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Re: On his morning jog, Charles runs through a park in the shape of a semi [#permalink]
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ans A..... along dia he takes 7 min and along circumference 11 mins
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Re: On his morning jog, Charles runs through a park in the shape of a semi [#permalink]
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chetan2u wrote:
ans A..... along dia he takes 7 min and along circumference 11 mins


Hey man, I'd go with A as well; how did you get 7 minutes? I am getting 6 minutes.

assume rate = R; distance_1=d; distance_2=22d/7; time(circumference) = t

\(22d/7R-d/R=4\) --> \(R=2,14/4\)

time taken=3,14(4)d/d2,14 which turns out to be roughly 6 minutes.

can you please elaborate? thanks
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Re: On his morning jog, Charles runs through a park in the shape of a semi [#permalink]
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gmat6nplus1 wrote:
chetan2u wrote:
ans A..... along dia he takes 7 min and along circumference 11 mins


Hey man, I'd go with A as well; how did you get 7 minutes? I am getting 6 minutes.

assume rate = R; distance_1=d; distance_2=22d/7; time(circumference) = t

\(22d/7R-d/R=4\) --> \(R=2,14/4\)

time taken=3,14(4)d/d2,14 which turns out to be roughly 6 minutes.

can you please elaborate? thanks


hi... let assume the same values as yours..
rate =R, dia(2r)=d,dist along circumference(Pi*r) will be =(d/2)*22/7=11d/7
given is ... (11d/7R)-d/R=4... 4d/7R=4 or d/R=7mins and we have to find d/R, which is 7 min
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Re: On his morning jog, Charles runs through a park in the shape of a semi [#permalink]
It's easy to forget to halve the circumference (the longer distance = 11D/7, and not 22D/7).

When solving, you can take advantage of the fact that the rate can be anything here, by plugging in R = 1. Given that Time = Distance / Rate, we then have D = T. From there you can take two approaches:

1)

T + 4 = 11T / 7
7T + 28 = 11T
28 = 4T
7 = T

2)

Assign variable 'x' for the time taken on the short path, and y as the time taken on the long path. Given that T can = R:

x + 4 = y
11x = 7y (based on the fact that y = x * 11 / 7)
Multiplying the first equation by 11 and subtracting the second equation, you get:
44 = 4y
y = 11
x = (11-4) 7

Answer is A.
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Re: On his morning jog, Charles runs through a park in the shape of a semi [#permalink]
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Bunuel wrote:
On his morning jog, Charles runs through a park in the shape of a semicircle, entering at point A and exiting at point B. If he runs the diameter from A to B, it takes him 4 minutes less than it takes him to run along the circumference of the semicircle from A to B. Assuming Charles runs at a constant rate regardless of which path, how long does it take him to run along the diameter? (Assume that π=227)

A. 7 minutes
B. 8 minutes
C. 9 minutes
D. 10 minutes
E. 11 minutes

Kudos for a correct solution.


Circumference of semi circle of radius R \(= \pi*R\)
Diameter of semi circle = 2R

Ratio of distance of Circumference:Diameter \(= \pi:2\)
If Speed is constant, ratio of time taken to run on Circumference: Diameter \(= \pi:2\)
On the ratio scale, this difference is 22/7 - 2 = 8/7 but it is actually 4 mins.
So time taken to run along the diameter is 2*4/(8/7) = 7 mins

Answer (A)
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Re: On his morning jog, Charles runs through a park in the shape of a semi [#permalink]
VeritasPrepKarishma wrote:
Bunuel wrote:
On his morning jog, Charles runs through a park in the shape of a semicircle, entering at point A and exiting at point B. If he runs the diameter from A to B, it takes him 4 minutes less than it takes him to run along the circumference of the semicircle from A to B. Assuming Charles runs at a constant rate regardless of which path, how long does it take him to run along the diameter? (Assume that π=227)

A. 7 minutes
B. 8 minutes
C. 9 minutes
D. 10 minutes
E. 11 minutes

Kudos for a correct solution.


Circumference of semi circle of radius R \(= \pi*R\)
Diameter of semi circle = 2R

Ratio of distance of Circumference:Diameter \(= \pi:2\)
If Speed is constant, ratio of time taken to run on Circumference: Diameter \(= \pi:2\)
On the ratio scale, this difference is 22/7 - 2 = 8/7 but it is actually 4 mins.
So time taken to run along the diameter is 2*4/(8/7) = 7 mins

Answer (A)


Would it possible to expand on your steps for the following part: 'On the ratio scale, this difference is 22/7 - 2 = 8/7 but it is actually 4 mins.
So time taken to run along the diameter is 2*4/(8/7) = 7 mins'

I understood the reasoning up to then but was confused by this part. Thanks.
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Re: On his morning jog, Charles runs through a park in the shape of a semi [#permalink]
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I did it like this:

Diameter = 2r --> This is the distance in one way. Time is 4 minutes less in this way (T-4)
Circumference = 2πr --> This is the distance in the other way. Time is T this way.

We can use the RTD chart, knowing that the Rate is the same, and solve the equation.

