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

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

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New post 11 Jan 2018, 14:53
Bunuel wrote:
The Official Guide For GMAT® Quantitative Review, 2ND Edition

Image
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


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

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New post 06 May 2018, 15:13
Bunuel wrote:
SOLUTION

Image
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.


Hi pushpitkc,

i have written the correct equation but simply got stuck in terms of solving the equation

\((x+12)x = 540\)

\(x^2+12x=540\)

\(x^2+12x-540 =0\) ok here i got stuck, i simply couldnt figure out the numbers that are muplitiplied by each other and result is -540 and when we add same namber we get +12

540 is kinda large number... how to tackle it

Any advice ? :)
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In the figure shown above, line segment QR has length 12, an  [#permalink]

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New post 07 May 2018, 02:32
1
dave13 wrote:
Bunuel wrote:
SOLUTION

Image
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.


Hi pushpitkc,

i have written the correct equation but simply got stuck in terms of solving the equation

\((x+12)x = 540\)

\(x^2+12x=540\)

\(x^2+12x-540 =0\) ok here i got stuck, i simply couldnt figure out the numbers that are muplitiplied by each other and result is -540 and when we add same namber we get +12

540 is kinda large number... how to tackle it

Any advice ? :)


Hi dave13


For ax2 + bx + c = 0, the values of x which are the solutions of the equation are given by:

\(\frac{-b + \sqrt{b^2 - 4ac}}{2a}\) , \(\frac{-b - \sqrt{b^2 - 4ac}}{2a}\)


In this equation \(x^2+12x-540 =0\), a=1, b=12,and c=-540

We can calculate the discriminant(\(\sqrt{b^2 - 4ac}\)) by substituting these value of a,b,and c

Solving, we get \(\sqrt{144 - 4(-540)} = \sqrt{144 + 2160} = \sqrt{2304} = 48\)

Therefore, the roots of the equation are \(\frac{-12 + 48}{2} = 18\) and \(\frac{-12 - 48}{2} = -30\)

When the numbers don't strike you, this is another method to find the roots of the quadratic equation

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

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New post 24 Feb 2019, 09:54
Imo, easiest way is to prime factor 540 and then pick something that fits.
540 --> 2*270 --> 2*3*90 --> 2*3^3*10 --> 2^2 * 3^3 *5
Our equation is x^2 + 12x - 540 = 0

My thought process: the middle term coefficient is positive which means the larger amount in (BIG)-(small) is positive and that 12 has a units digit of 2 which means the 5 is in the BIG amount (since 5 will either leave 0 or 5 as units digit). From here it's pretty easy to pick and check that with our remaining prime factors we can get 2*3*5 - 3*2*3 = 12, so (x+30)(x-18).
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In the figure shown above, line segment QR has length 12, an   [#permalink] 24 Feb 2019, 09:54

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