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# Greg and Brian are both at Point A (above). Starting at the

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Greg and Brian are both at Point A (above). Starting at the  [#permalink]

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Updated on: 12 Jul 2013, 01:36
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64% (02:12) correct 36% (02:00) wrong based on 461 sessions

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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 2/3 that of Brian's.
(2) Brian's average speed is 20 miles per hour greater than Greg's.

Originally posted by ghostdude on 22 May 2011, 11:28.
Last edited by Bunuel on 12 Jul 2013, 01:36, edited 1 time in total.
Edited the question and added the OA.
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Re: Greg and Brian  [#permalink]

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22 May 2011, 19:04
6
1
A

~20 sec method:

ratio AB:AC is fixed, so only ratio speed_G:speed_B matters.
1) we know ration speed_B:speed_G. sufficient
2) it can be 20.0001:0.0001 (Brian is first) or 10020:10000 (Greg is first). Insufficient
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Re: Greg and Brian  [#permalink]

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22 May 2011, 15:14
6
1
ghostdude wrote:
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 2/3 that of Brian's.
(2) Brian's average speed is 20 miles per hour greater than Greg's.

I am not sure of the OA

Greg's speed: $$s_g$$
Distance covered by Greg: $$D$$
Time taken by Greg: $$t_g=\frac{D}{s_g}$$

Brian's speed $$s_b$$
Distance covered by Greg $$D\sqrt{2}$$ (Hypotenuse of right isosceles triangle = root(2) times base)
Time taken by Brian $$t_b=\frac{D\sqrt{2}}{s_b}$$

1.
$$s_g=\frac{2}{3}s_b$$
$$t_g=\frac{D}{s_g}=\frac{D}{\frac{2}{3}s_b}=\frac{3D}{2s_b}$$-------------1
$$t_b=\frac{\sqrt{2}D}{s_b}$$----------------2

Comparing 1 and 2:
$$\frac{3}{2} > \sqrt{2}$$. Thus, Greg took relatively more time to reach his destination.
Sufficient.

2.
$$s_b=s_g+20$$
$$t_b=\frac{\sqrt{2}D}{s_b}=\frac{\sqrt{2}D}{s_g+20}$$
$$t_g=\frac{D}{s_g}$$

Let's prove opposite of st1:
$$t_b>t_g$$
$$\frac{\sqrt{2}D}{s_g+20}>\frac{D}{s_g}$$
$$\frac{\sqrt{2}}{s_g+20}>\frac{1}{s_g}$$
$$s_g\sqrt{2}>s_g+20$$
$$0.4s_g>20$$
$$s_g>50$$

Thus, if greg's speed is approx more than 50, brian would take more time, otherwise greg would take more time.

Not Sufficient.

Ans: "A"
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Re: Greg and Brian  [#permalink]

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22 May 2011, 22:48
1
We know distances for both as x and xroot2
A gives us ratio of speeds, if we know ratio of speeds and ratio of distance, we can find ratio of time, as variables for distance and speed will get eliminated/cancelled, and only the ratio will remain
B gives us comparison between teo speeds, but not as a ratio- in this case, the variables will not get eliminated- more specifically, the varibale for speed.
In other words, this would make the answer yes or no dependent upon the value of speed. Hence it is not sufficient.
A alone is sufficient
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Re: Greg and Brian  [#permalink]

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23 May 2011, 04:40
Vivesomnium,

The fact, that statement 2 doesn't fix speed ratio is not enough to say that it's not sufficient.
For example, if statement 2 were following:

"Greg's average speed is 20 miles per hour greater than Brian's."

statement 2 would be sufficient.
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Re: Greg and Brian  [#permalink]

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23 May 2011, 18:10
Let AB = D

then AC = D * root(2)

(1)

B's speed = v

G's Speed = 2v/3

D/v - Brian

D * root(2)/2v/3 - Greg

Now, D/v * 3root(2)/2 > D/v because 3root(2)/2 > 1

Sufficient

(2)

D/v - Brian

D * root(2)/(v-20) - Greg

D * root(2)/(v-20) may be < or > D/v depending on value of v

If v = 21, then D * root(2)/(v-20) is > D/v

If v = 40, then D * root(2)/(v-20) is < D/v

Insufficient.

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Re: Greg and Brian  [#permalink]

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24 May 2011, 02:26
a. the distances are 2^(1/2) and 1 units each.

B = 1 unit/interval then G = 2/3 unit/interval

thus 2^(1/2)/ 1 < 1/ (2/3). Hence sufficient.

b 2^(1/2)/ (g+20) < 1/g

for g = 1

LHS < RHS meaning B travels faster.

for g=100
LHS>RHS meaning G travels faster. not sufficient.

Hence A.
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Re: Greg and Brian are both at Point A (above). Starting at the  [#permalink]

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27 Mar 2017, 03:17
let AB = x

hence AC= sqrt2*x (using pythagorus)

assuming speed,distance and time taken by Brian = v1, d1, t1

and v2, d2, t2 be the time taken by Greg

1) V2 = 2V1/3

hence t1/t2 = 2*sqrt2/3

This is sufficent

2) V1 = V2 + 20

==sqrt2*x/t1 = x/t2 + 20

==sqrt2/t1 = 1/t2 + 20/x

since we dont know the value of x that is distance this is not sufficient

Hence, Answer is A
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Re: Greg and Brian are both at Point A (above). Starting at the  [#permalink]

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04 Jun 2018, 19:30
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# Greg and Brian are both at Point A (above). Starting at the

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