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# If |x| = |y|, x + y = ? (1) x - y = 4 (2) x/y = -1

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Joined: 02 Sep 2009
Posts: 62496
If |x| = |y|, x + y = ? (1) x - y = 4 (2) x/y = -1  [#permalink]

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12 Jul 2019, 07:00
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55% (hard)

Question Stats:

61% (01:28) correct 39% (01:26) wrong based on 325 sessions

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If $$|x| = |y|$$, $$x + y = ?$$

(1) $$x - y = 4$$

(2) $$\frac{x}{y} = -1$$

 This question was provided by Math Revolution for the Game of Timers Competition

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If |x| = |y|, x + y = ? (1) x - y = 4 (2) x/y = -1  [#permalink]

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Updated on: 13 Jul 2019, 08:30
4
1
Given: |x|=|y|
Question: x+y=?

Statement 1:
x−y=4
Since |x|=|y|
=> x = y or x=-y
x-y=4
x<>y => x=-y
x=2 & y=-2
x+y=0
SUFFICIENT

Statement 2:
x/y=−1
x=-y
x+y=0
SUFFICIENT

IMO D

Originally posted by Kinshook on 12 Jul 2019, 07:16.
Last edited by Kinshook on 13 Jul 2019, 08:30, edited 1 time in total.
##### General Discussion
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Joined: 06 Jun 2019
Posts: 138
If |x| = |y|, x + y = ? (1) x - y = 4 (2) x/y = -1  [#permalink]

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Updated on: 13 Jul 2019, 04:33
1
If |x|=|y|, x+y=?

Since x and y are equal (note: their absolute values are so), let's pick some easy numbers to work with.

ST1 x−y=4
Say x and y are 2 each. 2-2=0 (NO, must be 4). Does not work
Say x and y are (-2) each. (-2)-(-2)=0 (NO, must be 4). Does not work
Say x is (-2) and y is (2). (-2)-2=-4 (NO, must be 4). Does not work
Say x is 2 and y is (-2). 2-(-2)=4, yes it works. So 2+(-2)=0. We have an answer. Sufficient.

ST2 xy=−1
Let's do similar testing.
Say x and y are 2 each. 2/2=1(NO, must be (-1)). Does not work
Say x and y are (-2) each. (-2)/(-2)=1 (NO, must be (-1)). Does not work
Say x is (-2) and y is (2). (-2)/2=-1, yes it works. So, we have -2+2=0. We have an answer.
Say x is 2 and y is (-2). 2/(-2)=-1, yes it works. So 2+(-2)=0. Again, we have an answer. (Although two different sets of value works, both yield the same answer) Sufficient

Hence D
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Originally posted by JonShukhrat on 12 Jul 2019, 08:00.
Last edited by JonShukhrat on 13 Jul 2019, 04:33, edited 2 times in total.
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Re: If |x| = |y|, x + y = ? (1) x - y = 4 (2) x/y = -1  [#permalink]

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12 Jul 2019, 08:19
2
Given : |x|=|y|
Find: x+y? (single value should be available)
Analysis:
if x=y then, values are required
if x= -y then, x+y = 0
if -x = y then, x+y =0
(1)Statement 1:
x-y=4
x=y+4 => x not equal to y
we know |x|=|y| thus, we can have values like x(2), y(-2), implying that x+y = 0
Hence, sufficient

(ii) Statement 2:
x/y = -1
x=-y thus, x+y =0
Hence, Sufficient

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Re: If |x| = |y|, x + y = ? (1) x - y = 4 (2) x/y = -1  [#permalink]

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12 Jul 2019, 08:22
2
Quote:
If |x|=|y|, x+y=?

(1) x−y=4

(2) x/y=−1

It is given that |x|=|y|. Hence, either x = y or x = - y.
We need to find out what is x + y?
If x = y, then x + y is 2x or 2y.
If x = -y, then x + y = 0.

Let us analyze the statements separately first and see if it is required to analyze them together.

Statement 1:
x - y = 4.
From this statement we can conclude that x = y + 4, and x is not equal to y. From what we found in the beginning, it means that x = -y. In this case, we can substitute y by -x and write an equation as x + y = x - x = 0.
Statement is sufficient.

Statement 2:
x/y = -1.
From this stament, it can be calculated that x = -1 * y = -y. Here, we can solve it in the same way as we did above in analysis of the first statement.
Statement 2 is sufficient.

As we have both statements sufficient to solve the question, the Answer is D.
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If |x| = |y|, x + y = ? (1) x - y = 4 (2) x/y = -1  [#permalink]

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12 Jul 2019, 09:11
1
|x|=|y| --> x^2 = y^2 --> x^2 - y^2 = 0 --> (x-y)(x+y)=0

QUESTION: x+y=?

