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The length of a certain rectangle is 4 inches longer than its width. [#permalink]
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10 Sep 2015, 00:27
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74% (02:11) correct 26% (02:29) wrong based on 162 sessions
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The length of a certain rectangle is 4 inches longer than its width. [#permalink]
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10 Sep 2015, 01:38
l = w+4
\(l*w = 221\) \(w+4*w = 221\) \(w^2+4w221=0\)
\((w+17)(w13)\)
w is 17 or 13, only 13 is possible. Hence L = 13+4 = 17
\(13^2+17^2 = x^2\) \(169+289 = x^2\) \(458 = x^2\)
20*20 is 400, therefore x has to be greater 20. 22*22 (just tried that one) is 484, therefore it has to be below that. Hence C
(P.S.: No Idea how to do that task in <2 minutes, to much calculation involved)



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The length of a certain rectangle is 4 inches longer than its width. [#permalink]
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10 Sep 2015, 04:51
Solution: a(a+4) = 221. 221 is near to 15^2. So, look for two no. around 15 whose product will have 1 in its uits place.You dont have to look far, its 13 and 17. Length required = sqrt(13^2 + 17^2) = sqrt(169 + 289) = sqrt(458) 21^2 = 441 and 22^2 = 484.
Option C



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Re: The length of a certain rectangle is 4 inches longer than its width. [#permalink]
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10 Sep 2015, 07:10
Bunuel wrote: The length of a certain rectangle is 4 inches longer than its width. If the area of the rectangle is 221 square inches, then the length of the rectangle’s diagonal, in inches, is
(A) between 19 and 20 (B) between 20 and 21 (C) between 21 and 22 (D) between 22 and 23 (E) between 23 and 24
Kudos for a correct solution. Let the width of the rectangle be x,then the length will be x+4. The area is 221. so, x(x+4)=221 or \(x^2+4x221=0\) or \((x13)(x+17)=0\) or x=13 width is 13 so the lenght will 13+4= 17 So, The diagonal of the rectangle will be \(\sqrt{13^2 + 17^2}\) = \(\sqrt{458}\) Clearly, the sum lies between \(21^2\) i.e 441 and \(22^2\) i.e 484 Answer: C



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Re: The length of a certain rectangle is 4 inches longer than its width. [#permalink]
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11 Sep 2015, 00:34
L = 17, B = 13 169+289 = 458 is between 441 and 484 hence between 21 and 22. answer C



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Re: The length of a certain rectangle is 4 inches longer than its width. [#permalink]
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11 Sep 2015, 03:01
Bunuel wrote: The length of a certain rectangle is 4 inches longer than its width. If the area of the rectangle is 221 square inches, then the length of the rectangle’s diagonal, in inches, is
(A) between 19 and 20 (B) between 20 and 21 (C) between 21 and 22 (D) between 22 and 23 (E) between 23 and 24
Kudos for a correct solution. Width=x, Length=x+4 x*x+4=221 Solving for x, x=13,x+4=17 Therefore using pythagoras theorem, Diagonal^2=169+289=458 Diagonal=More than 21 but less than 22 Answer C



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Re: The length of a certain rectangle is 4 inches longer than its width. [#permalink]
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13 Sep 2015, 09:00
Bunuel wrote: The length of a certain rectangle is 4 inches longer than its width. If the area of the rectangle is 221 square inches, then the length of the rectangle’s diagonal, in inches, is
(A) between 19 and 20 (B) between 20 and 21 (C) between 21 and 22 (D) between 22 and 23 (E) between 23 and 24
Kudos for a correct solution. MANHATTAN GMAT OFFICIAL SOLUTION:First, set up an equation for the area of the rectangle. If x is the width, then we have x(x + 4) = 221 Note that if we put this into standard quadratic form and then try to factor, we wind up back where we started, in some sense: we are looking for two numbers that multiply to 221 and that differ by 4. x^2 + 4x – 221 = 0 Beyond pure trial and error, we can look for nearby squares. 221 is nearly 225, which equals 15^2. So we might try numbers near 15. As it turns out, 221 = 13 × 17. We might even get there by noticing a difference of squares: 221 = 225 – 4 = 15^2 – 2^2 = (15 – 2)(15 + 2) = 13 × 17. As a last resort, we could always use the quadratic formula, which gets us the roots of the equation as well. Now, the diagonal of the rectangle will be given by the Pythagorean Theorem: d^2 = 13^2 + 17^2 = 169 + 289 = 458. The square root of 458 is definitely larger than 20, since 20^2 = 400. Going up, we can compute 21^2 = 441 < 458, whereas 22^2 = 484 > 458. So the length of the diagonal must be between 21 and 22. The correct answer is C.
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Re: The length of a certain rectangle is 4 inches longer than its width. [#permalink]
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16 May 2017, 21:11
SHORTER METHOD
Let the dimensions be L & B Given: L  B= 4 & L.B= 221 Length of diagonal = (L^2 + B^2)^(1/2)
L^2 + B^2= √{(LB)^2 + 2L.B}=√{(4)^2 + 2.(221)}= √{16 + 442}= √(458)
Now 21^2=441 & 22^2= 484.
Hence 21< Diagonal length< 22. Hence C



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Re: The length of a certain rectangle is 4 inches longer than its width. [#permalink]
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17 May 2017, 11:45
Bunuel wrote: The length of a certain rectangle is 4 inches longer than its width. If the area of the rectangle is 221 square inches, then the length of the rectangle’s diagonal, in inches, is
(A) between 19 and 20 (B) between 20 and 21 (C) between 21 and 22 (D) between 22 and 23 (E) between 23 and 24
Kudos for a correct solution. W* (w+4)= 221 221 has four factors 1, 13, 17,and liself W= 13 length= 13 +4 = 17 According to , Pythagorean theoram C^2= a^2 + b^2 C^2= 13^2+ 17^2 thus Diagonal could be \sqrt{458} and could be Between 21 and 22.



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Re: The length of a certain rectangle is 4 inches longer than its width. [#permalink]
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15 Jun 2018, 22:23
Easier method:
Let width be x then length = x+4. Given: x(x+4) = 221 => x^2+4x = 221 (don't solve, just keep)...........(a)
We have to find diagonal. Using Pythagoras, x^2+(x+4)^2 = diagonal^2 => x^2 + x^2 + 16 + 8x = diagonal^2 => 2 (x^2+ 4x) +16 = diagonal^2 input value from (a) => 2(221) + 16 = 458.. so diagonal should be b/w 21 (441) and 22 (484)! No need to find roots of (a)! Cheers.




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