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The combined area of the two black squares is equal to 1000 square uni
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07 Feb 2012, 17:34
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82% (03:15) correct 18% (03:16) wrong based on 202 sessions
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The combined area of the two black squares is equal to 1000 square units. A side of the larger black square is 8 units longer than a side of the smaller black square. What is the combined area of the two white rectangles in square units? (A) 928 (B) 936 (C) 948 (D) 968 (E) 972 Attachment:
Square.PNG [ 1.52 KiB  Viewed 9249 times ]
Guys this is how I am trying to solve this: Let Side of smaller square be x Side of larger square will be x+8 Area of smaller square \(x^2\)Area of larger square \((x+8)^2\)So the sum of the areas = 1000 which brings me to the equation \(x^2\)+8x468 . I don't think we an simplify this equation any further, well I can't and therefore need help.
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The combined area of the two black squares is equal to 1000 square uni
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07 Feb 2012, 17:45
The combined area of the two black squares is equal to 1000 square units. A side of the larger black square is 8 units longer than a side of the smaller black square. What is the combined area of the two white rectangles in square units?(A) 928 (B) 936 (C) 948 (D) 968 (E) 972 I guess the black squares are the red squares on the diagram. The length of a smaller square  x; The length of a larger square  x+8; The area of entire square  x^2+(x+8)^2=1000 > 2x^2+16x=936; The combined area of the two white rectangles  x(x+8)+x(x+8)=2x^2+16x > look up: 2x^2+16x=936. Answer: B. Attachment:
Square.PNG [ 1.52 KiB  Viewed 9005 times ]
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Re: The combined area of the two black squares is equal to 1000 square uni
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27 May 2013, 23:21
enigma123 wrote: Attachment: Square.PNG The combined area of the two black squares is equal to 1000 square units. A side of the larger black square is 8 units longer than a side of the smaller black square. What is the combined area of the two white rectangles in square units? (A) 928 (B) 936 (C) 948 (D) 968 (E) 972 Let Side of smaller square be x Side of larger square will be x+8 Area of smaller square \(x^2\)Area of larger square \((x+8)^2\)So the sum of the areas = 1000 which brings me to the equation \(x^2\)+8x468 . I don't think we an simplify this equation any further, well I can't and therefore need help. X^2 +26x 18x468=0 ie X=18 and X= 26. Discard negative value thus X=18. Hope it helps...
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Re: The combined area of the two black squares is equal to 1000 square uni
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29 May 2013, 09:25
enigma123 wrote: Attachment: Square.PNG The combined area of the two black squares is equal to 1000 square units. A side of the larger black square is 8 units longer than a side of the smaller black square. What is the combined area of the two white rectangles in square units? (A) 928 (B) 936 (C) 948 (D) 968 (E) 972 Guys this is how I am trying to solve this:
Let
Side of smaller square be x Side of larger square will be x+8
Area of smaller square \(x^2\) Area of larger square \((x+8)^2\)
So the sum of the areas = 1000 which brings me to the equation \(x^2\)+8x468 . I don't think we an simplify this equation any further, well I can't and therefore need help. Let smaller square be of side x then bigger square has a side of x+8 (x)^2 +(x+8)^2=1000 2x^2+8x468=1000 (x+26)(x18)=0 x=18 Now,the entire figure is a square of side 18+26=44 Total are=44^2=1936 White are=total areacoloured area=19361000=936



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Re: The combined area of the two black squares is equal to 1000 square uni
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29 May 2013, 19:24
Let a be side of big square b be side of small square given a*a +b*b =1000 and a=8+b
so (8+b)^2+b^2=1000 on solving b = 18 and a=26
need to find total area of two rectangles given we get 44^21000 = 936
Sol:B



