# Math 800 Challenge– Can you crack it?

So far none of my students have been able to answer this problem (at least not without a little nudging on my part.) Let’s see if anybody can prove the correct answer and explanation.

You have 48 hours!

The sides of a rectangular box when added together equal 36.

Column A Column B
The length of the longest possible line
that can be drawn to connect two of the
vertices of rectangular box. A. The quantity in Column A is greater
B. The quantity in Column B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given

I posted the answer and an explanation in the comments, good luck!

## Author

• Chris Lele is the Principal Curriculum Manager (and vocabulary wizard) at Magoosh. Chris graduated from UCLA with a BA in Psychology and has 20 years of experience in the test prep industry. He's been quoted as a subject expert in many publications, including US News, GMAC, and Business Because. In his time at Magoosh, Chris has taught countless students how to tackle the GRE, GMAT, SAT, ACT, MCAT (CARS), and LSAT exams with confidence. Some of his students have even gone on to get near-perfect scores. You can find Chris on YouTube, LinkedIn, Twitter and Facebook!

### 15 Responses to Math 800 Challenge– Can you crack it?

1. Chris Lele August 9, 2011 at 4:48 pm #

Okay, let’s close the thread on this one, guys.

The answer is (D) because the longest possible line can either
be smaller or bigger than 8 depending on how you construct
the box. If you build a box with 5 x 2 x 2 (which will work to 36
for total perimeter), then you will have a line that is less than 8
(approximately 5.8).

If you construct a really skinny box with sides 8 x 1/2 x 1/2, then
the longest possible line will be greater than 8.

Column A cannot be both bigger and smaller than Column B. In
the GRE, on Quantitative Comparison, when you have more than
one answer (it could be A or B) the answer is (D).

Another way of looking at it – in QC when you arrive at an answer,
such as (A) can be bigger that doesn’t make it the final answer.
Your goal should be to disprove that, as we did here. If you can’t
disprove (A), that is (A) is always bigger given the constraints of
the problem, than the answer is (A). If you can come up with
one case where (A) is not the answer, say that one case is
(C), than the answer has to be (D).

2. inam August 9, 2011 at 9:19 am #

B)
4x+4y+4z=36
x+y+z=9
if we assume x=7 y=1 and x=1
then we get longest edge to be sq.root of 49+1+1
which gives something less than 8

3. Chris Lele August 8, 2011 at 9:41 pm #

True, it could be less than 8 as well if the dimensions of the box are more boxy. So it could be larger or smaller, depending on the dimensions of the box. (D).

Good eye! As nobody seems to have noticed this throughout the many posts. Everyone was proving that the box could be bigger.

4. venkatesh July 29, 2011 at 3:58 pm #

the ans is A.4(l+b+h)=36.so l+b+h =9 .the maximum value of line joining the vertices must be sqrt(8^2+0.5^2+0.5^2).so i think its A .

• venkatesh July 29, 2011 at 4:55 pm #

for ex sqrt(8^2+0.5^2+0.5^2)>8 so ans is A

• oldguy9999 August 8, 2011 at 1:50 pm #

formulas – area of base = L x W
perimeter of base = 2L + 2L
SURFACE AREA = (L x W) + (2L + 2W)H
diagonal rectangular block = sqrt(Lsqr+Wsqr+Hsqr)

if you think about the diagonal being only 8 then plug numbers that come close to that goal.

if L=8 you can imagine that a pencil that length could “almost” reach across to the furthest point.

box with L=8, W=2, H = 1 would have a surface area of 36 and a diagonal of sqrt(69) = approx 8.3, so a box with a surface area of 36 can have a longer diagonal than 8,

i think this is a fast way to answer the question wtihout going beyond what you need to do for an answer. the question does not ask you to solve the greatest possible length of the box’s diagonal. it only asks if it could be greater than 8.

• oldguy9999 August 8, 2011 at 2:03 pm #

well, on second thought, in gre format they ask you if it is bigger or not. the diagonal could also be smaller than 8. ( a cube is still a rectangular box). so answer should be D.

5. Giridhar Kumar July 26, 2011 at 10:42 pm #

Perimeter of cuboid = 36
4(L+B+H) = 36
L+B+H=9

Length of longest possible line connecting vertices = SQRT(L^2 + B^2 + H^2)

Case 1:
L=1, B=1, H=7
SQRT(L^2 + B^2 + H^2) = SQRT(1+1+49)=SQRT(51) 8
A is greater

• Giridhar Kumar July 26, 2011 at 10:47 pm #

Case 1: L=1, B=1, H=7
SQRT(L^2 + B^2 + H^2) = SQRT(1+1+49)=SQRT(51) 8; A is greater

• Giridhar Kumar July 26, 2011 at 10:48 pm #

Dont know why my answer is getting truncated

• Chris Lele July 27, 2011 at 10:47 am #

Hmm…that is interesting…The whole answer should usually post, even if it is several pages long, which yours clearly is not. I’ll check with the technical team on this.

• Giridhar Kumar August 3, 2011 at 4:04 pm #

L=1, B=1, H=7
Long diagnol = SQRT(51)8

6. Vivek Roongta July 12, 2011 at 12:12 am #

Ans: A

4L+4B+4H = 36

=> L+B+H = 9

now longest possible line joining vertices is diagonal of the box

which is equal to (L^2+B^2+H^2) …. for the minimum value of this expression L=B=H =3

so minimum value of diagonal is 27 > 8 … so A will always be bigger then B

7. Navee February 16, 2011 at 3:45 pm #

As length of a longest possible line in a rectangular box (cuboid) is Body diagonal .. which is Square root of (L^2 + B^2 + H^ ) and here L + B + H = 36 …. Even we take min L = 1 and B = 1 and H =34.. the body diagonal is Square root of ( 1 + 1 + 1156 ) …. this is greater than Col B which Square root of (64)…. So the final answer is : A

8. Navee February 16, 2011 at 3:40 pm #

ans is : A

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