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Two sides of a triangle have lengths x and y and meet at a r

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Two sides of a triangle have lengths x and y and meet at a r  [#permalink]

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New post Updated on: 08 Dec 2013, 08:02
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Two sides of a triangle have lengths x and y and meet at a right angle. If the perimeter of the triangle is 4x, what is the ratio of x to y ?

a) 2 : 3
b) 3 : 4
c) 4 : 3
d) 3 : 2
e) 2 : 1

Originally posted by xhimi on 08 Dec 2013, 01:57.
Last edited by xhimi on 08 Dec 2013, 08:02, edited 1 time in total.
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Re: Two sides of a triangle have lengths x and y and meet at a r  [#permalink]

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New post 08 Dec 2013, 03:13
7
E is correct. Here is my solution.
Hope it helps.
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Re: Two sides of a triangle have lengths x and y and meet at a r  [#permalink]

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New post 08 Dec 2013, 03:29
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pqhai wrote:
E is correct. Here is my solution.
Hope it helps.


There is a small typo there in your solution.

Actually B is the right answer.
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Re: Two sides of a triangle have lengths x and y and meet at a r  [#permalink]

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New post 08 Dec 2013, 05:14
Let's assume the simple set of sides for a right triangle as x=3, y=4 and hypotenuse= 5. Now the perimeter adds to 3+4+5= 12, which is equal to 3x. Therefore x:y= 3/4
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Re: Two sides of a triangle have lengths x and y and meet at a r  [#permalink]

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New post 08 Dec 2013, 11:47
1
Mau5
Ahhhh....my bad. Thank you, you're correct. 8x = 6y --> x/y = 3/4.
mau5 wrote:
pqhai wrote:
E is correct. Here is my solution.
Hope it helps.


There is a small typo there in your solution.

Actually B is the right answer.


Posted from my mobile device
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Re: Two sides of a triangle have lengths x and y and meet at a r  [#permalink]

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New post 06 Jan 2015, 11:36
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Hi All,

When right triangles show up on Test Day, they're rarely random - they're often one of the right triangles that are frequently tested: 3/4/5, 5/12/13, 30/60/90, 45/45/90.

In this prompt, the fact that the perimeter is equal to 4 times one of the shorter sides is an interesting (and rather specific) piece of information....MAYBE it matches one of the special right triangles I listed....?

reetskaur correctly realized that this description is a match for a 3/4/5 right triangle, so we can AVOID all of the crazy geometry/algebra and say that...

X = 3
Y = 4
Hypoteneuse = 5
Perimeter = 12

X:Y = 3:4

The GMAT is based heavily on established Quant and Verbal patterns - keep an eye open for pattern-based shortcuts throughout the Test - they're everywhere....

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Re: Two sides of a triangle have lengths x and y and meet at a r  [#permalink]

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New post 08 Jan 2015, 03:00
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Two equations

x^2+y^2=c^2

x+y+c=4x

we need x/y ratio, so should express c through x and y

c=4x-x-y=3x-y

substitute c in first equation

x^2+y^2=(3x-y)^2 => 8x=6y => x/y=6/8=3/4

B
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Re: Two sides of a triangle have lengths x and y and meet at a r  [#permalink]

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New post 04 May 2015, 01:32
EMPOWERgmatRichC wrote:
Hi All,

When right triangles show up on Test Day, they're rarely random - they're often one of the right triangles that are frequently tested: 3/4/5, 5/12/13, 30/60/90, 45/45/90.

In this prompt, the fact that the perimeter is equal to 4 times one of the shorter sides is an interesting (and rather specific) piece of information....MAYBE it matches one of the special right triangles I listed....?

reetskaur correctly realized that this description is a match for a 3/4/5 right triangle, so we can AVOID all of the crazy geometry/algebra and say that...

X = 3
Y = 4
Hypoteneuse = 5
Perimeter = 12

X:Y = 3:4

The GMAT is based heavily on established Quant and Verbal patterns - keep an eye open for pattern-based shortcuts throughout the Test - they're everywhere....

