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# Points A and B are 120 km apart. A motorcyclist starts from

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Points A and B are 120 km apart. A motorcyclist starts from [#permalink]  17 Apr 2010, 08:13
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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

[Reveal] Spoiler: OA
D, please provide workable method that can solve this in 2mins, i solved but it takes more than 3mins. can any one help.
thanks
[Reveal] Spoiler: OA

Last edited by einstein10 on 17 Apr 2010, 08:57, edited 1 time in total.
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Re: time distance problem [#permalink]  17 Apr 2010, 08:55
Starting distance is 120.

As the first motorcyclist moves east, the second motorcyclist moves north. This forms a right triangle. The distance between the two motorcyclists is the hypotenuse of the triangle.

I'd backsolve this one and plug in the answer choices. This minimum distance occurs at answer choice B. The value is the square root of 1380.
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Re: time distance problem [#permalink]  25 Apr 2010, 04:47
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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?
3 hours
3.4 hours
3.5 hours
3.6 hours
None

[Reveal] Spoiler: OA
D, please provide workable method that can solve this in 2mins, i solved but it takes more than 3mins. can any one help.
thanks

Attachment:

Untitled.png [ 2.5 KiB | Viewed 4254 times ]

The distance between two motorcyclists would be the length of the hypotenuse, which is square root of $$(120-30x)^2+(10x)^2=1000x^2-60*120x+120^2$$ (where x is the time in hours) . So we need to minimize the value of quadratic expression (function) $$1000x^2-60*120x+120^2$$.

Now quadratic function $$f(x)=ax^2+bx+c$$ reaches its minimum (or maximum when $$a$$ is negative - not our case), when $$x=-\frac{b}{2a}=\frac{60*120}{2*1000}=3.6$$

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Re: time distance problem [#permalink]  25 Apr 2010, 05:36
1
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Great Bunuel.

But, I think if you give some details on this equation, we can better understand.

I tried to solve this by plug-in values. Somehow, 1 hr before there is GC site problem and couldnt post my explanation.

As you move from A, the horizontal distance will reduce as per (120-30t). So, at t=4 hr A will reach at the origin of motoryst B. By refering the properties of triangle, you can figure it out.

1. At t=3.5hr, A will travel 105km (ramaining 15km to reach at origin) and B will travel 35km, so shortest distance = \sqrt{(15^2+35^2)} = \sqrt{(1450)}

2. At t=3.6hr, A will travel 108km (ramaining 12km to reach at origin) and B will travel 36km, so shortest distance = \sqrt{(12^2+36^2)} = \sqrt{(1440)}

So, OA should be D.
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Re: time distance problem [#permalink]  25 Apr 2010, 06:06
Hussain15 wrote:
I couldnt get the solution given by Bunuel.

Is it a GMAT level question?? What's the source?

this question is one of prep tests, i came across, i don't think it is of GMAT level question,
because, optimization of quadratic equation is not a subject matter of gmat,
unless, it involves very simple, integer based calculations or like that.
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Re: time distance problem [#permalink]  25 Apr 2010, 06:09
einstein10 wrote:
Hussain15 wrote:
I couldnt get the solution given by Bunuel.

Is it a GMAT level question?? What's the source?

this question is one of prep tests, i came across, i don't think it is of GMAT level question,
because, optimization of quadratic equation is not a subject matter of gmat,
unless, it involves very simple, integer based calculations or like that.

You mean GMAT prep tests or one of many like Kaplan, MGMAT etc?
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Re: time distance problem [#permalink]  25 Apr 2010, 06:11
Hussain15 wrote:
einstein10 wrote:
Hussain15 wrote:
I couldnt get the solution given by Bunuel.

Is it a GMAT level question?? What's the source?

this question is one of prep tests, i came across, i don't think it is of GMAT level question,
because, optimization of quadratic equation is not a subject matter of gmat,
unless, it involves very simple, integer based calculations or like that.

You mean GMAT prep tests or one of many like Kaplan, MGMAT etc?

sorry for confusion. it is from MGMAT prep test.
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Re: time distance problem [#permalink]  25 Apr 2010, 07:41
msand wrote:
Bunuel wrote:

Minimum value of quadratic function $$f(x)=ax^2+bx+c$$ is the y-coordinate of the vertex of this parabola. -- did not get what do you mean by this?

