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Learn how Kamakshi achieved a GMAT 675 with an impressive 96th %ile in Data Insights. Discover the unique methods and exam strategies that helped her excel in DI along with other sections for a balanced and high score.
24 Jun 2021, 03:15
Kudos
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
52% (02:25) correct
48%
(02:24)
wrong
based on 79
sessions
History
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Square 1 is inscribed in circle 1 and then another circle 2 is inscribed in square 1. Then another square 2 is inscribed in circle 2 and so on. What is the diameter of the circle 5 if the radius of circle 1 is 50 cm?
(A) 25 cm
(B) 25√2 cm
(C) 50 cm
(D) 50√2 cm
(E) None of these
(A) 25 cm
(B) 25√2 cm
(C) 50 cm
(D) 50√2 cm
(E) None of these
19 Aug 2021, 22:39
Kudos
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Circle 1 radius = 50 cm
Square 1 side = 50*√2 = 50√2 cm
Circle 2 radius = 50√2/2 = 25√2 cm
Square 2 side = 25√2*√2 = 50 cm
Circle 3 radius = 50/2 = 25 cm
Square 3 side = 25*√2 = 25√2 cm
Circle 4 radius = 25√2/2 = 12.5√2 cm
Square 4 side = 12.5√2*√2 = 25 cm
Circle 5 radius = 25/2 = 12.5 cm
Diameter of circle 5 = 2r = 25 cm
Hence, OA is (A).
Square 1 side = 50*√2 = 50√2 cm
Circle 2 radius = 50√2/2 = 25√2 cm
Square 2 side = 25√2*√2 = 50 cm
Circle 3 radius = 50/2 = 25 cm
Square 3 side = 25*√2 = 25√2 cm
Circle 4 radius = 25√2/2 = 12.5√2 cm
Square 4 side = 12.5√2*√2 = 25 cm
Circle 5 radius = 25/2 = 12.5 cm
Diameter of circle 5 = 2r = 25 cm
Hence, OA is (A).
20 Aug 2021, 03:37
Kudos
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Solution:
We have tried to make 2 layers of the said diagram which you can have a look at:
circlelayers.png [ 27.21 KiB | Viewed 2024 times ]
Now, starting from \(1st\) layer:
We know radius of \(C1 = 50cm\). So, diameter of \(C1 = 100cm\)
So, the sides of square \(S1\) will be \(= 50\sqrt2 cm\) (Pythagoras Theorem)
\(2nd\) Layer:
Now this \(50\sqrt2 cm\) will be diameter for C2.
Thus sides of \(S2 = 50cm\)
\(3rd\) Layer:
Now this \(50 cm\) will be diameter for C3.
Thus sides of \(S3 = 25\sqrt2 cm\)
\(4th\) Layer:
Now this \(25\sqrt2 cm\) will be diameter for C4.
Thus sides of \(S4 = 25cm\)
\(5th\) Layer:
Now this \(25cm\) will be diameter for C5.
Hence the right answer is Option A.
We have tried to make 2 layers of the said diagram which you can have a look at:
Attachment:
circlelayers.png [ 27.21 KiB | Viewed 2024 times ]
Now, starting from \(1st\) layer:
We know radius of \(C1 = 50cm\). So, diameter of \(C1 = 100cm\)
So, the sides of square \(S1\) will be \(= 50\sqrt2 cm\) (Pythagoras Theorem)
\(2nd\) Layer:
Now this \(50\sqrt2 cm\) will be diameter for C2.
Thus sides of \(S2 = 50cm\)
\(3rd\) Layer:
Now this \(50 cm\) will be diameter for C3.
Thus sides of \(S3 = 25\sqrt2 cm\)
\(4th\) Layer:
Now this \(25\sqrt2 cm\) will be diameter for C4.
Thus sides of \(S4 = 25cm\)
\(5th\) Layer:
Now this \(25cm\) will be diameter for C5.
Hence the right answer is Option A.
