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

It is currently 21 Oct 2019, 01:04

Close

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
Your Progress

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.

Close

Request Expert Reply

Confirm Cancel

For how many ordered pairs (x , y) that are solutions of the equation

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

Find Similar Topics 
Manager
Manager
avatar
S
Joined: 24 Jan 2019
Posts: 107
Location: India
Concentration: Strategy, Finance
GMAT 1: 730 Q51 V38
GPA: 4
Re: For how many ordered pairs (x , y) that are solutions of the equation  [#permalink]

Show Tags

New post 16 Jul 2019, 22:09
1
(x−y)^2+2(y^2)=18

For how many ordered pairs (x , y) that are solutions of the equation above are x and y both integers?


Both terms have minimum value of zero. if x, y are integers then (x-y) is also an integer.

Here, we need to divide 18 in such a way that it is of the form : (square of some integer) + 2* (square of other integer). Then we need to check for the values of x, y.

18 = 16 + 2*(1) = 0 + 2*(9)

first case, |x-y|=4 and |y|=1 ................possible (x,y) pairs: (3,-1),(-3,1),(5,1),(-5,-1).
second case, |x-y|=0 and |y| = 3...........possible (x,y) pairs: (0,-3),(0,3).

These are the only possibilities for given expression. (total 6)

ANSWER : C
Manager
Manager
User avatar
G
Joined: 12 Jan 2018
Posts: 117
Re: For how many ordered pairs (x , y) that are solutions of the equation  [#permalink]

Show Tags

New post 17 Jul 2019, 00:29
1
\((x−y)^{2}\)+2\(y^{2}\)=18

As we can see y can vary between -3 and +3.

Case 1: @y=+/-3, x is also +/-3.
So we have 2 solutions: (3,3) and (-3,-3)

Case 2:@y=+/-2 , no integer value of x satisfies the equation.

Case 3:@y=+1, x=5 or x=-3
So we have 2 solutions (-3,1) (5,1)

Case 4: @y=-1, x=3, x=-5
Sp we have 2 solutions (-5,-1) (3,-1)

Case 5: @y=0, no integer value of x satisfies the equation.

Therefore solutions are,
(3,3)
(-3,3)
(-3,1)
(5,1)
(-5,-1)
(3,-1)
Total 6 solutions.

Ans: C


_________________
"Remember that guy that gave up?
Neither does anybody else"
Manager
Manager
User avatar
G
Status: Not Applying
Joined: 27 Apr 2009
Posts: 179
Location: India
Schools: HBS '14 (A)
GMAT 1: 730 Q51 V36
Re: For how many ordered pairs (x , y) that are solutions of the equation  [#permalink]

Show Tags

New post 17 Jul 2019, 00:41
There are 6 integer solutions to this equation.
x = -5, y = -1
x = 5, y = 1
x = 3, y = -1
x = -3, y = 1
x = -3, y = -3
x = 3, y = 3
_________________
http://www.wizius.in
Better Prep. Better Scores. Better Schools

Guaranteed Admission to Top-50 MBA Programs
You either get-in or get your money-back.
Senior Manager
Senior Manager
User avatar
P
Joined: 20 Mar 2018
Posts: 379
Location: Ghana
Concentration: Finance, Real Estate
Re: For how many ordered pairs (x , y) that are solutions of the equation  [#permalink]

Show Tags

New post 17 Jul 2019, 01:40
(x−y)^(2)+2y^(2)=18
(x-y)^(2) + 2y^(2)= (9•2)
(x-y)^(2) + 2y^(2)=3^2•2

(x-y)^2 =3^2 , 2y^2=2
x-y= +/-3, y= +/-1

Four instances (1) +and + ,(2) -and + ,(3)+ and - (4)- and -

Instance 1 : x-y =3 ,y=1 —> x=4
.: (4,1)
Instance 2: x-y=-3,y=1 —> x=-2
.: (-2,1)
Instance 3: x-y=3,y=-1 —> x=2
.: (2,-1)
Instance 4: x-y=-3,y=-1 —> x=-4
.: (-4,-1)

And yes x and y are integers since we have an integer on the RHS of the equation and sum of squares on the LHS

Answer B

Posted from my mobile device
Intern
Intern
avatar
B
Joined: 09 Jul 2019
Posts: 38
Re: For how many ordered pairs (x , y) that are solutions of the equation  [#permalink]

Show Tags

New post 17 Jul 2019, 02:06
I manually counted and tried different numbers and got only four pairs of integers, therefore B
Intern
Intern
avatar
B
Joined: 11 Jun 2014
Posts: 23
CAT Tests
Re: For how many ordered pairs (x , y) that are solutions of the equation  [#permalink]

