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Re: Points A and B are 120 km apart. A motorcyclist starts from
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03 Nov 2013, 20:17
the official answer is D, but I am confused in 3.6 hours the motorcyclist covers (3.6*30) 108 km & the cyclist covers (3.6*10) 36 km, now the hypotenuse is sqrt(120108)^2 + sqrt(36)^2 = 48 km
after 4 hours motorcyclist reaches point b and the cyclist covers 40 km, as the road is perpendicular, the distance between two will be 40 km which is less than 48 km at 3.6 hours. so the answer should be E.



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Re: Points A and B are 120 km apart. A motorcyclist starts from
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03 Nov 2013, 22:17



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Re: Points A and B are 120 km apart. A motorcyclist starts from
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04 Nov 2013, 00:34
it is clear that the closer to 4 hours the less the distance. So, it is D



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Re: Points A and B are 120 km apart. A motorcyclist starts from
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04 Nov 2013, 21:21
I don't think this problems needs calculations of any sort. Simply visualize a variable right triangle with A toward B forming the base and perpendicularly away from B as the height. The resulting hypotenuse represents the shortest distance between the bikers at any point. The variable, A towards B (Base) diminishes at a much faster rate than does perpendicularly away from B (The height) because of the significant difference in the speeds of the bikers. Now since the hypotenuse (or the distance between the bikers) is the sum of the squares of the sides, it's least when their combination is minimum ( which obviously happens just before 4 hours or closest to 4 hours).
Guess, visualizing this concept takes much lesser time than reading it (say under a minute!!)



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Re: Points A and B are 120 km apart. A motorcyclist starts from
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22 Nov 2013, 16:19
correct me if I'm wrong, but wouldn't the hypotenuse be at it's shortest when one of the legs is minimized, thus 3.6 would have to be the answer here? The cyclist going north is only going at 10mph, so the leg of the triangle from the cyclist traveling from A to B is getting shorter at a MUCH faster rate than the cyclist traveling north's leg is getting longer. By this logic the longer you wait, the shorter the AB cyclists leg is going to get, without nearly as much increase in the northward leg. Based on this I said D



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Re: Points A and B are 120 km apart. A motorcyclist starts from
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23 Nov 2013, 06:09
AccipiterQ wrote: correct me if I'm wrong, but wouldn't the hypotenuse be at it's shortest when one of the legs is minimized, thus 3.6 would have to be the answer here? The cyclist going north is only going at 10mph, so the leg of the triangle from the cyclist traveling from A to B is getting shorter at a MUCH faster rate than the cyclist traveling north's leg is getting longer. By this logic the longer you wait, the shorter the AB cyclists leg is going to get, without nearly as much increase in the northward leg. Based on this I said D No, that's not correct. Why not the answer then 4 hours (E. None)? In this case one of the legs is 0 and the distance is 40 km, which is more than ~38 km in case of 3.6 hours. Correct solution is here: pointsaandbare120kmapartamotorcycliststartsfrom92837.html#p717460
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Re: Points A and B are 120 km apart. A motorcyclist starts from
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05 May 2014, 09:45
prasannajeet wrote: Hi Bunuel
Please explain the below part.....
Now quadratic function f(x)=ax^2+bx+c reaches its minimum (or maximum when a is negative  not our case), when x=\frac{b}{2a}=\frac{60*120}{2*1000}=3.6
Rgds Prasannajeet I dont think differential / integral calculus must be getting tested in GMAT. If yes , then a single differential is just a proof of point of inflection. The equation would need to be differentiated again to confirm that the point was inflection was maxima or minima



