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In the above figure, ABCD is a rectangle with a circle inscribed insid
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Updated on: 29 Jul 2019, 09:19
Question Stats:
38% (02:42) correct 62% (01:58) wrong based on 34 sessions
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In the above figure, ABCD is a rectangle with a circle inscribed inside it. There is another rectangle with an inscribed circle inside it. The colored regions are marked in the figure above. if a dart is fired towards the block, and it hits the block in colored region, what is the probability that the dart hits the region which is redcolored? A) 1/2 B) 1/3 C) 1/4 D) 1/5 E) 1/6 GMATbuster's Collection Attachment:
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Originally posted by gmatbusters on 29 Jul 2019, 07:05.
Last edited by gmatbusters on 29 Jul 2019, 09:19, edited 1 time in total.



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Re: In the above figure, ABCD is a rectangle with a circle inscribed insid
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29 Jul 2019, 09:06
The GMAT won't test you on technicalities around the definition of "inscribed"  unless you know precisely how that word is defined, you wouldn't know if it's possible to inscribe a nonsquare rectangle in a circle (touching at just two points). You can't actually do that, because of the definition of "inscribed", so the two rectangles here must be squares, but if this were a real GMAT question, it would just tell you both of the quadrilaterals are squares. It's a pure ratio problem, so we can invent a number  say the edge of the big square is 4. Then the area of that square is 16, and the area of the big circle, which has a radius of 2, is 4π. Subtracting the circle from the square, we get the total area of the four corners (one of which is red), so that total area is 16  4π, and the area of one corner is 1/4 of that, or 4  π. So that's the redshaded area. The diameter 4 of the big circle is the diagonal of the small square. Since the diagonal of a square is √2 times an edge of that square, each edge of the small square is 4/√2 = 2√2. So the area of the small square is (2√2)^2 = 8. The small circle has a radius half the length of a side of the small square, so a radius of √2 .Its area is thus 2π. We get the total yellow area by subtracting the area of the small circle from the area of the small square, so the yellow area is 8  2π. Notice now that the yellow area is exactly twice as big as the red area, so 1/3 of the coloured area is red, so that's the probability, if we pick a random point from the coloured regions, that the point is in the red region.
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Re: In the above figure, ABCD is a rectangle with a circle inscribed insid
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29 Jul 2019, 09:30
gmatbusters wrote: The rectangle here indeed is a Square, but it is not mentioned explicitly to increase the level of question a bit.
I understood that, but that's not how they increase the difficulty level of real GMAT questions.
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Re: In the above figure, ABCD is a rectangle with a circle inscribed insid
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29 Jul 2019, 09:14
The rectangle here indeed is a Square, but it is not mentioned explicitly to increase the level of question a bit. IanStewart wrote: The GMAT won't test you on technicalities around the definition of "inscribed"  unless you know precisely how that word is defined, you wouldn't know if it's possible to inscribe a nonsquare rectangle in a circle (touching at just two points). You can't actually do that, because of the definition of "inscribed", so the two rectangles here must be squares, but if this were a real GMAT question, it would just tell you both of the quadrilaterals are squares.
It's a pure ratio problem, so we can invent a number  say the edge of the big square is 4. Then the area of that square is 16, and the area of the big circle, which has a radius of 2, is 4π. Subtracting the circle from the square, we get the total area of the four corners (one of which is red), so that total area is 16  4π, and the area of one corner is 1/4 of that, or 4  π. So that's the redshaded area.
The diameter 4 of the big circle is the diagonal of the small square. Since the diagonal of a square is √2 times an edge of that square, each edge of the small square is 4/√2 = 2√2. So the area of the small square is (2√2)^2 = 8. The small circle has a radius half the length of a side of the small square, so a radius of √2 .Its area is thus 2π. We get the total yellow area by subtracting the area of the small circle from the area of the small square, so the yellow area is 8  2π.
Notice now that the yellow area is exactly twice as big as the red area, so 1/3 of the coloured area is red, so that's the probability, if we pick a random point from the coloured regions, that the point is in the red region.
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Re: In the above figure, ABCD is a rectangle with a circle inscribed insid
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30 Jul 2019, 04:57
Lets assume first of all that this is a square bcuz the circle inscribed touches every side. Let side be 4
Area of bigger square = 4*4 = 12
Area of bigger circle = 22/7*2*2 = 88/7 Radius is half of the side of sq
Area of the red portion = (1288/7)/4 = 24/28
Side of inner sq = 4/√2
Area of inner sq = 8
Area of inner circle = 22/7*2/√2*2/√2 = 44/7
Area of yellow region = 844/7 = 12/7
Probability = favourable outcome / all outcomes
= Area of red region / area of all colored region = (24/28)/(12/7+24/28) = 1/3
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In the above figure, ABCD is a rectangle with a circle inscribed insid
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03 Aug 2019, 19:43
All we need to find is Area \(\frac{(Colored Red)}{Area (Colored Red) + Area (Colored Yellow)}\) > (a)We need to only find the ratios so we can assume any size for the given figure. Since a circle is inscribed in the rectangle, the rectangle has to be a square. Consider Figure A:
Let, Side of smaller Square RQ = 2 Therefore, Radius of smaller circle OT = 1 Radius of bigger circle OS = \(\sqrt{2}\) Side of bigger square BD = \(2\sqrt{2}\) Finding Area of Yellow Regions:Consider OHSJ: Area of Yellow Portion in OSHJ = Area (OSHJ) – Area of Sector (OHJ) = (\(1 * 1) – (\frac{1}{4} * \pi * 1^2) = \frac{4 – \pi}{4}\) Now there are four such yellow regions. So total area of Yellow shaded region = \(\frac{4 – \pi}{4} * 4 = 4 – \pi\)> (b) Finding Area of Red Region:Consider OIAK: Area of Red Portion on OIAK = Area (OIAK) – Area of Sector (OIK) = \((\sqrt{2} * \sqrt{2}) – (\frac{1}{4} * \pi * (\sqrt{2})^2) = \frac{4 – \pi}{2}\) > (c)Hence From (a), the required probability = \(\frac{(c)}{(b) + (c)} = \frac{4 – \pi}{\frac{4 – \pi}{2} + 4 – \pi} = \frac{1}{3}\) Answer B
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In the above figure, ABCD is a rectangle with a circle inscribed insid
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03 Aug 2019, 19:43






