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

 It is currently 20 Oct 2018, 21:57

### 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

# For a display, identical cubic boxes are stacked in square

Author Message
TAGS:

### Hide Tags

Senior Manager
Joined: 07 Sep 2010
Posts: 266
For a display, identical cubic boxes are stacked in square  [#permalink]

### Show Tags

Updated on: 23 Sep 2013, 06:02
18
63
00:00

Difficulty:

25% (medium)

Question Stats:

76% (02:09) correct 24% (02:20) wrong based on 869 sessions

### HideShow timer Statistics

For a display, identical cubic boxes are stacked in square layers. Each layer consists of cubic boxes arranged in rows that form a square, and each layer has 1 fewer row and 1 fewer box in each remaining row than the layer directly below it. If the bottom of the layer has 81 boxes and the top of the layer has only 1 box, how many boxes are in display?

A. 236
B. 260
C. 269
D. 276
E. 285

Originally posted by imhimanshu on 23 Sep 2013, 05:10.
Last edited by Bunuel on 23 Sep 2013, 06:02, edited 1 time in total.
Renamed the topic and edited the question.
Math Expert
Joined: 02 Sep 2009
Posts: 50007
For a display, identical cubic boxes are stacked in square  [#permalink]

### Show Tags

23 Sep 2013, 06:23
39
33
imhimanshu wrote:
For a display, identical cubic boxes are stacked in square layers. Each layer consists of cubic boxes arranged in rows that form a square, and each layer has 1 fewer row and 1 fewer box in each remaining row than the layer directly below it. If the bottom of the layer has 81 boxes and the top of the layer has only 1 box, how many boxes are in display?

A. 236
B. 260
C. 269
D. 276
E. 285

Basically we have a 9-layer pyramid as shown below:

(Actually this pyramid 8-layer, couldn't find 9-layer one image)

The number of boxes would be: $$9^2 + 8^2 + 7^2 + 6^2 + 5^2 + 4^2 + 3^2 + 2^2 + 1 = 285$$.

You can use the sum of the first n perfect squares formula to calculate: $$\frac{n(n+1)(2n+1)}{6}=\frac{9*(9+1)(2*9+1)}{6}=285$$.

Hope it's clear.

Attachment:

pyramid_with_corner_cube_from_istock.jpg [ 9.94 KiB | Viewed 49611 times ]

_________________
Intern
Joined: 25 Mar 2014
Posts: 31
Re: For a display, identical cubic boxes are stacked in square  [#permalink]

### Show Tags

28 Jul 2014, 12:55
9
3
Easy calculation:

1) Sum the UNIT digits.
1 + 4 + 9 + 16 + 25 +36+ 49+ 64 +81

2) Agrupate in 10's
1+9
4+6
6+4
9+1
5+ ....

The unit digit must be 5, therefore the correct answer is E.
##### General Discussion
Manager
Joined: 01 Sep 2012
Posts: 122
Re: For a display, identical cubic boxes are stacked in square  [#permalink]

### Show Tags

23 Sep 2013, 14:46
3
Thanks bunuel but how did you manage to understand that?
I read the question again and cannot imagine the picture you uploaded in my head at all.
Thanks!
_________________

If my answer helped, dont forget KUDOS!

IMPOSSIBLE IS NOTHING

Math Expert
Joined: 02 Sep 2009
Posts: 50007
Re: For a display, identical cubic boxes are stacked in square  [#permalink]

### Show Tags

24 Sep 2013, 00:56
26
1
roygush wrote:
Thanks bunuel but how did you manage to understand that?
I read the question again and cannot imagine the picture you uploaded in my head at all.
Thanks!

I read the stem carefully. We are told that:
Cubic boxes are stacked in square layers --> each layer is a square;
The bottom of the layer has 81 boxes --> the bottom layer has 9 rows and each row has 9 boxes.
Each layer has 1 fewer row and 1 fewer box in each remaining row than the layer directly below it --> the second layer has 8 rows and each row has 8 boxes.
...

