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ds rate (m08q34)

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tinki
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Re: ds rate (m08q34) [#permalink] New post   02 Feb 2011, 13:58
GMAT TIGER wrote:
bmwhype2 wrote:
A plane takes off from a hill at 750 meters above sea level and lands some time later in a town located at 50 meters below sea level. During the first part of its flight, the plane gained height at a rate of 50 meters per minute but then it started to descend at a rate of 20 meters per minute. The duration of the first part of the flight was what percent of the total flight time?

1. The duration of the descent is known.
2. The total flight time is known.



mandb wrote:
walker wrote:
D

general equation: 750-50=20*td-50ta ==> 2*td-5*ta=70
where ta - time of ascent, td - time of descent

1. we know td. we can find ta from general equation. SUFF.
2. we know ta+td. we can solve system of 2 equations. SUFF.


Thx... I was wondering why it was 750+50 in explanation..


It is.

Present height + (+ accent) - (- level below the sea) = 20d
750 + 50a - (-50) = 20d
75 + 5a + 5 = 2d
80 + 5a = 2d

St. 1) The duration of the descent (i.e. d) is known.
Say d = 50, a = 4. SUFF...

St. 2) The total flight (i.e. a+d) time is known.
For simplicity, say a+d = 54
a = 54 - d

Then, 80 + 5(54-d) = 2d.
d = 50. SUFF.

D



i am still confused with the above highlighted part. can somebody explain?
i cant visualize how the distance without considering slope is obtained. something like drawing would be highly appreciated.
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snkrhed
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Re: ds rate (m08q34) [#permalink] New post   Updated on: 02 Nov 2011, 09:38
Takeaway: Essentially this is a system of eqn or substitution problem

20y = 800 + 50x

or

20y - 50x = 800

1) we know y; sufficient
2) a+Y=given; sufficient

Try to get one variable.

Edit: oops mistype; lol i give up typing on my phone...
Posted from my mobile device

Originally posted by snkrhed on 02 Nov 2011, 09:25.
Last edited by snkrhed on 02 Nov 2011, 09:38, edited 2 times in total.
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Re: ds rate (m08q34) [#permalink] New post   02 Nov 2011, 09:33
tinki wrote:
GMAT TIGER wrote:

Present height + (+ accent) - (- level below the sea) = 20d
750 + 50a - (-50) = 20d
75 + 5a + 5 = 2d
80 + 5a = 2d

St. 1) The duration of the descent (i.e. d) is known.
Say d = 50, a = 4. SUFF...

St. 2) The total flight (i.e. a+d) time is known.
For simplicity, say a+d = 54
a = 54 - d

Then, 80 + 5(54-d) = 2d.
d = 50. SUFF.

D



i am still confused with the above highlighted part. can somebody explain?
i cant visualize how the distance without considering slope is obtained. something like drawing would be highly appreciated.
[/quote]

Descent = beginning + ascend / min - ending location

Or

20y = 750+50x-(-50)

Posted from my mobile device

Posted from my mobile device
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Re: ds rate (m08q34) [#permalink] New post   02 Nov 2011, 10:24
Thanks for explaining.
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Re: ds rate (m08q34) [#permalink] New post   18 Nov 2011, 01:41
bmwhype2 wrote:
A plane takes off from a hill at 750 meters above sea level and lands some time later in a town located at 50 meters below sea level. During the first part of its flight, the plane gained height at a rate of 50 meters per minute but then it started to descend at a rate of 20 meters per minute. The duration of the first part of the flight was what percent of the total flight time?

1. The duration of the descent is known.
2. The total flight time is known.

Show SpoilerOA
D

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IMO the question is rongly worded; If quations ask for actual numbers then we have to answer wether we can calculate the number based on the statements . However, the statement povided does not provide enough information to calulate the exact percentage, therefore both statements are insufficient.

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Re: ds rate (m08q34) [#permalink]
18 Nov 2011, 01:41
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