ds rate (m08q34)

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

Oh I see one is above sea level and other one is below

let duration of ascend = u mins [up]
let duration of descend = d mins [down]

from the information we can write the below equation
50u+20d=800

(1) if d is known, we can calculate u's value
(2) if u+d is known, then we'll have two equations and we can solve them to get u's value

I donot understand why vertical height is being considered here. A plane will never travel vertically down. The path followed is the hypotenuse of the triangle formed by the ground distance covered and the vertical height. The descent distance will definitely be greater than the vertical distance. Shouldnt we try to form a triangle to solve this? Plz help me out here.

Say the plane ascended by X meter @ 50mpm and descended (X+750+50) = (X+800) meter @ 20 mpm.
Time to go up, Tu = X/50 min ------------------------ eq1
Time to go down, Td = (X+800)/20 min. -------- eq2
Total time, Tt = Tu + Td = (7X+4000)/100 ----------- eq3

1) We have Td. So we can find out X from eq2 and plug in eq1 to get Tu. Sufficient.
2) We have Tt. So we can find out X from eq3 and plug in eq1 to get Tu. Sufficient.

D

If the plane took T mins to reach from height X to ground, no matter what horizontal distance it covered, it would still have taken T mins to reach from hieght X to height 0.
Consider this on coordinate plane XY. Say the plane was initially at (1,10) and it took T min to reach (X,0). Now it matters not what the value of X is, since we know that displacement of body from Y=10 from Y=0 happened in T mins.

Makes sense??

Thats the beauty of DS.
Even if you'd incorrectly assumed both were ABOVE sea level, you'd still have been able to answer the question.

Yes.. now I get it.. The slope of the path taken upward and downward are perpendicular. Assuming this, we need not consider the horizontal distance covered.

IMO this question is wrongly worded. By asking for the %, the answer should be sufficient if only we are able to give a number back....
therefore IMO rephrase the question or give some numbers to (1) and (2)

the question allows the answer to be different values...different values mean different answers therefore insufficient statements

I beg to differ.
DS questions do not ask us to find the answer, they just ask us whether or not it is possible to find the answer. Agreed, for different values of Descent Time and Total Time we'd get different values, but answer to DS question would always be same "Yes, we can find the answer" or "No, we can not find answer", whichever the case may be.

: ) i disagree - the DS questions ask if the info is enought to answer the question! the statements are not enough to answer the question! if the statements were sufficient to answer the question, then our answer should be a defined % !

anyone else who would like to contribute to this discusion?

Thanks for the explaination guys.

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.

This qn is already discussed . But I still dont undesrtand the logic.

I have translated the qn to a figure(attached with this prob),as per my understanding.

b = distance travelled while gaining height at a rate of 50mps
c+d= distance travelled while descending at a rate of 20mps
Qn asks for T(b)/T(c+d)
where Tb= Time to ascent ;ie time to travel distance of b.
T=Total time.time to cover b+c+d

Is there anything wrong in my understanding ? I am not sure how to proceed further .

No response even after a day !!
Wonder ..is the qn so difficult or is the qn so simple ..that nobody is answering !

Don't bother about distance traveled towards x-axis. Consider only the y-axis, i.e. going up and coming down. That is what is discussed and the rate (50 and 20) mentioned is in terms of height gained or lost.


So, say if t1 is time for going up and t2 is time taken while going down :

20t2 = 750+50t1-(-50)
=> 20 t2 = 800 + 50 t1 ---1

We need to find : t1/(t1+t2) * 100

Statement 1: If t2 is known, we can calculate t1, hence the % can be calculated - Sufficient

Statement 2: If (t1+t2) is known, say t1+t2 = K (some constant) --- 2

Then we have two independent equations in terms of t1 and t2. So, again we can calculate the % - Sufficient

Hence the answer should be D.

Hope, it helps.

Thanks ! I have spent more than 1 hour ! Just one word "height" made all the difference !