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by alltimeacheiver » Tue Apr 05, 2011 9:48 pm
If circle A has 3 times the area of circle B, and circle B has one-sixth of the area of square WXYZ, what is the ratio of the area of a circle inscribed within WXYZ to that of the area of circle A?


a pie/12

b pie/6

c pie/2

d 2/pie

e squareroot of 12pie

ques 2 If 125% of j is equal to 25% of k, 150% of k is equal to 50% of l, and 175% of l is equal to 75% of m, then 20% of m is equal to what percent of 200% of j ?
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by Anurag@Gurome » Tue Apr 05, 2011 10:02 pm
alltimeacheiver wrote:If circle A has 3 times the area of circle B, and circle B has one-sixth of the area of square WXYZ, what is the ratio of the area of a circle inscribed within WXYZ to that of the area of circle A?


a pie/12

b pie/6

c pie/2

d 2/pie

e squareroot of 12pie
Let area of circle A = a
Area of circle B = b
Area of square WXYZ = x and let 's' is the side of square. The side 's' of the square will be the diameter of the circle inscribed in square WXYZ. So, radius of inscribed circle = s/2.
Then area of inscribed circle = (pi)(s/2)² = (pi)(s²/4)
Also, a = 3b and b = x/6
Now, a = 3b = 3*x/6 = x/2 implies a = x/2
Also, x = s² or s = √x
So, area of inscribed circle: area of circle A = (pi)x/4 : x/2 = (pi)/2

The correct answer is C.
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by manpsingh87 » Tue Apr 05, 2011 10:03 pm
alltimeacheiver wrote:If circle A has 3 times the area of circle B, and circle B has one-sixth of the area of square WXYZ, what is the ratio of the area of a circle inscribed within WXYZ to that of the area of circle A?


a pie/12

b pie/6

c pie/2

d 2/pie

e squareroot of 12pie
circle A area= 3pir^2; where r is the radius of circle b;......1
also pir^2=a^2/6 where a is the side of the square. ..........2)
and the radius of the circle inscribed in the square = a/2, therefore its area= pi*a^2/4;
subsituting value of a^2 from 2 we have..
=((pi)^2*6*r^2)/4;..........3)
therefore required ratio is obtained by dividing 3 by 1;
=pi/2;
hence C
Last edited by manpsingh87 on Tue Apr 05, 2011 10:13 pm, edited 1 time in total.
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by Anurag@Gurome » Tue Apr 05, 2011 10:11 pm
alltimeacheiver wrote: ques 2 If 125% of j is equal to 25% of k, 150% of k is equal to 50% of l, and 175% of l is equal to 75% of m, then 20% of m is equal to what percent of 200% of j ?
1.25j = 0.25k implies 5j = k
1.5k = l/2 implies 3k = l
1.75l = 0.75m implies 7l = 3m
Let 20% of m is equal to N% of 200% of j
0.2m = 0.0N * 2j implies N = 10m/j = 10(7l/3)/(k/5) = 10*(7l/3)*(5/l/3) = 350

Hence N = [spoiler]350%[/spoiler]
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by force5 » Wed Apr 06, 2011 12:23 am
1.25j = 0.25k implies 5j = k
1.5k = l/2 implies 3k = l
1.75l = 0.75m implies 7l = 3m
Let 20% of m is equal to N% of 200% of j
0.2m = 0.0N * 2j implies N = 10m/j = 10(7l/3)/(k/5) = 10*(7l/3)*(5/l/3) = 350
Good one..
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