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Re: The combined area of the two black squares is equal to 1000 square uni
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29 May 2013, 21:32
sum of area of black squares(1000) + sum of area of white rectangles(X) = area of the larger square (Y) From the options, X is an integer, hence Y (= X + 1000) is an integer too. As Y is a square number, From the properties of square numbers, Y can only end with 0,1,4,6,9 or 25. From the given options, only B (ending 6) is a possible solution, as +1000 doesn't change unit digit value. Number ending 2 or 8 + 1000 (as in all other options) can never be a square number. Besides, all of the above ways are equally correct.
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Re: The combined area of the two black squares is equal to 1000 square uni
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07 Dec 2015, 05:33
hi Any quick way to solve and get factor for x2+8x468



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Re: The combined area of the two black squares is equal to 1000 square uni
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30 Dec 2016, 05:17
Assume the larger red square has a side of length x + 4 units and the smaller red square has a side of length x  4 units. This satisfies the condition that the side length of the larger square is 8 more than that of the smaller square. Therefore, the area of the larger square is (x + 4)2 or x^2 + 8x + 16. Likewise, the area of the smaller square is (x  4)^2 or x^2  8x + 16. Set up the following equation to represent the combined area: (x^2 + 8x +16) + (x^2  8x +16) = 1000 2x^2 + 32 = 1000 2x^2 = 968 It is possible, but not necessary, to solve for the variable x here. The two white rectangles, which are congruent to each other, are each x + 4 units long and x  4 units high. Therefore, the area of either rectangle is (x + 4)(x  4), or x^2  16. Their combined area is 2(x^2  16), or 2x^2  32. Since we know that 2x^2 = 968, the combined area of the two white rectangles is 968  32, or 936 square units. The correct answer is B.
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Re: The combined area of the two black squares is equal to 1000 square uni
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17 Feb 2017, 08:25
I solved it in under 2 mins using the below approach: Let X and x be the sides of big and small squares respectively.
Xx=8 X^2 +x^2= 1000 or (Xx)^2 +2Xx=1000 or 64 +2Xx= 1000 or 2Xx = 936
Now. area of the two white squares are Xx +Xx = 2Xx = 936



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Re: The combined area of the two black squares is equal to 1000 square uni
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22 Jan 2018, 16:46
Hi All, If you don't see the 'elegant' approach to this question, then you can still solve it with some 'brute force' math (and a bunch of multiplication)  it's not 'pretty', but if you're comfortable doing math by hand then you can still get to the correct answer in 2 minutes: We're told that the sum of two squares is 1000 and that the two numbers (before they are squared) differ by 8. Since the answer choices to the question are integers, it's likely that the two side lengths of the two squares are also integers. Let's list out some perfect squares and get a sense of how big the numbers can get... 20^2 = 400 30^2 = 900 Since the larger square is clearly bigger, it's area will be more than half of the 1000, so it's side length will fall into the range of 2030. The side length of the smaller triangle will be 8 LESS... Now let's try to narrow things down a bit... 15^2 = 225 25^2 = 625 17^2 = does NOT end in a 5, so there's no way the sum would be 1000, but this example probably isn't too far away from the correct answer... 26^2 = 676 18^2 = 324 This IS the sum that we're looking for, so these MUST be the dimensions of the two squares. By extension, the dimensions of the two white rectangles are 26x18 and the area of each is 468. Since there are two of those rectangles, their total area = 468(2) = 936. Final Answer: GMAT assassins aren't born, they're made, Rich
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Re: The combined area of the two black squares is equal to 1000 square uni
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22 Jun 2018, 05:46
gultrage wrote: enigma123 wrote: Attachment: Square.PNG The combined area of the two black squares is equal to 1000 square units. A side of the larger black square is 8 units longer than a side of the smaller black square. What is the combined area of the two white rectangles in square units? (A) 928 (B) 936 (C) 948 (D) 968 (E) 972 Guys this is how I am trying to solve this:
Let
Side of smaller square be x Side of larger square will be x+8
Area of smaller square \(x^2\) Area of larger square \((x+8)^2\)
So the sum of the areas = 1000 which brings me to the equation \(x^2\)+8x468 . I don't think we an simplify this equation any further, well I can't and therefore need help. Let smaller square be of side x then bigger square has a side of x+8 (x)^2 +(x+8)^2=1000 2x^2+8x468=1000 (x+26)(x18)=0 x=18 Now,the entire figure is a square of side 18+26=44 Total are=44^2=1936 White are=total areacoloured area=19361000=936 How did you get "468"? Posted from my mobile device



