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# In the rectangle shown, if d = 10, what is x? (1) The perimeter of

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Joined: 02 Sep 2009
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In the rectangle shown, if d = 10, what is x? (1) The perimeter of  [#permalink]

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18 Sep 2018, 21:04
00:00

Difficulty:

95% (hard)

Question Stats:

39% (02:12) correct 61% (01:49) wrong based on 59 sessions

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In the rectangle shown, if d = 10, what is x?

(1) The perimeter of the rectangle is 28.
(2) The area of the rectangle is 48.

Attachment:

image006.jpg [ 4.76 KiB | Viewed 489 times ]

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Re: In the rectangle shown, if d = 10, what is x? (1) The perimeter of  [#permalink]

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18 Sep 2018, 21:24
Bunuel wrote:

In the rectangle shown, if d = 10, what is x?

(1) The perimeter of the rectangle is 28.
(2) The area of the rectangle is 48.

Attachment:
image006.jpg

St1:- Perimeter of the rectangle=2(x+w)=28 or, x+w=14.....(1)

We have two symmetrical right angled triangles about the diagonal of the rectangle.

So , $$x^2+w^2=10^2$$---(2)
From (1) & (2), we have x=6 or 8.
No unique value of x.
Therefore insufficient.

St2:- Area of rectangle =x*w=48---(3)
From (1)&(3), we have
$$(x+w)^2=x^2+w^2+2xw$$=100+96=196
Or, x+w=14.....(4)
From (2) & (4), we have x=6 or 8
No unique value of x.
Insufficient.

Combined, no new information to restrict the value of x.
Hence insufficient.

Ans. (E)
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Joined: 04 Jan 2015
Posts: 2466
Re: In the rectangle shown, if d = 10, what is x? (1) The perimeter of  [#permalink]

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19 Sep 2018, 01:29

Solution

Given:
• The diagonal of a rectangle, d = 10
• x, w are the length and breadth of the rectangle

To find:
• The value of x

Approach and Working:
• We can write, $$d^2 = w^2 + x^2$$
• Implies, $$x^2 + w^2 = 10^2 = 100$$ …………. (1)

With this understanding, let’s analyse the given statements

Analysing Statement 1
“Perimeter of the rectangle = 28”
• Implies, 2(x + w) = 28
o x + w = 14 => w = 14 – x ……. (2)
• Substituting (2) in (1), we get,
o $$x^2 + (14 - x)^2 = 100$$
o $$2x^2 – 28x + 196 = 100$$
o $$x^2 – 14x + 48 = 0$$
o By factorization, we get, (x – 6) * (x – 8) =0
• Thus, x = 6 or 8

Therefore, Statement 1 is NOT sufficient to arrive at a unique answer

Analysing Statement 2
“Area of the rectangle = 48”
• Implies, x*w = 48
o $$w = \frac{48}{x}$$ ……... (3)
• Substituting (3) in (1), we get,
o $$x^2 + (\frac{48}{x})^2 = 100$$
o $$(x + \frac{48}{x})^2 – 2*x*\frac{48}{x} = 100$$
o $$(x + \frac{48}{x})^2 = 196$$
o Implies, $$(x + \frac{48}{x}) = 14$$
o $$x^2 - 14x + 48 = 0$$
• Thus, x = 6 or 8

Therefore, Statement 2 is NOT sufficient to arrive at a unique answer

Combining Both Statements
• From Statement 1, we got, x = 6 or 8
• From Statement 2, we got, x = 6 or 8
• Combining both, we get, x = 6 or 8

Therefore, both statements TOGETHER are NOT sufficient to arrive at a unique answer

Hence, the correct answer is option E.

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In the rectangle shown, if d = 10, what is x? (1) The perimeter of  [#permalink]

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19 Sep 2018, 01:56
Hi,

Its a 30-sec question, if we know the common pythogorean triplets asked in GMAT

Question: What is x ?

Given, it is a rectangle, so each vertex angle is 90 degree and also it is given that diagonal d = 10.

Since the questions is a Value DS question, we no need to do the algebra here, we just need to know the common pythogorean triplets.

6 – 8 – 10 is a very common pythogorean triplet, similarly 3 – 4 -5 , 12 – 5 -13 are common pythogorean triplets asked in GMAT.

So here,

x can be 6 and w can be 8 or

x can be 8 and w can be 6.

So there are two values of x.

Statement I is insufficient:

The perimeter of the rectangle is 28.

i.e., 2*(x+w) = 28

x+w = 14

Same reasoning,

x can be 6 and w can be 8 or

x can be 8 and w can be 6.

So not sufficient.

Statement II is insufficient:

The area of the rectangle is 48

i.e., x * w = 48

Same reasoning here too,

x can be 6 and w can be 8 or

x can be 8 and w can be 6.

Together also, same thing x can be 6 or 8. So not sufficient.

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Joined: 12 Sep 2018
Posts: 15
Re: In the rectangle shown, if d = 10, what is x? (1) The perimeter of  [#permalink]

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21 Sep 2018, 03:51
So we can't just assume x being the highest value of two because in the figure it's length seems to be almost double?
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Re: In the rectangle shown, if d = 10, what is x? (1) The perimeter of  [#permalink]

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21 Sep 2018, 04:06
1
artiom01 wrote:
So we can't just assume x being the highest value of two because in the figure it's length seems to be almost double?

Hi artiom01,

We can't assume anything in GMAT with naked eye unless and until it is specifically mentioned.
Problem Solving
Figures: All figures accompanying problem solving questions are intended to provide information useful in solving the problems. Figures are drawn as accurately as possible. Exceptions will be clearly noted. Lines shown as straight are straight, and lines that appear jagged are also straight. The positions of points, angles, regions, etc., exist in the order shown, and angle measures are greater than zero. All figures lie in a plane unless otherwise indicated.

Data Sufficiency:
Figures:
• Figures conform to the information given in the question, but will not necessarily conform to the additional information given in statements (1) and (2).
• Lines shown as straight are straight, and lines that appear jagged are also straight.
• The positions of points, angles, regions, etc., exist in the order shown, and angle measures are greater than zero.
• All figures lie in a plane unless otherwise indicated.

You may go thru below link for more details and drills:-
https://gmatclub.com/forum/gmat-trick-r ... 01412.html
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PKN

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Joined: 21 Jun 2017
Posts: 212
Concentration: Finance, Economics
WE: Corporate Finance (Commercial Banking)
Re: In the rectangle shown, if d = 10, what is x? (1) The perimeter of  [#permalink]

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21 Sep 2018, 08:41
I got E by

Statement I : x^2 +w^2 =100 and after certain applications we get w*x=48. Now, w and x can be any irrational squares whose sum can add up to be 100, since it has not been mentioned whether x and y are integers.

Statement II : Like above.

Experts please guide. Am i correct with this approach ?
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Re: In the rectangle shown, if d = 10, what is x? (1) The perimeter of &nbs [#permalink] 21 Sep 2018, 08:41
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