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# Alex and Brenda both stand at point X. Alex begins to walk a

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Re: Alex and Brenda both stand at point X. Alex begins to walk a [#permalink]

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26 Nov 2013, 07:01
GmatPrepK wrote:
Hi,

I tried using this method:

let T be the time Brenda has to travel in order to cover twice the distance that Alex does:

So,

Distance (Brenda) = 2 Distance(Alex)
Speed of Brenda * T = 2 [ 4*T + 4 ] (the additional 4 miles covered in initial hour)
R * T = 8T + 8
T (R-8) = 8
T = 8 / (R-8) (C)

Your approach is correct but you stopped early like me If you re-read the question it is asking for the time of Alex and not that of Brenda. T is time of Brenda in hours. Alex started 1 hour earlier so Alex's time is T+1 which gives...

$$T+1 = \frac{8}{(R-8)} + 1$$

$$= \frac{8+R-8}{(R-8)}$$

$$= \frac{R}{R-8}$$

Hope this helps.
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Re: Alex and Brenda both stand at point X. Alex begins to walk a [#permalink]

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23 Jun 2014, 04:30
Bunuel wrote:
enigma123 wrote:
Alex and Brenda both stand at point X. Alex begins to walk away from Brenda in a straight line at a rate of 4 miles per hour. One hour later, Brenda begins to ride a bicycle in a straight line in the opposite direction at a rate of R miles per hour. If R > 8, which of the following represents the amount of time, in terms of R, that Alex will have been walking when Brenda has covered twice as much distance as Alex?

A. R-4
B. R/R+4
C. R/R-8
D. 8/R-8
E. 2R - 4

Guys - I don't have an OA for this. Can you please help in terms of how this can be solved?

Let T be the time that Alex will have been walking when Brenda has covered twice as much distance as Alex.

In T hours Alex will cover 4T miles;
Since Brenda begins her journey 1 hour later than Alex then total time for her will be T-1 hours, and the distance covered in that time will be R(T-1);

We want the distance covered by Brenda to be twice as much as that of Alex: 2*4T=R(T-1) --> 8T=RT-R --> T=R/(R-8).

Very clear and concise explanation, thank you.
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Re: Alex and Brenda both stand at point X. Alex begins to walk away from B [#permalink]

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23 Nov 2014, 03:03
Gmat1008 wrote:
Alex and Brenda both stand at point X. Alex begins to walk away from Brenda in a straight line at a rate of 4 miles per hour. One hour later, Brenda begins to ride a bicycle in a straight line in the opposite direction at a rate of R miles per hour. If R > 8, which of the following represents the amount of time, in terms of R, that Alex will have been walking when Brenda has covered twice as much distance as Alex?
1)R – 4
2)R / (R + 4)
3)R / (R – 8)
4)8 / (R – 8)
5)R^2 – 4

Now I have this doubt. I can approach this question by following 2 approaches:
1) R*T = 2*4(T+1) -> T = 8/(R-8)
2) R*(T-1) = 2*4T -> T = R/(R-8)

Can somebody help me why 1st is wrong and 2nd is correct. I can assume time either ways.

Thanks!!

hi first is not wrong. as per your equation. T+1 is the total time for which alex have been walking and T is the time for which brenda have been riding on a bicycle. so, we need to add 1 in the final answer.
i.e. T+1= [8/(R-8)] +1
= R/R-8
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Re: Alex and Brenda both stand at point X. Alex begins to walk a [#permalink]

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27 Nov 2014, 14:21
VeritasPrepKarishma wrote:
enigma123 wrote:
Alex and Brenda both stand at point X. Alex begins to walk away from Brenda in a straight line at a rate of 4 miles per hour. One hour later, Brenda begins to ride a bicycle in a straight line in the opposite direction at a rate of R miles per hour. If R > 8, which of the following represents the amount of time, in terms of R, that Alex will have been walking when Brenda has covered twice as much distance as Alex?

A. R-4
B. R/R+4
C. R/R-8
D. 8/R-8
E. 2R - 4

Guys - I don't have an OA for this. Can you please help in terms of how this can be solved?

