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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?

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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.

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

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06 Mar 2014, 02:51

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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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06 Mar 2014, 08:50

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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

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

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08 Mar 2014, 17:21

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

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

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08 Mar 2014, 19:45

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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08 Mar 2014, 21:12

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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 algebra-101576.html#p787276

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
_________________

“If you can't fly then run, if you can't run then walk, if you can't walk then crawl, but whatever you do you have to keep moving forward.”

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

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09 Mar 2014, 00:11

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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.

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

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20 Jun 2015, 10:29

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.

Answer: B.

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

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.

Answer: B.

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

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

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22 Jun 2015, 04:43

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 algebra-101576.html#p787276

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...

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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06 Nov 2015, 11:41

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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Share some Kudos, if my posts help you. Thank you !

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

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13 Jun 2016, 09:54

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.

Answer B

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.

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

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04 Dec 2017, 11:22

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

Answer is B
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