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Math Expert V
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In the figure shown above, line segment QR has length 12, an  [#permalink]

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Difficulty:   45% (medium)

Question Stats: 74% (03:09) correct 26% (03:08) wrong based on 889 sessions

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The Official Guide For GMAT® Quantitative Review, 2ND Edition In the figure shown above, line segment QR has length 12, and rectangle MPQT is a square. If the area of rectangular region MPRS is 540, what is the area of rectangular region TQRS?

(A) 144
(B) 216
(C) 324
(D) 360
(E) 396

Problem Solving
Question: 135
Category: Geometry; Algebra Area; Second-degree equations
Page: 79
Difficulty: 600

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Math Expert V
Joined: 02 Sep 2009
Posts: 59674
Re: In the figure shown above, line segment QR has length 12, an  [#permalink]

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SOLUTION In the figure shown above, line segment QR has length 12, and rectangle MPQT is a square. If the area of rectangular region MPRS is 540, what is the area of rectangular region TQRS?

(A) 144
(B) 216
(C) 324
(D) 360
(E) 396

The area of MPRS = the area of MPQT + the area of TQRS.

540 = x^2 + 12x --> x = 18.

The area of TQRS = 12*18 = 216.

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Re: In the figure shown above, line segment QR has length 12, an  [#permalink]

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22
3
Ans B
Since the area of the total Fig is given as 540.
We can plug ans choices as follows
A) 540-144=396( not a perfect square, should be a perfect square because fig MPQT is given as a square)
B)540-216=324 (perfect square, thus sides are 18)
C)540-324= not a perfect square
D)540-360= not a perfect square
E)540-396=144 ( perfect square, but if both are 12 then the total area will come out to be 144*2 and not 540)

Just another simpler approach .
Let me know if i missed something ##### General Discussion
Director  Joined: 25 Apr 2012
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Re: In the figure shown above, line segment QR has length 12, an  [#permalink]

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1 In the figure shown above, line segment QR has length 12, and rectangle MPQT is a square. If the area of rectangular region MPRS is 540, what is the area of rectangular region TQRS?

(A) 144
(B) 216
(C) 324
(D) 360
(E) 396

Let the side of square MPQT be "X"

Then area of rectangular region MRPS will be (12+X)*X= 540

Solving for Quadratic Eqn, we get X^2+12X-540=0

or X= {-12 +/- \sqrt{$$(12)^2-4*540*1$$} }/ 2

or X = (-12 +/- 48)/2 or X=18 or -30 because X cannot be negative

So Area of rectangular region TQRS= 18*12 = 216

Ans is B

Difficulty level 600 is okay.
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Re: In the figure shown above, line segment QR has length 12, an  [#permalink]

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1
Refer fig below:

$$x^2 + 12x = 540$$

$$x^2 + 12x - 540 = 0$$

x = 18 (Ignore -ve value)

Area TQRS = 18 x 12 = 216 = Answer = B
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Posts: 35
Re: In the figure shown above, line segment QR has length 12, an  [#permalink]

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manusingh3 wrote:
Ans B
Since the area of the total Fig is given as 540.
We can plug ans choices as follows
A) 540-144=396( not a perfect square, should be a perfect square because fig MPQT is given as a square)
B)540-216=324 (perfect square, thus sides are 18)
C)540-324= not a perfect square
D)540-360= not a perfect square
E)540-396=144 ( perfect square, but if both are 12 then the total area will come out to be 144*2 and not 540)

Just another simpler approach .
Let me know if i missed something How did you know 324 was a perfect square? you skipped that part Manager  Joined: 20 Dec 2013
Posts: 222
Location: India
Re: In the figure shown above, line segment QR has length 12, an  [#permalink]

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Option B.
Let each side of square=x
Given that,
x(x+12)=540
Therefore x=18
We have to find,
12x=12*18=216

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Re: In the figure shown above, line segment QR has length 12, an  [#permalink]

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I'm confused on this one... I set up the equation correctly as 540 = x * (12 + x) (or 540 = x^2 + 12x) but how can I simplify from there? I don't know a good way on knowing that x = 18. Any help would be greatly appreciated. Thanks.
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Re: In the figure shown above, line segment QR has length 12, an  [#permalink]

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dbiersdo wrote:
I'm confused on this one... I set up the equation correctly as 540 = x * (12 + x) (or 540 = x^2 + 12x) but how can I simplify from there? I don't know a good way on knowing that x = 18. Any help would be greatly appreciated. Thanks.