So, according to the RTD chart: R=D/T. We make the 2 Rates equal:

[(2πr) / T] / [(2r) / (T-4)]
= [2πr (T-4)] / [2πrT]
= (2πrT - 8πr) / (2rT)
= π - [(8πr) / rT]
= π - [(8π) / T]
= (πΤ - 8π) / Τ
= - 7π

Which led me to ANS A.

I am sorry if there are any mistakes in the calculation and feel free to point out my mistakes!
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Re: On his morning jog, Charles runs through a park in the shape of a semi [#permalink]
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Bunuel wrote:
On his morning jog, Charles runs through a park in the shape of a semicircle, entering at point A and exiting at point B. If he runs the diameter from A to B, it takes him 4 minutes less than it takes him to run along the circumference of the semicircle from A to B. Assuming Charles runs at a constant rate regardless of which path, how long does it take him to run along the diameter? (Assume that π=227)

A. 7 minutes
B. 8 minutes
C. 9 minutes
D. 10 minutes
E. 11 minutes

Kudos for a correct solution.


VERITAS PREP OFFICIAL SOLUTION:

Since Rate = Distance/Time, and you know that the rate is the same in each case, you can use that to set up the equation:

Circumference Distance/Circumference Time = Diameter Distance/Diameter Time

And since you're solving for Diameter Time, you can call that time T and the Circumference Time (T + 4).

The relationship for distance is that Circumference=π(Diameter)=22/7*D, so half the circumference would be 11/7*D This sets up the equation:

\(\frac{\frac{11}{7}*D}{(T+4)}=\frac{D}{T}\)
And your goal is to solve for T, so cross-multiply to get:

\(\frac{11}{7}*DT=DT+4D\)
Then combine like terms by subtracting DT from both sides:

\(\frac{4}{7}*DT=4D\)
Then divide both sides by D:

\(\frac{4}{7}*T=4\)
And you should see that T=7, leading to answer choice A.
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Re: On his morning jog, Charles runs through a park in the shape of a semi [#permalink]
Hi ,
I solved in underneath way using table method.
Let me know is this approach correct?
Diameter=2r
Circumference of semi cirlce=pi*r

Distance Speed Time
Diameter 2r 2r/x-4 x-4
Circumference pi*r pi*r/x x
Total

Speed is constant
So,2r/x-4=pi*r/x
x=11
and so
x-4 =7
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Re: On his morning jog, Charles runs through a park in the shape of a semi [#permalink]
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Bunuel wrote:
On his morning jog, Charles runs through a park in the shape of a semicircle, entering at point A and exiting at point B. If he runs the diameter from A to B, it takes him 4 minutes less than it takes him to run along the circumference of the semicircle from A to B. Assuming Charles runs at a constant rate regardless of which path, how long does it take him to run along the diameter? (Assume that π=227)

A. 7 minutes
B. 8 minutes
C. 9 minutes
D. 10 minutes
E. 11 minutes

Kudos for a correct solution.


Hi Bunuel, Kindly correct the typo \(\pi = \frac{22}{7}\)
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Re: On his morning jog, Charles runs through a park in the shape of a semi [#permalink]
Answer = A = 7

Let the diameter of semicircle = 1, time required = t

Perimeter of circular path \(= 2 * \frac{22}{7} * \frac{1}{2} * \frac{1}{2} = \frac{22}{14}\), time required = t+4

Setting up the speed equation given that speed is constant

\(\frac{1}{t} = \frac{22}{14(t+4)}\)

t = 7 minutes
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Re: On his morning jog, Charles runs through a park in the shape of a semi [#permalink]
23a2012 wrote:
On his morning jog, Charles runs through a park in the shape of a semicircle, entering at point A and exiting at point B. If he runs the diameter from A to B, it takes him 4 minutes less than it takes him to run along the circumference of the semicircle from A to B. Assuming Charles runs at a constant rate regardless of which path, how long does it take him to run along the diameter? (Assume that =22/7)



7 minutes

8 minutes

9 minutes

10 minutes

11 minutes


Let radius =r, speed of charles =s

we need to find 2r/s

as per the question we have,

pi(r)/s - 2r/s =4

r/s(22/7 -2) =4

r/s(8/7) =4

r/s = 7/2

2r/s =7

hence A
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Re: On his morning jog, Charles runs through a park in the shape of a semi [#permalink]
(⫪d/2)/d=⫪/2
⫪/2=t/(t-4)
t=11 minutes
t-4=7 minutes
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Re: On his morning jog, Charles runs through a park in the shape of a semi [#permalink]
Bunuel wrote:
On his morning jog, Charles runs through a park in the shape of a semicircle, entering at point A and exiting at point B. If he runs the diameter from A to B, it takes him 4 minutes less than it takes him to run along the circumference of the semicircle from A to B. Assuming Charles runs at a constant rate regardless of which path, how long does it take him to run along the diameter? (Assume that π=227)

A. 7 minutes
B. 8 minutes
C. 9 minutes
D. 10 minutes
E. 11 minutes

Kudos for a correct solution.


t = pi*r/s
t-4 = 2r/s
=> pi*r/s = 2r/s + 4
=> r(pi - 2) = 4s
or 2r(pi-2) = 8s
=> 2r/s = 8/(pi-2) = 7
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