Let me preface this problem: you will have to be doing really well on the GRE to even see a problem of this difficulty. Basically even if you score 800 you may only see one problem of this difficulty level.

For problems that are this difficult make sure you are mapping them out. Do not stare at the computer screen hoping that the answer will fall out of the screen. Redraw the figure on scratch paper (if relevant to the problem) and begin writing down information that will help you get to the answer. I recommend doing the following for the next challenge problem.

One final thing–and perhaps the most important part–whoever can crack this problem first, providing me with both the answer and the explanation, will win a free access to Magoosh GRE. That’s hundreds upon hundreds of free questions!

Okay, so here it is:

**How many squares, each with a side of 1, can fit inside a circle with an area of 50π, such that none of the squares fall outside the circle?**

**(A) 100**

**(B) 124**

**(C) 132**

**(D) 138**

**(E) 157**

Hi Chris,

Sorry for digging up this post from a year back! Can you provide an elaborate explanation for this problem? I don’t get the explanation you have given in the comments.

Hi Praveen,

Actually this problem is not a GRE problem. It was meant as a challenge to win a free subscription, so I made it super hard. To be frank, focusing on a question such as this one is a waste of time, from a GRE perspective. There are far more relevant questions to tackle – in fact, I have a perfect score math challenge coming in a few days 🙂

radius of circle = sqrt[(50 * pi) / pi] = sqrt(50) = 7.071

Consider a semicircle of radius 7.071. Divide them into 7 horizontal bands

of width 1. Remaining space of 0.071 at the very top is not enough to fit

a square of size 1 and therefore there is no 8th band.

In the first band which starts in the horizontal diameter of the circle,

pack squares adjacent to each other leaving little space in the very left

and the very right. This space will chosen(more on this later) to be

fractional portion of whatever length of chord we have over which

we are arranging the squares divided by 2. The bottom portion of first band

has length 14.142, we should be able to pack 14 squares, leaving

0.142 / 2 = 0.071 each in the left and right end.

Half the length of bottom portion of second band = sqrt(7.071^2 – 1^2) = 6.999.

Length of bottom portion of second band = 6.999 * 2 = 13.998. We should be

able to pack 13 squares in the second band.

Half the length of bottom portion of third band = sqrt(7.071^2 – 2^2) = 6.782.

Length of bottom portion of third band = 6.782 * 2 = 13.564. We should be

able to pack 13 squares in the third band.

Half the length of bottom portion of fourth band = sqrt(7.071^2 – 3^2) = 6.403.

Length of bottom portion of fouth band = 6.403 * 2 = 12.806. We should be able

to pack 12 squares in the fourth band.

Half the length of bottom portion of fifth band = sqrt(7.071^2 – 4^2) = 5.830.

Length of bottom portion of fifth band = 5.830 * 2 = 11.66. We should be able

to pack 11 squares in the fifth band.

Half the length of bottom portion of sixth band = sqrt(7.071^2 – 5^2) = 4.999.

Length of bottom portion of sixth band = 4.999 * 2 = 9.998. We should be able

to pack 9 squares in the sixth band.

Half the length of bottom portion of seventh band = sqrt(7.071^2 – 6^2) = 3.741.

Length of bottom portion of seventh band = 3.741 * 2 = 7.482. We should be

able to pack 7 squares in the seventh band.

Maximum number of squares that could be packed in semicircle

= 14 + 13 + 13 + 12 + 11 + 9 + 7 = 79

Number of squares that could be packed in circle = 79 * 2 = 158

We have assumed that whatever space we leave(by distributing evenly fractional

part of the length of the chord) before the first square

and the after the last square, is enough to prevent the top left of the

first square and top right of the last square from overshooting the circle

circumference. For the sixth band, the assumption will most likely not hold,

because fractional part is too small. I don’t know how to calculate the minimum

distance needed on either side to keep squares inside the circle.