(1) x−y=4
If (x−y)(x+y)=0 and (x−y)=4, then (x+y) must be 0
(1) is SUFFICIENT

(2) x/y=−1
x/y=−1 --> x = −y --> x+y = 0
(2) is SUFFICIENT

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If |x| = |y|, x + y = ? (1) x - y = 4 (2) x/y = -1  [#permalink]

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12 Jul 2019, 12:24
1
1
Given: |x|=|y|

Question : x+y=0

St 1: x-y = 4 ==> x = 2, y = -2 ==> 2+(-2)=0 ===> sufficient since we can't find any other set of numbers that satisfies St 1 and |x|=|y|.
St 2: $$\frac{x}{y}$$=-1 ==> x=-y ==> x+y=0 ==> sufficient

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Re: If |x| = |y|, x + y = ? (1) x - y = 4 (2) x/y = -1  [#permalink]

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12 Jul 2019, 17:26
1
Statement 1
|x|=|y| and x-y=4
It implies that x and y gas opposite signs
Hence x+y=0
Sufficient

Statement 2
|x|=|y| and xy=-1
Again we get that x and y are equal in magnitude, but have opposite signs
x+y=0
sufficient

IMO D
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Re: If |x| = |y|, x + y = ? (1) x - y = 4 (2) x/y = -1  [#permalink]

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12 Jul 2019, 19:22
1
|x|=|y|
--> x = ± y
--> x = y (or) x = -y

(1) x − y = 4
--> x & y can have ONLY opposite sign; x is positive & y is negative
--> x = -y
-y - y = 4
--> -2y = 4
--> y = -2
--> x = -(-2) = 2

x + y = 2 - 2 = 0

Sufficient

(2) x/y=−1
--> x & y are of same value, BUT opposite sign
--> x = -y

x + y = -y + y = 0

Sufficient

IMO Option D

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Re: If |x| = |y|, x + y = ? (1) x - y = 4 (2) x/y = -1  [#permalink]

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12 Jul 2019, 22:22
1

If |x|=|y|, x+y=?

(1) x−y=4, from this we can get that x and y are opposite signs.
If they both were negative, zero: -2--2=-2+2=0.
If they both were positive aslo x-y=0.
So we infer that both of them need to be opposite signs and x need to be positive. eg: 2--2=2+2=4

(2) x/y=−1 from this we get x and y different signs, hence x+y=0, sufficient.

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If |x| = |y|, x + y = ? (1) x - y = 4 (2) x/y = -1  [#permalink]

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12 Jul 2019, 22:42
1
If |x|=|y| , x+y=?

$$|x|^2$$ = $$|y|^2$$ (squaring both sides)
$$x^2$$ = $$y^2$$
($$x^2$$-$$y^2$$) = 0
(x-y)(x+y) = 0
(x-y)=0 or (x+y) = 0

STATEMENT (1) x−y=4
this statement tells us that (x-y) 0
then definitely (x+y) = 0 SUFFICIENT

STATEMENT (2) $$\frac{x}{y}$$=−1
$$\frac{x}{y}$$+1 = 0
$$\frac{(x+y)}{y}$$=0
(x+y)=0 (since y cant be equal to zero)
SUFFICIENT

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Re: If |x| = |y|, x + y = ? (1) x - y = 4 (2) x/y = -1  [#permalink]

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13 Jul 2019, 00:12
1
If |x|=|y|, then x=y or x=-y
What is x+y?

1. x-y=6
The condition is that x=y or x=-y.
So x-y=6
The values of x and y that satisfy this condition are x=3 and y=-3 and this satisfies the condition x=-y
3-(-3)=6. Hence we can say that x+y=0
Statement 1 alone is sufficient.

Statement 2: x/y=-1
We know that both x and y can be both positive or they both negative or one can be positive while the other is negative.
So we mutiply both sides by y^2 to ensure the signs don’t change.
y^2+xy=0
y(y+x)=0
either y=0, which makes x=0, or y+x=0 which means that x=-y
And this is true for any pairs of x=-y, such as {1,-1; 2,-2; 3,-3; -4,4}
Hence we can say that x+y=0
Statement 2 is therefore sufficient.

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Re: If |x| = |y|, x + y = ? (1) x - y = 4 (2) x/y = -1  [#permalink]

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13 Jul 2019, 03:22
1
If |x|=|y|, x+y=?
Possibilities:
x and y can be pos and neg:
'+' + '+' = 2x or 2y
'-' + '-' = 2x or 2y
'+' + '-' = 0
'-' + '+' = 0

(1) x−y=4
if x and y:
'+' '+' = no sence, coz |x|=|y|
'-' '-' = no sence, coz |x|=|y|
'+' '-' = can be x = 2 y = -2
'-' '+' = no sense, coz result need to be negative
SUFFICIENT: '+' + '-' = 0

(2) xy=−1
if x and y:
'+' '+' = no sence, multiple must be positive
'-' '-' = no sence, multiple must be positive
'+' '-' = can be
'-' '+' = can be
SUFFICIENT:
'+' + '-' = 0
'-' + '+' = 0

Answ D
Re: If |x| = |y|, x + y = ? (1) x - y = 4 (2) x/y = -1   [#permalink] 13 Jul 2019, 03:22
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