Since your answer is in terms of R, you have the flexibility of putting in any value for R.
Say R = 16
Alex has already covered 4 miles in the first hour. In the next hour, Alex covers another 4 miles while Brenda covers 16, thereby covering twice the distance covered by Alex. (I chose R = 16 because it is twice of 8) Alex has been walking for 2 hrs so that's the answer we are looking for.
Plug R = 16 in the options. Only option C gives you 2. Answer (C)

LEARNING: Assuming the right number can save a lot of time. I chose R=9...so it took some time to decipher that only in after 9 hrs of Alex's travelling time will Brenda's distance be double of Alex [i could have done LCM but did not]. Had i chosen 16 which double of 8, i could have saved time.
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Re: Alex and Brenda both stand at point X. Alex begins to walk a [#permalink]

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28 Nov 2014, 21:16
If we take R=12 miles/hr then in 2 hrs brenda would have travelled 24 miles and by that same time a;ex would have travelled 8 miles as his speed is 4m/h now Alex started ahead of brenda 1 hour earlier so total distance travelled by Alex =12 miles which is half of what brenda has travelled . Now put value of R =12 in the options and see which option gives 3 as answer as total time traveled by Alex is 3 hrs . Hope this explanation helps. Happy preparing
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Re: Alex and Brenda both stand at point X. Alex begins to walk a [#permalink]

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23 Feb 2015, 01:43
GmatPrepK wrote:
Hi,

I tried using this method:

let T be the time Brenda has to travel in order to cover twice the distance that Alex does:

So,

Distance (Brenda) = 2 Distance(Alex)
Speed of Brenda * T = 2 [ 4*T + 4 ] (the additional 4 miles covered in initial hour)
R * T = 8T + 8
T (R-8) = 8
T = 8 / (R-8) (C)

This logic works very well I think you did a good job with it. Can you explain your question further?
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Re: Alex and Brenda both stand at point X. Alex begins to walk a [#permalink]

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23 Feb 2015, 12:20
distance= speed *time.
speed of Alex is Sa 4 miles per hour
speed of brenda is Sb R>8. take R as 9miles per hour for convenience.
Sb *t=2(Sa*t+4) => twice distance covered by brenda and 4km travelled early by alex in one hour.
9t=2(4t+4)

9t=8t+8

9t-8t=8

t= 8. Sb 9*8=72.
Sa 4*8 +4=36. Total time is 8hrs+1 hr earlier. 9hrs.
By plugging options C satisfy. R/R-8 = 9/9-8=9hrs.
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Re: Alex and Brenda both stand at point X. Alex begins to walk a [#permalink]

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12 Nov 2015, 00:11
Tom and Linda stand at point A. Linda begins to walk in a straight line away from Tom at a constant rate of 2 miles per hour. One hour later, Tom begins to jog in a straight line in the exact opposite direction at a constant rate of 6 miles per hour. If both Tom and Linda travel indefinitely, what is the positive difference, in minutes, between the amount of time it takes Tom to cover half of the distance that Linda has covered and the amount of time it takes Tom to cover twice the distance that Linda has covered?
When Tom has covered 1/2 Linda's distance, the following equation will hold: 6T = 0.5(2(T + 1)). We can solve for T:
6T = 0.5(2(T + 1))
6T = 0.5(2T + 2)
6T = T+1
5T = 1
T = 1/5

So it will take Tom 1/5 hour, or 12 minutes, to cover 1/2 Linda's distance. When Tom has covered twice Linda's distance, the following equation will hold: 6T = 2(2(T + 1)). We can solve for T:
6T = 2(2(T + 1))
6T = 2(2T + 2)
6T = 4T + 4
2T = 4
T = 2

If i apply same logic to following question
Alex and Brenda both stand at point X. Alex begins to walk away from Brenda in a straight line at a rate of 4 miles per hour. One hour later, Brenda begins to ride a bicycle in a straight line in the opposite direction at a rate of R miles per hour. If R > 8, which of the following represents the amount of time, in terms of R, that Alex will have been walking when Brenda has covered twice as much distance as Alex?

then 2*4*(t+1)= Rt
which gives me as option D ---- > 8/R-1
Can you clarify me
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Alex and Brenda both stand at point X. [#permalink]

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10 Feb 2016, 19:02
1
Alex and Brenda both stand at point X. Alex begins to walk away from Brenda in a straight line at a rate of 4
miles per hour. One hour later, Brenda begins to ride a bicycle in a straight line in the opposite direction at a
rate of R miles per hour. If R > 8, which of the following represents the amount of time, in terms of R, that
Alex will have been walking when Brenda has covered twice as much distance as Alex?
A R – 4
B R / (R + 4)
C R / (R – 8)
D 8 / (R – 8)
E R2 – 4

