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13 May 2019, 23:16
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
49% (03:31) correct
51%
(02:51)
wrong
based on 45
sessions
History
Date
Time
Result
Not Attempted Yet
A closed box with a square base is to be wrapped with a square sheet of wrapping paper. The box is centered on the wrapping paper with the vertices of the base lying on the midlines of the square sheet of paper, as shown in the figure on the left. The four corners of the wrapping paper are to be folded up over the sides and brought together to meet at the center of the top of the box, point A in the figure on the right. The box has base length w and height h. What is the area of the sheet of wrapping paper?
(A) \(2(w+h)^2\)
(B) \(\frac{(w+h)^2}{2}\)
(C) \(2w^2+4wh\)
(D) \(2w^2\)
(E) \(w^2h\)
307dceed58be3002607d62631a356808a8841be4.png [ 26.96 KiB | Viewed 3677 times ]
(A) \(2(w+h)^2\)
(B) \(\frac{(w+h)^2}{2}\)
(C) \(2w^2+4wh\)
(D) \(2w^2\)
(E) \(w^2h\)
Attachment:
307dceed58be3002607d62631a356808a8841be4.png [ 26.96 KiB | Viewed 3677 times ]
22 May 2019, 06:35
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Probably one of the first things that comes to our mind when we look at this question – ‘Whoa, really? Am I supposed to solve this question in 2 minutes?’
Well, there’s no other option, really!
In reality though, this is a question that looks difficult but, IS NOT difficult. You only have to be able to break it down into smaller geometrical entities for which you know the formulas for area calculation.
When you look at it in detail, you need to know the following formulas:
What is actually difficult though, is to visualize how the wrapping paper will wrap around the box. This is where your visual reasoning skills are tested and this is also where GMAT tells you that it is not testing you on your knowledge of Geometry alone. It’s actually not that hard if you follow the dotted lines.
To help you understand how the wrapping paper will get wrapped eventually, we have made some constructions in the diagram below:
22nd May 2019 - Reply 3.JPG [ 81.67 KiB | Viewed 3283 times ]
We have marked some points with alphabets. We have connected point F to point D. You can now see that rectangle DFGH corresponds to one of the lateral (bounding) surfaces of the cube. The area of this rectangle is w * h.
There will be 4 such bounding surfaces, so the total area of these bounding surfaces will be 4wh.
Triangle DEF, when folded will cover the triangle HKG. HK represents half the diagonal of the square.
The diagonal of the square = w√2. Therefore, HK = ½ * w√2 = \(\frac{w}{√2}\)
.
The diagonals of a square bisect each other at right angles. Hence, KG will also be \(\frac{w}{√2}\). Also, HKG will be a right angled triangle.
So, area of triangle DEF = area of triangle HKG = ½ * \(\frac{w}{√2}\) * \(\frac{w}{√2}\). = \(\frac{w^2}{4}\).
We have 8 such triangles (4 at the top and 4 at the bottom). Therefore, the total area of these bounding surfaces will be 2*\(w^2\).
Triangle FGI is a right angled triangle, which represents the part of the wrapping paper, which will remain unused.
Area of triangle FGI = ½ * h * h = \(\frac{h^2}{2}\). There are 4 such triangles, so the total area of these triangles will be 2*\(h^2\).
Hence, the total area of the wrapping paper = 4wh + 2 * \(w^2\) + 2 * \(h^2\), which simplifies to 2\((w + h)^2\).
An ideal time to solve this question could be anywhere between 2 to 3 minutes. However, if you are stuck with this question beyond 3 minutes, it’s a good idea to eliminate answer options like B, D and E, based on the area of the rectangles (4wh), because none of these answer options give 4wh. Beyond this stage, option C can be eliminated because it does not take care of the unused portion of the wrapping paper.
So, even when you are stuck, you have a way out to zero in on the right answer, if you have practiced eliminating options based on simple mathematical logic.
I hope this helped!
Thanks
Well, there’s no other option, really!
In reality though, this is a question that looks difficult but, IS NOT difficult. You only have to be able to break it down into smaller geometrical entities for which you know the formulas for area calculation.
When you look at it in detail, you need to know the following formulas:
- Area of a rectangle = length * breadth
Area of a right angled triangle = ½ * (product of perpendicular sides)
Length of a diagonal of a square in terms of its side = √2 * side
What is actually difficult though, is to visualize how the wrapping paper will wrap around the box. This is where your visual reasoning skills are tested and this is also where GMAT tells you that it is not testing you on your knowledge of Geometry alone. It’s actually not that hard if you follow the dotted lines.
To help you understand how the wrapping paper will get wrapped eventually, we have made some constructions in the diagram below:
Attachment:
22nd May 2019 - Reply 3.JPG [ 81.67 KiB | Viewed 3283 times ]
We have marked some points with alphabets. We have connected point F to point D. You can now see that rectangle DFGH corresponds to one of the lateral (bounding) surfaces of the cube. The area of this rectangle is w * h.
There will be 4 such bounding surfaces, so the total area of these bounding surfaces will be 4wh.
Triangle DEF, when folded will cover the triangle HKG. HK represents half the diagonal of the square.
The diagonal of the square = w√2. Therefore, HK = ½ * w√2 = \(\frac{w}{√2}\)
.
The diagonals of a square bisect each other at right angles. Hence, KG will also be \(\frac{w}{√2}\). Also, HKG will be a right angled triangle.
So, area of triangle DEF = area of triangle HKG = ½ * \(\frac{w}{√2}\) * \(\frac{w}{√2}\). = \(\frac{w^2}{4}\).
We have 8 such triangles (4 at the top and 4 at the bottom). Therefore, the total area of these bounding surfaces will be 2*\(w^2\).
Triangle FGI is a right angled triangle, which represents the part of the wrapping paper, which will remain unused.
Area of triangle FGI = ½ * h * h = \(\frac{h^2}{2}\). There are 4 such triangles, so the total area of these triangles will be 2*\(h^2\).
Hence, the total area of the wrapping paper = 4wh + 2 * \(w^2\) + 2 * \(h^2\), which simplifies to 2\((w + h)^2\).
An ideal time to solve this question could be anywhere between 2 to 3 minutes. However, if you are stuck with this question beyond 3 minutes, it’s a good idea to eliminate answer options like B, D and E, based on the area of the rectangles (4wh), because none of these answer options give 4wh. Beyond this stage, option C can be eliminated because it does not take care of the unused portion of the wrapping paper.
So, even when you are stuck, you have a way out to zero in on the right answer, if you have practiced eliminating options based on simple mathematical logic.
I hope this helped!
Thanks
25 May 2022, 13:11
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The Diagonal of the sheet can be visualised as 2* (w/2 + h +w/2) = 2(w+h)
Let the side of square sheet = S
S^2 +S^2 = (2(w+h))^2
S^2 = 2(w+h)^2
Answer : A
Let the side of square sheet = S
S^2 +S^2 = (2(w+h))^2
S^2 = 2(w+h)^2
Answer : A
