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Math Challenge 800– Win Free Access to Magoosh GRE!

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

By the way, students who use Magoosh GRE improve their scores by an average of 8 points on the new scale (150 points on the old scale.) Click here to learn more.

20 Responses to Math Challenge 800– Win Free Access to Magoosh GRE!

  1. Praveen April 19, 2012 at 7:35 am #

    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.

    • Chris Lele
      Chris April 19, 2012 at 12:07 pm #

      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 🙂

  2. Ravi September 2, 2011 at 11:24 am #

    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.

  3. Faeze August 25, 2011 at 1:03 pm #

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


    would you please give the explanation


  4. inam August 9, 2011 at 9:06 am #


  5. Chris Lele
    Chris Lele July 29, 2011 at 3:29 pm #

    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!

  6. satdharsan July 29, 2011 at 11:56 am #

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

  7. satdharsan July 29, 2011 at 11:54 am #

    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

  8. Ahmed July 27, 2011 at 2:05 am #

    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


    • Chris Lele
      Chris Lele July 27, 2011 at 10:42 am #

      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.

  9. Durga June 7, 2011 at 1:51 am #

    Hi Chris,

    Yeah, the problem was interesting.
    Thanks for the solution reply.

  10. Durga May 30, 2011 at 1:38 am #


    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 !!



    • Chris Lele
      chris June 1, 2011 at 5:15 pm #

      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.

      • Marlie August 5, 2011 at 8:51 pm #

        Way to use the inrentet to help people solve problems!

  11. Durga Prasana May 28, 2011 at 8:01 am #


    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. 🙂

  12. jaismin May 26, 2011 at 9:11 pm #

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

  13. Yogesh Pate April 10, 2011 at 11:53 am #

    Ans: 2500 squares

  14. Chrislele February 22, 2011 at 8:06 pm #

    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!

    • anirudha May 18, 2011 at 8:34 am #

      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!

  15. Navee February 19, 2011 at 2:27 pm #

    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 .

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