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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 Top Member 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: 10197 messages Followed by: 497 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: 2950 messages Followed by: 19 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 scott@targettestprep.com See why Target Test Prep is rated 5 out of 5 stars on BEAT the GMAT. Read our reviews Top Member Legendary Member Joined 02 Mar 2018 Posted: 1171 messages Followed by: 2 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 • FREE GMAT Exam Know how you'd score today for$0

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