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# The length of rectangle B is 10 percent less than the length

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The length of rectangle B is 10 percent less than the length [#permalink]  15 Feb 2017, 08:41
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Question Stats:

69% (00:44) correct 30% (01:20) wrong based on 144 sessions

The length of rectangle B is 10 percent less than the length of rectangle A, and the width of rectangle B is 10 percent greater than the width of rectangle A.

 Quantity A Quantity B The area of rectangle A The area of rectangle B

A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
[Reveal] Spoiler: OA

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Re: The length of rectangle B is 10 percent less than the length [#permalink]  21 Feb 2017, 16:56
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Explanation

Lets assume length of rectangle A is x and width of rectangle A is y.

Clearly area of rectangle A = $$x \times y$$

Now, length of rectangle B is 10 percent less than the length of rectangle A

length of rectangle B = 0.9 x

width of rectangle B is 10 percent greater than the width of rectangle A

width of rectangle B = 1.1 y

Clearly area of rectangle B $$= 0.9 x \times 1.1 y = 0.99 \times x \times y$$.

Irrespective of values of x and y (which can only be positive real numbers), Quantity A is always greater.

Hence option A is correct.
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Re: The length of rectangle B is 10 percent less than the length [#permalink]  28 Apr 2019, 21:12
but if we assume area of B is xy then what happened?
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Re: The length of rectangle B is 10 percent less than the length [#permalink]  29 Apr 2019, 05:24
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Not sure what did you mean .........

regards
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Re: The length of rectangle B is 10 percent less than the length [#permalink]  22 Jan 2020, 06:28
how much time we should take to solve these kind of questions
i took 45 secs to solved it correctly
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Re: The length of rectangle B is 10 percent less than the length [#permalink]  29 Jan 2020, 07:02
In a rectangle length is greater than width

10%(width of A) is less than 10%(length of A)

So, even though width of B is 10% more than A since its length is 10% less

A still has a greater area
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Re: The length of rectangle B is 10 percent less than the length [#permalink]  29 Jan 2020, 11:49
Carcass wrote:

The length of rectangle B is 10 percent less than the length of rectangle A, and the width of rectangle B is 10 percent greater than the width of rectangle A.

 Quantity A Quantity B The area of rectangle A The area of rectangle B

A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.

We know:

Length of rectangle B = 90% of the length of rectangle A
$$=> L(B) = 0.9 * L(A)$$ ... (i)

Width of rectangle B = 110% of the width of rectangle A
$$=> W(B) = 1.1 * W(A)$$ ... (ii)

Multiplying (i) and (ii):

$$L(B) * W(B) = 0.9 * 1.1 * L(A) * W(A)$$

$$=> Area(B) = 0.99 * Area(A)$$

Thus, Area of rectangle A is greater

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Re: The length of rectangle B is 10 percent less than the length   [#permalink] 29 Jan 2020, 11:49
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