Hope it helps.
_________________
Intern
Joined: 24 Jan 2012
Posts: 9
Re: For a display, identical cubic boxes are stacked in square  [#permalink]

### Show Tags

24 Sep 2013, 08:58
2
Bunuel wrote:
roygush wrote:
Thanks bunuel but how did you manage to understand that?
I read the question again and cannot imagine the picture you uploaded in my head at all.
Thanks!

I read the stem carefully. We are told that:
Cubic boxes are stacked in square layers --> each layer is a square;
The bottom of the layer has 81 boxes --> the bottom layer has 9 rows and each row has 9 boxes.
Each layer has 1 fewer row and 1 fewer box in each remaining row than the layer directly below it --> the second layer has 8 rows and each row has 8 boxes.
...

Hope it helps.

I got confused (and I still am) by the line which says "Each layer has 1 fewer row and 1 fewer box in each remaining row than the layer directly below it"

I got that the bottom layer will have 9 x 9 boxes
I also understand that the next level up will have 8 rows of boxes and since this layer also has to form a square hence it needs 8 boxes in the column as well.

What is elusive for me is "and 1 fewer box in each remaining row[/color] than the layer directly below it". Can you please explain again.
Math Expert
Joined: 02 Sep 2009
Posts: 50007
Re: For a display, identical cubic boxes are stacked in square  [#permalink]

### Show Tags

24 Sep 2013, 15:04
3
violetsplash wrote:
Bunuel wrote:
roygush wrote:
Thanks bunuel but how did you manage to understand that?
I read the question again and cannot imagine the picture you uploaded in my head at all.
Thanks!

I read the stem carefully. We are told that:
Cubic boxes are stacked in square layers --> each layer is a square;
The bottom of the layer has 81 boxes --> the bottom layer has 9 rows and each row has 9 boxes.
Each layer has 1 fewer row and 1 fewer box in each remaining row than the layer directly below it --> the second layer has 8 rows and each row has 8 boxes.
...

Hope it helps.

I got confused (and I still am) by the line which says "Each layer has 1 fewer row and 1 fewer box in each remaining row than the layer directly below it"

I got that the bottom layer will have 9 x 9 boxes
I also understand that the next level up will have 8 rows of boxes and since this layer also has to form a square hence it needs 8 boxes in the column as well.

What is elusive for me is "and 1 fewer box in each remaining row[/color] than the layer directly below it". Can you please explain again.

Each layer has 1 fewer row and 1 fewer box in each remaining row than the layer directly below it:

1st layer has 9 rows and 9 boxes in each of them. 2nd row has 1 fewer, so 8 rows and each of the remaining 8 rows has 1 fewer box, so 8 boxes in it.

Hope it's clear.
_________________
Manager
Joined: 14 Jan 2013
Posts: 145
Concentration: Strategy, Technology
GMAT Date: 08-01-2013
GPA: 3.7
WE: Consulting (Consulting)
Re: For a display, identical cubic boxes are stacked in square  [#permalink]

### Show Tags

30 Mar 2014, 16:02
Bunuel,

Do you have more examples like these questions? I am finding it difficult to understand after reading question.

Thanks

~M14
_________________

"Where are my Kudos" ............ Good Question = kudos

"Start enjoying all phases" & all Sections

__________________________________________________________________
http://gmatclub.com/forum/collection-of-articles-on-critical-reasoning-159959.html

http://gmatclub.com/forum/percentages-700-800-level-questions-130588.html

http://gmatclub.com/forum/700-to-800-level-quant-question-with-detail-soluition-143321.html

Manager
Joined: 20 Oct 2013
Posts: 54
Re: For a display, identical cubic boxes are stacked in square  [#permalink]

### Show Tags

24 May 2014, 06:30
superb explanation Bunnel! i couldnt understand the qs...
_________________

Hope to clear it this time!!
GMAT 1: 540
Preparing again

Senior Manager
Joined: 13 Jan 2012
Posts: 286
Weight: 170lbs
GMAT 1: 740 Q48 V42
GMAT 2: 760 Q50 V42
WE: Analyst (Other)
Re: For a display, identical cubic boxes are stacked in square  [#permalink]

### Show Tags

04 Apr 2015, 17:42
Bunuel wrote:
imhimanshu wrote:
For a display, identical cubic boxes are stacked in square layers. Each layer consists of cubic boxes arranged in rows that form a square, and each layer has 1 fewer row and 1 fewer box in each remaining row than the layer directly below it. If the bottom of the layer has 81 boxes and the top of the layer has only 1 box, how many boxes are in display?

