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M16-22

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Re: M16-22  [#permalink]

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New post 21 Sep 2017, 20:44
Simba9 wrote:
Hi Bunuel

Since you say it's better to solve (when we get an equation like a^2 − 32a −32=0, as we cannot assume that one value will be positive and the other neg.) Would you mind posting a complete resolution to this problem? i.e. including your calculations to get to the positive and negative (invalid) solution.

I am asking this as I think you might be doing the calculus in a much more straightforward way than I am.

Thanks in advance!


Factoring Quadratics: http://www.purplemath.com/modules/factquad.htm
Solving Quadratic Equations: http://www.purplemath.com/modules/solvquad.htm
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Re: M16-22  [#permalink]

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New post 06 Nov 2017, 20:02
Can someone correct where am i going wrong?

let diameter = d
long side: d-2
short side: d /3

using Pythagorean theorem
d^2 = (d-2)^2 + (d/3)^2 or
d^2 = d^2 + 4 - 4d + d^2/9 subtracting d^2 on both sides
0 = 4 - 4d + d^2/9
multiplying both sides by 9
d^2 - 36d - 36 = 0

this is a different equation that derived in the original expert ans. What am I missing here?
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New post 06 Nov 2017, 20:49
sevenplusplus wrote:
Can someone correct where am i going wrong?

let diameter = d
long side: d-2
short side: d /3

using Pythagorean theorem
d^2 = (d-2)^2 + (d/3)^2 or
d^2 = d^2 + 4 - 4d + d^2/9 subtracting d^2 on both sides
0 = 4 - 4d + d^2/9
multiplying both sides by 9
d^2 - 36d - 36 = 0

this is a different equation that derived in the original expert ans. What am I missing here?


You have: d^2 = (d-2)^2 + (d/3)^2
Original solution: a^2 -32a-32=0

Do you see the difference? You have d, while the OE has a (longer side).
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M16-22  [#permalink]

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New post 11 Feb 2018, 11:52
We know the ratio of the length of the right triangle (1: \sqrt{3}: 2) and since the larger side is 2 meters less than the diagonal.
Can't we just assume that d is equal to 2x = x\sqrt{3} + 2 ?
And so we just need statement A to figure out the perimeter ?

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New post 11 Feb 2018, 12:11
alvarezy wrote:
We know the ratio of the length of the right triangle (1: \sqrt{3}: 2) and since the larger side is 2 meters less than the diagonal.
Can't we just assume that d is equal to 2x = x\sqrt{3} + 2 ?
And so we just need statement A to figure out the perimeter ?

Thank you


Do we know that the diagonal will divide the rectangle into 30-60-90 triangles? No. Why it cannot be say 1-89-90? Or 7.5-82.5-90?
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New post 11 Feb 2018, 15:06
Understood. Very clear. Thank you for the quick reply.


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Re: M16-22  [#permalink]

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New post 11 Aug 2018, 01:44
Simba9 wrote:
Hi Bunuel

Since you say it's better to solve (when we get an equation like a^2 − 32a −32=0, as we cannot assume that one value will be positive and the other neg.) Would you mind posting a complete resolution to this problem? i.e. including your calculations to get to the positive and negative (invalid) solution.

I am asking this as I think you might be doing the calculus in a much more straightforward way than I am.

Thanks in advance!


Here you go!! I know it's a bit late but it might be useful for others as well.

\(a^2 -32a-32=0\). You can't solve this using factorization. So we should use other method

\(D = \sqrt{b^2-4*a*c}\)

\(D = \sqrt{(-32)^2-4*1*(-32)}\)

\(D = \sqrt{(32 * 32) + (4*32)}\)

\(D = \sqrt{32 * (32 + 4)}\)

\(D = \sqrt{32 * 36}\)

\(D = 24 \sqrt{2}\)

\(D = 34\) approx.

\(x = \frac{(-b + D)}{(2*a)}\)

\(a =\frac{(- - 32 + 34)}{(2*1)}\)

\(a = \frac{(32 + 34)}{2}\)

\(a = 33\)

\(x = \frac{(-b - D)}{(2* a)}\)
\(a = \frac{(- - 32 - 34)}{(2*1)}\)
\(a = \frac{(32 - 34)}{2}\)
\(a = \frac{-2}{2}\) = -1 Not Applicable.

so a = 33, we can calculate value of b. And Perimeter of Rectangle as well.
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Re M16-22  [#permalink]

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New post 28 Aug 2018, 01:52
I think this is a high-quality question and the explanation isn't clear enough, please elaborate. If longer side has two value ,then shorter side will have two value then how will we come to solution the value of perimeter?
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New post 28 Aug 2018, 05:15
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Re: M16-22  [#permalink]

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New post 02 Oct 2018, 21:33
if shorter side is 2, diagonal is 6 and longer side is 4 (Meet conditions of both statements)
then perimeter is=2(2+4)=12

if shorter side is 10, diagonal is 30 and longer side is 28 (Meet conditions of both statements)
then perimeter is=2(10+28)=76

two different answers so should't the correct option be E?
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New post 03 Oct 2018, 00:28
HasnainAfxal wrote:
if shorter side is 2, diagonal is 6 and longer side is 4 (Meet conditions of both statements)
then perimeter is=2(2+4)=12

if shorter side is 10, diagonal is 30 and longer side is 28 (Meet conditions of both statements)
then perimeter is=2(10+28)=76

two different answers so should't the correct option be E?


In a rectangle two adjacent sides and the diagonal form a right triangle, so the lengths of the sides must satisfy side^2 + side^2 = diagonal^2. Do your examples satisfy this?
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M16-22  [#permalink]

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New post 03 Oct 2018, 01:13
Bunuel wrote:
HasnainAfxal wrote:
if shorter side is 2, diagonal is 6 and longer side is 4 (Meet conditions of both statements)
then perimeter is=2(2+4)=12

if shorter side is 10, diagonal is 30 and longer side is 28 (Meet conditions of both statements)
then perimeter is=2(10+28)=76

two different answers so should't the correct option be E?


In a rectangle two adjacent sides and the diagonal form a right triangle, so the lengths of the sides must satisfy side^2 + side^2 = diagonal^2. Do your examples satisfy this?

Thanks got it :)

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New post 09 Feb 2019, 11:01
I think this is a high-quality question and the explanation isn't clear enough, please elaborate. It would be better if there is a rectangle showing what's each letter written on the solution.
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New post 22 Mar 2019, 09:17
gmatdemolisher1234 wrote:
Bunuel

when we get an equation like a^2 − 32a −32=0 , would you recommend solving the equation during the official test.. isnt it really time consuming..
Since GMAT isnt a calulation based exam.. Can we assume that one value will be positive and the other neg.

what would u personally suggest ?

Thank in advance



Generally, in such situations, I prefer to quickly use the concept of product of roots. In a quadratic equation, ax^2 + bx + c, product of roots is c/a. If the product of roots is negative, one of the roots is negative else both are positive.
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Re: M16-22   [#permalink] 22 Mar 2019, 09:17

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