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Re: Points A and B are 120 km apart. A motorcyclist starts from
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05 May 2014, 21:11
himanshujovi wrote: prasannajeet wrote: Hi Bunuel
Please explain the below part.....
Now quadratic function f(x)=ax^2+bx+c reaches its minimum (or maximum when a is negative  not our case), when x=\frac{b}{2a}=\frac{60*120}{2*1000}=3.6
Rgds Prasannajeet I dont think differential / integral calculus must be getting tested in GMAT. If yes , then a single differential is just a proof of point of inflection. The equation would need to be differentiated again to confirm that the point was inflection was maxima or minima I would say this question is borderline. GMAT expects you to know how to find the roots of the quadratic using the formula (though it never expects you to actually use it). Also, knowing what the graph of a quadratic looks like is a useful skill on GMAT. We know that when the coefficient of the x^2 term is positive, it is upward facing so it will have a minimum value. By symmetry, the minimum value will lie right in the center of the roots i.e. it will be the average of the two roots. The two roots are given by \(\frac{b + \sqrt{b^2  4ac}}{2a} and \frac{b  \sqrt{b^2  4ac}}{2a}\). Their average is b/2a. So you don't actually need to use differential calculus. I wouldn't recommend my students to ignore this question if they were looking at 750.
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Re: Points A and B are 120 km apart. A motorcyclist starts from
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12 Sep 2014, 05:06
VeritasPrepKarishma wrote: himanshujovi wrote: prasannajeet wrote: Hi Bunuel
Please explain the below part.....
Now quadratic function f(x)=ax^2+bx+c reaches its minimum (or maximum when a is negative  not our case), when x=\frac{b}{2a}=\frac{60*120}{2*1000}=3.6
Rgds Prasannajeet I dont think differential / integral calculus must be getting tested in GMAT. If yes , then a single differential is just a proof of point of inflection. The equation would need to be differentiated again to confirm that the point was inflection was maxima or minima I would say this question is borderline. GMAT expects you to know how to find the roots of the quadratic using the formula (though it never expects you to actually use it). Also, knowing what the graph of a quadratic looks like is a useful skill on GMAT. We know that when the coefficient of the x^2 term is positive, it is upward facing so it will have a minimum value. By symmetry, the minimum value will lie right in the center of the roots i.e. it will be the average of the two roots. The two roots are given by \(\frac{b + \sqrt{b^2  4ac}}{2a} and \frac{b  \sqrt{b^2  4ac}}{2a}\). Their average is b/2a. So you don't actually need to use differential calculus. I wouldn't recommend my students to ignore this question if they were looking at 750. Hi Karishma Am I correct in understanding from your explanation that assuming the more time passes the less the distance? One closes the distance very quickly so the more time passes the less the distance. This is wrong? If the bicycle would start at the middle do the same straight line, the more time passes the less the distance. Why is this different?



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Re: Points A and B are 120 km apart. A motorcyclist starts from
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28 Jul 2015, 23:20
jjack0310 wrote: WholeLottaLove wrote: Points A and B are 120 km apart. A motorcyclist starts from A to B along straight road AB with speed 30 kmph. At the same time a cyclist starts from B along a road perpendicular to road AB, with a speed of 10 kmph. After how many hours will the distance between them be the least? See attached graphThis problem is basically looking for the shortest hypotenuse in a triangle comprised of biker on road AB and cyclist moving directly perpendicular to AB from point B. Using the graph, we can plug in numbers into the quadratic formula to see what length C is the shortest. a^2+b^2=c^2 1 Hour: Biker has traveled 30 km and cyclist has traveled 10 km. 90^2 + 10^2 = c^2 8200 = c^2 We don't need to find the square  we can compare values of c^22 Hour: Biker has traveled 60 km and cyclist has traveled 20 km. 60^2+20^ = c^2 4000 = c^2 3 hour: Biker has traveled 90 km and cyclist has traveled 30 km. 30^2 + 30^2 = c^2 1800 = c^2 For each increasing hour, the cyclist and biker move closer together. Out of all the possible answer choices, 3.6 reduces the bikers time the most (and thus the square of a larger speed i.e. 30 km/h) It's also possible to just look at the graph and recognize that the more hours that pass, the shorter the red line between the biker and cyclist is. ANSWER: D. 3.6 hoursOne other thing...technically none of the answer choices are correct. Distance would be minimized when the biker hit B and was 40km away from the cyclist traveling perpendicular from line AB: a^2+b^2=c^2 ===> 0^2+40^2=c^2 ===> 1600 = c^2 so wouldn't E actually be the right answer? This. I thought the answer would be E, just because the distance will be the least when the motorcyclist reaches point B, because then the distance will be only 40 km. No, because even after 3.5 hours, the hypotenuse would be 5root58 < 5*8 = 40.



