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
SIM Motion
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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.
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!!
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and by the way ..this question should go under the problem solving thread
The powers of two are bloody impolite!!
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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
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
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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.
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.
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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)
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)
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okay let B took T & A took T+1kumadil2011 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)
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
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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
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
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Let R = 16mph.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
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.
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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].
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