SIM Motion

This topic has expert replies
Master | Next Rank: 500 Posts
Posts: 328
Joined: Thu Aug 07, 2008 5:25 pm
Location: Philadelphia
Thanked: 4 times
GMAT Score:550

SIM Motion

by Abdulla » Sun Jun 14, 2009 2:42 pm
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) R^2 – 4

OA is C
Abdulla

Legendary Member
Posts: 752
Joined: Sun May 17, 2009 11:04 pm
Location: Tokyo
Thanked: 81 times
GMAT Score:680

by tohellandback » Sun Jun 14, 2009 10:49 pm
just plugin any value for R
I plugged in 16 miles per hour
so in one hour
if Brenda take time t to cover twice the distance traveled by alex in t1 time
16t=2*4*t1
t1=2t
if t=1
t1=2
now use the answer options
A)R-4=12 no
B)R/R+4=16/20 no
C)R/R-8=16/16-8=2 yes
D)8/R-8=8/8=1 No
E)Nooo

of course if you get same values for two options you need to substitute some other values.
The powers of two are bloody impolite!!

Legendary Member
Posts: 752
Joined: Sun May 17, 2009 11:04 pm
Location: Tokyo
Thanked: 81 times
GMAT Score:680

by tohellandback » Sun Jun 14, 2009 10:51 pm
and by the way ..this question should go under the problem solving thread
The powers of two are bloody impolite!!

Senior | Next Rank: 100 Posts
Posts: 32
Joined: Sat Jun 13, 2009 12:33 pm
Location: Mumbai, India

by pg850 » Sun Jun 14, 2009 11:49 pm
Another way to solve the problem:

IF 'x' is the amount of time for Brenda to cover twice the distance of Alex,
then (4+4x)*2 = Rx [Equatinf the distance travelled by both]

Solving for x would give, x = 8/(R-8)

To find total time for Alex, add 1 to this giving you the answer,

R/(R-8)

-PG

User avatar
Legendary Member
Posts: 682
Joined: Fri Jan 16, 2009 2:40 am
Thanked: 32 times
Followed by:1 members

by Vemuri » Mon Jun 15, 2009 4:06 am
If none of the above approaches click on the D day, just assume a value for R, i.e. R=9 & then check when the distance travelled by Brenda becomes double the value of Alex's distance.

Hr ---- Alex ---- Brenda
1 ---- 4 miles 0 miles
2 ---- 8 miles 9 miles
3 ---- 12 miles 18 miles
4 ---- 16 miles 27 miles
5 ---- 20 miles 36 miles
6 ---- 24 miles 45 miles
7 ---- 28 miles 54 miles
8 ---- 32 miles 63 miles
9 ---- 36 miles 72 miles

Plug in the value of R & you will get C as the answer.

Junior | Next Rank: 30 Posts
Posts: 17
Joined: Sat Nov 12, 2011 10:20 pm

by kumadil2011 » Wed Nov 30, 2011 9:09 am
Why cant i solve by following method

let time taken by Bre be T, and Alex be T+1..

I get the solution if i take T-1 for Bre and T for Alex..can some one guide me when to pick pair like (T and T+1) and (T and T-1)

Master | Next Rank: 500 Posts
Posts: 385
Joined: Fri Sep 23, 2011 9:02 pm
Thanked: 62 times
Followed by:6 members

by user123321 » Wed Nov 30, 2011 11:55 am
kumadil2011 wrote:Why cant i solve by following method

let time taken by Bre be T, and Alex be T+1..

I get the solution if i take T-1 for Bre and T for Alex..can some one guide me when to pick pair like (T and T+1) and (T and T-1)
okay let B took T & A took T+1

so from question
distance travelled by A * 2 = distance travelled by B
=> 4(T+1)*2 = R*T
=> 8T+8 = RT
=> T(R-8) = 8
=> T = 8/(R-8)
since we need to tell the time from Alex's point of view, we need to find T+1 from above
=> T+1 = 8/(R-8) + 1
=> T+1 = R/(R-8)

hope this helps

user123321
Just started my preparation :D
Want to do it right the first time.

User avatar
Master | Next Rank: 500 Posts
Posts: 425
Joined: Wed Dec 08, 2010 9:00 am
Thanked: 56 times
Followed by:7 members
GMAT Score:690

by LalaB » Thu Dec 01, 2011 9:29 am
4t=S
r(t-1)=2S

8t=r(t-1)
rt-8t=r
t=r/(r-8)

User avatar
Master | Next Rank: 500 Posts
Posts: 103
Joined: Mon Jun 23, 2014 11:31 pm
Thanked: 25 times
Followed by:12 members
GMAT Score:770

by BestGMATEliza » Wed Jul 09, 2014 9:49 pm
So lets make t= the time (in hours) that alex has walked. Then, the time Brenda has biked would be t-1, because she started an hour after alex.

We know that the Distance = Rate*Time, so we want to find the time when Brenda's Distance is twice that of Alex.

For Alex D=4t and for Brenda D=R(t-1), thus we are looking for when 8t=R(t-1)

let's multiply out the R, so 8t=Rt -R

now let's put all the ts on one side thus, 8t-Rt=-R

take out the t, so (8-R)t=-R

Now, divide both sides by (8-R) and we have t=-R/(8-R)

this exact equation isn't an answer choice, but we can multiply the top and the bottom of the fraction by -1 to get: R/(R-8), which is C
Eliza Chute
Best GMAT Prep Courses
GMAT course comparison and reviews
Your one stop for all your GMAT studying needs!

User avatar
GMAT Instructor
Posts: 15539
Joined: Tue May 25, 2010 12:04 pm
Location: New York, NY
Thanked: 13060 times
Followed by:1906 members
GMAT Score:790

by GMATGuruNY » Thu Jul 10, 2014 4:05 am
Abdulla 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) R^2 - 4
Let R = 16mph.
We can PLUG IN THE ANSWERS, which represents Alex's time.

Answer choice C: R/(R-8) = 16/(16-8) = 2 hours.
At a rate of 4mph, the distance traveled by Alex in 2 hours = 4*2 = 8 miles.
Since Brenda starts to bike 1 hour after Alex starts to walk, Brenda bikes for 1 hour.
At a rate of 16mph, the distance traveled by Brenda in 1 hour = 16*1 = 16 miles.
Success!
Brenda's distance is twice Alex's distance.

The correct answer is C.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3