5-Day Free Trial
5-day free, full-access trial TTP
Available with Beat the GMAT members only code
MORE DETAILS
Triangle Problem 1
This topic has expert replies
-
aditiniyer
- Senior | Next Rank: 100 Posts
- Posts: 45
- Joined: Fri Aug 19, 2016 1:42 am
The figure is made up of a series of inscribed equilateral triangles.If the pattern continues until the length of a side of the largest triangle( entire figure) is exactly 128 times that of the smallest triangle, what fraction of the total figure will be shaded.
- Attachments
-
GMAT/MBA Expert
-
[email protected]
- Elite Legendary Member
- Posts: 10392
- Joined: Sun Jun 23, 2013 6:38 pm
- Location: Palo Alto, CA
- Thanked: 2867 times
- Followed by:511 members
- GMAT Score:800
Hi aditiniyer,
When posting questions, you should make sure to post them in the proper sub-Forum. This is the DS Forum, but the prompt is clearly NOT a DS question. In addition, it does not include the answer choices, which means that we essentially have just one way to approach this question - algebraically.
I'm going to give you a hint so that you can retry this question on your own: Notice in the first 'layer' of equilateral triangles that 3 out of the 4 are shaded... what ratio of equilateral triangles in the second 'layer' are shaded? How about in the third 'layer'? With each new layer that is added, how does that impact the fraction of the total that is shaded?
GMAT assassins aren't born, they're made,
Rich
When posting questions, you should make sure to post them in the proper sub-Forum. This is the DS Forum, but the prompt is clearly NOT a DS question. In addition, it does not include the answer choices, which means that we essentially have just one way to approach this question - algebraically.
I'm going to give you a hint so that you can retry this question on your own: Notice in the first 'layer' of equilateral triangles that 3 out of the 4 are shaded... what ratio of equilateral triangles in the second 'layer' are shaded? How about in the third 'layer'? With each new layer that is added, how does that impact the fraction of the total that is shaded?
GMAT assassins aren't born, they're made,
Rich
-
GMATGuruNY
- GMAT Instructor
- Posts: 15539
- Joined: Tue May 25, 2010 12:04 pm
- Location: New York, NY
- Thanked: 13060 times
- Followed by:1906 members
- GMAT Score:790
Here is the complete problem, with answer choices:
In the figure above, each side of the large outer triangle (2) is DOUBLE each side of the four small interior triangles (1).
Implication:
As the triangles increase in size, each successive triangle must have a side that is DOUBLE that of the next smallest triangle.
Thus, from smallest to biggest, the lengths of the sides must be proportioned as follows:
2�, 2¹, 2². 2³, 2�, 2�, 2�, 2�.
Because all of the triangles have the same combination of angles -- 60, 60, 60 -- they are all SIMILAR.
RULE:
If two triangles are similar, and each side of the larger triangle is X TIMES the corresponding side in the smaller triangle, then the AREA of the larger triangle is X² TIMES the area of the smaller triangle.
In the problem here, since the sides from smallest to biggest keep DOUBLING, the areas from smallest to biggest must be QUADRUPLING.
Thus, from smallest to biggest, the areas must be proportioned as follows:
2�, 2², 2�, 2�, 2�, 2¹�, 2¹², 2¹�.
The figure above indicates that -- for each successive pair of triangles -- 3/4 of the larger triangle will be shaded.
The blue values in the list above represent the larger triangle in each successive pair.
Since 3/4 of these areas will be shaded, we get:
Total shaded = (3/4)(2² + 2� + 2¹� + 2¹�).
The total AREA is equal to the area of the largest triangle, represented by the greatest blue value (2¹�).
Thus:
(total shaded)/(total area) = [(3/4)(2² + 2� + 2¹� + 2¹�)]/2¹� = (3/4)(2¯¹² + 2¯� + 2¯� + 2�).
The correct answer is C.
Consider the following case:The figure is made up of a series of inscribed equilateral triangles.If the pattern continues until the length of a side of the largest triangle( entire figure) is exactly 128 times that of the smallest triangle, what fraction of the total figure will be shaded.
