In the figure shown above, line segment QR has length 12, and rectangl

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

It's easy to solve this equation without using the formula (which is cumbersome in my opinion)
x^2 + 12x - 540 = 0

We try to split 540 into two factors such that their difference is 12
540 = 54*10 = 18*3*10
You can see immediately that the 2 factors will be 18 and 30.

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

Here is a post discussing how to split the middle term quickly: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2013/12 ... equations/
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Note: Not drawn to scale.

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.
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A very good contender for back solving, lets start with option C (Always start with C)

Area of rectangle.........Side QT.......Area of square.....Total area of the figure
(A) 144
(B) 216
(C) 324....................324/12=27.... 27*27 =729........... 729 + 324 = 1053
(D) 360
(E) 396

The total area was supposed to be 540 however it is 1053 hence we move up (go for smaller values). Now we know that answer is either A or B so we try any of them.


Area of rectangle.........Side QT.......Area of square.....Total area of the figure
(A) 144
(B) 216....................216/12 =18.....18*18 =324.......... 324 + 216 =540 (BINGO!)
(C) 324....................324/12=27.... 27*27 =729........... 729 + 324 = 1053
(D) 360
(E) 396

Answer is B.

Add the given information...



Let x = the length of each side of square MPQT


GIVEN: The area of rectangular region MPRS is 540

In other words: (base)(height) = 540
Substitute values to get: (x + 12)(x) = 540
Expand: x² + 12x = 540
Set equal to zero: x² + 12x - 540 = 0
Factor: (x + 30)(x - 18) = 0
So, EITHER x = -30 OR x = 18

Since the length cannot be negative, it must be the case that x = 18

What is the area of rectangular region TQRS?
Area = (base)(height) = (12)(18) = 216

Answer: B

Cheers,
Brent

Since PQMT is a square, we can let PQ = PM = n. Since QR = 12, PR = n + 12. Finally, since the area of rectangle MPRS is 540, we have:
n(n + 12) = 540

n^2 + 12n = 540

n^2 + 12n - 540 = 0

(n + 30)(n - 18) = 0

n = -30 or n = 18

Since n can’t be negative, n = 18, and the area of square MPQT is 18 x 18 = 324. Thus, the area of rectangle TQRS is 540 - 324 = 216,

Answer: B
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Thank you for your answer, but still : how do you find 48 quickly? you don't find the root of 12^2-4*540 so easily, do you?

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 \(b^2=4ac\)

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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
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Hi Karishma, obviously another awesome post! The blog post helped me a lot.

Would you say though that this technique would be even faster than simply back-solving when it comes to quadratic equations?

Thanks

Franco

Backsolving is a useful technique but its relevance in GMAT is decreasing. The options given are such that it is harder to back solve now. For example, here you will need to divide the option by 12 (the length of QR) and then try to plug in what you get in the equation. Notice that all options are divisible by 12 (which will be true for all good options) so it might be some time before you arrive at the answer using backsolving. During the exam, use whatever comes to your mind - you wouldn't have the time to consciously decide to use one method over another.
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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.

Factoring Quadratics: https://www.purplemath.com/modules/factquad.htm
Solving Quadratic Equations: https://www.purplemath.com/modules/solvquad.htm
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well I found the correct equation within seconds but then I could not find the roots fast enough

Is there any shortcut/trick to find the roots ?

Check the links below:

Factoring Quadratics: https://www.purplemath.com/modules/factquad.htm
Solving Quadratic Equations: https://www.purplemath.com/modules/solvquad.htm

Theory on Algebra: https://gmatclub.com/forum/algebra-101576.html
Algebra - Tips and hints: https://gmatclub.com/forum/algebra-tips- ... 75003.html

DS Algebra Questions to practice: https://gmatclub.com/forum/search.php?se ... &tag_id=29
PS Algebra Questions to practice: https://gmatclub.com/forum/search.php?se ... &tag_id=50

Hope it helps.
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why do you always start with C? Also can you please explain your strategy? Thanks.

ayas7 , a comment and links that might help.

For questions where the answer choices are listed in ascending order, and we want to backsolve, we start with C because it is the middle value. C gives a benchmark.

If C yields an answer that is too large? Toss out Answers D and E, which will be greater than C. Then we only have to pick between Answers A and B.

There are two really good explanations for starting with C, given by a GMATclub expert, HERE
and HERE

THIS POST
by Bunuel includes the link I gave above, and a lot more. Scroll down slightly to "2. Strategies and Tactics."

One more. The post immediately above, which I just found rather accidentally, is part of what looks to be a phenomenal collection, Ultimate GMAT Quantitative Megathread, by Bunuel, composed of GMAT Quant . . . everything. MEGATHREAD HERE

Hope that helps.
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Here are the links:
Backsolving on GMAT Math
How to Plug in Numbers on GMAT Math Questions
Why Approximate?
GMAT Math Strategies — Estimation, Rounding and other Shortcuts
The Power of Estimation for GMAT Quant
The 4 Math Strategies Everyone Must Master, Part 1 (1. Test Cases and 2. Choose Smart Numbers.)
The 4 Math Strategies Everyone Must Master, part 2 (3. Work Backwards and 4. Estimate)
Intelligent Guessing on GMAT
How to Avoid Tedious Calculations on the Quantitative Section of the GMAT
GMAT Tip of the Week: No Calculator? No Problem.
The Importance of Sorting Answer Choices on the GMAT
Identifying the Correct Answer on GMAT Quant Questions
How to Manage Unmanageable Numbers on the GMAT
Should You Double Check Your Answer Choices on the GMAT?
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