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Greg and Brian

by ronaldramlan » Sun Aug 07, 2011 4:26 am
Image
Greg and Brian are both at Point A (above). Starting at the same time, Greg drives to point B while Brian drives to point C. Who arrives at his destination first?

(1) Greg's average speed is 2/3rd that of Brian's.

(2) Brian's average speed is 20 miles per hour greater than Greg's.

(A) Statement (1) alone is sufficient, but statement (2) alone is not sufficient.
(B) Statement (2) alone is sufficient, but statement (1) alone is not sufficient.
(C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
(D) Each statement ALONE is sufficient.
(E) Statements (1) and (2) TOGETHER are NOT sufficient.
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Source: — Data Sufficiency |
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by GmatKiss » Sun Aug 07, 2011 5:27 am
is AC=AB, attachment not visible for me :(

if AC=AB, then IMO: D
if AC is not equal to AB then, IMO:E
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by akhilsuhag » Sun Aug 07, 2011 5:37 am
Given: Distance to be traveled is the same = D
Let the speeds be: G and B respectively and the time taken Tg and Tb.

Now,
D= G * Tg or D= B * Tb; This gives us the equation : G * Tg = B * Tb

Statement 1:

G = 2/3*B (Given)
So our equation becomes: (2/3)B * Tg = B * Tb or (2/3)* Tg = Tb

Clearly Tb is two-thirds of Tg and so B or brian gets their first.

Hence Sufficient

Statement 2:

B = G + 20; So our equation becomes: G * Tg = (G+20) * Tb

or, (G/(G+20))* Tg = Tb

Now G + 20 will always be greater than G and so Tb will always be smaller than Tg.

So, Sufficient

The answer seems to be C. [spoiler]Also instead of all the work, both statements tall us who is faster. Since we know that the Distance is the same if we know who is faster we know who gets their first. So you really don't have to work in this problem.[/spoiler]

I hope I am correct tho. I am really not confident about my quant abilities!!
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by akhilsuhag » Sun Aug 07, 2011 5:39 am
I meant D I wrote C by mistake. Both are sufficient. Can we have the OA.
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by ronaldramlan » Sun Aug 07, 2011 7:33 am
OA is A
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by akhilsuhag » Sun Aug 07, 2011 7:56 am
Stupid of me I took AC=AB and not AB=BC..
Need to ream questions properly!!
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by gmatboost » Sun Aug 07, 2011 11:16 pm
I always enjoy going from A to B.
Keep in mind that the one who arrives first is the one who took less time.

If AB is x, then AC is x*root(2), since it is a 45-45-90 triangle.

If we draw a RTD chart here, before the statements:
Greg: Rg * Tg = x
Brian: Rb * Tb = x*root(2)

R = Rate, T = Time, g, = greg, b = brian
We want to compare Tg and Tb

Statement 1:
Greg: (2/3)Rb * Tg = x
Brian: Rb * Tb = x*root(2)

Tg = x / [2Rb/3] = x * [3/(2Rb)] = 3x/(2Rb)
Tb = x*root(2)/Rb

Tg = (x/Rb) * (3/2)
Tb = (x/Rb) * root(2)

Since 3/2 > root(2), [spoiler]Greg's time was longer, so Greg gets there later. Sufficient. Disappointing for me (Greg).[/spoiler]

We cannot make the same conclusion using Statement 2. The speeds could be 21 and 1, in which case Brian would get there first easily because he goes so much faster. Or the speeds could be 9920 and 9900, in which case Greg would get there first easily because the speeds are very similar and Greg's distance is much less. Insufficient.
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by smackmartine » Sun Aug 07, 2011 11:32 pm
IMO A
Let equal sides be d(AB=BC=d) and ,thus AC = rt(2)d
If g=Greg and b= Brian
RgTg = d ---- I
RbTb=rt(2)d ----II

St 1.

Rg=(2/3) Rb

From above we know that solving I and II will give me the ratio Tg/Tb (just wo of these variables) = 3/2rt(2) --suff

St 2.
[RgTg/(Rg+20)][Tg/Tb]=1/rt(2) -- we will end up in three variables after solving. as we do not know Rg ratio Tg/Tb is not possible. ---Insuff
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by GMATGuruNY » Mon Aug 08, 2011 4:07 am
ronaldramlan wrote:Image
Greg and Brian are both at Point A (above). Starting at the same time, Greg drives to point B while Brian drives to point C. Who arrives at his destination first?

(1) Greg's average speed is 2/3rd that of Brian's.

(2) Brian's average speed is 20 miles per hour greater than Greg's.
In an isosceles right triangle, the sides are proportioned x : x : x√2.

Statement 1: Greg's average speed is 2/3rd that of Brian's.

Let Greg's distance AB = 30.
Then Brian's distance AC = 30√2 ≈ 42.

Let Brian's speed = 42 miles per hour.
Then Greg's speed = (2/3)*42 = 28 miles per hour.

Time for Brian to arrive = d/r = 42/42 = 1 hour.
Time for Greg to arrive = d/r = 30/28 = more than 1 hour.
Brian arrives first.

The case above illustrates the following:
For every 2 miles that Greg travels, Brian travels 3 miles.
Thus, Brian is traveling 1.5 times as fast as Greg.
Brian's distance is √2 times Greg's distance.
Since √2 < 1.5, Brian's distance is less than 1.5 times Greg's distance.
Since Brian is traveling 1.5 times as fast as Greg, and his distance is less than 1.5 times Greg's distance, Brian arrives first.
Sufficient.

Statement 2: Brian's average speed is 20 miles per hour greater than Greg's.

Case 1:
Let Greg's distance AB = 30.
Then Brian's distance AC = 30√2 ≈ 42.

Let Greg's speed = 22 miles per hour.
Then Brian's speed = 22+20 = 42 miles per hour.

Time for Brian to arrive = d/r = 42/42 = 1 hour.
Time for Greg to arrive = d/r = 30/22 = more than 1 hour.
Brian arrives first.

Case 2:
Let Greg's distance AB = 300.
Then Brian's distance AC = 300√2 ≈ 420.

Let Greg's speed = 300 miles per hour.
Then Brian's speed = 300+20 = 320 miles per hour.

Time for Greg to arrive = d/r = 300/300 = 1 hour.
Time for Brian to arrive = d/r = 420/320 = more than 1 hour.
Greg arrives first.

Since in the first case Brian arrives first and in the second case Greg arrives first, insufficient.

The correct answer is A.
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