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The combined area of the two black squares is equal to 1000 square uni
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23 Jun 2018, 08:40
MrJglass wrote: gultrage wrote: enigma123 wrote: Attachment: Square.PNG The combined area of the two black squares is equal to 1000 square units. A side of the larger black square is 8 units longer than a side of the smaller black square. What is the combined area of the two white rectangles in square units? (A) 928 (B) 936 (C) 948 (D) 968 (E) 972 Guys this is how I am trying to solve this:
Let Side of smaller square be x Side of larger square will be x+8
Area of smaller square \(x^2\) Area of larger square \((x+8)^2\)
So the sum of the areas = 1000 which brings me to the equation \(x^2\)+8x468 . I don't think we an simplify this equation any further, well I can't and therefore need help. Let smaller square be of side x then bigger square has a side of x+8 (x)^2 +(x+8)^2=1000 2x^2+8x468=1000 (x+26)(x18)=0 x=18 Now,the entire figure is a square of side 18+26=44 Total are=44^2=1936 White are=total areacoloured area=19361000=936 How did you get "468"? MrJglass , welcome to GMAT club! That post was awhile ago, so getting an answer from the author is unlikely. I agree: (468) in that post is confusing. This equation is not selfevident, and seems to me to be insensible: \(2x^2+8x468=1000\) That said, "468" comes from dividing one form of the equation by 2 and subtracting RHS from both sides. Thus: \(x^2 +(x+8)^2=1,000\) > \(2x^2+16x+64=1,000\) Divide by 2, subtract 500 from both sides \(x^2+8x+32=500\) \(x^2+8x468=0\) Hope that helps.
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The combined area of the two black squares is equal to 1000 square uni
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24 Jun 2018, 14:17
generis wrote: Welcome to GMAT club! That post was awhile ago, so getting an answer from the author is unlikely. I agree: (468) in that post is confusing.
This equation is not selfevident, and seems to me to be insensible: \(2x^2+8x468=1000\)
That said, "468" comes from dividing one form of the equation by 2 and subtracting RHS from both sides.
Thus: \(x^2 +(x+8)^2=1,000\) >
\(2x^2+16x+64=1,000\) Divide by 2, subtract 500 from both sides \(x^2+8x+32=500\) \(x^2+8x468=0\)
Hope that helps.
Yea, thanks! Posted from my mobile device



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Re: The combined area of the two black squares is equal to 1000 square uni
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25 May 2019, 00:44
although a straight fwd question but time intensive because of calculations given square 1 + square 2 area = 1000 so x^2+(x+8)^2=1000 solve for x we get 18 & 26 since ve value wont be correct so x= 18 and other side of square = 18+8 ; 26 so rectangle white side area ; 18*26 * 2; 936 IMO B enigma123 wrote: The combined area of the two black squares is equal to 1000 square units. A side of the larger black square is 8 units longer than a side of the smaller black square. What is the combined area of the two white rectangles in square units? (A) 928 (B) 936 (C) 948 (D) 968 (E) 972 Attachment: Square.PNG Guys this is how I am trying to solve this: Let Side of smaller square be x Side of larger square will be x+8 Area of smaller square \(x^2\)Area of larger square \((x+8)^2\)So the sum of the areas = 1000 which brings me to the equation \(x^2\)+8x468 . I don't think we an simplify this equation any further, well I can't and therefore need help.
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Re: The combined area of the two black squares is equal to 1000 square uni
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