Show Tags

New post 17 Jul 2019, 03:43
Answer C
(x-y)^2+2y^2=18
putting values as follow
x=3,y=3
x=5,y=1
x=-3,y=1
x=3,y=-1
x=-5,y=-1
x=-3,y=-3
Manager
Manager
avatar
S
Joined: 07 Dec 2018
Posts: 113
Location: India
Concentration: Technology, Finance
GMAT 1: 670 Q49 V32
Re: For how many ordered pairs (x , y) that are solutions of the equation  [#permalink]

Show Tags

New post 17 Jul 2019, 04:22
1
\((x−y)^2\)+\(2y^2\)=18

Possible values for \(y^2\) can be from 1 to 9.
But to make y an integer, \(y^2\) can only be 1, 4 or 9.
If \(y^2\)=1 =>y=1 or -1
If \(y^2\)=4 =>y=2 or -2
If \(y^2\)=9 =>y=3 or -3

Now, if y=1,
\((x−y)^2\)+\(2y^2\)=18
\((x−1)^2\)+2=18
\((x−1)^2\) =16
So, x=5 or x=-3

Thus we get two points : (5,1), (-3,1)

Similarly if y=-1,
\((x−y)^2\)+\(2y^2\)=18
We will get two values of x=3 & -5

Thus we get two more points : (-5,-1), (3,-1)

If y=-2 or 2, X will not be an integer.

If y=-3 or 3, we will get two values of x : -3 & 3 respectively.

Thus we get two more points : (-3,-3), (3,-3)

So, ans should be 6.
Option (C)
Senior Manager
Senior Manager
User avatar
V
Joined: 25 Sep 2018
Posts: 428
Location: United States (CA)
Concentration: Finance, Strategy
GMAT 1: 640 Q47 V30
GPA: 3.97
WE: Investment Banking (Investment Banking)
CAT Tests
Re: For how many ordered pairs (x , y) that are solutions of the equation  [#permalink]

Show Tags

New post 17 Jul 2019, 04:25
1
Solution:
\((x - y)^2\) + \(2y^2\) = 18

Let's take x-y = a
& y=b
We get,
\(a^2\) + \(2b^2\) = 18...(i)
Substituting values in equation (i),

we can see that, if a=0 & x-y =0
b = 3 then & y=3. Therefore x= 3
(3,3) satisfying
(-3,-3) satisfying

Substituting values in equation (i)
a=4,-4,
b=1,-1

x-y = 4 & y=1
only (5,1) , (-5,-1) ,(3,-1) ,(-3,1) are satisfying,

So in total we have integer 6 pairs which are the solutions for x & y

Hence the answer is C.
_________________
Why do we fall?...So we can learn to pick ourselves up again
Senior Manager
Senior Manager
User avatar
P
Status: eternal student
Joined: 06 Mar 2018
Posts: 292
Location: Kazakhstan
GPA: 3.87
GMAT ToolKit User Premium Member
Re: For how many ordered pairs (x , y) that are solutions of the equation  [#permalink]

Show Tags

New post 17 Jul 2019, 05:08
(x−y)^2+2y^2=18

For how many ordered pairs (x , y) that are solutions of the equation above are x and y both integers?

I've tried to put different numbers to this equation and there are 4 (x,y) pairs possible.

A. 2
B. 4
C. 6
D. 8
E. 10


B is the answer. :heart
_________________
My SC approach flowchart

(no one is ideal, please correct if you see any mistakes or gaps in my explanation, it will be helpful for both of us, thank you)

___________________
"Nothing in this life is to be feared, it is only to be understood"
~ Marie Curie
Intern
Intern
avatar
B
Joined: 10 Aug 2017
Posts: 29
Re: For how many ordered pairs (x , y) that are solutions of the equation  [#permalink]

Show Tags

New post 17 Jul 2019, 05:27
For how many ordered pairs (x , y) that are solutions of the equation above are x and y both integers?

A. 2
B. 4
C. 6 (Answer)
D. 8
E. 10

Posted from my mobile device
Manager
Manager
avatar
S
Joined: 30 Nov 2016
Posts: 103
Location: India
Concentration: Finance, Strategy
Re: For how many ordered pairs (x , y) that are solutions of the equation  [#permalink]

Show Tags

New post 17 Jul 2019, 05:38
\((x-y)^2\)=18-\(2y^2\)
the only values of y which gives us a integer value for x are 1,-1
so the corresponding pair(x,y) would be:(5,1),(-5,-1)
Intern
Intern
User avatar
S
Joined: 24 Mar 2018
Posts: 48
Location: India
Concentration: Operations, Strategy
Schools: ISB '21
WE: Project Management (Energy and Utilities)
Re: For how many ordered pairs (x , y) that are solutions of the equation  [#permalink]