Solution - Lets assume Alex travels X miles when Brenda begins to ride. Then,

Time taken by Alex = Time taken by Brenda
X/4 = [(X+4)2]/R --- since Brenda covers twice as distance as Alex in total thus Brenda distance is (X+4)2

which gives X = 32/(R-8)

Now time taken by Alex = X/4 = [32/(R-8)]/4 = 8/(R-8)

WHERE DID I GO WRONG ? Answer is C, not D
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Re: Alex and Brenda both stand at point X. [#permalink]

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10 Feb 2016, 20:54
ravi007shankar wrote:
Alex and Brenda both stand at point X. Alex begins to walk away from Brenda in a straight line at a rate of 4
miles per hour. One hour later, Brenda begins to ride a bicycle in a straight line in the opposite direction at a
rate of R miles per hour. If R > 8, which of the following represents the amount of time, in terms of R, that
Alex will have been walking when Brenda has covered twice as much distance as Alex?
A R – 4
B R / (R + 4)
C R / (R – 8)
D 8 / (R – 8)
E R2 – 4

Solution - Lets assume Alex travels X miles when Brenda begins to ride. Then,

Time taken by Alex = Time taken by Brenda
X/4 = [(X+4)2]/R --- since Brenda covers twice as distance as Alex in total thus Brenda distance is (X+4)2

which gives X = 32/(R-8)

Now time taken by Alex = X/4 = [32/(R-8)]/4 = 8/(R-8)

WHERE DID I GO WRONG ? Answer is C, not D

Hi,
The Q seems more on lines of IIM CAT than GMAT..
your solution is correct, except the last step..
x/4 is the time, both have travelled ..
so Alex time is x/4 +1= 8/(R-8) +1..
{8 + (R-8)}/(R-8) =R/(R-8)..

choice D is given to force students to make errors..

Hope it helped
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Re: Alex and Brenda both stand at point X. [#permalink]

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11 Feb 2016, 00:02
HI ravi007shankar,

This question should probably be posted in the PS Forum here:

gmat-problem-solving-ps-140/

Here's how you can solve this question by TESTing VALUES:

We're told that Alex and Brenda both stand at point X. Alex begins to walk away from Brenda in a straight line at a rate of 4
miles per hour. One hour later, Brenda begins to ride a bicycle in a straight line in the opposite direction at a
rate of R miles per hour (and that R > 8). We're asked which of the following represents the amount of time, in terms of R, that
Alex will have been walking when Brenda has covered TWICE as much distance as Alex.

Let's TEST R = 12

After 1 hour....
Alex = 4 miles walked
Brenda = 0 miles biked

After 2 hours...
Alex = 8 miles walked
Brenda = 12 miles biked

After 3 hours...
Alex = 12 miles walked
Brenda = 24 miles biked
This is a MATCH for what we were told, so we're looking for an answer that equals 3 when R=12. There's only one answer that matches...

