GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 27 Jun 2019, 03:29 GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  An open-empty rectangular box with negligible wall thickness is 6 feet

Author Message
TAGS:

Hide Tags

Manager  Joined: 10 Feb 2014
Posts: 85
GMAT 1: 690 Q50 V33 An open-empty rectangular box with negligible wall thickness is 6 feet  [#permalink]

Show Tags

2
4 00:00

Difficulty:   95% (hard)

Question Stats: 39% (02:35) correct 61% (02:08) wrong based on 96 sessions

HideShow timer Statistics Attachment: box.PNG [ 1.98 KiB | Viewed 2439 times ]
An open-empty rectangular box with negligible wall thickness is 6 feet wide, 6 feet long, and 2 feet high. An ant starts crawling from one corner (point A) to reach the diametrically opposite corner (point B). If the ant takes the shortest route possible to reach the destination by crawling on the inner surface of the box, how many feet does the ant travel?

A. $$6\sqrt{2}$$

B. $$\sqrt{10}$$

C. $$2\sqrt{19}$$

D. $$10$$

E. $$14$$

Originally posted by itzmyzone911 on 30 Apr 2015, 01:53.
Last edited by Bunuel on 30 Apr 2015, 03:44, edited 3 times in total.
Edited the question.
Manager  Joined: 17 Mar 2015
Posts: 116
Re: An open-empty rectangular box with negligible wall thickness is 6 feet  [#permalink]

Show Tags

If I understood the question correctly I really wonder if thats actually a gmat question. Not only my answer is different from what is proposed (my answer is 9 feet or approximately 9.54 feet) but also the way to solve it required me to use things which are most likely beyond the bounadires of gmat requirements.
Basically the way I approached this task was evaluating 2 scenarios: and The first one gave me the lowest answer after conducting analysis on the distance function (sum of lengths of AC and CB)
Math Expert V
Joined: 02 Sep 2009
Posts: 55804
Re: An open-empty rectangular box with negligible wall thickness is 6 feet  [#permalink]

Show Tags

1
itzmyzone911 wrote:
Attachment:
box.PNG
An open-empty rectangular box with negligible wall thickness is 6 feet wide, 6 feet long, and 2 feet high. An ant starts crawling from one corner (point A) to reach the diametrically opposite corner (point B). If the ant takes the shortest route possible to reach the destination by crawling on the inner surface of the box, how many feet does the ant travel?

A. $$6\sqrt{2}$$

B. $$\sqrt{10}$$

C. $$2\sqrt{19}$$

D. $$10$$

E. $$14$$

Similar questions to practice:
an-ant-is-clinging-to-one-corner-of-a-box-in-the-shape-of-a-135055.html
s97-184708.html
an-ant-crawls-from-one-corner-of-a-room-to-the-diagonally-134454.html

Hope it helps.
_________________
Manager  Joined: 10 Feb 2014
Posts: 85
GMAT 1: 690 Q50 V33 Re: An open-empty rectangular box with negligible wall thickness is 6 feet  [#permalink]

Show Tags

1
Zhenek wrote:
If I understood the question correctly I really wonder if thats actually a gmat question. Not only my answer is different from what is proposed (my answer is 9 feet or approximately 9.54 feet) but also the way to solve it required me to use things which are most likely beyond the bounadires of gmat requirements.
Basically the way I approached this task was evaluating 2 scenarios: and The first one gave me the lowest answer after conducting analysis on the distance function (sum of lengths of AC and CB)

How did you arrive at your ans?? Explain and then we shall see whether this is within or beyond GMAT scope
Manager  Joined: 10 Feb 2014
Posts: 85
GMAT 1: 690 Q50 V33 Re: An open-empty rectangular box with negligible wall thickness is 6 feet  [#permalink]

Show Tags

Bunuel wrote:
itzmyzone911 wrote:
Attachment:
box.PNG
An open-empty rectangular box with negligible wall thickness is 6 feet wide, 6 feet long, and 2 feet high. An ant starts crawling from one corner (point A) to reach the diametrically opposite corner (point B). If the ant takes the shortest route possible to reach the destination by crawling on the inner surface of the box, how many feet does the ant travel?

A. $$6\sqrt{2}$$

B. $$\sqrt{10}$$

C. $$2\sqrt{19}$$

D. $$10$$

E. $$14$$

Similar questions to practice:
an-ant-is-clinging-to-one-corner-of-a-box-in-the-shape-of-a-135055.html
s97-184708.html
an-ant-crawls-from-one-corner-of-a-room-to-the-diagonally-134454.html

Hope it helps.

Bunuel you rock ...you seem to have virtually every GMAT problem on earth on your finger tips..
Manager  Joined: 17 Mar 2015
Posts: 116
An open-empty rectangular box with negligible wall thickness is 6 feet  [#permalink]

