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) = 20d750 + 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.