The figure is made up of a series of inscribed equilateral triangles.

The Answer is C.


To get a triangle 128 times(2^7) smaller than the main triangle, you will have to partition it 7 times.
For each Odd partition you will get 3 Shaded regions.
For each Even partition you will get 1 shaded region the will again be partitions into regions by subsequent odd partition.So, we will no consider these partitions.

Partition 1 2 3 4 5 6 7
Shaded Regions 3 1 3 1 3 1 3

Consider the area of main triangle be A.
So the shaded region will be (3)*(1/4^7)A + (3)*(1/4^5)A + (3)*(1/4^3)A + (3)*(1/4^1)A

= 3/4 (2^-12 + 2^-8 + 2^-4 + 2^0 ) A

To get the fraction of shaded region , (3/4 (2^-12 + 2^-8 + 2^-4 + 2^0 ) A) / A
= 3/4 (2^-12 + 2^-8 + 2^-4 + 2^0 )

may i ask which source it is from? i dont think real test will have such kinda of question

IMO C.

I did not calculate it completely. Just calculated the area of first two partitions and established a pattern and then used the pattern to guess the answer. That's what I am going to do if something like this comes in real test as well

Can you please explain why u skipped certain sides?

Excuse me but would you explain why you skipped the even partitions?

Cheers!
J

I believe the even portions are skipped because those are considered shaded in the next odd portions.

I didn't get into details, and solved it with logic.
we have 4 equal triangles. we know for sure that the shaded area will be greater than 3/4 of the total figure.
A and B are right away out.

i started with E.
3/4 *(1+1/2 + 1/4 + 1/8) = 3/4 * (8+4+2+1)/8 = 3/4 * 15/8 ->result is greater than 1. so out.

D
3/4 * (1+1/4 +1/16 + 1/64) = 3/4* (64+16+4+1)/64 = 3/4 * 85/64 = 255/256 - clearly no!


only left - C

Just a glance at the image shows that the shaded area would be around 75%. Calculating all the options till two terms only as rest are too small, we get C as the answer.

Though i marked the wrong answer, Lets try to solve easy way..
1/128 = 1/2^7 . So center triangle will go upto 8th triangle in the center and will be unshaded.
Let A be the area of bigger triangle. Lets number biggest triangle as 1 and then smaller center triangles as 2 and next smaller center triangle as 3 and so on.

So, First focusing on 1st bigger triangle no. 1..
Shaded region = 3/4 A

Now , In 2nd smaller triangle no. 2 , we have only shaded region from 1/4th of the part i.e from triangle no.3. So we will consider it in the next calculation in triangle no.3

In triangle 3,
Shaded region = 3/4 * triangle 3 = 3/4 *(1/4)^2 A

In triangle no. 4 , we have only shaded region from 1/4th of the part i.e from triangle no.5. So we will consider it in the next calculation in triangle no.5

In triangle 5,
Shaded region = 3/4 * triangle 5 = 3/4* 1/4 *1/4 * triangle 3 = 3/4 *(1/4)^2 * (1/4)^2 A

In triangle 6, we have only shaded region from 1/4th of the part i.e from triangle no.7. So we will consider it in the next calculation in triangle no.7
Shaded region = 3/4 * (1/4)^2 * (1/4)^2 * (1/4)^2 A


Sum of shaded regions = 3/4 A + 3/4 *(1/4)^2 A + 3/4 *(1/4)^2 * (1/4)^2 A + 3/4 * (1/4)^2 * (1/4)^2 * (1/4)^2 A
= 3/4 (1/2^0 + 1/2^4 + 1/2^8 + 1/2^12) A

Answer C

I solved it as per the image attached, hope it is clear and helps.

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