Show Tags

1
Haha, I'm glad I actually looked at it again, anyway, I made a mistake in my calculations and it is indeed 10 in the end, damn. I'll just explain my way of thinking then:
My 2nd picture, the point C is on the line and X is a variable. Length function is $$\sqrt{36+x^2} + \sqrt{36+(2-x)^2}$$.
With the help of derivatives I found that its lowest value is $$2*\sqrt{37}$$ when X = 1 (although one could also apply some common sense there since if u try drawing the thing abit differently - as a double-rectangular figure omitting the folding, the shortest distance ends up being a straight line and it happens that since the rectangles have equal lengths, the line will cross the folding in its middle thus X = 1). $$2*\sqrt{37}$$ > 12.
First picture is similar. Length function is $$F(x) = \sqrt{x^2+4} + \sqrt{36+(6-x)^2}$$.
In order to find minimum value I went with the rule that this $$X_0$$ is pretty much a root of an equation $$F'(X_0) = 0$$
$$\frac{2*x}{\sqrt{x^2+4}} - \frac{2*(6-x)}{\sqrt{36+(6-x)^2}} = 0$$
$$x* \sqrt{36+(6-x)^2} = (6-x)* \sqrt{4+x^2}$$
solving this you get x = 1,5 as the only fitting answer, y ends up being 10 which is the answer to the question
Manager  Joined: 10 Feb 2014
Posts: 85
GMAT 1: 690 Q50 V33 Re: An open-empty rectangular box with negligible wall thickness is 6 feet  [#permalink]

Show Tags

Zhenek wrote:
Haha, I'm glad I actually looked at it again, anyway, I made a mistake in my calculations and it is indeed 10 in the end, damn. I'll just explain my way of thinking then:
My 2nd picture, the point C is on the line and X is a variable. Length function is $$\sqrt{36+x^2} + \sqrt{36+(2-x)^2}$$.
With the help of derivatives I found that its lowest value is $$2*\sqrt{37}$$ when X = 1 (although one could also apply some common sense there since if u try drawing the thing abit differently - as a double-rectangular figure omitting the folding, the shortest distance ends up being a straight line and it happens that since the rectangles have equal lengths, the line will cross the folding in its middle thus X = 1). $$2*\sqrt{37}$$ > 12.
First picture is similar. Length function is $$F(x) = \sqrt{x^2+4} + \sqrt{36+(6-x)^2}$$.
In order to find minimum value I went with the rule that this $$X_0$$ is pretty much a root of an equation $$F'(X_0) = 0$$
$$\frac{2*x}{\sqrt{x^2+4}} - \frac{2*(6-x)}{\sqrt{36+(6-x)^2}} = 0$$
$$x* \sqrt{36+(6-x)^2} = (6-x)* \sqrt{4+x^2}$$
solving this you get x = 1,5 as the only fitting answer, y ends up being 10 which is the answer to the question

Well dear, I appreciate your insights...very thorough indeed.

However, the question can be solved using basics of Pythagoras theorem...do visit the link suggested by Bunuel at

an-ant-crawls-from-one-corner-of-a-room-to-the-diagonally-134454.html

Relevant discussions have taken place along with diagrams to make the point better. Do revert if any doubts..
Manager  Joined: 17 Mar 2015
Posts: 116
An open-empty rectangular box with negligible wall thickness is 6 feet  [#permalink]

Show Tags

3
Yea I did understand the approach displayed there and I find it very clever. I guess I will elaborate with the changed picture: AB' is the shortest path the ant has to take and its the straight line, thus we got a triangle ADB' where AB' = hypotenuse and is easily found to be equal to $$\sqrt{36+64} = 10$$
Manager  Joined: 08 Sep 2010
Posts: 58
An open-empty rectangular box with negligible wall thickness is 6 feet  [#permalink]

Show Tags

itzmyzone911 wrote:
Attachment:
box.PNG
An open-empty rectangular box with negligible wall thickness is 6 feet wide, 6 feet long, and 2 feet high. An ant starts crawling from one corner (point A) to reach the diametrically opposite corner (point B). If the ant takes the shortest route possible to reach the destination by crawling on the inner surface of the box, how many feet does the ant travel?

A. $$6\sqrt{2}$$

B. $$\sqrt{10}$$

C. $$2\sqrt{19}$$

D. $$10$$

E. $$14$$

Hi-
I think there is a typo in the question. I believe either width or length needs to be 2, if the answer is indeed the correct answer.
Using the data given the question:
Basically if we unfold the box, this will give us two rectangles, 1 rectangle with height 2 and base 6, 2nd rectangle with height 2 and base 6. The two rectangles have point A and B as diametrically opposite points. Now to find AB, it's sqrt( (6+6)^2 + 2^2) and that's not the correct answer.
Bunuel - I have followed the other posts that you have mentioned, but not sure what I am missing here in this question. Any inputs will be very helpful.

Thanks,
Chanakya
Manager  G
Joined: 25 Jul 2018
Posts: 67
Location: Uzbekistan
Concentration: Finance, Organizational Behavior
GRE 1: Q168 V167 GPA: 3.85
WE: Project Management (Investment Banking)
An open-empty rectangular box with negligible wall thickness is 6 feet  [#permalink]

Show Tags

4
Hola amigos What would I do if I had this box and ant in real life? In order to show this little buddy the shortest path from $$A$$ to $$B$$, I would definitely UNFOLD the box. Now the shortest route from $$A$$ to $$B$$ is the hypotenuse of the right triangle ABC. Thus $$AB = 10$$.
Attachments photo_2019-04-28_20-44-58.jpg [ 45.61 KiB | Viewed 226 times ]

_________________
Let's help each other with kudos. Kudos will be payed back  An open-empty rectangular box with negligible wall thickness is 6 feet   [#permalink] 28 Apr 2019, 08:55
Display posts from previous: Sort by

An open-empty rectangular box with negligible wall thickness is 6 feet  