If quadrilaterals ABCD and ADEF have the same areas, is AF28?

Question

GMATH  practice exercise (Quant Class 15)

https://i.postimg.cc/hGP6RWVW/15-Mar19-3w.gif

If quadrilaterals ABCD and ADEF have the same areas, is AF > 28 ?

(1) EF = 10
(2) CE = 24

Edited: 
01. The contest related to this post is finished!
02. We have presented the official solution.
03. We have added the dotted line, to avoid the interpretation that F could be "below" line AB.

Shobhit7 · Answer

IMO, Ans C

Considering both statements together and a rectangle with base CE and height FE, resultant area 240.

So, area of each quad is less than / equal to 240/2= 120 max.

In trapezium / rectangle ADEF, FE is 10, so average of FE+AD is less than or equal to 10. So, max possible value of DE is 12.

In triangle DEF with base 12 and height 10, Hyp FD is root 244, approx 15.something.
As AF is smaller than FD, AF cannot be greater than 28.

siyeezy · Answer

(1) EF = 10

We don't know the length of DE. You can stretch DE very far apart (could be to a ridiculous number, say to 1 million) then clearly AF, which forms the hypotenuse of the right triangle with base the length of DE & height < 10 would be greater than 28. However, if DE is very short, such as length of 1, then AF is the hypotenuse of a right triangle with other sides 1 & a side less than 10, so AF is less than 28.

Insufficient

(2) CE = 24

A similar thought exercise can be performed here, where if EF is a ridiculously long length, then you can construct AF to be >28, but if EF is sufficiently short enough, then AF < 28.

Insufficient

(Both 1 & 2)

Knowing EF is 10 and CE is 24, we can try to maximize the length of AF to see if it exceeds 28. If it doesn't, then we know the answer is sufficient.

We can set BA and AD to be very small, almost 0, such that AF forms the hypotenuse of a right triangle with sides 10 and 24, which a quick calculation shows that at most AF can be 26. So AF is definitely less than 28.

Answer is C

fskilnik · Answer

The contest is finished  from this moment on!

We thank  Shobhit7  and  siyeezy  for your beautiful contributions!

We will start commenting on both solutions (separately) based on our judge´s considerations:

Shobhit:

01. You did not prove each statement alone is insufficient ("it´s obvious" is not a good argument, at least in a prize-winning contest)...
02. Area of each quadrilateral must be LESS than 240/2, because from the figure given points B, A, and F should be considered NOT collinear in Data Sufficient situations like this one!
03. Nice arguments using the "middle-base" of the trapezoid (although, again, DE must be less than 12, because 10=FE>AD) and hypotenuse FD, offering an upper limit to AF <<< 28 !!

siyeezy:

01. Very nice viable "geometric bifurcations" (using the term we have created in our method) to guarantee the insufficiency of each statement alone. 
02. Saw the important right triangle 10,24,26 (we consider "5m,12m,13m" a GMAT-shortcut!) but did not mention *explicitly* that the "equal areas restriction" would be temporarily ignored for that. The fact that the upper limit (26) is already less than 28 (our FOCUS) makes this approach excellent because (according to our method) the FOCUS is the most important guide to our decisions!

All that put, the judge (and I) believe both contributions were REALLY great, therefore we will consider... a DRAW! Both of you will receive the prize we had in mind: our "Lightning" course!

I hope you will be able to strengthen even more your powerful skills through our method!

Congratulations to both of you and thanks for joining!!
Fabio.

P.S.: I will wait a private message from each one of you, to give the step-by-step instructions for collecting the prize! Cheers!

fskilnik · Answer

https://i.postimg.cc/pL6zR4S5/15-Mar19-10j.gif {S_{{\mathop{ m ABCD} olimits} }} = {S_{{ m{ADEF}}}} = {S_{{ m{ADEG}}}} + {S_{{ m{AGF}}}}\,\,\,\,\,\left( * ight) AF\,\,\mathop > \limits^? \,\,28 \left( 1 ight)\,\,\left\{ \matrix{ \,{ m{image}}\,\,{ m{left}}:\,\,\left( * ight)\,\,4 \cdot 6 = 3 \cdot 6 + {{3 \cdot 4} \over 2}\,\,\,\,\, \Rightarrow \,\,\,\,{ m{viable}}\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{ m{NO}}} ight angle \hfill \cr \,{ m{image}}\,\,{ m{right}}:\,\,\left( * ight)\,\,x \cdot 6 = 28 \cdot 6 + {{28 \cdot 4} \over 2}\,\,\,\,\, \Rightarrow \,\,\,\,{ m{viable}}\,\,\,\left( {x = {{28 \cdot 8} \over 6} > 0} ight)\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{ m{YES}}} ight angle \hfill \cr} ight. \left( 2 ight)\,\,\left\{ \matrix{ \,{ m{image}}\,\,{ m{left}}:\,\,\left( * ight)\,\,18 \cdot y = 6 \cdot y + {{6 \cdot 4} \over 2}\,\,\,\,\, \Rightarrow \,\,\,\,{ m{viable}}\,\,\,\,\left( {y = 1 > 0} ight) \Rightarrow \,\,\,\,\left\langle {{ m{NO}}} ight angle \hfill \cr \,{ m{image}}\,\,{ m{right}}:\,\,\left( * ight)\,\,18 \cdot y = 6 \cdot y + {{6 \cdot 28} \over 2}\,\,\,\,\, \Rightarrow \,\,\,\,{ m{viable}}\,\,\,\left( {y = 7 > 0} ight)\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{ m{YES}}} ight angle \hfill \cr} ight. \left( {1 + 2} ight)\,\,{ m{Extremal}}\,\,{ m{Scenario}}\,\,\,\left( {\left( * ight)\,\,{ m{ignored}}} ight)\,\,:\,\,A{F_{\,{ m{ExtScen}}}}\,\,\mathop < \limits^{{ m{near}}} \,\,13 \cdot 2 = 26 \Rightarrow \,\,AF < \,\,A{F_{\,{ m{ExtScen}}}} < 28\,\,\,\, \Rightarrow \,\,\,\,{ m{SUFF}}. The correct answer is (C). We follow the notations and rationale taught in the GMATH method. Regards, Fabio.

bumpbot · Answer

Automated notice from GMAT Club BumpBot: 

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

 This post was generated automatically.

If quadrilaterals ABCD and ADEF have the same areas, is AF>28?

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Data Sufficiency (DS) Questions