A. 236
B. 260
C. 269
D. 276
E. 285

Basically we have a 9-layer pyramid as shown below:
Attachment:
pyramid_with_corner_cube_from_istock.jpg
(Actually this pyramid 8-layer, couldn't find 9-layer one image)

The number of boxes would be: 9^2 + 8^2 + 7^2 + 6^2 + 5^2 + 4^2 + 3^2 + 2^2 + 1 = 285.

You can use the sum of the first n perfect squares formula to calculate: $$\frac{n(n+1)(2n+1)}{6}=\frac{9*(9+1)(2*9+1)}{6}=285$$.

Hope it's clear.

Is there any chance you can apply why that 6 is there? I want to make sure I can apply this formula in more complicated cases.
Intern
Joined: 26 Jul 2014
Posts: 3
Re: For a display, identical cubic boxes are stacked in square  [#permalink]

### Show Tags

11 Oct 2015, 19:42
Here's where non-native speakers could have trouble. By display I kept thinking of a computer display and I tried to visualize boxes arranged within the TV, and jumped into the conclusion that this was similar to a problem in the OG (13th Ed. PS 124).
Now, if you don't know the formula for the sum of the first n perfect squares (I actually forgot it on a second attempt), it is just nonsense to sum each square result. plaverbach's approach is the appropriate one. After taking a look at the answers and noticing that only two of them have the same units number, you pray that those are wrong and go ahead and find that unit.
Intern
Joined: 12 Nov 2014
Posts: 19
Re: For a display, identical cubic boxes are stacked in square  [#permalink]

### Show Tags

03 Jan 2016, 10:17
I liked plaverbach approach, as I couldn't understand the question in the first place and choose random wrong answer.
But when I saw the picture posted, I could use the plaverbach approach.
Intern
Joined: 26 May 2014
Posts: 40
Location: India
Concentration: General Management, Technology
WE: Information Technology (Computer Software)
Re: For a display, identical cubic boxes are stacked in square  [#permalink]

### Show Tags

29 Aug 2016, 13:01
Attachment:
File comment: Visualize the Question this Way

IMG_20160830_012335.jpg [ 1.04 MiB | Viewed 26117 times ]

In the above figures ,small circles are akin to cubic boxes and lines are rows. So bottom most layer has 9*9 = 81 boxes . Now this figure has more layers stacked on top of it and each layer has 1 less box(small circle) and 1 less row( line). If you follow this theory then you will notice that no of rows = no of boxes in each row.

So 2nd layer from the top will have 2 rows with 2 boxes each. Similarly top most layer will have 1 row and 1 box.

So the total no of boxes will be : 9* 9 + 8*8 +.....+ 1*1 = 285 .
Intern
Joined: 10 Aug 2016
Posts: 5
Re: For a display, identical cubic boxes are stacked in square  [#permalink]

### Show Tags

20 Oct 2016, 02:00
2
possibly the worst worded question i've ever read
Math Expert
Joined: 02 Aug 2009
Posts: 6970
Re: For a display, identical cubic boxes are stacked in square layers  [#permalink]

### Show Tags

16 Jan 2017, 19:39
1
ashikaverma13 wrote:
For a display, identical cubic boxes are stacked in square layers. Each layer consists of cubic boxes arranged in rows that form a square, and each layer has 1 fewer row and 1 fewer box in each remaining row than the layer directly below it. If the bottom layer has 21 boxes and the top layer has only 1 box, how many boxes are in the display?

A. 236
B. 260
C. 269
D. 276
E. 285

Hi,
The Q seems to be flawed as 21 cannot be the number of boxes . It is either 81 or 25.

You can imagine this as a huge cube from which steps are made in 2 sides by removing a row of boxes in each layer.
In numerical value,all the layers will be square of integers starting from 1 on top to 5 or 9 in lowermost.

