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Senior Manager  G
Joined: 04 Sep 2017
Posts: 291
When opened and lying flat, a birthday card is in the shape of a regul  [#permalink]

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8 00:00

Difficulty:   75% (hard)

Question Stats: 54% (02:21) correct 46% (02:32) wrong based on 156 sessions

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When opened and lying flat, a birthday card is in the shape of a regular hexagon. The card must be folded in half along 1 of its diagonals before being placed in an envelope for mailing. Assuming that the thickness of the folded card will not be an issue, will the birthday card fit inside a rectangular envelope that is 4 inches by 9 inches?

(1) Each side of the regular hexagon is 4 inches long.
(2) The area of the top surface (which is the same as the area of the bottom surface) of the folded birthday card is less than 36 square inches.

DS61791.01
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Status: As cheeks from my insta feed say: soon...
Joined: 17 Jan 2016
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Re: When opened and lying flat, a birthday card is in the shape of a regul  [#permalink]

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gmatt1476 wrote:
When opened and lying flat, a birthday card is in the shape of a regular hexagon. The card must be folded in half along 1 of its diagonals before being placed in an envelope for mailing. Assuming that the thickness of the folded card will not be an issue, will the birthday card fit inside a rectangular envelope that is 4 inches by 9 inches?

(1) Each side of the regular hexagon is 4 inches long.
(2) The area of the top surface (which is the same as the area of the bottom surface) of the folded birthday card is less than 36 square inches.

DS61791.01

won't statement 2 give us the sizes of the card?
Intern  B
Joined: 05 Mar 2018
Posts: 22
Re: When opened and lying flat, a birthday card is in the shape of a regul  [#permalink]

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I think given the area you can decide the length of the sides, like a right triangle.
Manager  B
Joined: 19 Jan 2018
Posts: 84
When opened and lying flat, a birthday card is in the shape of a regul  [#permalink]

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gmatt1476 wrote:
When opened and lying flat, a birthday card is in the shape of a regular hexagon. The card must be folded in half along 1 of its diagonals before being placed in an envelope for mailing. Assuming that the thickness of the folded card will not be an issue, will the birthday card fit inside a rectangular envelope that is 4 inches by 9 inches?

(1) Each side of the regular hexagon is 4 inches long.
(2) The area of the top surface (which is the same as the area of the bottom surface) of the folded birthday card is less than 36 square inches.

DS61791.01

You can actually solve this question by using very little math and by using some logic:

If you fold a hexagon from one of it's diagonal, you end up with a trapezoid with three sides as X (because a regular hexagon has equal sides), and Base 2 as Y. Now you need to know the values of X or Y to determine the dimensions of the figure. (Refer to pic attachment for a visual)

1.) Gives you the value of X, so you have enough information to determine a specific value of Y which will lead to your answer! SUFFICIENT

2.) The area is less than 36 square inches, but we don't know the values of either x or y. You can have a trapezoid that is super long but very small in height (assuming the area of it is 35.999999) or a trapezoid that's super tall but the base is super small. Or have a very small area (0.000001) and the right dimensions would be enough for the folded hexagon to fit in the envelope. NOT SUFFICIENT

Attachments

File comment: Hexagon Folded in Half Trap.png [ 14 KiB | Viewed 1240 times ]

Manager  S
Joined: 01 Dec 2018
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Re: When opened and lying flat, a birthday card is in the shape of a regul  [#permalink]

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gmatt1476 wrote:
When opened and lying flat, a birthday card is in the shape of a regular hexagon. The card must be folded in half along 1 of its diagonals before being placed in an envelope for mailing. Assuming that the thickness of the folded card will not be an issue, will the birthday card fit inside a rectangular envelope that is 4 inches by 9 inches?

(1) Each side of the regular hexagon is 4 inches long.
(2) The area of the top surface (which is the same as the area of the bottom surface) of the folded birthday card is less than 36 square inches.