Final Answer:

GMAT assassins aren't born, they're made,
Rich


How can you assume x to be 3 and not 4.
Could it not be x 4 y3 z 5? I mean its basically the same triangle just a different distributionß

Going with the algebraic approach x^2+y^2=(3x-y)^2 => 8x=6y => x/y=6/8=3/4
I don't get how x^2+y^2=(3x-y)^2 becomes => 8x=6y. Could someone add some more steps?
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Two sides of a triangle have lengths x and y and meet at a r  [#permalink]

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New post 04 May 2015, 03:17
xhimi wrote:
Two sides of a triangle have lengths x and y and meet at a right angle. If the perimeter of the triangle is 4x, what is the ratio of x to y ?

a) 2 : 3
b) 3 : 4
c) 4 : 3
d) 3 : 2
e) 2 : 1



Let \(y=K*x\)
Then the hypotnuse will be \(4x- x - Kx = (3-K)x\)
So,
\(1+ K^2 = (3-K)^2\)
\(K= \frac{4}{3}\)
so \(y= \frac{4}{3} * x\)
Answer : B
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Re: Two sides of a triangle have lengths x and y and meet at a r  [#permalink]

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New post 04 May 2015, 14:13
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Hi ZLukeZ,

In the prompt, we're told that X and Y are the two "legs" of the right triangle and that the PERIMETER = 4X

IF.....
the legs are X=3 and Y=4, then the perimeter is 4(3) = 12.
A 3/4/5 triangle has a perimeter of 3+4+5 = 12, so this "matches" what we were told.

IF....
the legs are X=4 and Y=3, then the perimeter is 4(4) = 16
This does NOT match the perimeter of a 3/4/5 triangle, so X CANNOT be 4.

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Re: Two sides of a triangle have lengths x and y and meet at a r  [#permalink]

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New post 06 May 2015, 14:20
Hi ZuleZ,

I also had to do a double take on that part of the question

So x^2 + y^2 = (3x-y)^2
Let us focus on the (3x-y)^2 part of the equation
FOIL it and you get (3x-y)(3x-y) ------> 9x^2-6xy+y^2, lets put this back into the main equation

x^2+y^2 = 9x^2 - 6xy +y^2 ---------> y^2-y^2 + 6xy = 9^2- x^2 --------------> 6xy = 8x^2 ---divide by x -----> 6y=8x

Hope this provides some clarification
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Re: Two sides of a triangle have lengths x and y and meet at a r  [#permalink]

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New post 27 Feb 2017, 22:27
The question can very easily be solved using the answer options, in my opinion. It would take lesser time than the algebra would.
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Re: Two sides of a triangle have lengths x and y and meet at a r  [#permalink]

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New post 22 Aug 2018, 05:51
EMPOWERgmatRichC wrote:
Hi ZLukeZ,

In the prompt, we're told that X and Y are the two "legs" of the right triangle and that the PERIMETER = 4X

IF.....
the legs are X=3 and Y=4, then the perimeter is 4(3) = 12.
A 3/4/5 triangle has a perimeter of 3+4+5 = 12, so this "matches" what we were told.


IF....
the legs are X=4 and Y=3, then the perimeter is 4(4) = 16
This does NOT match the perimeter of a 3/4/5 triangle, so X CANNOT be 4.

GMAT assassins aren't born, they're made,
Rich


Why are you assuming it's a 3/4/5 triangle right off the bat? Could be a 45, 45, 90? I understand it works if we make it a 3,4,5 and make x=3 and y=4, but how do we know that's the only possible solution?

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Re: Two sides of a triangle have lengths x and y and meet at a r  [#permalink]

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New post 22 Aug 2018, 06:12
I think the empower provide method should ideally be avoided since it is not always applicable and at some times if you apply it wrongly it will lead you to wrong answers
Its best to follow the basics here!

