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655-705 (Hard)|   Algebra|      
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Bunuel
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Does (x − y)^2 = 100 ? (1) x^2 + y^2 = 52 (2) x + y = 2 04 Aug 2020, 22:55
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

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55% (hard)

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64% (01:59) correct 36% (02:01) wrong based on 161 sessions

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Does (x − y)^2 = 100 ?

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


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C
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Does (x − y)^2 = 100 ? (1) x^2 + y^2 = 52 (2) x + y = 2 04 Aug 2020, 23:57
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Kindly see the attachment.

DONOT Fall under the trap of assuming (x-y)^2 = 100 for the first Option.
INS - Insufficient
IMO C
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1.jpeg [ 63.19 KiB | Viewed 3574 times ]

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Does (x − y)^2 = 100 ? (1) x^2 + y^2 = 52 (2) x + y = 2 05 Aug 2020, 00:16
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(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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Does (x − y)^2 = 100 ? (1) x^2 + y^2 = 52 (2) x + y = 2 05 Aug 2020, 00:25
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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).
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Does (x − y)^2 = 100 ? (1) x^2 + y^2 = 52 (2) x + y = 2 05 Aug 2020, 01:19
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Bunuel
Does (x − y)^2 = 100 ?

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

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.
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Does (x − y)^2 = 100 ? (1) x^2 + y^2 = 52 (2) x + y = 2 05 Aug 2020, 10:22
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Bunuel
Does (x − y)^2 = 100 ?

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



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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
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Does (x − y)^2 = 100 ? (1) x^2 + y^2 = 52 (2) x + y = 2 05 Aug 2020, 23:05
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(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
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Does (x − y)^2 = 100 ? (1) x^2 + y^2 = 52 (2) x + y = 2 05 Nov 2023, 08:32
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\(x^2 - 2xy + y^2 = 100 \)
\(x^2 + y^2 = 100 + 2xy\)..... [1]

St. 1:\( x^2 + y^2 = 52\)
Substituting in [1]
\(52 = 100 + 2xy\)
\(-2xy = 48\)
\(-xy = 24\) or \(xy = -24\).... [2]

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.
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