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17 Dec 2015, 21:58
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x + y = ?
(1) y=2x−1
(2) y^2=−|1−2x|
As this is a two variable equation with two unknowns there should be two equations to solve it.
From eq1 alone we cant find x+y so using eq2 we should see wether x+y can be found or not
From eq2 for x<1/2 |1-2x| is positive so y^2=2x-1 this happens only when y=y^2 so Y=1 => X=1
for x>1/2 Y^2=1-2X and Y=2x-1 this happens when Y=-1 => X=0
for x=1/2 Y=0 so, by using the above two conditions we can determine x+y
(1) y=2x−1
(2) y^2=−|1−2x|
As this is a two variable equation with two unknowns there should be two equations to solve it.
From eq1 alone we cant find x+y so using eq2 we should see wether x+y can be found or not
From eq2 for x<1/2 |1-2x| is positive so y^2=2x-1 this happens only when y=y^2 so Y=1 => X=1
for x>1/2 Y^2=1-2X and Y=2x-1 this happens when Y=-1 => X=0
for x=1/2 Y=0 so, by using the above two conditions we can determine x+y
19 Dec 2015, 02:57
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x+y = ?
1) y = 2x -1 clearly INSUFFICIENT. two unknown one equation.
for x = 0, y = -1
for x = 1/2, y = 0
for x = 1, y = 1
2) y^2 = - |1-2x|
y^2 + |1-2x| = 0, As we know that square of a number is always +ve and absolute value is always positive too. sum of two +ve entities can be zero only when their individual value is zero
Hence, y = 0 and |1-2x| will only be zero when x = 1/2
x+y = 1/2+0 = 1/2 --- definite, SUFFICIENT.
Option B is the correct answer.
1) y = 2x -1 clearly INSUFFICIENT. two unknown one equation.
for x = 0, y = -1
for x = 1/2, y = 0
for x = 1, y = 1
2) y^2 = - |1-2x|
y^2 + |1-2x| = 0, As we know that square of a number is always +ve and absolute value is always positive too. sum of two +ve entities can be zero only when their individual value is zero
Hence, y = 0 and |1-2x| will only be zero when x = 1/2
x+y = 1/2+0 = 1/2 --- definite, SUFFICIENT.
Option B is the correct answer.
19 Dec 2015, 04:30
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x + y = ?
Taking choice 1
(1) y=2x−1
x, y can be anything. When x =0, y=-1 or when x=1,y=1.
Therefore this choice is not sufficient.
(2) y2=−|1−2x|
x, y can be anything. When x =-1, y=- (3^0.5) or when x=1/2,y=0.
Therefore this choice is not sufficient.
Combining choice 1 and choice 2.
y=2x−1 , y2=−|1−2x|
Squaring choice 1. y^2 = (2x-1)^2 and substituting in choice 2 in place of y^2.
(2x-1)^2 = - |1-2x|
|1-2x| can also be written as (1-2x)^2 which can also be written as ((-1)(2x-1))^2 which is (2x-1)^2.
Therefore (2x-1)^2 = - (2x-1)^2
(2x-1)^2 + (2x-1)^2 =0
2((2x-1)^2)=0
x=1/2.
Substituting x value in choice 1 y=2x-1 gives y=0.
Therefore x+y= 1/2.
So, both statements (1) and (2) TOGETHER are sufficient to answer the question and answer choice C.
Taking choice 1
(1) y=2x−1
x, y can be anything. When x =0, y=-1 or when x=1,y=1.
Therefore this choice is not sufficient.
(2) y2=−|1−2x|
x, y can be anything. When x =-1, y=- (3^0.5) or when x=1/2,y=0.
Therefore this choice is not sufficient.
Combining choice 1 and choice 2.
y=2x−1 , y2=−|1−2x|
Squaring choice 1. y^2 = (2x-1)^2 and substituting in choice 2 in place of y^2.
(2x-1)^2 = - |1-2x|
|1-2x| can also be written as (1-2x)^2 which can also be written as ((-1)(2x-1))^2 which is (2x-1)^2.
Therefore (2x-1)^2 = - (2x-1)^2
(2x-1)^2 + (2x-1)^2 =0
2((2x-1)^2)=0
x=1/2.
Substituting x value in choice 1 y=2x-1 gives y=0.
Therefore x+y= 1/2.
So, both statements (1) and (2) TOGETHER are sufficient to answer the question and answer choice C.
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