GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 17 Jun 2019, 00:01 GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  Two sides of a triangle have lengths x and y and meet at a r

Author Message
TAGS:

Hide Tags

Intern  Status: Hulk
Joined: 17 Jul 2013
Posts: 11
Location: Singapore
Concentration: Entrepreneurship, Finance
GMAT 1: 600 Q44 V30 GMAT 2: 650 Q45 V34 GMAT 3: 690 Q49 V34 Two sides of a triangle have lengths x and y and meet at a r  [#permalink]

Show Tags

3
23 00:00

Difficulty:   45% (medium)

Question Stats: 67% (02:12) correct 33% (02:21) wrong based on 480 sessions

HideShow timer Statistics

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.
Retired Moderator Joined: 16 Jun 2012
Posts: 1006
Location: United States
Re: Two sides of a triangle have lengths x and y and meet at a r  [#permalink]

Show Tags

7
E is correct. Here is my solution.
Hope it helps.
Attachments math.png [ 9.17 KiB | Viewed 9987 times ]

_________________
Please +1 KUDO if my post helps. Thank you.

"Designing cars consumes you; it has a hold on your spirit which is incredibly powerful. It's not something you can do part time, you have do it with all your heart and soul or you're going to get it wrong."

Chris Bangle - Former BMW Chief of Design.
Verbal Forum Moderator B
Joined: 10 Oct 2012
Posts: 608
Re: Two sides of a triangle have lengths x and y and meet at a r  [#permalink]

Show Tags

1
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.
_________________
Intern  Status: FTW
Joined: 01 Apr 2012
Posts: 9
Location: India
GMAT Date: 09-27-2014
GPA: 3.5
WE: Consulting (Venture Capital)
Re: Two sides of a triangle have lengths x and y and meet at a r  [#permalink]

Show Tags

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
Retired Moderator Joined: 16 Jun 2012
Posts: 1006
Location: United States
Re: Two sides of a triangle have lengths x and y and meet at a r  [#permalink]

Show Tags

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
_________________
Please +1 KUDO if my post helps. Thank you.

"Designing cars consumes you; it has a hold on your spirit which is incredibly powerful. It's not something you can do part time, you have do it with all your heart and soul or you're going to get it wrong."

Chris Bangle - Former BMW Chief of Design.
EMPOWERgmat Instructor V
Status: GMAT Assassin/Co-Founder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 14343
Location: United States (CA)
GMAT 1: 800 Q51 V49 GRE 1: Q170 V170 Re: Two sides of a triangle have lengths x and y and meet at a r  [#permalink]

Show Tags

2
1
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....

GMAT assassins aren't born, they're made,
Rich
_________________
760+: Learn What GMAT Assassins Do to Score at the Highest Levels
Contact Rich at: Rich.C@empowergmat.com

*****Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!*****

Rich Cohen

Co-Founder & GMAT Assassin Follow
Special Offer: Save $75 + GMAT Club Tests Free Official GMAT Exam Packs + 70 Pt. Improvement Guarantee www.empowergmat.com/ Director  G Joined: 23 Jan 2013 Posts: 547 Schools: Cambridge'16 Re: Two sides of a triangle have lengths x and y and meet at a r [#permalink] Show Tags 1 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 Intern  Joined: 06 Mar 2015 Posts: 7 Re: Two sides of a triangle have lengths x and y and meet at a r [#permalink] Show Tags 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? Director  Joined: 07 Aug 2011 Posts: 518 Concentration: International Business, Technology GMAT 1: 630 Q49 V27 Two sides of a triangle have lengths x and y and meet at a r [#permalink] Show Tags 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 EMPOWERgmat Instructor V Status: GMAT Assassin/Co-Founder Affiliations: EMPOWERgmat Joined: 19 Dec 2014 Posts: 14343 Location: United States (CA) GMAT 1: 800 Q51 V49 GRE 1: Q170 V170 Re: Two sides of a triangle have lengths x and y and meet at a r [#permalink] Show Tags 2 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 _________________ 760+: Learn What GMAT Assassins Do to Score at the Highest Levels Contact Rich at: Rich.C@empowergmat.com *****Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!***** Rich Cohen Co-Founder & GMAT Assassin Follow Special Offer: Save$75 + GMAT Club Tests Free
Official GMAT Exam Packs + 70 Pt. Improvement Guarantee
www.empowergmat.com/ B
Joined: 13 Jan 2015
Posts: 97
Location: United Kingdom
Concentration: Other, General Management
Schools: LBS '19 (WL)
GMAT 1: 690 Q48 V36 Re: Two sides of a triangle have lengths x and y and meet at a r  [#permalink]

Show Tags

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
Manager  S
Joined: 24 Dec 2016
Posts: 96
Location: India
Concentration: Finance, General Management
WE: Information Technology (Computer Software)
Re: Two sides of a triangle have lengths x and y and meet at a r  [#permalink]

Show Tags

The question can very easily be solved using the answer options, in my opinion. It would take lesser time than the algebra would.
SC Moderator P
Status: GMAT - Pulling Quant and Verbal together
Joined: 04 Sep 2017
Posts: 201
Location: United States (OH)
GPA: 3.6
WE: Sales (Computer Software)
Re: Two sides of a triangle have lengths x and y and meet at a r  [#permalink]

Show Tags

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
_________________
Would I rather be feared or loved? Easy. Both. I want people to be afraid of how much they love me.

How to sort questions by Topic, Difficulty, and Source:
https://gmatclub.com/forum/search.php?view=search_tags
Intern  Joined: 14 Sep 2014
Posts: 9
Re: Two sides of a triangle have lengths x and y and meet at a r  [#permalink]

Show Tags

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.

Posted from my mobile device
EMPOWERgmat Instructor V
Status: GMAT Assassin/Co-Founder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 14343
Location: United States (CA)
GMAT 1: 800 Q51 V49 GRE 1: Q170 V170 Re: Two sides of a triangle have lengths x and y and meet at a r  [#permalink]

Show Tags

1
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.

GMAT assassins aren't born, they're made,
Rich
_________________
760+: Learn What GMAT Assassins Do to Score at the Highest Levels
Contact Rich at: Rich.C@empowergmat.com

*****Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!*****

Rich Cohen

Co-Founder & GMAT Assassin Follow
Special Offer: Save \$75 + GMAT Club Tests Free
Official GMAT Exam Packs + 70 Pt. Improvement Guarantee
www.empowergmat.com/
Intern  B
Joined: 02 Feb 2018
Posts: 3
GMAT 1: 710 Q42 V46 Re: Two sides of a triangle have lengths x and y and meet at a r  [#permalink]

Show Tags

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!
Manager  S
Joined: 22 Sep 2018
Posts: 249
Re: Two sides of a triangle have lengths x and y and meet at a r  [#permalink]

Show Tags

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 $$\frac{x}{y} = \frac{3}{4}$$
Senior Manager  P
Joined: 15 Feb 2018
Posts: 257
Re: Two sides of a triangle have lengths x and y and meet at a r  [#permalink]

Show Tags

1
Please don't skip heaps of steps when posting. It can be much too hard to follow.

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

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