Hi,

Well you can solve the equation in second degree by factorizing the same

x^2+12x-540=0

Solving equations of degree 2 : QUADRATIC

The general form of a quadratic equation is $$ax^2+bx+c=0$$

This equation has 2 solutions given by $$\frac{-b \pm \sqrt{b^2-4ac}}{2a}$$ if $$b^2>4ac$$

The equation has no solution if $$b^2<4ac$$

The equation has exactly one solution if $$b^2=4ac$$

[/color]

When you solve the equation using the above , we get 2 values of x which are -30 and 18...Now x can't be negative so you take the positive value which is 18.

You may also want to refer to GMAT Club Math Book where you can find information on GMAT Math.
For Algebra, you may refer to the the link

Another way to solve the quadratic as in the above case to plug in the answer choices and see which one fits. Plugin probably will be the quickest method. I missed that one

Hope it helps
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Intern  Joined: 08 Jan 2014
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Re: In the figure shown above, line segment QR has length 12, an  [#permalink]

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2
TroyfontaineMacon wrote:
manusingh3 wrote:
Ans B
Since the area of the total Fig is given as 540.
We can plug ans choices as follows
A) 540-144=396( not a perfect square, should be a perfect square because fig MPQT is given as a square)
B)540-216=324 (perfect square, thus sides are 18)
C)540-324= not a perfect square
D)540-360= not a perfect square
E)540-396=144 ( perfect square, but if both are 12 then the total area will come out to be 144*2 and not 540)

Just another simpler approach .
Let me know if i missed something How did you know 324 was a perfect square? you skipped that part Its just something I memorized. All squares up to 25. Its easy and will help you on the exam day- With equations and Right angle triangles(Pythagorean triples etc. Intern  B
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GMAT 1: 470 Q37 V18 GMAT 2: 570 Q36 V32 GMAT 3: 560 Q37 V30 GMAT 4: 610 Q41 V34 Re: In the figure shown above, line segment QR has length 12, an  [#permalink]

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Total area = 540

Totale area = Total Length * Total Width
Possible Length * Width:
60*9
30*18
15*36

We know that the sides of PQMT are equal ( = square )
60-12=48 and 9 ( not equal )
30-12=18 and 18 ( EQUAL !)

Total Area - Area Square = Area Rectangle = 540 - (18*18) = 216 . ANSWER B
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Re: In the figure shown above, line segment QR has length 12, an  [#permalink]

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Bunuel wrote:
SOLUTION In the figure shown above, line segment QR has length 12, and rectangle MPQT is a square. If the area of rectangular region MPRS is 540, what is the area of rectangular region TQRS?

(A) 144
(B) 216
(C) 324
(D) 360
(E) 396

The area of MPRS = the area of MPQT + the area of TQRS.

540 = x^2 + 12x --> x = 18.

The area of TQRS = 12*18 = 216.

What is the best way to solve complex quadratics like this one? I couldn't figure out x^2 + 12x - 540 = 0 quick enough.

Thanks.
Math Expert V
Joined: 02 Sep 2009
Posts: 59674
Re: In the figure shown above, line segment QR has length 12, an  [#permalink]

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Bunuel wrote:
SOLUTION In the figure shown above, line segment QR has length 12, and rectangle MPQT is a square. If the area of rectangular region MPRS is 540, what is the area of rectangular region TQRS?

(A) 144
(B) 216
(C) 324
(D) 360
(E) 396

The area of MPRS = the area of MPQT + the area of TQRS.

540 = x^2 + 12x --> x = 18.

The area of TQRS = 12*18 = 216.

What is the best way to solve complex quadratics like this one? I couldn't figure out x^2 + 12x - 540 = 0 quick enough.

Thanks.

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Posts: 152
Re: In the figure shown above, line segment QR has length 12, an  [#permalink]

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WoundedTiger wrote:
dbiersdo wrote:
I'm confused on this one... I set up the equation correctly as 540 = x * (12 + x) (or 540 = x^2 + 12x) but how can I simplify from there? I don't know a good way on knowing that x = 18. Any help would be greatly appreciated. Thanks.

