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# What is the area of the rectangular region above?

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What is the area of the rectangular region above? [#permalink]

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24 Feb 2011, 11:39
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What is the area of the rectangular region above?

(1) l + w = 6. Not sufficient to get the value of .
(2) d^2 = 20
[Reveal] Spoiler: OA

Last edited by Bunuel on 01 Feb 2012, 14:10, edited 1 time in total.
Edited the question
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Re: OG Quant review book question [#permalink]

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24 Feb 2011, 11:46
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(1)
$$l+w = 6$$
$$(l+w)^2 = l^2+w^2+2w*l=36$$
$$w*l=\frac{36-l^2-w^2}{2}$$
Not Sufficient.

(2)
$$d^2=l^2+w^2=20$$
Not Sufficient.

Combining both;
$$w*l=\frac{36-(l^2+w^2)}{2}$$
$$w*l=\frac{36-20}{2}$$
$$w*l=\frac{16}{2}$$
$$w*l=8$$(Area is w*l)

Sufficient.

Ans: "C"
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Re: OG Quant review book question [#permalink]

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24 Feb 2011, 11:49
Thanks!
Actually, can we not use the pythagorean theorem and 45-45-90 triangle rule to get l and w using (2)?
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Re: OG Quant review book question [#permalink]

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24 Feb 2011, 12:10
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heygirl wrote:
What is the area of a rectangle with length l and width w?
(1)l+w=6
(2)d^2=20(d is the diagonal)

Make sure you type the question in exactly as it was stated from the source. Yuo should not reword or/and shorten the questions.

Original question:

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What is the area of the rectangular region above?

$$area=lw=?$$

(1) l + w = 6. Not sufficient to get the value of $$lw$$.
(2) d^2 = 20 --> $$l^2 +w^2 = 20$$. Not sufficient to get the value of $$lw$$.

(1)+(2) Square (1): $$l^2+2lw+w^2=36$$, as from (2) $$l^2 +w^2 = 20$$ then $$2lw+20=36$$ --> $$lw=8$$. Sufficient.

heygirl wrote:
Thanks!
Actually, can we not use the pythagorean theorem and 45-45-90 triangle rule to get l and w using (2)?

Usually the diagonal does not divide a rectangle into two 45-45-90 triangles (it'll be correct only for squares, so when l=w).

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Re: OG Quant review book question [#permalink]

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24 Feb 2011, 12:12
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Thanks for letting me know Bunuel. I shall follow the rules henceforth!!
I actually thought b could be the right answer here. I made a wrong assumption : diag of a rect make 45 degrees!
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Re: OG Quant review book question [#permalink]

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24 Feb 2011, 17:37
Both Bunuel and fluke approaches were great. Thanks for that!
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09 Mar 2011, 00:19
Lolaergasheva wrote:
What is the area of the rectangular region ?
(1) l + w = 6
(2) d^2 = 20

Area of rectangular region is given by $$l*w$$.

Statement 1 gives us,$$l+w=6$$, but l and w can take any values, so insufficient

Statement 2 gives us $$d^2 = 20$$ . Assuming that d refers to length of diagonal, we have $$d^2 = l^2 + w^2 = 20$$. Again l and w can take multiple values, so insufficient.

Combining 1 and 2,

we get $$l+w=6$$ ... (1)

and $$l^2 + w^2 = 20$$.... (2)

Squaring both sides of (1), we get

so, $$l^2 + w^2 + 2l*w = 36$$

Putting $$l^2 + w^2 = 20$$ in above we get $$2l*w = 16$$ or $$l*w = 8$$

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Re: What is the area of the rectangular region above? [#permalink]

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What is the area of the rectangular region above? [#permalink]

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07 Sep 2015, 07:28
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Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and equations ensures a solution.

What is the area of the rectangular region above?

(1) l + w = 6. Not sufficient to get the value of .
(2) d^2 = 20

In the original condition we have 2 variables for the rectangle(width, length) and thus we need 2 variables to match the number of variables and equations. Since there is 1 each in 1) and 2), C is likely the answer.
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What is the area of the rectangular region above?   [#permalink] 07 Sep 2015, 07:28
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