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100 points for $49 worth of Veritas practice GMATs FREE VERITAS PRACTICE GMAT EXAMS Earn 10 Points Per Post Earn 10 Points Per Thanks Earn 10 Points Per Upvote ## The area of one square is x^2 + 10x + 25 and the area tagged by: M7MBA ##### This topic has 3 expert replies and 1 member reply ## The area of one square is x^2 + 10x + 25 and the area The area of one square is x^2 + 10x + 25 and the area of another square is 4x^2 - 12x + 9. If the sum of the perimeters of both squares is 32, what is the value of x? A. 0 B. 2 C. 2.5 D. 4.67 E. 10 The OA is the option B. What are the equations that I should use here? Experts, may you tell me how to use the quadratic equations? I need your help here. ### GMAT/MBA Expert GMAT Instructor Joined 04 Oct 2017 Posted: 551 messages Followed by: 11 members Upvotes: 180 Top Reply Quote: The area of one square is x^2 + 10x + 25 and the area of another square is 4x^2 - 12x + 9. If the sum of the perimeters of both squares is 32, what is the value of x? A. 0 B. 2 C. 2.5 D. 4.67 E. 10 The OA is the option B. What are the equations that I should use here? Experts, may you tell me how to use the quadratic equations? I need your help here. Hi M7MBA, Let's take a look at your question. We will first find the length of side for both the squares and then find the perimeters of both. $$Area\ of\ first\ square\ =\ x^2+10x+25$$ Let length of side of the square is 'a', then, $$a\times a=\ x^2+10x+25$$ $$a^2=\ x^2+10x+25$$ $$a^2=\ x^2+2\left(x\right)\left(5\right)+\left(5\right)^2$$ $$a^2=\ \left(x+5\right)^2$$ $$a=x+5$$ We know that perimeter of a square is 4 times the length of a side, therefore, $$Perimeter\ of\ first\ square\ =\ 4\left(x+5\right)$$ $$Area\ of\ second\ square=4x^2-12x+9$$ Let length of side of the square is 'b', then, $$b\times b=4x^2-12x+9$$ $$b^2=4x^2-12x+9$$ $$b^2=\left(2x\right)^2-2\left(2x\right)\left(3\right)+\left(3\right)^2$$ $$b^2=\left(2x-3\right)^2$$ $$b=2x-3$$ $$Perimeter\ of\ second\ square\ =\ 4\left(2x-3\right)$$ We know that the sum of the perimeters of both squares is 32, therefore, $$4\left(x+5\right)+\ 4\left(2x-3\right)=32$$ $$4x+20+\ 8x-12=32$$ $$12x+8=32$$ $$12x=32-8$$ $$12x=24$$ $$x=\frac{24}{12}=2$$ Therefore, Option B is correct. Hope it helps. I am available if you'd like any help. _________________ GMAT Prep From The Economist We offer 70+ point score improvement money back guarantee. Our average student improves 98 points. Free 7-Day Test Prep with Economist GMAT Tutor - Receive free access to the top-rated GMAT prep course including a 1-on-1 strategy session, 2 full-length tests, and 5 ask-a-tutor messages. Get started now. ### GMAT/MBA Expert Elite Legendary Member Joined 23 Jun 2013 Posted: 9942 messages Followed by: 492 members Upvotes: 2867 GMAT Score: 800 Hi M7MBA, We're told that the area of one square is X^2 + 10X + 25, the area of another square is 4X^2 - 12X + 9 and the sum of the perimeters of both squares is 32. We're asked for the value of X. This question is based around standard Quadratic rules and a bit of Geometry. To start, since we're dealing with two SQUARES, we know that the the reverse-FOIL of each area will be the same 'term' twice (since those terms represent the lengths of the sides of each square). X^2 + 10X + 25 = (X+5)(X+5) 4X^2 - 12X + 9 = (2X - 3)(2X - 3) The PERIMETER of a square = 4(side), so the two perimeters are....4(X+5) and 4(2X - 3). We're told that the SUM of those two perimeters is 32, so we can set up an equation and solve for X: 4(X+5) + 4(2X - 3) = 32 4X + 20 + 8X - 12 = 32 12X + 8 = 32 12X = 24 X = 2 Final Answer: B GMAT assassins aren't born, they're made, Rich _________________ Contact Rich at Rich.C@empowergmat.com ### GMAT/MBA Expert GMAT Instructor Joined 25 Apr 2015 Posted: 1776 messages Followed by: 14 members Upvotes: 43 M7MBA wrote: The area of one square is x^2 + 10x + 25 and the area of another square is 4x^2 - 12x + 9. If the sum of the perimeters of both squares is 32, what is the value of x? A. 0 B. 2 C. 2.5 D. 4.67 E. 10 We have to determine the side length of each square. Since the area of the first square is x^2 + 10x + 25 = (x + 5)^2, its side is x + 5. Similarly, since the area of the second square is 4x^2 - 12x + 9 = (2x - 3)^2, its side is 2x - 3. Furthermore, the perimeter of the first square is 4(x + 5), and that of the second square is 4(2x - 3). Since the sum of the perimeters of the two squares is 32, we can create the following equation: 4(x + 5) + 4(2x - 3) = 32 4x + 20 + 8x - 12 = 32 12x + 8 = 32 12x = 24 x = 2 Answer: B _________________ Scott Woodbury-Stewart Founder and CEO ### Top Member Legendary Member Joined 02 Mar 2018 Posted: 721 messages Followed by: 1 members $$A1=x^2+10x+25$$ $$A2=4x^2-12x+9$$ $$let\ the\ perimeter\ be\ represented\ by\ p1\ \&\ p2\ for\ A1\ and\ A2\ respectively$$ $$Therefore,\ p1+p2=32$$ $$A1=\left(x^2+5x\right)+\left(5x+25\right)$$ $$A1=x\left(x+5\right)+5\left(x+5\right)$$ $$A1=\left(x+5\right)\left(x+5\right)$$ $$A2=4x^2-12x+9$$ $$A2=\left(4x^2-6x\right)-\left(6x+9\right)$$ $$A2=2x\left(2x^{ }-3\right)-3\left(2x-3\right)$$ $$A2=\left(2x^{ }-3\right)\left(2x-3\right)$$ $$\sin ce\ perimeter\ =4l,\ \ \ therefore\ 4l1+4l2=32$$ $$4\left(2x-3\right)+4\left(2x+5\right)=32$$ $$8x-12+4x+20=32$$ $$12x+8=32$$ $$12x=24$$ $$x=\frac{24}{12}=2$$ Hence the correct answer is option B • Award-winning private GMAT tutoring Register now and save up to$200

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