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Re: Points A and B are 120 km apart. A motorcyclist starts from
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09 Sep 2016, 11:15
wakk0 wrote: rrsnathan wrote: gurpreetsingh wrote: IMO D
Distance traveled by 1st = 30t in t time distance traveled by 2nd = 10t in t time
now at any instance a right angled triangle is formed whose sides are as follow
1. The distance traveled by 2nd 2. The distance left between the 1st and B 3. Present distance between both the bikers This is the hypotenuse.
Thus the equation of their distance becomes
d^2 = (10t)^2 + (12030t)^2
Which is maximum when its differential is 0 => 20t = 2(12030t) *3 => t = 3.6 Hi, Can anyone can explain this part of this solution ""Which is maximum when its differential is 0 => 20t = 2(12030t) *3 => t = 3.6 "" Thanks in advance, Rrsnathan gurpreetsingh is drawing on a point from first semester calculus. Let me explain: In calculus you can take a derivative of a function to calculate the min or max of the function. When you are applying a min or max to a parabola what you are actually finding is the vertex of the parabola (the min or max as the case may be). In order to find this you take the first derivative of the function, set it equal to zero, and then solve for your variable. This solution will produce the answer (the min or max of the parabola) for the problem. If you would like an explanation of this process let me know and I will post it. However, if you are a bit rusty on calculus Bunuel's answer is probably the better course to take. Hi, I am not good at calculus. But I know a little about differentiation. Would you, please, tell me in respect of what you are differentiating?



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Re: Points A and B are 120 km apart. A motorcyclist starts from
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23 Oct 2016, 02:52
NaeemHasan wrote: Hi, I am not good at calculus. But I know a little about differentiation. Would you, please, tell me in respect of what you are differentiating? We are differentiating W.R.T. \(t\). You can solve this question without Calculus. Please check previous posts.



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Re: Points A and B are 120 km apart. A motorcyclist starts from
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02 May 2017, 20:42
Shouldn't the answer be E? For example at 3.7 hours the distance would be lesser than what it was in 3.6 hrs. Please help me understand. Thanks, Jat
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Points A and B are 120 km apart. A motorcyclist starts from
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23 May 2018, 00:08
Hi, To be honest, when I saw the question for the first time, I had no idea to solve it. ( I roughly remember that calculus might work, thus I tried it. ) Here is my method to solve the question T  in hour D  distance between cars
I started with forming a distance equation (12030T)^2 + (10T)^2 = D^2
Then, I expanded terms
120^2  2*30T*120 + 900T^2 + 100T^2 = D^2
I knew that, in this case, I did not have to care about D value, so I just applied differentiation: *right side is zero because it's the vertex(might be max or mix value). In this case, I was sure that it gives time at minimum distance. 2*30*120 +2000T=0
T=3.6 hrs
Best regards, G.



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Re: Points A and B are 120 km apart. A motorcyclist starts from
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20 Oct 2018, 20:45
einstein10 wrote: Points A and B are 120 km apart. A motorcyclist starts from A to B along straight road AB with speed 30 kmph. At the same time a cyclist starts from B along a road perpendicular to road AB, with a speed of 10 kmph. After how many hours will the distance between them be the least? A. 3 hours B. 3.4 hours C. 3.5 hours D. 3.6 hours E. None D, please provide workable method that can solve this in 2mins, i solved but it takes more than 3mins. can any one help. thanks Distance traveled by 1st = 30t in t time distance traveled by 2nd = 10t in t time now at any instance a right angled triangle is formed whose sides are as follow 1. The distance traveled by 2nd 2. The distance left between the 1st and B 3. Present distance between both the bikers This is the hypotenuse. Thus the equation of their distance becomes d^2 = (10t)^2 + (12030t)^2 Which is maximum when its differential is 0 => 20t = 2(12030t) *3 => t = 3.6 Can someone please explain how the last equation has come from first.I was trying to do first differential but I am no getting the same result.



