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The sides of rectangle A are each multiplied by x to form rectangle B

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The sides of rectangle A are each multiplied by x to form rectangle B  [#permalink]

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New post 15 Jan 2018, 21:52
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The sides of rectangle A are each multiplied by x to form rectangle B and by y to form rectangle C. When multiplied by x, the area of rectangle A equals 10, and when multiplied by y, the area of rectangle A equals 5. If the difference in area between rectangle B and C is 300, what is x−y?

A. 5
B. 20
C. 30
D. 50
E. 60

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Re: The sides of rectangle A are each multiplied by x to form rectangle B  [#permalink]

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New post 15 Jan 2018, 22:01
B

Let area of rectangle A be lb.

Hence area of rectangle B is lb*x^2 and rectangle C is lb*y^2.
Also lb*x=10 and lb*y=5.
x=10/lb and y=5/lb.

Also lb(x^2-y^2)=300
Solving we get lb=1/4 and x=40 y=20
Hence x-y=20


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Re: The sides of rectangle A are each multiplied by x to form rectangle B  [#permalink]

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New post 15 Jan 2018, 22:44
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Bunuel wrote:
The sides of rectangle A are each multiplied by x to form rectangle B and by y to form rectangle C. When multiplied by x, the area of rectangle A equals 10, and when multiplied by y, the area of rectangle A equals 5. If the difference in area between rectangle B and C is 300, what is x−y?

A. 5
B. 20
C. 30
D. 50
E. 60


Length = \(l\), Width = \(w\).
Area of rectangle A = \(lw\).
Area of rectangle B = \(lx*wx = (x^2)lw\).
Area of rectangle C = \(ly*wy = (y^2)lw\).

\(lwx = 10, lwy = 5\).
\(x = \frac{10}{(lw)}, y = \frac{5}{(lw)}\)
\(lw(x-y) = 10-5 = 5\).

\(lw(x^2-y^2) = 300\)
\(lw(x+y)(x-y) = 300\)
\(5(x+y) = 300\)
\((x+y) = 60\).

\(\frac{10}{(lw)} + \frac{5}{(lw)} = 60\)
\(15lw = 60(lw)^2\)
\(lw = \frac{15}{60} = \frac{1}{4}.\)

\((x-y) = 5*4 = 20\).
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Re: The sides of rectangle A are each multiplied by x to form rectangle B  [#permalink]

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New post 15 Jan 2018, 23:11
1
2
Assumption : Let area of rectangle A be a.b where a and b are sides of rectangle

Given : Area of B = x^2(a.b) , Area of C = y^2(a.b)
x*(a.b) = 10 ------------------ 1
y*(a.b)=5 ------------------ 2
ab(x^2-y^2) = 300 ------------------ 3

Required : x-y = ?

Solution : Add eq 2 and eq 1, we get a.b(x+y) = 15 ------------------ 4
Divide eq 3 with eq 4, we get x-y=20

Hence, B is the answer.
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Re: The sides of rectangle A are each multiplied by x to form rectangle B  [#permalink]

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New post 10 Aug 2019, 10:47
1
If a is the area of rect A,
then xa = 10
ya = 5
=> x = 2y

Now (x^2)*a - (y^2)*a = 300
4(y^2)a - (y^2)*a = 300
(y^2)*a = 100
a y y =100
5y = 100
y =20
And hence x = 40.
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Re: The sides of rectangle A are each multiplied by x to form rectangle B  [#permalink]

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New post 12 Aug 2019, 18:14
So we have three rectangles:
Let, Rectangle A = a * b
then, Rectangle B = ax * bx = \(x^2 * a * b\)
then, Rectangle C = ay * by = \(y^2 * a * b\)

Given,
When multiplied by x, the area of rectangle A equals 10 = x * a * b = 10 -> (a)
when multiplied by y, the area of rectangle A equals 5 = y * a * b = 5 -> (b)

The difference in the area between rectangle B and C is 300 = \(x^2*a*b - y^2*a*b = 300\) => \((x^2 - y^2) (a * b) = 300\) => \((x + y)(x - y)(a * b) = 300\) ->(c)

Adding (a) + (b)
(x + y)(a * b) = 15 -> (d)

Diving (c) by (d)
\(\frac{(x + y)(x - y)(a * b)}{(x + y)(a * b)}\) = \(\frac{300}{15}\)
=> \((x - y) = 20\)

Answer B
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Re: The sides of rectangle A are each multiplied by x to form rectangle B   [#permalink] 12 Aug 2019, 18:14
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