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A and B form 1/2 of the square, while C and D form the other 1/2 of th

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Bunuel
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A and B form 1/2 of the square, while C and D form the other 1/2 of th [#permalink] New post   17 Sep 2021, 01:30
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C
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E

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A and B form 1/2 of the square, while C and D form the other 1/2 of the square. The area of A is 1/2 the area of B, while the area of C is 2/5 the area of D. What fraction of the square is the area of C?

A. 1/9
B. 1/8
C. 1/7
D. 1/6
E. 1/5



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C
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Re: A and B form 1/2 of the square, while C and D form the other 1/2 of th [#permalink] New post   17 Sep 2021, 03:07
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Let area of the entire square = 420
ATQ,
A+B =210
C+D =210

Also given,
C = 2/5 *D
D =150 and C = 60

(Area of C)/(Total area of square) = 60/ 210 = 1/7

IMO C
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Re: A and B form 1/2 of the square, while C and D form the other 1/2 of th [#permalink] New post   17 Sep 2021, 03:48
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2(C+D) = Total Area

If C = (2/5)D then D = 5C/2

Substituting:

2(C + 5C/2) = Total Area
7C = Total Area

You need 7 times the area of C to cover the total area. This means that C is 1/7 of the Total Area.

ANSWER C
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Re: A and B form 1/2 of the square, while C and D form the other 1/2 of th [#permalink] New post   17 Sep 2021, 08:18
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A and B form 1/2 of the square, while C and D form the other 1/2 of the square. The area of A is 1/2 the area of B, while the area of C is 2/5 the area of D. What fraction of the square is the area of C?

We have:
\(C = \frac{2}{5} D\)
=> \(\frac{C}{D}\) = \(\frac{2}{5}\)
<=> \(\frac{C}{(D+C)}\) = \(\frac{2}{(5+2)}\)
<=> \(\frac{C}{((1/2)ABCD)}\) = \(\frac{2}{7}\)
<=> \(\frac{C}{ABCD}\) = \(\frac{2}{7} * \frac{1}{2}\)
<=> \(\frac{C}{ABCD}\) = \(\frac{ 1}{7}\)

IMO: C
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Re: A and B form 1/2 of the square, while C and D form the other 1/2 of th [#permalink] New post   17 Sep 2021, 09:57
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let total area of square = 100
so AB is 50 and CD is 50
given C = 2/5 of D
x+2x/5 = 50
x= 250/7
C= 250/7 * 2/5 ; 100/7
fraction of the square is the area of C
100/100/7 ; 1/7
option C


Bunuel wrote:

A and B form 1/2 of the square, while C and D form the other 1/2 of the square. The area of A is 1/2 the area of B, while the area of C is 2/5 the area of D. What fraction of the square is the area of C?

A. 1/9
B. 1/8
C. 1/7
D. 1/6
E. 1/5


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Re: A and B form 1/2 of the square, while C and D form the other 1/2 of th [#permalink] New post   17 Sep 2021, 10:41
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An easy one I would say

C is 2/5 of D
So C = 2/5D
Now we know (C + D) = 1/2 area of square = A / 2
Therefore D + 2D/5 = A / 2
Therefore D = 5A / 14
Hence C = D * 2 / 5 = A * 5/14 * 2/5 = A / 7

So answer is C

Bunuel wrote:

A and B form 1/2 of the square, while C and D form the other 1/2 of the square. The area of A is 1/2 the area of B, while the area of C is 2/5 the area of D. What fraction of the square is the area of C?

A. 1/9
B. 1/8
C. 1/7
D. 1/6
E. 1/5


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Re: A and B form 1/2 of the square, while C and D form the other 1/2 of th [#permalink] New post   18 Sep 2021, 00:20
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We need, Area of c / area of square = \((\frac{2}{5} *D)/(A+B+C+D) \)= \((\frac{2}{5} *D)/(C+D+C+D)\)

= \((\frac{1}{5} *D)/(\frac{2}{5} *D + D)\) = \(\frac{1}{7}\).

So, I think C. :)
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Re: A and B form 1/2 of the square, while C and D form the other 1/2 of th [#permalink]
18 Sep 2021, 00:20
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