🎉 Ace XAT 2025 with 2IIM's Crash Course - Enrol now


🎉Effortlessly calculate your score with the 2IIM Score Calculator!🎉

IIFT 2018 Question Paper | Quants

IIFT Previous Year Paper | IIFT Quants Questions | Question 10

The best way to boost your IIFT prep is to practice the actual IIFT Question Papers. 2IIM offers you exactly that, in a student friendly format to take value from this. In the 2018 IIFT, Quants was a mixed bag of questions of varying difficulty, with some routine questions and the others were very demanding. Some beautiful questions that laid emphasis on Learning ideas from basics and being able to comprehend more than remembering gazillion formulae and shortcuts.

Question 10 : Let \S_{1}) be a square of side 4 cm. Circle \C_{1}) circumscribes the square \S_{1}) such that all its corners are on \C_{1}). Another square \S_{2}) circumscribes the circle \C_{1}). Circle \C_{2}) circumscribes the square \S_{2}), and square \S_{3}) circumscribes circle \C_{2}), & so on. If \A_{N}) is the area between the square \S_{N}) and the circle \C_{N}), where N is the natural number. then the ratio of sum of all \A_{N}) to \A_{1}) is

  1. 1
  2. \\frac{\pi}{2}) - 1
  3. Can't be determined
  4. None of the above

🎉 Ace XAT 2025 with 2IIM's Crash Course - Enrol now


Best CAT Online Coaching
Try upto 40 hours for free
Learn from the best!


2IIM : Best Online CAT Coaching.

🎉Effortlessly calculate your score with the 2IIM Score Calculator!🎉


2IIM : Best Online CAT Coaching.


Best CAT Coaching in Chennai


CAT Coaching in Chennai - CAT 2022
Limited Seats Available - Register Now!


Explanatory Answer


Sq circle 1
The side of the square is 4 cm.
When a circle inscribes a square, the ratio of the radius of the circle and the side of the square is 1 : \\sqrt{2})
Therefore the radius of C1 = \\frac{4}{\sqrt{2}}) = \2 \sqrt{2})
sq circle 2
The ratio of the radius of the first circle C1 to second square is 1 : 2
The ratio of the side of the second square S2 to the radius of C2 is \\sqrt{2}: 1),
So, finally the ratio of radius of C1 to C2 will be 1 : \\sqrt{2})
Similarly the ratio of radii of C2 and C3 will also be 1 : \\sqrt{2})
So, the radii of the circles follows a geometric progression with common ratio \\sqrt{2})
Radius of C1 = \2 \sqrt{2})
Radius of C2 = 4
Radius of C3 = 4\\sqrt{2})
So radius of Cn = 2\\times \sqrt{2}^{n})
square in a circle 3
The ratio of the side of the Square S1 to the radius of the Circle C1 = \\sqrt{2}: 1)
The ratio of radius of the circle C1 to the side of the square S2 = 2 : 1.
So the ratio of the slides of S1 and S2 is 1 : \\sqrt{2})
Similarly the ratio of radii of S2 and S3 will also be 1 : \\sqrt{2})
So, the squares follow a geometric progression with common ratio \\sqrt{2})
Side of S1 = 4
Side of S2 = 4\\sqrt{2})
Side of S3 = 8
So Side of Sn = 4\\times \sqrt{2}^{n-1})
Now, the area between Cn and Sn = \A_{n})= (\\Pi \times r^{2})) - \a^{2})
= (\\Pi \times\left(2 \times \sqrt{2}^{n}\right)^{2})) - \\left(4 \times \sqrt{2}^{n-1}\right)^{2})
= \\left(\Pi \times 4 \times 2^{n}\right)-\left(16 \times 2^{n-1}\right))
= \\left(\Pi \times 4 \times 2^{n}\right)-\left(8 \times 2^{n}\right))
= \(\Pi-2)\left(4 \times 2^{n}\right))
\A_{n})= \(\Pi-2)\left(4 \times 2^{n}\right))
Area between S1 and C1 = \A_{1})= \(\Pi-2)\left(4 \times 2^{1}\right)) = \8(\Pi-2))
Sum of such n areas = \\sum 4(\Pi-2)\left(2^{n}\right)) = \4(\Pi-2) \times \Sigma\left(2^{n}\right)) = \8(\Pi-2)\left(2^{n}-1\right))
The question asks us the ratio between the Sum of all the intermediate areas to that of the a-rea between S1 and C1, which is
\\frac{8(\Pi-2)\left(2^{n}-1\right)}{8(\Pi-2)}) = \\left(2^{n}-1\right))


The answer is 'None of the above'

Choice D is the correct answer.

CAT Questions | CAT Quantitative Aptitude

CAT Questions | Verbal Ability for CAT


Where is 2IIM located?

2IIM Online CAT Coaching
A Fermat Education Initiative,
58/16, Indira Gandhi Street,
Kaveri Rangan Nagar, Saligramam, Chennai 600 093

How to reach 2IIM?

Mobile: (91) 99626 48484 / 94459 38484
WhatsApp: WhatsApp Now
Email: info@2iim.com