Hi,

Well you can solve the equation in second degree by factorizing the same

x^2+12x-540=0

Solving equations of degree 2 : QUADRATIC

The general form of a quadratic equation is $$ax^2+bx+c=0$$

This equation has 2 solutions given by $$\frac{-b \pm \sqrt{b^2-4ac}}{2a}$$ if $$b^2>4ac$$

The equation has no solution if $$b^2<4ac$$

The equation has exactly one solution if $$b^2=4ac$$

[/color]

When you solve the equation using the above , we get 2 values of x which are -30 and 18...Now x can't be negative so you take the positive value which is 18.

You may also want to refer to GMAT Club Math Book where you can find information on GMAT Math.
For Algebra, you may refer to the the link

Another way to solve the quadratic as in the above case to plug in the answer choices and see which one fits. Plugin probably will be the quickest method. I missed that one

Hope it helps

How is that an answer to his question? When you use this approach you will have to know the square root of 576 which is no easier to solve for...
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Re: In the figure shown above, line segment QR has length 12, an  [#permalink]

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noTh1ng wrote:
Well you can solve the equation in second degree by factorizing the same

x^2+12x-540=0

How is that an answer to his question? When you use this approach you will have to know the square root of 576 which is no easier to solve for...

Hi noTh1ng,

Best way to solve the quadratic as in the above case is to plug in the answer choices and see which one fits. Plugin probably will be the quickest method.

Second possibility is to prime factorise the number 540

i.e. 540 = 54 x 10 = 2x3x3x3x2x5 = now see if any combination of two parts of these numbers make sum or difference of the two as 12 (co-efficient of x)

18 x 30 = 540
and 30-18 = 12

so x^2+12x-540=0 can be re-written as

x^2 + 30x - 18x + 540 = 0

i.e. x(x+30) -18(x+30) = 0

i.e. (x+30)*(x-18) = 0

i.e. x= 18 or -30

Dimensions are always positive numbers hence -30 can be rejected

I hope this helps!
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In the figure shown above, line segment QR has length 12, an  [#permalink]

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dbiersdo wrote:
I'm confused on this one... I set up the equation correctly as 540 = x * (12 + x) (or 540 = x^2 + 12x) but how can I simplify from there? I don't know a good way on knowing that x = 18. Any help would be greatly appreciated. Thanks.

We can complete a square here: divide $$(12/2)^2=36$$ -> and add 36 to both sides $$x^2+12x+36=540+36$$ -> $$(x+6)^2=576$$
x+6=+/-24, x=18,-30 Answer (B) we can use only +ve values

It's by far not a 600 Level --> I would say 650 (regarding 55%)
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Re: In the figure shown above, line segment QR has length 12, an  [#permalink]

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I got frustrated trying to figure out X^2+12x-540 so I picked random numbers and plugged in X.

20^2+12(20)=640 Pretty close!

15^2+12(15)=400something... Close too but opposite direction

So X is between 20 and 15.

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Re: In the figure shown above, line segment QR has length 12, an  [#permalink]

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DJ1986 wrote:
I got frustrated trying to figure out X^2+12x-540 so I picked random numbers and plugged in X.

20^2+12(20)=640 Pretty close!

15^2+12(15)=400something... Close too but opposite direction

So X is between 20 and 15.

Sorry to bump the topic!

But it was the same thing for me when I faced X^2 + 12x - 540. For me what works the best in cases like this is to do this: x(x + 12) = 540. Think about a number that multiplied by (itself + 12) will result in 540. Hope it helps.
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Re: In the figure shown above, line segment QR has length 12, an  [#permalink]

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PM*PR=540
pm*(PQ+12) =540
PQ(PQ+12) =540 ( MPQT is a square)
PQ^2 +12PQ-540 =0
PQ^2 +30PQ-18PQ-540=0
PQ=18=PM=QT
area of TQRS = 12*18=216
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Re: In the figure shown above, line segment QR has length 12, an  [#permalink]

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Quick way to solve this is to work backward using answer choices:
from question stem it is clear that the area of TQRS is 12*PQ

try few smart values and you will find the way easily

If PQ=10 then PQ^2 =100 and 12*PQ=120
the sum would be more than 220
therefore PQ must be more than 10

If PQ=20 then PQ^2 =400 and 12*PQ=240
the sum would be more than 540
therefore PQ is less than 20

A- TQRS = 244 ------> PQ = 12 ------> total area = 388 <540 (so out)
B- TQRS = 216------> PQ = 18 ------> total area = 540
C- TQRS = 324------> PQ = 26 ------> (out)
D- PQ >20 --> out
E- PQ >20 --> out

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