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23 Nov 2015, 12:53
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E
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Difficulty:
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85% (01:27) correct
15%
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A plane was originally flying at an altitude of x feet when it ascended 2,000 feet and then descended 5,000 feet. If the plane's altitude after these two changes was 1/3 its original altitude, then the solution of which of the following equations gives the plane's original altitude, in feet ?
(A) x + 2000 = 1/3*(x - 3000)
(B) 1/3*(x - 3000) = x
(C) x + 3000 = 1/3*x
(D) x - 7000 = 1/3*x
(E) x - 3000 = 1/3*x
(A) x + 2000 = 1/3*(x - 3000)
(B) 1/3*(x - 3000) = x
(C) x + 3000 = 1/3*x
(D) x - 7000 = 1/3*x
(E) x - 3000 = 1/3*x
23 Nov 2015, 20:03
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Bunuel
Plane's original altitude = x
Plane's new altitude after ascending 2000 ft = x+2000
Plane's new altitude after descending 5000 ft from previous altitude = x+2000-5000 = x-3000
So, after two changes plane is at 1/3 its original altitude => x-3000 = x/3
Answer (E)
19 Oct 2019, 20:53
Kudos
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Plane starts at altitude x.
The altitude then increases 2000 ft (x+2000).
It then decreases 5000 ft from there ((x+2000)-5000)). Net change= x-3000
This is equal to 1/3 of x. Set both equations equal:
x-3000=1/3x
Ans=E
The altitude then increases 2000 ft (x+2000).
It then decreases 5000 ft from there ((x+2000)-5000)). Net change= x-3000
This is equal to 1/3 of x. Set both equations equal:
x-3000=1/3x
Ans=E