Show Tags

New post 17 Jul 2019, 06:09
1
\((x−y)^2+2y^2=18\)

Let x-y = z

\(z^2+2y^2=18\)

Both z^2 and y^2 are perfect squares. Following equations are possible with perfect square values:

16+2.(1) = 18 ---> eqn 1
0+2.(9) = 18 ---> eqn 2

For eqn 1, z=4, y=1 or z=-4, y=-1 or z=4, y=-1 or z=-4,y=1
x-y=4, y=1 or x-y=-4, y=-1 or x-y=4, y=-1 or x-y=-4,y=1
x=5,y=1 or x=-5,y=-1 or x=3,y=-1 or x=-3,y=1
There are 4 ordered pairs

For eqn 2, z=0, y=3 or z=0, y=-3
x-y=0, y=3 or x-y=0, y=-3
x=3,y=3 or x=-3,y=-3
There are 2 ordered pairs

In total, there are 6 ordered pairs.

Option (C).
Senior Manager
Senior Manager
avatar
G
Joined: 25 Jul 2018
Posts: 273
Re: For how many ordered pairs (x , y) that are solutions of the equation  [#permalink]

Show Tags

New post 17 Jul 2019, 06:54
1
(x−y)^2+2y^2=18 For how many ordered pairs (x , y) that are solutions of the equation above are x and y both integers?

Both two parts of the equation cannot be negative :
"(x-y)^2" and "2y^2" could be positive integer or 0(zero) to satisfy the equation.
We have only two options:
1) (x-y)^2=16 and 2y^2=2---> 16+2=18
2) (x-y)^2=0 and 2y^2=18---> 0+18=18.

1) 2y^2=2 --> y=1 --->(x-1)^2=16
(x-1)=4 --> x=5
(x-1)=-4 --> x=-3 (That means we have 2 pairs for y=1: (5;1) (-3;1)

y=-1 --> (x+1)^2=16
(x+1)=4 --> x=3
(x+1)=-4 --> x=-5 (That means we have 2 pairs for y=-1: (3;-1) (-5;-1)

2) 2y^2=18 y=3 --->(x-3)^2=0
(x-3)=0 --> x=3 (That means we have 1 pair for y=3: (3;3)

y=-3 --> (x+3)^2=0
(x+3)=0 --> x=-3 (That means we have 1 pair for y=-3: (-3;-3)
In total, 4 pairs for Case1 + 2 pairs for Case2 =6

The answer choice is C.
Director
Director
avatar
D
Joined: 20 Jul 2017
Posts: 930
Location: India
Concentration: Entrepreneurship, Marketing
WE: Education (Education)
Re: For how many ordered pairs (x , y) that are solutions of the equation  [#permalink]

Show Tags

New post 17 Jul 2019, 07:27
1
\((x−y)^2+2y^2=18\)

Possible values
Case 1: 0^2 + 2*3^2 = 18
--> (x, y) = (3, 3), (-3, -3)

Case 2: 4^2 + 2*1^2 = 18
--> (x, y) = (5, 1), (-3, 1), (3, -1), (-5, -1)

So, 6 pairs are possible

IMO Option C

Pls Hit Kudos if you like the solution
Manager
Manager
avatar
S
Joined: 10 Nov 2014
Posts: 63
Location: United States
Concentration: Marketing, Strategy
Re: For how many ordered pairs (x , y) that are solutions of the equation  [#permalink]

Show Tags

New post 17 Jul 2019, 21:39
Square + even no = 18 - therefore the square should also be even or 0.

The only possibilities are
0 + 18
4 + 14
16 + 2

First scenario, 0 + 18
(x-y)^2 = 0 and 2y^2=18
x=y and y=+/-3
(x,y) = (3,3) or (-3,-3)

Second scenario, 4 + 14
2y^2 =14, y^2 =7 which is not possible since y is an integer

Third scenario, 16 + 2
(x-y)^2=16 and 2y^2=2
x-y=+/-4 and y=+/-1

if y=1 and x-y=4 , we get y=1 and x=5
if y=1 and x-y=-4 we get y=1 and x=-3
if y=-1 and x-y=4 , we get y=-1 and x=3
if y=-1 and x-y =-4 , we get y=-1 and x=-5



(3,3)
(3,-3)
(5,1)
(3,-1)
(3,-1)
(-5,-1)

6 pairs
Ans choice C
GMAT Club Bot
Re: For how many ordered pairs (x , y) that are solutions of the equation   [#permalink] 17 Jul 2019, 21:39

Go to page   Previous    1   2   3   4   [ 75 posts ] 

Display posts from previous: Sort by

For how many ordered pairs (x , y) that are solutions of the equation

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  





Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne