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At 7:57 am, Flight 501 is at an altitude of 6 miles above the ground a

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liranmaymoni
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At 7:57 am, Flight 501 is at an altitude of 6 miles above the ground a [#permalink] At 7:57 am, Flight 501 is at an altitude of 6 miles above the ground a   11 Feb 2014, 06:15
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At 7:57 am, Flight 501 is at an altitude of 6 miles above the ground and is on a direct approach (i.e., flying in a direct line to the runway) towards The Airport, which is located exactly 8 miles due north of the plane’s current position. Flight 501 is scheduled to land at The Airport at 8:00 am, but, at 7:57 am, the control tower radios the plane and changes the landing location to an airport 15 miles directly due east of The Airport. Assuming a direct approach (and negligible time to shift direction), by how many miles per hour does the pilot have to increase her speed in order to arrive at the new location on time?


A. \(5\sqrt{13} - 10\) miles/hr

B. 100 miles/hr

C. \(100\sqrt{13} - 200\) miles/hr

D. 200 miles/hr

E. \(100\sqrt{13}\) miles/hr
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Re: At 7:57 am, Flight 501 is at an altitude of 6 miles above the ground a [#permalink] At 7:57 am, Flight 501 is at an altitude of 6 miles above the ground a   14 Feb 2014, 03:51
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Temurkhon wrote:
what if like this:
R*T=D
8/3m/min*3min=8miles
5m/min*3min=15miles
5-8/3=7/3m/min or 140m/hour


Given the flight has an altitude of 6 miles and the airport is exactly 8 miles due north. The flight has to descend to the airport along the hypotenuse of a right angled triangle with legs 6 & 8 miles. hypotenuse = 10 miles. after that we can apply your approach => 3 mins per 10 miles => 60 mins is 60*10/3 = 200 mph. ---original speed.

Now since the flight got a new destination exactly 15 miles east of the current airport. we have 2nd right angled triangle with 8 & 15 as legs. hypotenuse is 17. Further, now the flight has to descend to the new airport along the hypotenuse of a third triangle with legs 6 miles(altitude of the plane) & 17 miles which is \(\sqrt{6^2+17^2} = 5\sqrt{13}\)

The plane has to descend \(5\sqrt{13}\) miles in 3 mins. hence in 60 mins it has to have a new speed of \(60*\frac{5}{3}\sqrt{13}\)

\(= 20*5\sqrt{13}\) --- new speed

Increase in speed is \(20*5\sqrt{13}-20*10\)

\(20(5\sqrt{13}-10)\)
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Re: At 7:57 am, Flight 501 is at an altitude of 6 miles above the ground a [#permalink] At 7:57 am, Flight 501 is at an altitude of 6 miles above the ground a   11 Feb 2014, 06:40
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Assume Euclidean geometry. If the original airport is 8 miles due north, and the new airport is 15 miles due east of the original airport, then the new airport is 17 miles away. (17² = 8² + 15²)

Now the plane is 6 miles above ground, and the original airport is 8 miles away. Then the Euclidean distance from the plane to the original airport is 10 miles (10² = 6² + 8²). Because it would have taken 3 minutes to travel the 10 miles, We can say it would have taken 10/3 miles/min.

The new airport is 17 miles away and the plane is 6 miles above ground. So the distance the plane must now travel is 5√13 miles( [5√13]² = 17² + 6²). To do this in 3 minutes, the plane must travel at 5√13/3 miles/min.

The difference in speed is thus 5√13/3 - 10/3 miles/min. But we need to find the answer in miles/hour, so we need to multiply by 60 to get 60(5√13/3 - 10/3) miles/hour or 20(5√13 - 10) miles/hour.
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Re: At 7:57 am, Flight 501 is at an altitude of 6 miles above the ground a [#permalink] At 7:57 am, Flight 501 is at an altitude of 6 miles above the ground a   14 Feb 2014, 01:08
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what if like this:
R*T=D
8/3m/min*3min=8miles
5m/min*3min=15miles
5-8/3=7/3m/min or 140m/hour
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Re: At 7:57 am, Flight 501 is at an altitude of 6 miles above the ground a [#permalink] At 7:57 am, Flight 501 is at an altitude of 6 miles above the ground a   21 Nov 2017, 04:02
Temurkhon wrote:
what if like this:
R*T=D
8/3m/min*3min=8miles
5m/min*3min=15miles
5-8/3=7/3m/min or 140m/hour



If you get 140m/hr, thats good enough since 100 √ 13−200 miles/hr is approximately 140 miles an hour.