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Special Offer: Save $75 + GMAT Club Tests Free Official GMAT Exam Packs + 70 Pt. Improvement Guarantee www.empowergmat.com/ ***********************Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!*********************** Board of Directors Joined: 17 Jul 2014 Posts: 2726 Location: United States (IL) Concentration: Finance, Economics GMAT 1: 650 Q49 V30 GPA: 3.92 WE: General Management (Transportation) Alex and Brenda both stand at point X. [#permalink] ### Show Tags 11 Feb 2016, 08:22 1 ravi007shankar wrote: Alex and Brenda both stand at point X. Alex begins to walk away from Brenda in a straight line at a rate of 4 miles per hour. One hour later, Brenda begins to ride a bicycle in a straight line in the opposite direction at a rate of R miles per hour. If R > 8, which of the following represents the amount of time, in terms of R, that Alex will have been walking when Brenda has covered twice as much distance as Alex? A R – 4 B R / (R + 4) C R / (R – 8) D 8 / (R – 8) E R2 – 4 Solution - Lets assume Alex travels X miles when Brenda begins to ride. Then, Time taken by Alex = Time taken by Brenda X/4 = [(X+4)2]/R --- since Brenda covers twice as distance as Alex in total thus Brenda distance is (X+4)2 which gives X = 32/(R-8) Now time taken by Alex = X/4 = [32/(R-8)]/4 = 8/(R-8) WHERE DID I GO WRONG ? Answer is C, not D the wording is indeed confusing...and one might get messy with all the VIC's... My approach, assign variables: we need to find the time of A, when B covered twice the distance A covered. since R>8, suppose R=9. when alex covered 4 miles, brenda covered 0 - 1hour when alex covered 8 miles, brenda - 9 miles - 2 hours when alex covered 12 miles, brenda - 18 miles - 3 hours when alex covered 16 miles, brenda covered - 27 miles - 4 hours when alex covered 20 miles, brenda - 36 miles - 5 hours when alex covered 24 miles, brenda - 45 miles - 6 hours when alex covered 28 miles, brenda - 54 miles - 7 hours when alex covered 32 miles, brenda - 63 miles - 8 hours when alex covered 36 miles, brenda - 72 miles - AHA, we have 2X the distance. time - 9 hours. ok, so when R=9, time must be 9. now plug in values: A R – 4 = 9-4=5 - out B R / (R + 4) - 9/13 - out C R / (R – 8) = 9/1 = 9 works, let's see other options D 8 / (R – 8) = 8/-1 = -8 E R2 – 4 = 18-4=14 - nope. C alone works. thus, C is the answer. p.s. I see Rich is more experienced with testing values..he picked the numbers way better than me, and saved a lot of time EMPOWERgmat Instructor Status: GMAT Assassin/Co-Founder Affiliations: EMPOWERgmat Joined: 19 Dec 2014 Posts: 11957 Location: United States (CA) GMAT 1: 800 Q51 V49 GRE 1: 340 Q170 V170 Re: Alex and Brenda both stand at point X. [#permalink] ### Show Tags 11 Feb 2016, 10:59 1 Hi mvictor, You saw the opportunity to TEST VALUES, which is good. The next 'step' in your training should be to spot the 'clues' in the prompt that can help you to pick the most efficient VALUES to TEST. In the answer choices, you should notice all of the '4s' and '8s' - implying that multiples of 4 will likely be useful here. When combined with the fact that Alex is walking 4 miles/hour and Brenda's speed has to be GREATER than 8, the number 12 seems like a really logical choice. GMAT assassins aren't born, they're made, Rich _________________ 760+: Learn What GMAT Assassins Do to Score at the Highest Levels Contact Rich at: Rich.C@empowergmat.com # Rich Cohen Co-Founder & GMAT Assassin Special Offer: Save$75 + GMAT Club Tests Free
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05 Aug 2016, 10:23
Alex and Brenda both stand at point X. Alex begins to walk away from Brenda in a straight line at a rate of 4 miles per hour. One hour later, Brenda begins to ride a bicycle in a straight line in the opposite direction at a rate of R miles per hour. If R > 8, which of the following represents the amount of time, in terms of R, that Alex will have been walking when Brenda has covered twice as much distance as Alex?
A. R – 4 B. R / (R + 4) C. R / (R – 8) D. 8 / (R – 8) E. R2 – 4

Hi, I have solved this to arrive at Option D as below:

4 + (4* X) = (R*X)/2 -------------> X = 8/(R-8)

But the author has solved and has arrived at Option C as below:

(4*X) = (R*(X-1))/2 ----------------> X = R/(R-8)

Please get me clarified where i am going wrong.
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05 Aug 2016, 13:24
smineshk wrote:
Alex and Brenda both stand at point X. Alex begins to walk away from Brenda in a straight line at a rate of 4 miles per hour. One hour later, Brenda begins to ride a bicycle in a straight line in the opposite direction at a rate of R miles per hour. If R > 8, which of the following represents the amount of time, in terms of R, that Alex will have been walking when Brenda has covered twice as much distance as Alex?
A. R – 4 B. R / (R + 4) C. R / (R – 8) D. 8 / (R – 8) E. R2 – 4

Hi, I have solved this to arrive at Option D as below:

4 + (4* X) = (R*X)/2 -------------> X = 8/(R-8)

But the author has solved and has arrived at Option C as below:

(4*X) = (R*(X-1))/2 ----------------> X = R/(R-8)

Please get me clarified where i am going wrong.

You didn't include the hour that Alex took before Brenda started.