A. (1\4)(2^0 + 2^-4 + 2^-8 + 2^-12)
B. (1\4)(2^0 + 2^-2 + 2^-4 + 2^-6)
C. (3\4)(2^0 + 2^-4 + 2^-8 + 2^-12)
D. (3\4)(2^0 + 2^-2 + 2^-4 + 2^-6)
E. (3\4)(2^0 + 2^-1 + 2^-2 + 2^-3)
In the figure above, each side of the large outer triangle (2) is DOUBLE each side of the four small interior triangles (1).
Implication:
As the triangles increase in size, each successive triangle must have a side that is DOUBLE that of the next smallest triangle.
Thus, from smallest to biggest, the lengths of the sides must be proportioned as follows:
2�, 2¹, 2². 2³, 2�, 2�, 2�, 2�.
Because all of the triangles have the same combination of angles -- 60, 60, 60 -- they are all SIMILAR.
RULE:
If two triangles are similar, and each side of the larger triangle is X TIMES the corresponding side in the smaller triangle, then the AREA of the larger triangle is X² TIMES the area of the smaller triangle.
In the problem here, since the sides from smallest to biggest keep DOUBLING, the areas from smallest to biggest must be QUADRUPLING.
Thus, from smallest to biggest, the areas must be proportioned as follows:
2�, 2², 2�, 2�, 2�, 2¹�, 2¹², 2¹�.
The figure above indicates that -- for each successive pair of triangles -- 3/4 of the larger triangle will be shaded.
The blue values in the list above represent the larger triangle in each successive pair.
Since 3/4 of these areas will be shaded, we get:
Total shaded = (3/4)(2² + 2� + 2¹� + 2¹�).
The total AREA is equal to the area of the largest triangle, represented by the greatest blue value (2¹�).
Thus:
(total shaded)/(total area) = [(3/4)(2² + 2� + 2¹� + 2¹�)]/2¹� = (3/4)(2¯¹² + 2¯� + 2¯� + 2�).
The correct answer is C.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
-
alanforde800Maximus
- Master | Next Rank: 500 Posts
- Posts: 187
- Joined: Tue Sep 13, 2016 12:46 am
Thank You Mitch. This explanation was very easy to understand.GMATGuruNY wrote:Here is the complete problem, with answer choices:
Consider the following case:The figure is made up of a series of inscribed equilateral triangles.If the pattern continues until the length of a side of the largest triangle( entire figure) is exactly 128 times that of the smallest triangle, what fraction of the total figure will be shaded.
A. (1\4)(2^0 + 2^-4 + 2^-8 + 2^-12)
B. (1\4)(2^0 + 2^-2 + 2^-4 + 2^-6)
C. (3\4)(2^0 + 2^-4 + 2^-8 + 2^-12)
D. (3\4)(2^0 + 2^-2 + 2^-4 + 2^-6)
E. (3\4)(2^0 + 2^-1 + 2^-2 + 2^-3)
In the figure above, each side of the large outer triangle (2) is DOUBLE each side of the four small interior triangles (1).
Implication:
As the triangles increase in size, each successive triangle must have a side that is DOUBLE that of the next smallest triangle.
Thus, from smallest to biggest, the lengths of the sides must be proportioned as follows:
2�, 2¹, 2². 2³, 2�, 2�, 2�, 2�.
Because all of the triangles have the same combination of angles -- 60, 60, 60 -- they are all SIMILAR.
RULE:
If two triangles are similar, and each side of the larger triangle is X TIMES the corresponding side in the smaller triangle, then the AREA of the larger triangle is X² TIMES the area of the smaller triangle.
In the problem here, since the sides from smallest to biggest keep DOUBLING, the areas from smallest to biggest must be QUADRUPLING.
Thus, from smallest to biggest, the areas must be proportioned as follows:
2�, 2², 2�, 2�, 2�, 2¹�, 2¹², 2¹�.
The figure above indicates that -- for each successive pair of triangles -- 3/4 of the larger triangle will be shaded.
The blue values in the list above represent the larger triangle in each successive pair.
Since 3/4 of these areas will be shaded, we get:
Total shaded = (3/4)(2² + 2� + 2¹� + 2¹�).
The total AREA is equal to the area of the largest triangle, represented by the greatest blue value (2¹�).
Thus:
(total shaded)/(total area) = [(3/4)(2² + 2� + 2¹� + 2¹�)]/2¹� = (3/4)(2¯¹² + 2¯� + 2¯� + 2�).
The correct answer is C.
