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.
The area of one square is x^2 + 10x + 25 and the area
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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
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Rich
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
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Hi M7MBA,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.
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.
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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: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
4(x + 5) + 4(2x - 3) = 32
4x + 20 + 8x - 12 = 32
12x + 8 = 32
12x = 24
x = 2
Answer: B
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$$A1=x^2+10x+25$$
$$A2=4x^2-12x+9$$
$$let\ the\ perimeter\ be\ represented\ by\ p1\ \&\ p2\ for\ A1\ and\ A2\ respectively$$
$$Therefore,\ p1+p2=32$$
$$A1=\left(x^2+5x\right)+\left(5x+25\right)$$
$$A1=x\left(x+5\right)+5\left(x+5\right)$$
$$A1=\left(x+5\right)\left(x+5\right)$$
$$A2=4x^2-12x+9$$
$$A2=\left(4x^2-6x\right)-\left(6x+9\right)$$
$$A2=2x\left(2x^{ }-3\right)-3\left(2x-3\right)$$
$$A2=\left(2x^{ }-3\right)\left(2x-3\right)$$
$$\sin ce\ perimeter\ =4l,\ \ \ therefore\ 4l1+4l2=32$$
$$4\left(2x-3\right)+4\left(2x+5\right)=32$$
$$8x-12+4x+20=32$$
$$12x+8=32$$
$$12x=24$$
$$x=\frac{24}{12}=2$$
Hence the correct answer is option B
$$A2=4x^2-12x+9$$
$$let\ the\ perimeter\ be\ represented\ by\ p1\ \&\ p2\ for\ A1\ and\ A2\ respectively$$
$$Therefore,\ p1+p2=32$$
$$A1=\left(x^2+5x\right)+\left(5x+25\right)$$
$$A1=x\left(x+5\right)+5\left(x+5\right)$$
$$A1=\left(x+5\right)\left(x+5\right)$$
$$A2=4x^2-12x+9$$
$$A2=\left(4x^2-6x\right)-\left(6x+9\right)$$
$$A2=2x\left(2x^{ }-3\right)-3\left(2x-3\right)$$
$$A2=\left(2x^{ }-3\right)\left(2x-3\right)$$
$$\sin ce\ perimeter\ =4l,\ \ \ therefore\ 4l1+4l2=32$$
$$4\left(2x-3\right)+4\left(2x+5\right)=32$$
$$8x-12+4x+20=32$$
$$12x+8=32$$
$$12x=24$$
$$x=\frac{24}{12}=2$$
Hence the correct answer is option B