When we add these, it becomes $$1^2+2^2+3^2+4^2+5^2+6^2+7^2+8^2+9^2$$
Either add all of them or use formula
SUM=$$\frac{n(n+1)(2n+1)}{6}$$=9*(9+1)*(2*9+1)/6=9*10*19/6=15*19=285
E
_________________

1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html

GMAT online Tutor

Intern
Joined: 07 Dec 2016
Posts: 41
Re: For a display, identical cubic boxes are stacked in square  [#permalink]

### Show Tags

27 Mar 2017, 15:14
I had a tough time understanding the question as well
_________________

Cheers!
If u like my post..... payback in Kudos!!

Director
Joined: 26 Aug 2016
Posts: 642
Location: India
Concentration: Strategy, Marketing
GMAT 1: 690 Q50 V33
GMAT 2: 700 Q50 V33
GMAT 3: 730 Q51 V38
GPA: 4
WE: Information Technology (Consulting)
Re: For a display, identical cubic boxes are stacked in square  [#permalink]

### Show Tags

06 Jul 2017, 06:57
1
Yeah i thought from each square floor, one block is removed.
solved like-
9^2 -1 + 8^2 -1 + .... + 2^2 -1 + 1 ( since it is said from below floor one block is removed) .
And perplexed over where is the answer.
Now i understood what does it mean. It simply means a square box is missing from row in each subsequent floor.
Intern
Joined: 15 May 2017
Posts: 9
Re: For a display, identical cubic boxes are stacked in square  [#permalink]

### Show Tags

13 Feb 2018, 11:03
As we know the series is 9^2+8^2.....+2^2+1
we know there are 5 odd+4 even=odd hence striking off Answer choices A,B,and D now we can sum only the unit digits of the squares

EMPOWERgmat Instructor
Status: GMAT Assassin/Co-Founder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 12687
Location: United States (CA)
GMAT 1: 800 Q51 V49
GRE 1: Q170 V170
Re: For a display, identical cubic boxes are stacked in square  [#permalink]

### Show Tags

03 Mar 2018, 16:06
2
Hi All,

The math behind this question is not difficult - it's just about adding the perfect squares from 1^2 to 9^2, inclusive. While it might take a little time to "visualize" what this question describes, the actual "work" is not too challenging. If a child can do the math, then you should be able to (and faster).

There is an interesting shortcut built into the math though. From the answer choices, you'll notice that almost all of the units digits are unique (6, 0, 9 and 5). This provides a likely shortcut that can help us to avoid a bit of the math.

1
4
9
16
25
36
49
64
81
---
?

The nine numbers above can be "paired up" to create 4 values that end in a 0:
1+ 9 --> ends in 0
4 + 16 --> ends in 0
36 + 64 --> ends in 0
49 + 81 --> ends in 0
and 25

The sum MUST end in a 5 and there's only one answer that matches…

GMAT assassins aren't born, they're made,
Rich
_________________

760+: Learn What GMAT Assassins Do to Score at the Highest Levels
Contact Rich at: Rich.C@empowergmat.com

# Rich Cohen

Co-Founder & GMAT Assassin

Special Offer: Save \$75 + GMAT Club Tests Free
Official GMAT Exam Packs + 70 Pt. Improvement Guarantee
www.empowergmat.com/

*****Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!*****

Manager
Joined: 05 Dec 2016
Posts: 115
For a display, identical cubic boxes are stacked in square  [#permalink]

### Show Tags

22 Mar 2018, 09:55
1
Bunuel wrote:

You can use the sum of the first n perfect squares formula to calculate: $$\frac{n(n+1)(2n+1)}{6}=\frac{9*(9+1)(2*9+1)}{6}=285$$.

Hope it's clear.

Attachment:
pyramid_with_corner_cube_from_istock.jpg

Or you can just multiply the units of the numbers and add them all together you will get a result of 5. The only answer ending in 5 is the correct one
_________________

lets all unite to reach our target together

For a display, identical cubic boxes are stacked in square &nbs [#permalink] 22 Mar 2018, 09:55
Display posts from previous: Sort by