When Brenda sets off, Alex is 4 miles away from Brenda.
$$4\text{mph} \times 1\text{ hour} = 4 \text{ miles}$$

Calculating $$X$$ as the time from when Brenda starts to when she has covered double the distance.
$$4 + 4X = \frac{RX}{2} \implies X = \frac{8}{R-8}$$

Calculating $$Y$$ as the time that Alex has been walking when Brenda has covered double the distance.

$$Y = X + 1 = \frac{8}{R-8} + \frac{R-8}{R-8} = \frac{R}{R-8}$$
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Re: Alex and Brenda both stand at point X. Alex begins to walk a [#permalink]

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10 Nov 2016, 01:00
enigma123 wrote:
Alex and Brenda both stand at point X. Alex begins to walk away from Brenda in a straight line at a rate of 4 miles per hour. One hour later, Brenda begins to ride a bicycle in a straight line in the opposite direction at a rate of R miles per hour. If R > 8, which of the following represents the amount of time, in terms of R, that Alex will have been walking when Brenda has covered twice as much distance as Alex?

A. R-4
B. R/(R+4)
C. R/(R-8)
D. 8/(R-8)
E. 2R - 4

Guys - I don't have an OA for this. Can you please help in terms of how this can be solved?

Check the solution in attachment
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Re: Alex and Brenda both stand at point X. Alex begins to walk a [#permalink]

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15 May 2017, 10:36
Bunuel wrote:
enigma123 wrote:
Alex and Brenda both stand at point X. Alex begins to walk away from Brenda in a straight line at a rate of 4 miles per hour. One hour later, Brenda begins to ride a bicycle in a straight line in the opposite direction at a rate of R miles per hour. If R > 8, which of the following represents the amount of time, in terms of R, that Alex will have been walking when Brenda has covered twice as much distance as Alex?

A. R-4
B. R/R+4
C. R/R-8
D. 8/R-8
E. 2R - 4

Guys - I don't have an OA for this. Can you please help in terms of how this can be solved?

Let T be the time that Alex will have been walking when Brenda has covered twice as much distance as Alex.

In T hours Alex will cover 4T miles;
Since Brenda begins her journey 1 hour later than Alex then total time for her will be T-1 hours, and the distance covered in that time will be R(T-1);

We want the distance covered by Brenda to be twice as much as that of Alex: 2*4T=R(T-1) --> 8T=RT-R --> T=R/(R-8).

Hi bunuel..!
In this question why are we not considering it from Alex's point of view. I mean why not
2[4(t+1)]=Rt.?
I will be thankful if you clear my doubt.

Posted from my mobile device
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Re: Alex and Brenda both stand at point X. Alex begins to walk a [#permalink]

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15 May 2017, 10:43
Bunuel wrote:
enigma123 wrote:
Alex and Brenda both stand at point X. Alex begins to walk away from Brenda in a straight line at a rate of 4 miles per hour. One hour later, Brenda begins to ride a bicycle in a straight line in the opposite direction at a rate of R miles per hour. If R > 8, which of the following represents the amount of time, in terms of R, that Alex will have been walking when Brenda has covered twice as much distance as Alex?

A. R-4
B. R/R+4
C. R/R-8
D. 8/R-8
E. 2R - 4

Guys - I don't have an OA for this. Can you please help in terms of how this can be solved?

Let T be the time that Alex will have been walking when Brenda has covered twice as much distance as Alex.

In T hours Alex will cover 4T miles;
Since Brenda begins her journey 1 hour later than Alex then total time for her will be T-1 hours, and the distance covered in that time will be R(T-1);

We want the distance covered by Brenda to be twice as much as that of Alex: 2*4T=R(T-1) --> 8T=RT-R --> T=R/(R-8).

Hi bunuel..!
In this question why are we not considering it from Alex's point of view. I mean why not
2[4(t+1)]=Rt.?
I will be thankful if you clear my doubt.

Posted from my mobile device

T in my solution is the time that Alex will have been walking when Brenda has covered twice as much distance as Alex because the question asks which of the following represents the amount of time, in terms of R, that Alex will have been walking when Brenda has covered twice as much distance as Alex?

What is T in your solution?
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Re: Alex and Brenda both stand at point X. Alex begins to walk a [#permalink]

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31 May 2017, 04:25
I did it this way and got the answer in less than a min.

Let's assume the value of R as 10.

So when Alex reaches 20 miles, Brenda will cover 40 miles which is what question asks.

For Alex to reach 20 mils, he needs to travel for 5 hours.

Now put the value of R in answer options and only option C will give the value as 5 hours.

This method may not work every time but sometimes it makes life easy
Re: Alex and Brenda both stand at point X. Alex begins to walk a   [#permalink] 31 May 2017, 04:25

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