DS61791.01

IF WE USE THE BASIC CONCEPTS THIS PROBLEM CAN BE DONE IN SECONDS .

INFO GIVEN - the envelope has B=4 AND L=9.
The hexagon has been folded into half along one of its longest diagonal . note longest diagonal of hexagon is 2(side)

1)we know the side of Hexagon which gives us longest side of folded shape / longest diagonal of hexagon as 2*4 = 8 .
and we know the longest side of rectangular envelope is 9 . so we can easily fit the card into envelope .
SUFFICIENT .

2)Please note just by having same area of two different objects doesn't mean it will fit into each other . The shape matters (the sides matter ). NOT SUFFICIENT
Intern  B
Joined: 29 Aug 2019
Posts: 15
Re: When opened and lying flat, a birthday card is in the shape of a regul  [#permalink]

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The actual issue with this problem is at the core of data sufficiency.

You are mostly led to believe that statement II is sufficient because you have it in your head that each side is still 4. Given that the side is 4 and the area is 36 sure it is sufficient. But you have to get statement I completely out of your head and realize that there are a number of ways in which base 2 is in fact more than 9 yet still leads to an area of 36.

As a matter of fact you can calculate the area without being told the area just by knowing that side and chopping it up - but that's a waste of time for a problem like this. Simple logic will serve you best here.

Forget trying to prove why statement II doesn't work - just imagine statement I and II are flipped around and do the problem again and you will see that just having an area of 36 is not sufficient.
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Joined: 23 Nov 2018
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Re: When opened and lying flat, a birthday card is in the shape of a regul  [#permalink]

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Bunuel can you help out with this please?
Director  V
Joined: 24 Oct 2016
Posts: 580
GMAT 1: 670 Q46 V36 GMAT 2: 690 Q47 V38 When opened and lying flat, a birthday card is in the shape of a regul  [#permalink]

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Bunuel can you help out with this please?

gmatt1476 wrote:
When opened and lying flat, a birthday card is in the shape of a regular hexagon. The card must be folded in half along 1 of its diagonals before being placed in an envelope for mailing. Assuming that the thickness of the folded card will not be an issue, will the birthday card fit inside a rectangular envelope that is 4 inches by 9 inches?

(1) Each side of the regular hexagon is 4 inches long.
(2) The area of the top surface (which is the same as the area of the bottom surface) of the folded birthday card is less than 36 square inches.

DS61791.01

When hexagon is folded, it forms an isosceles trapezium as shown in the attached picture (Credit: jsound996)
Let's say the smaller parallel side is B1, longer parallel side is B2 and Height of trapezium is H.

Since regular hexagon divides into 6 equilateral triangles, we know:
1) B2 = 2B1
2) Height of equilateral triangle = √3*side/2

NEED: To make sure that we can satisfy 2 conditions: 1) H <=4 and 2) B2 <= 9.

1) B1 = 4; B2 = 8
Condition 2 met.
Let's calculate H.

Since regular hexagon divides into 6 equilateral triangles, we can calculate the height.
Height of equilateral triangle = √3*side/2 = 2√3 = 2 * 1.73 = 3.4 => Condition 1 met.

Since both conditions met, statement 1 is sufficient.

2) Area of trapezium = 0.5 (B1 + B2) * H
Statement says: Area < 36
0.5 (B1 + B2) * H < 36
0.5 (3B1) * H < 36

H = √3*side/2 = √3*B1/2

0.5 * (3B1) * √3*B1/2 < 36
(B1^2)*3√3/4 < 36
B1^2 < (4*36)/3√3
B1^2 < 4*12/√3
B1^2 < 4*4*3/√3
B1^2 < 4*4*√3
B1 < 4*3(1/4)
B1 < 4 * 1.3
B1 < 5.26

B2 < 2B1 => B2 < 10.52

Hence B2 could be less than or more than 9. Condition 2 not met.

Not sufficient.