Here is an alternative method of doing it.
X and y are the right angled sides so the third side is the sqrt of their squares.
Since the perimeter is 4x

That is. X + y + z = 4x
X^2 + y^2 = z^2

Now for this sort of equation to solve either x y and z all have to be rational numbers or a rational multiple of same irrational number.

In either case their ratio will be rational over another rational.
And since the third side is the square root of their squares sum the ratio of x to y in pure number form needs to yield a rational number .
That is if a/b is the ratio of x and y then a^2 + b^2. = a rational number squared.

Only option b satisfies this.

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Re: Two sides of a triangle have lengths x and y and meet at a r  [#permalink]

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New post 22 Aug 2018, 21:32
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MikeScarn wrote:
EMPOWERgmatRichC wrote:
Hi ZLukeZ,

In the prompt, we're told that X and Y are the two "legs" of the right triangle and that the PERIMETER = 4X

IF.....
the legs are X=3 and Y=4, then the perimeter is 4(3) = 12.
A 3/4/5 triangle has a perimeter of 3+4+5 = 12, so this "matches" what we were told.


IF....
the legs are X=4 and Y=3, then the perimeter is 4(4) = 16
This does NOT match the perimeter of a 3/4/5 triangle, so X CANNOT be 4.

GMAT assassins aren't born, they're made,
Rich


Why are you assuming it's a 3/4/5 triangle right off the bat? Could be a 45, 45, 90? I understand it works if we make it a 3,4,5 and make x=3 and y=4, but how do we know that's the only possible solution?

CC: Bunuel


Hi MikeScarn,

You ask some really good questions. To answer them, you have to remember a few things:

1) This is the GMAT, not a general "math test." The GMAT is built around patterns and is predictable (NOTHING in a question is there by chance - the wording is always really specific, the concepts tested are really specific and even the answer choices are specific), so you can use the patterns used by the question-writers to your advantage.
2) The answer choices can often provide a clue as to how to approach the question. Here, they help to define what is NOT possible. If we were dealing with a 45/45/90 triangle, then the ratio of the legs would be 1:1. That's clearly not an option here, so the triangle cannot be a 45/45/90.

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Re: Two sides of a triangle have lengths x and y and meet at a r  [#permalink]

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New post 20 Jan 2019, 07:59
Thank you for these great explanations - I finally understand how to get to 6y = 8x. Then, as I understand it, I am looking for the ratio of x:y, or x/y. In that case, I simply divide both sides by y to get 6 = 8x/y and then divide both sides by 8 to get 6/8 = x/y. Obviously somehow I have reversed the two. Could someone help me with where I went wrong?

Thanks a ton!
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Re: Two sides of a triangle have lengths x and y and meet at a r  [#permalink]

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New post 24 Jan 2019, 14:09
xhimi wrote:
Two sides of a triangle have lengths x and y and meet at a right angle. If the perimeter of the triangle is 4x, what is the ratio of x to y ?

a) 2 : 3
b) 3 : 4
c) 4 : 3
d) 3 : 2
e) 2 : 1


Since it is a right trangle:
\(x^2 + y^2 = c^2\)
\(c = sqrt(x^2 + y^2)\)

Perimeter = 4x so \(x + y + sqrt(x^2 + y^2) = 4x\)

Simplify this down and we get \(\frac{x}{y} = \frac{3}{4}\)
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Re: Two sides of a triangle have lengths x and y and meet at a r  [#permalink]

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New post 24 Jan 2019, 23:12
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Please don't skip heaps of steps when posting. It can be much too hard to follow.

\(x+y+√(x^2+y^2)=4x\)
\(y+√(x^2+y^2)=3x\)
\(√(x^2+y^2)=3x-y\)
\(x^2+y^2=(3x-y)^2\)
\(x^2+y^2=9x^2-6xy+y^2\)
\(0=8x^2-6xy\)
\(6xy=8x^2\)
\(6y=8x\)
\(3y=4x\)
\(3y/4=x\)
\(3/4=x/y\)
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Re: Two sides of a triangle have lengths x and y and meet at a r   [#permalink] 24 Jan 2019, 23:12
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