Does (x − y)^2 = 100 ? (1) x^2 + y^2 = 52 (2) x + y = 2

Question

Does (x − y)^2 = 100 ?

(1) x^2 + y^2 = 52 
(2) x + y = 2

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sambitspm · Answer

Kindly see the attachment.

DONOT Fall under the trap of assuming (x-y)^2 = 100 for the first Option.    
INS - Insufficient
IMO C

yashikaaggarwal · Answer

(x − y)^2 = 100 
To prove: X^2+Y^2-2XY = 100

(1) x^2 + y^2 = 52
We need -2xy value to determine whether X^2+Y^2-2XY = 100 or not (insufficient)

(2) x + y = 2
Squaring both sides
X^2+y^2+2xy = 4
We need -2xy value not +2xy to determine whether X^2+Y^2-2XY = 100 or not (insufficient)

Combining both.
X^2+y^2-4 = -2xy
52-4 = -2xy
48 = -2xy

Putting value of -2xy in x^2+y^2 = 52
Subtracting -2xy from both sides
X^2+y^2-2xy = 52-2xy
(X-y)^2 = 52+48
(X-y)^2 = 100
(Sufficient)

Answer is C

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KaramveerBakshi · Answer

We need to find if (x-y)^2 = 100
Or x-y=+-10

Statements:

(1) x^2 + y^2 = 52 
We don't have another statement to solve for x and y.
 Insufficient.

(2) x+y=2
We don't have another statement to solve for x and y.
 Insufficient.

Combining, we have x^2 + y^2 = 52 and x+y=2
Solving we get (x,y) = (6,-4) or (-4,6)

We have x-y = 10 or -10.
In both cases, (x-y)^2 = 100
Hence,  Sufficient.

Answer is Option (C).

GMATWhizTeam · Answer

Solution 
 Step 1: Analyse Question Stem 
 • We need to find if  (x-y)^2 = 100  
So basically, we need to find the value of  (x-y)^2

Step 2: Analyse Statements Independently (And eliminate options) – AD/BCE 
 Statement 1:    x^2 + y^2 = 52 
 •  (x-y)^2 = x^2 + y^2 – 2xy ⟹ (x-y)^2 = 52 – 2xy 
• However, we don’t know the value of 2xy. 
Hence, statement 1 is NOT sufficient and we can eliminate answer Options A and D.

Statement 2:   x+y = 2 
 • With this statement, we cannot find the value of x-y, therefore we cannot find  (x-y)^2  
Hence, statement 2 is also NOT sufficient and we can eliminate answer Option B.

Step 3: Analyse Statements by combining. 
 • From statement 1:  x^2 + y^2 = 52….Eq.(i) 
 o And  (x-y)^2 = 52 – 2xy……Eq.(ii)  
• From statement 2:   x+ y = 2 
 o   (x+y)^2 = 4 ⟹ x^2 + y^2 + 2xy = 4…….Eq.(iii)  
• On combining both statements:
 o Subtracting Eq.(i) from Eq.(iii) we get,
   2xy = 4-52 = -48  
o By substituting the value of 2xy from the above equation into Eq.(ii), we can easily find the value of  (x-y)^2    
Thus, the correct answer is  Option C.

Archit3110 · Answer

target is (x − y)^2 = 100
or say x-y=10
#1
x^2 + y^2 = 52 
only possible when x,y (6,4)(-6,-4) ( -6,4) & ( 6,-4)
insufficient
#2
x+y=2
many possiblities insufficient
from 1 &2
6-4 ; possible pair
OPTION C

sood1596 · Answer

(x-y)^2 = 100?
Or x-y=+-10

(1) x^2 + y^2 = 52
(x-y)^2 = 100?
x^2 + y^2 + 2xy
=> 52 + 2xy 
We don't have values for x and y.
Insufficient.

(2) x+y=2
We don't have values for x and y.
Insufficient.

Combining, we have x^2 + y^2 = 52 and x+y=2
Solving we get x=6 and y =-4 or 
x=-4 and y=6

x-y = 10 or -10.
Sufficient

Answer -C

youcouldsailaway · Answer

x^2 - 2xy + y^2 = 100  
 x^2 + y^2 = 100 + 2xy .....

St. 1:  x^2 + y^2 = 52 
Substituting in  
 52 = 100 + 2xy 
 -2xy = 48 
 -xy = 24  or  xy = -24 ....

But -24 has many factors such as, -12 & 2, 6 & -4, -3 & 8, and -24 & 1. More pairings are also possible. Not sufficient.

St. 2:  x + y = 2 
x and y could be any number of values such as 1 & 1, 8 and -7, 6 and -5 etc. Not sufficient.

Combining St. 1 and 2:
From the factors of 24 above, the only possible way the combination works is if one of the values is 6 and the other is -4. Thus, C is the answer.

Does (x − y)^2 = 100 ? (1) x^2 + y^2 = 52 (2) x + y = 2

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