Sir how do you jump from 10 to 7 and 7 ta 2 either

2*(14+13+12+11+10+7+2)

would you please give the explanation

thnx

c)

The answer is (D) 138. A full explanation is above. Unfortunately, satdharsan, the contest ended a long time ago. It was only for the first person to supply a correct answer, and someone answered months ago. Don’t worry – we will have more Magoosh challenges!

sir pls respond for the answer given waiting for the responseand if possible to get a free account at magoosh so kindly response

the total diameter of the circle is 14.1 cm approximately

on the diameter the number of squares possibly can be placed is 13

followeby 13 and 12,11,10,8

totally 67 for the semi circle and altogether 134 for whole circle

so the near option is C 132

The answer is 2*(14+13+12+11+10+7+2)= 2*69 (69 is the number of squares in half circle) = 138

so the answer is

D

Yes the answer is 138. I’d posted above the answer is (C) 138 but I mixed up the letters. So it should be (D) 138.

Hi Chris,

Yeah, the problem was interesting.

Thanks for the solution reply.

Hi,

A few days back, I posted a solution to this problem, no feedback was replied back against it..rather the solution post had been deleted !!

:O

Thanks,

Durga

Hi Durga,

Sorry about that– I’m not quite sure why your well thought out example was erased. So this is a tricky–nay, diabolical–problem. You were so close. The twist, however, is to maximize the number of small squares you do not have to put them into one large 10×10 square. Instead, see how many little squares you can maximize in each column. For the centermost columns you can fit 14, then, moving outwards, you can fit 13 on each side, then 12, then 11, then 10, then 7, then 2. Add those up and you should get (C) 138.

Way to use the inrentet to help people solve problems!

Given,

A circle of area 50π, inside which to fit maximum possible number of squares(each of side 1).

Radius of circle = sqrt(50).

It is possible to get maximum number of squares into the circle, imagining one large square in the centre with diagonal = diameter of the circle = 2*sqrt(50). & then the remaining 4 arced spaces with more squares.

So, side of the square = sqrt(50 + 50) = 10.

Hence, making it possible for 100 squares(each of side 1) to fit in, symmetrically from the centre.

Now, we have to find out the number of squares that can be accommodated in the remaining arced gaps (on four sides of the square), formed by the sides of the square & the perimeter of the circle.

The maximum additional number of rows of squares that can be placed :

= flr(sqrt(50) – 5)

= flr(7.071067812 – 5)

= 2.

Now, for each additional row, we need to find the maximum no. of squares that can lie :

a- For row at perpendicular dist. of 6 from centre,

= flr(2 * sqrt(50 – 6*6))

= flr(2 * sqrt(14))

= 7

b- For row at perpendicular dist. of 7 from centre,

= flr(2 * sqrt(50 – 7*7))

= flr(2 * sqrt(1))

= 2

Hence, from (a) & (b) cases,

Total additional squares in each curved space:

= 9

Thus, the total number of squares, being :

= 100 + 4*9

= 136 (ANSWER)

But, to my surprise, I can’t find this answer in options, However, I’ll stick to my answer, since I don’t like bluffing. Guess, I’m missing something more intricate, incase I just missed the right answer. 🙂

answer is 157 as the area is50n and n is 3.14 multiplying both we get 157

Ans: 2500 squares

Actually, we can fit more than that. I'm still taking answer/explanations. Remember the first correct answer/explanation will get to use Magoosh for free!

Answer is C

the area of circle is 50*pi; i.e; 50*3.14= 155.70(approx)

area of 1 square :1

hence: D is eliminated as 157 is more than the area of the circle

therefore, the next highest number , i.e. C: 138 fits the bill accurately!

Its (A) 100…. as diameter of circle is : square root of 200 . And diameter of the circle is the diagonal of big square. So big square side becomes 10 and area is 100. If divide this big square of side 10 with a square of side 1 unit , we get 100 small squares .