Bunuel VeritasKarishma I was wondering whether there is a shorter way to do this Q. Thanks!
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Veritas Prep GMAT Instructor V
Joined: 16 Oct 2010
Posts: 9848
Location: Pune, India
Re: When opened and lying flat, a birthday card is in the shape of a regul  [#permalink]

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1
dabaobao wrote:
Bunuel can you help out with this please?

gmatt1476 wrote:
When opened and lying flat, a birthday card is in the shape of a regular hexagon. The card must be folded in half along 1 of its diagonals before being placed in an envelope for mailing. Assuming that the thickness of the folded card will not be an issue, will the birthday card fit inside a rectangular envelope that is 4 inches by 9 inches?

(1) Each side of the regular hexagon is 4 inches long.
(2) The area of the top surface (which is the same as the area of the bottom surface) of the folded birthday card is less than 36 square inches.

DS61791.01

When hexagon is folded, it forms an isosceles trapezium as shown in the attached picture (Credit: jsound996)
Let's say the smaller parallel side is B1, longer parallel side is B2 and Height of trapezium is H.

Since regular hexagon divides into 6 equilateral triangles, we know:
1) B2 = 2B1
2) Height of equilateral triangle = √3*side/2

NEED: To make sure that we can satisfy 2 conditions: 1) H <=4 and 2) B2 <= 9.

1) B1 = 4; B2 = 8
Condition 2 met.
Let's calculate H.

Since regular hexagon divides into 6 equilateral triangles, we can calculate the height.
Height of equilateral triangle = √3*side/2 = 2√3 = 2 * 1.73 = 3.4 => Condition 1 met.

Since both conditions met, statement 1 is sufficient.

2) Area of trapezium = 0.5 (B1 + B2) * H
Statement says: Area < 36
0.5 (B1 + B2) * H < 36
0.5 (3B1) * H < 36

H = √3*side/2 = √3*B1/2

0.5 * (3B1) * √3*B1/2 < 36
(B1^2)*3√3/4 < 36
B1^2 < (4*36)/3√3
B1^2 < 4*12/√3
B1^2 < 4*4*3/√3
B1^2 < 4*4*√3
B1 < 4*3(1/4)
B1 < 4 * 1.3
B1 < 5.26

B2 < 2B1 => B2 < 10.52

Hence B2 could be less than or more than 9. Condition 2 not met.

Not sufficient.

Bunuel VeritasKarishma I was wondering whether there is a shorter way to do this Q. Thanks!

For stmnt 1, you don't need to calculate anything. Since it is a regular hexagon, knowing a side will give all other measurements so you will be able to say whether the card will fit in the envelope.
For stmnt 2, you will need to find the maximum value of the side to see if it may or may not fit in the envelope.
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Re: When opened and lying flat, a birthday card is in the shape of a regul  [#permalink]

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1

Hello Karishma,

The question says it is a regular hexagon.

The topside of the hexagon will have 3 equilateral triangles.
Let it side be s.
Then according to the question Area of a top hexagon is less than 36

Solving it we get s<24 * root 3

Therefore insufficient as we can have both yes and no case here.
Is this approach ok?
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Joined: 16 Oct 2010
Posts: 9848
Location: Pune, India
Re: When opened and lying flat, a birthday card is in the shape of a regul  [#permalink]

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akash7gupta11 wrote:

Hello Karishma,

The question says it is a regular hexagon.

The topside of the hexagon will have 3 equilateral triangles.
Let it side be s.
Then according to the question Area of a top hexagon is less than 36

Solving it we get s<24 * root 3

Therefore insufficient as we can have both yes and no case here.
Is this approach ok?

Certainly you can use the concept of equilateral triangles. The diagonal along the fold will be made up of two sides of equilateral triangles so its length will be 2s.
You will get that s < 5.something
So 2s < 10.something

Hence not sufficient since 2s may be less than 9 or may not be.
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Karishma
Veritas Prep GMAT Instructor Re: When opened and lying flat, a birthday card is in the shape of a regul   [#permalink] 22 Nov 2019, 00:40
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