So, basically we are looking for the x-coordinate of its vertex (to get min value of the function) --> $$x_{vertex}=-\frac{b}{2a}=\frac{60*120}{2*1000}=3.6$$.

I believe what he means is Equation of a parabola can be written as

y = (x-h)^2 + K

Now the above value will be minimum when x = h (since x-h need to be 0)

So for an equation y = ax^2+bx+c the value of h is given by h = -b/2a.
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Re: time distance problem [#permalink]  25 Apr 2010, 07:50
3
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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 + (120-30t)^2

Which is maximum when its differential is 0
=> 20t = 2(120-30t) *3 => t = 3.6
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Re: Points A and B are 120 km apart. A motorcyclist starts from [#permalink]  27 Jun 2013, 01:25
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Re: Points A and B are 120 km apart. A motorcyclist starts from [#permalink]  27 Jun 2013, 05:37
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Consider two moving objects, one starting from A and moving at a speed of s1 towards B and the other starting from B and moving perpendicular to AB at a speed of s2. The least distance at any point in time between them is the hypotenuse. The distance between them is given by $$\sqrt{(s1*t) ^ 2 + (s2*t) ^ 2}$$

As the objects move, the length of the hypotenuse will keep changing. The least length of the hypotenuse is given by,

$$\sqrt{(s1*tmin) ^ 2 + (s2*tmin ) ^ 2}$$

where $$tmin= s1*d / (s1^2 + s2^2)$$and is the time taken when the least length of the hypotenuse is reached.

and d is the distance between A and B.

For this problem $$tmin= 30 * 120 / (30 ^2 + 10 ^ 2) = 3.6 hrs.$$

Note:$$tmin$$ can be derived as done in Gurpreet Singh's post.
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Re: Points A and B are 120 km apart. A motorcyclist starts from [#permalink]  05 Aug 2013, 13:30
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?

Total distance = 120
Total distance = d1+d2
120 = (r*t) + (r*t)
120 = (30*3) + (10*3)
120 = 90+30
120 = 120

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Re: Points A and B are 120 km apart. A motorcyclist starts from [#permalink]  05 Aug 2013, 13:54
1
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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.

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?
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Re: time distance problem [#permalink]  18 Aug 2013, 00:33
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 + (120-30t)^2

Which is maximum when its differential is 0
=> 20t = 2(120-30t) *3 => t = 3.6

Hi,

Can anyone can explain this part of this solution

""Which is maximum when its differential is 0
=> 20t = 2(120-30t) *3 => t = 3.6 ""

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Re: Points A and B are 120 km apart. A motorcyclist starts from [#permalink]  29 Aug 2013, 20:13
Hi Bunuel

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
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Re: time distance problem [#permalink]  29 Aug 2013, 21:43
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 + (120-30t)^2

Which is maximum when its differential is 0
=> 20t = 2(120-30t) *3 => t = 3.6

Hi,

Can anyone can explain this part of this solution

""Which is maximum when its differential is 0
=> 20t = 2(120-30t) *3 => t = 3.6 ""

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.
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Re: Points A and B are 120 km apart. A motorcyclist starts from [#permalink]  29 Aug 2013, 21:48
prasannajeet wrote:
Hi Bunuel

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

Bunuel is drawing on the fact that the max or min of a parabola is the vertex. The formula for calculating the vertex of the parabola is given by:
$$x = - \frac{b}{2a}$$ where $$a, b$$ are found in the equation $$f(x) = ax^2+bx+c$$ In our case our values are:

$$b=60*120$$ and $$a=1000$$

and plugging those into our vertex equation yields the answer of 3.6.

I hope this helps!
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Re: Points A and B are 120 km apart. A motorcyclist starts from [#permalink]  31 Aug 2013, 09:33
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.

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.
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Re: Points A and B are 120 km apart. A motorcyclist starts from [#permalink]  20 Sep 2013, 15:27
Isn't the distance minimized at a 45/45/90 triangle?
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Re: Points A and B are 120 km apart. A motorcyclist starts from [#permalink]  20 Sep 2013, 20:08
Devon wrote:
Isn't the distance minimized at a 45/45/90 triangle?

It's correct for a fixed Perimeter. In our case mortocyclist is approaching faster than cyclist is going away, so overal triangle in 3.6 h will be smaller than traingle in 3 h, when it is 45/45/90 traingle.

Hope it helps.
Re: Points A and B are 120 km apart. A motorcyclist starts from   [#permalink] 20 Sep 2013, 20:08

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