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On his morning jog, Charles runs through a park in the shape of a semi
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27 Jan 2015, 06:48
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Re: On his morning jog, Charles runs through a park in the shape of a semi
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28 Jan 2015, 04:47
ElCorazon wrote: 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. It is a Ratios concept. Say two numbers are in the ratio 5:3. If the first number is 25, the second one will be 15, right? Yes, because the multiplier is 5. The multiplier is the number with which you multiply the ratio to get the actual values. So 25/5 = 5 gives you the multiplier. Now multiply 3 by 5 to get 15, the actual value of the second number. If instead, you are given that the actual difference between these numbers is 10, can you find the individual numbers? Yes, because on ratio scale, the difference between the numbers is 2 (obtained by 53) but actually the difference is 10 so you can again say that the multiplier is 10/2 = 5. Check out this post for more on this ratios concept: http://www.veritasprep.com/blog/2011/03 ... ofratios/Here, on ratio scale the difference is 22/7  2 = 8/7. This difference is actually 4 mins. So the multiplier is 4/(8/7) = 7/2 So the actual value of 22/7 term is (22/7)*(7/2) = 11 mins And actual value of 2 is 2*7/2 = 7 mins
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Re: On his morning jog, Charles runs through a park in the shape of a semi
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27 Jan 2015, 07:43
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
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27 Jan 2015, 09:07
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 \(22d/7Rd/R=4\) > \(R=2,14/4\) 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
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27 Jan 2015, 09:14
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 \(22d/7Rd/R=4\) > \(R=2,14/4\) 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
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27 Jan 2015, 14:34
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 = (114) 7
Answer is A.



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Re: On his morning jog, Charles runs through a park in the shape of a semi
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27 Jan 2015, 20:59
Hi All, While the answer choices appear "close together", the way that the prompt is written allows for a great estimation "shortcut" that you can take advantage of. You do still need to do some algebra though.... We're asked to compare HALF the circumference of a circle to the diameter of a circle, so let's deal with THAT comparison first.... Circumference = 2(pi)R HALF a Circumference = (pi)R Diameter = 2R We're told that a runner moves at a constant speed and that the difference between running (pi)R units and 2R units is 4 minutes (we're never told what the units are, but that doesn't matter). Here we have the equation: pi(R)  2R = 4 minutes We're asked to figure out how long it takes to run the DIAMETER, so we're looking for the value of 2R. I'm going to 'factor out' the R from the above equation: R(pi  2) = 4 Since pi = approximately 3.14, we now have..... R(3.14  2) = 4 R(1.14) = 4 Without doing the exact calculation, I can see that R is BETWEEN 3 (which is too small) and 4 (which is too big). If R = 3 EXACTLY, then 2R = 6 minutes If R = 4 EXACTLY, then 2R = 8 minutes I know that 6 is too SMALL and 8 is too BIG, so the answer has to be BETWEEN 6 and 8. There's only one answer that fits... Final Answer: GMAT assassins aren't born, they're made, Rich
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Re: On his morning jog, Charles runs through a park in the shape of a semi
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27 Jan 2015, 21:03
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
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28 Jan 2015, 04:28
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
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29 Jan 2015, 05:57
I did it like this:
Diameter = 2r > This is the distance in one way. Time is 4 minutes less in this way (T4) 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) / (T4)] = [2πr (T4)] / [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
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02 Feb 2015, 01:59
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 crossmultiply 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
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02 Feb 2015, 22:16
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/x4 x4 Circumference pi*r pi*r/x x Total
Speed is constant So,2r/x4=pi*r/x x=11 and so x4 =7



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Re: On his morning jog, Charles runs through a park in the shape of a semi
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03 Feb 2015, 23:12
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
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03 Feb 2015, 23:26
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
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20 Feb 2015, 07:27
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
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05 Jan 2016, 17:31
(⫪d/2)/d=⫪/2 ⫪/2=t/(t4) t=11 minutes t4=7 minutes



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Re: On his morning jog, Charles runs through a park in the shape of a semi
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06 Jan 2016, 09:52
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 t4 = 2r/s => pi*r/s = 2r/s + 4 => r(pi  2) = 4s or 2r(pi2) = 8s => 2r/s = 8/(pi2) = 7
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Re: On his morning jog, Charles runs through a park in the shape of a semi
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