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Re: Points A and B are 120 km apart. A motorcyclist starts from [#permalink]
21 Sep 2013, 22:59

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 graph

This 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^2

2 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 hours

One 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.

Hey the distance between the 2 bikers will be approximately 38 Kms after 3.6 hrs which is less than 40 Kms after 4 hours.

After 3.6 hrs Biker A will have moved 3.6 * 30 = 108 Kms i.e. 12 Kms remaining and Biker B will have moved by 3.6 * 10 = 36 Kms

Re: Points A and B are 120 km apart. A motorcyclist starts from [#permalink]
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(120-108)^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.

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

Expert's post

khosru wrote:

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(120-108)^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.

Check your math: D. \(\sqrt{(120-30*3.6)^2 + (10*3.6)^2} = \sqrt{1,440}\). E. \(\sqrt{(120-30*4.6)^2 + (10*4.6)^2} = \sqrt{2,440}\). _________________

Re: Points A and B are 120 km apart. A motorcyclist starts from [#permalink]
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!!)

Re: Points A and B are 120 km apart. A motorcyclist starts from [#permalink]
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 A-B cyclists leg is going to get, without nearly as much increase in the northward leg. Based on this I said D

Re: Points A and B are 120 km apart. A motorcyclist starts from [#permalink]
23 Nov 2013, 06:09

Expert's post

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 A-B 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.

Re: Points A and B are 120 km apart. A motorcyclist starts from [#permalink]
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

Re: Points A and B are 120 km apart. A motorcyclist starts from [#permalink]
05 May 2014, 21:11

1

This post received KUDOS

Expert's post

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 co-efficient 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. _________________

Re: Points A and B are 120 km apart. A motorcyclist starts from [#permalink]
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 co-efficient 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?

Re: Points A and B are 120 km apart. A motorcyclist starts from [#permalink]
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 graph

This 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^2

2 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 hours

One 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.

gmatclubot

Re: Points A and B are 120 km apart. A motorcyclist starts from
[#permalink]
28 Jul 2015, 23:20

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