*d
From work to home: 60 = t2*d

The distance d is the same for both trips, and also t1 + t2 = 2

So t1 = d/45 and t2 = d/60.

Substitute away t1 and t2 in the total time equation: d/45 + d/60 = 2.

Now solving for d: (60d + 45d)/2700 = 2 --> 105*d = 5,400 --> d = 360/7.
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Re: GRE Math Challenge #7-David drove to work at an average [#permalink]
speed = distance/ time

We will get 2 equation for 1. To office and 2. Back from office.

Solving for 2 variables from 2 equations gives the answer.
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Re: GRE Math Challenge #7-David drove to work at an average [#permalink]
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the equation is x/45 + x/60 = 2 ==> x = 360/7

where X is the distance between home and office.
Answer -option C
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Re: GRE Math Challenge #7-David drove to work at an average [#permalink]
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soumya1989 wrote:
David drove to work at an average (arithmetic mean) speed of 45 miles per hour. After work, David drove home at an average speed of 60 miles per hour. If David spent a total of 2 hours commuting to and from work, how many miles does David drive to work?

A)48
B)\(\frac{256}{5}\)
C)\(\frac{360}{7}\)
D)\(\frac{105}{2}\)
E)\(\frac{160}{3}\)


We can let d = the distance each way and create the equation for time:

d/45 + d/60 = 2

Multiplying by 180, we have:

4d + 3d = 360

7d = 360

d = 360/7

Answer: C
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Re: GRE Math Challenge #7-David drove to work at an average [#permalink]
soumya1989 wrote:
David drove to work at an average (arithmetic mean) speed of 45 miles per hour. After work, David drove home at an average speed of 60 miles per hour. If David spent a total of 2 hours commuting to and from work, how many miles does David drive to work?

A)48
B)\(\frac{256}{5}\)
C)\(\frac{360}{7}\)
D)\(\frac{105}{2}\)
E)\(\frac{160}{3}\)



Let's start with a "word equation"
We can write: (David's travel time TO work) + (David's travel time FROM work) = 2 hours
time = distance/speed

Let d = the distance each way

We can write: d/45 + d/60 = 2
To eliminate the fractions, multiply both sides by 180, the LCM of 45 and 60

We get: 4d + 3d = 360
Simplify: 7d = 360
Solve: d = 360/7

Answer: C

Cheers,
Brent
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Re: GRE Math Challenge #7-David drove to work at an average [#permalink]
I did D/2 = 45 + 60( equating speed) but got a different answer, what's wrong with this approach?

GreenlightTestPrep wrote:
soumya1989 wrote:
David drove to work at an average (arithmetic mean) speed of 45 miles per hour. After work, David drove home at an average speed of 60 miles per hour. If David spent a total of 2 hours commuting to and from work, how many miles does David drive to work?

A)48
B)\(\frac{256}{5}\)
C)\(\frac{360}{7}\)
D)\(\frac{105}{2}\)
E)\(\frac{160}{3}\)



Let's start with a "word equation"
We can write: (David's travel time TO work) + (David's travel time FROM work) = 2 hours
time = distance/speed

Let d = the distance each way

We can write: d/45 + d/60 = 2
To eliminate the fractions, multiply both sides by 180, the LCM of 45 and 60

We get: 4d + 3d = 360
Simplify: 7d = 360
Solve: d = 360/7

Answer: C

Cheers,
Brent
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Re: GRE Math Challenge #7-David drove to work at an average [#permalink]
Chaithraln2499 wrote:
I did D/2 = 45 + 60( equating speed) but got a different answer, what's wrong with this approach?


d/2 = David's average speed for the entire trip.
Conversely, 45 + 60 is the SUM of his speeds driving to and from work.
So, you are not equating to equivalent values.
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Re: GRE Math Challenge #7-David drove to work at an average [#permalink]
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