√ 13 = between 3 (3*3=9) and 4(4*4=16). take 3.5 for example.

you'd get 100*3.5 - 200

you get a difference of approximately 150 m/hr which is the closest to the answer you got.
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Re: At 7:57 am, Flight 501 is at an altitude of 6 miles above the ground a [#permalink] At 7:57 am, Flight 501 is at an altitude of 6 miles above the ground a   28 Oct 2019, 07:37
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Just a quick question. Why are we taking the legs 8 and 15 rather than 6 and 15?

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Re: At 7:57 am, Flight 501 is at an altitude of 6 miles above the ground a [#permalink] At 7:57 am, Flight 501 is at an altitude of 6 miles above the ground a   02 Nov 2019, 01:30
gmatprav wrote:
Temurkhon wrote:
what if like this:
R*T=D
8/3m/min*3min=8miles
5m/min*3min=15miles
5-8/3=7/3m/min or 140m/hour


Given the flight has an altitude of 6 miles and the airport is exactly 8 miles due north. The flight has to descend to the airport along the hypotenuse of a right angled triangle with legs 6 & 8 miles. hypotenuse = 10 miles. after that we can apply your approach => 3 mins per 10 miles => 60 mins is 60*10/3 = 200 mph. ---original speed.

Now since the flight got a new destination exactly 15 miles east of the current airport. we have 2nd right angled triangle with 8 & 15 as legs. hypotenuse is 17. Further, now the flight has to descend to the new airport along the hypotenuse of a third triangle with legs 6 miles(altitude of the plane) & 17 miles which is \(\sqrt{6^2+17^2} = 5\sqrt{13}\)

The plane has to descend \(5\sqrt{13}\) miles in 3 mins. hence in 60 mins it has to have a new speed of \(60*\frac{5}{3}\sqrt{13}\)

\(= 20*5\sqrt{13}\) --- new speed

Increase in speed is \(20*5\sqrt{13}-20*10\)

\(20(5\sqrt{13}-10)\)




Can you sketch a drawing for this problem? I am unable to imagine this.
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Re: At 7:57 am, Flight 501 is at an altitude of 6 miles above the ground a [#permalink] At 7:57 am, Flight 501 is at an altitude of 6 miles above the ground a   19 Nov 2019, 05:36
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The trick here to consider the hypotenuse as the distance like mentioned in the above answers. Good Question!
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Re: At 7:57 am, Flight 501 is at an altitude of 6 miles above the ground a [#permalink] At 7:57 am, Flight 501 is at an altitude of 6 miles above the ground a   01 Mar 2021, 05:55
VeritasKarishma plz solve this question
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At 7:57 am, Flight 501 is at an altitude of 6 miles above the ground a [#permalink] At 7:57 am, Flight 501 is at an altitude of 6 miles above the ground a   01 Mar 2021, 07:22
Joyal Thomas wrote:
gmatprav wrote:
Temurkhon wrote:
what if like this:
R*T=D
8/3m/min*3min=8miles
5m/min*3min=15miles
5-8/3=7/3m/min or 140m/hour


Given the flight has an altitude of 6 miles and the airport is exactly 8 miles due north. The flight has to descend to the airport along the hypotenuse of a right angled triangle with legs 6 & 8 miles. hypotenuse = 10 miles. after that we can apply your approach => 3 mins per 10 miles => 60 mins is 60*10/3 = 200 mph. ---original speed.

Now since the flight got a new destination exactly 15 miles east of the current airport. we have 2nd right angled triangle with 8 & 15 as legs. hypotenuse is 17. Further, now the flight has to descend to the new airport along the hypotenuse of a third triangle with legs 6 miles(altitude of the plane) & 17 miles which is \(\sqrt{6^2+17^2} = 5\sqrt{13}\)

The plane has to descend \(5\sqrt{13}\) miles in 3 mins. hence in 60 mins it has to have a new speed of \(60*\frac{5}{3}\sqrt{13}\)

\(= 20*5\sqrt{13}\) --- new speed

Increase in speed is \(20*5\sqrt{13}-20*10\)

\(20(5\sqrt{13}-10)\)




Can you sketch a drawing for this problem? I am unable to imagine this.




I also got confused with this scenario.

Imagine the plan moving in one direction and original destination is 8 miles away.
Now changed target is 15 miles east from the original destination not the plane.