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Re: Points A and B are 120 km apart. A motorcyclist starts from
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20 Oct 2018, 20:46
prabsahi wrote: einstein10 wrote: Points A and B are 120 km apart. A motorcyclist starts from A to B along straight road AB with speed 30 kmph. At the same time a cyclist starts from B along a road perpendicular to road AB, with a speed of 10 kmph. After how many hours will the distance between them be the least? A. 3 hours B. 3.4 hours C. 3.5 hours D. 3.6 hours E. None D, please provide workable method that can solve this in 2mins, i solved but it takes more than 3mins. can any one help. thanks Distance traveled by 1st = 30t in t time distance traveled by 2nd = 10t in t time now at any instance a right angled triangle is formed whose sides are as follow 1. The distance traveled by 2nd 2. The distance left between the 1st and B 3. Present distance between both the bikers This is the hypotenuse. Thus the equation of their distance becomes d^2 = (10t)^2 + (12030t)^2 Which is maximum when its differential is 0 => 20t = 2(12030t) *3 => t = 3.6 Can someone please explain how the last equation has come from first.I was trying to do first differential but I am no getting the same result. chetan sir request your help.



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Re: Points A and B are 120 km apart. A motorcyclist starts from
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20 Oct 2018, 22:25
Hi prabsahi, Quote: Thus the equation of their distance becomes
d^2 = (10t)^2 + (12030t)^2
Which is maximum when its differential is 0
=> 20t = 2(12030t) *3 => t = 3.6
After differentiating with 't' and equating to zero, we get following: \(2*10t*10 + 2*(12030t)*(30) \Rightarrow 200t + 1800t  120*60 = 0 \Rightarrow 2000t = 120*60 \Rightarrow t = 3.6\) Alternatively, you can first expand the expression and then differentiate. i.e. \(d^2 = 100t^2 + 900t^2  2*30t*120 + 120^2 = 1000t^2  2*30t*120 + 120^2\) If we differentiate above expression w.r.t. 't' and equate it equal to zero, we get \(2000t  2*30*120 =0 => t = 3.6\) In general, it is better to avoid calculus for GMAT question. You can use the following results. These results can be obtained using the method of completing the square. Minimum or Maximum Value of a Quadratic Equation The minimum and maximum value of a quadratic equation \(f(x) = ax^2 + bx + c ; (a \neq 0)\) occurs at \(x = \frac{b}{2a}\)  if \(a > 0\), then the minimum value is \(f(\frac{b}{2a})\)
 if \(a < 0\), then the maximum value is \(f(\frac{b}{2a})\)
Hope this helps.



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Re: Points A and B are 120 km apart. A motorcyclist starts from
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20 Oct 2018, 22:31
ganand wrote: Hi prabsahi, Quote: Thus the equation of their distance becomes
d^2 = (10t)^2 + (12030t)^2
Which is maximum when its differential is 0
=> 20t = 2(12030t) *3 => t = 3.6
After differentiating with 't' and equating to zero, we get following: \(2*10t*10 + 2*(12030t)*(30) \Rightarrow 200t + 1800t  120*60 = 0 \Rightarrow 2000t = 120*60 \Rightarrow t = 3.6\) Alternatively, you can first expand the expression and then differentiate. i.e. \(d^2 = 100t^2 + 900t^2  2*30t*120 + 120^2 = 1000t^2  2*30t*120 + 120^2\) If we differentiate above expression w.r.t. 't' and equate it equal to zero, we get \(2000t  2*30*120 =0 => t = 3.6\) In general, it is better to avoid calculus for GMAT question. You can use the following results. These results can be obtained using the method of completing the square. Minimum or Maximum Value of a Quadratic Equation The minimum and maximum value of a quadratic equation \(f(x) = ax^2 + bx + c ; (a \neq 0)\) occurs at \(x = \frac{b}{2a}\)  if \(a > 0\), then the minimum value is \(f(\frac{b}{2a})\)
 if \(a < 0\), then the maximum value is \(f(\frac{b}{2a})\)
Hope this helps. Indeed this is very helpful .Thanks a lot!!




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