So first we need to determine on ground distance of new target from the plane, with two sides of right angle triangle known as 8 and 15 -> 17
Once this is obtained, we need to find distance to be covered by plane as it is having altitude of 6 miles .

Thus next right angle triangle will have two sides as 6 and 17.

Thus this last obtained distance is to be covered in 3 mins and then we determine per hour.
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Re: At 7:57 am, Flight 501 is at an altitude of 6 miles above the ground a [#permalink] At 7:57 am, Flight 501 is at an altitude of 6 miles above the ground a   02 Mar 2021, 01:48
Hard part is drawing the 3-D shape.

Original come in was the hypotenuse of 6 and 8 ———> 10 miles which he had to make in 3 minutes

Now, the approach is the main diagonal through a 3 D box with dimensions 6 - 15 - 8

Sqrt(325)

Then make sure to confer to miles per hour, not miles per minute


Sqrt(325) / (3/60)

-

10 / (3/60)
——————

= 100 sqrt(13) - 200

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Re: At 7:57 am, Flight 501 is at an altitude of 6 miles above the ground a [#permalink] At 7:57 am, Flight 501 is at an altitude of 6 miles above the ground a   02 Mar 2021, 02:40
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liranmaymoni wrote:
At 7:57 am, Flight 501 is at an altitude of 6 miles above the ground and is on a direct approach (i.e., flying in a direct line to the runway) towards The Airport, which is located exactly 8 miles due north of the plane’s current position. Flight 501 is scheduled to land at The Airport at 8:00 am, but, at 7:57 am, the control tower radios the plane and changes the landing location to an airport 15 miles directly due east of The Airport. Assuming a direct approach (and negligible time to shift direction), by how many miles per hour does the pilot have to increase her speed in order to arrive at the new location on time?


A. \(5\sqrt{13} - 10\) miles/hr

B. 100 miles/hr

C. \(100\sqrt{13} - 200\) miles/hr

D. 200 miles/hr

E. \(100\sqrt{13}\) miles/hr



There are 3 right triangles here in different planes. On the ground, we have the two airports 15 miles apart (due east) and the plane is 8 miles to the south of A1 (point G). So this is the 8-15-17 triangle.

Attachment:
Screenshot 2021-03-02 at 14.54.50.png
Screenshot 2021-03-02 at 14.54.50.png [ 25.07 KiB | Viewed 9759 times ]



The points P represent the current location of the plane at height 6. The green triangles are in the air with base on the ground. The plane needs to travel from P to A1 or A2 in the two situations.

P to A1 - Distance is 10
Time taken = 3 mins
\(Speed = \frac{10}{3/60} miles/hr = 200 miles/hr\)

P to A2 - Distance is \(\sqrt{17^2 + 6^2} = 5\sqrt{13}\)
Time taken = 3 mins
\(Speed = \frac{5\sqrt{13}}{(3/60)} = 100\sqrt{13}\) miles/hr

\(Increase = (100\sqrt{13} - 200)\) miles/hr

Answer (C)
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Re: At 7:57 am, Flight 501 is at an altitude of 6 miles above the ground a [#permalink] At 7:57 am, Flight 501 is at an altitude of 6 miles above the ground a   16 Aug 2022, 05:08
Be clever. The question asks how many miles per hour should the speed increase. All the options give a substraction. From here we can intuitively see the second term refers to the current speed! So when you find the current speed, you can choose the correct answer with a high % accuracy and then move on!

liranmaymoni wrote:
At 7:57 am, Flight 501 is at an altitude of 6 miles above the ground and is on a direct approach (i.e., flying in a direct line to the runway) towards The Airport, which is located exactly 8 miles due north of the plane’s current position. Flight 501 is scheduled to land at The Airport at 8:00 am, but, at 7:57 am, the control tower radios the plane and changes the landing location to an airport 15 miles directly due east of The Airport. Assuming a direct approach (and negligible time to shift direction), by how many miles per hour does the pilot have to increase her speed in order to arrive at the new location on time?


A. \(5\sqrt{13} - 10\) miles/hr

B. 100 miles/hr

C. \(100\sqrt{13} - 200\) miles/hr

D. 200 miles/hr

E. \(100\sqrt{13}\) miles/hr
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Re: At 7:57 am, Flight 501 is at an altitude of 6 miles above the ground a [#permalink] At 7:57 am, Flight 501 is at an altitude of 6 miles above the ground a   14 Sep 2023, 17:14
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