Math Revolution and GMAT Club Contest! x + y = ?

Guys,

The first statement doesnt give us any value for x+y and the 2nd statement is against the math basics being that the absolute value of (1-2x) is always positive for all x and y^2 cannot be a negative number so answer is E right?

I picked E, although i have my doubts.


x + y = ?

(1) y=2x−1
knowing this, we have x+2x-1 = 3x-1, not sufficient.

(2) y^2=−|1−2x|
ok, how the hell can a squared number be negative? impossible.
let's multiply everything by -1.
-y^2 = |1-2x|
ok, now:
-y^2 = 1-2x or -y^2 = 2x-1

doesn't tell us much:

combine 1+2.

y=2x-1
-y^2 = either 1-2x or 2x-1.

square first equation to get 4x^2 - 4x + 1

now, negative of this, will equal either 1-2x or 2x-1, let's test both cases:
-(4x^2 - 4x + 1) = 1-2x
-4x^2 +4x - 1 = 1-2x
-4x^2 +6x -2 = 0 | divide everything by 2
-2x^2 +3x - 2 = 0
-2 = 2x^2 - 3x
-2 = x(2x-3)
doesn't tell much. so this should raise red flags that C is not sufficient.

-4x^2 +4x-1 = 2x-1
-4x^2 -2x=0
-4x^2 = 2x | divide by 2x
-2x = 1
x = -1/2
ok, now if x=-1/2, then y should be 0.

nevertheless, since 2 outcomes are possible. C is insufficient, and the answer is E.

We're asked whether, given the information in the two statements we can calculate the one value of x+y.

Statement 1) is obviously not sufficient to tie down \(x+y\) as \(y\) increases with \(x\) and therefore \(x+y\) can take an infinite number of values. For example if \(x=1\), \(y=1\), and \(x+y=2\); and if \(x=2\), \(y=3\), and \(x+y=5\). We have two different answers and therefore 1) is not sufficient.

Statement 2) is a little more complicated. We know that in GMAT no imaginary numbers are allowed, therefore a square number must not be \(<0\), and the left hand side of the equation is \(>=0\). The right hand side of the equation is a negative sign next to a modulus, and since a modulus is \(>=0\), the negative of a modulus is \(<=0\). The only value which each side of the equation can take is \(0\), and this gives us \(y=0\) and \((1-2x)=0\) which gives us \(x=1/2\). Therefore statement 2) is sufficient and the answer is B.

Find x+y

1) y = 2x - 1 -> no way to find what x+y is. Insufficient

2) y^2 = -|1-2x| -> this can be written as follows.

y^2 = 2x-1 (x can be any positive integer greater than 0)
y^2 = 1-2x (x can be any negative integer <= 0)

In either case, we cannot find x+y.

1&2 together -

when y^2 = 2x-1 & y = 2x-1 -> (x, y) = (1/2, 0) or (1, 1)
when Y^2 = 1-2x & y=2x-1 -> (x, y) = (1/2, 0) or (0, -1)

here (1/2, 0) satisfies both forms of y^2 = -|1-2x|. Hence x+y - 1/2. or Answer is C

(1) Insufficient.
(2) \(y^2 = -|1-2x| => y = 0 & x = 1/2 => x+y = 1/2\) Sufficient
Answer B

x + y =?

St1: y = 2x−1
x = 1 --> y = 1 --> x + y = 2
x = 2 --> y = 3 --> x + y = 5
St1 does not result in a unique solution. So St1 is not sufficient.

St2: y^2 = -|1 - 2x|
x = 1 --> y^2 = -1 --> y^2 can never be negative. So this answer is not possible.
x = -1 --> y^2 = -3 --> Again it is not possible
x = 1/2 --> y^2 = 0 --> y = 0 --> Possible

Thus x = 1/2 and y = 0 --> x + y = 1/2
St2 is sufficient

Answer: B

x + y = ?

(1) y=2x−1
(2) y 2 =−|1−2x|
Solution-
1)y+x=3x-1 ,hence not sufficient.
2)y 2 =−|1−2x| -->y2=0 (since y2 can't be negative)-->1-2x=0--->x=1/2------>x+y=1/2 Hence sufficient.

Correct Choice-B

Statement 1: x+y=x+2x-1=3x-1, not suficient
Statement 2:Y^2=-mod(1-2x)
the only possibility is 1-2x should be 0 because square of a number cannot be negative
=>x=1/2, y=0, x+y=1/2, sufficient
Answer=B

With Just I, we have 1 equation and 2 variables. Hence Not Sufficient

with II, we can construct 2 equations for 2 variables
\(y^2 = 1 - 2x\) and \(y^2 = 2x - 1\)
we can solve for x and y and find the value of x + y

Answer is B

With Just I, we have 1 equation and 2 variables. Hence Not Sufficient

with II, we can construct 2 equations for 2 variables
\(y^2 = 1 - 2x\) and \(y^2 = 2x - 1\)
we can solve for x and y and find the value of x + y

Answer is B

1) y=2x-1

2) y^2= -|1-2x|

now x+y=?

1) insufficient

2) as LHS will always be positive, we can say |1-2x| will be positive , so its equal to (1-2x)

thus, y^2 = -( 1-2x)
y^2 = 2x-1

again In sufficient

1+2.. gives (1/2,0) and (1,1).. but not unique soln

Thus ans E

QUESTION #9:
x + y = ?

(1) y=2x−1
(2) y2=−|1−2x|

Solution:

Statement: (A): not sufficient
(B): Not Sufficient
H'ever, taking both the statements together we got:
y^2=−|1−2x|
or, y^2=-(1-2x)
or, (2x-1)^2=-1+2x
or, (2x-1)(2x-1)=2x-1
or, (2x-1)(2x-1)-(2x-1)=0
or, (2x-1)((2x-1-1)=0
so, 2x-1=0; 2x-1-1=0
so, x=1/2; x-1
so, x=1/2, 1

when x=1, y=(2x2)-1=3
when x=1/2, y=(2.1/2)-1=0

Answer: "C"

QUESTION #9:

x + y = ?

(1) \(y=2x−1\)
(2) \(y^2=−|1−2x|\)

Answer:

(1) \(x+y=x+2x-1=3x-1\) - unknown x --> insufficient
(2) \(y^2=−|1−2x|\): can find x as function of \(y^2\) and plug in \(x+y\) but still y is unknown --> insufficient

(1) + (2):
\(y=2x-1\) --> \(x=\frac{y+1}{2}\) --> \(x+y= \frac{3y+1}{2}\) (*)
\(y^2=−|1−2x|\)
When \(x =<1/2\) --> \(y^2=2x−1\) --> \(x = \frac{y^2+1}{2}\) --> \(x+y=\frac{y^2+2y+1}{2}\) (**)
from (*) and (**) --> \(\frac{3y+1}{2}=\frac{y^2+2y+1}{2}\) --> \(y=0\) or \(y=1\) --> \(x+y\) has 2 solution --> insufficient

Answer E

from statement 1
y=2x-1. we cant find the values of x and y


from statement 2

y^2=-mod1-2x

square of any number is always>=0. of non negative.

mod always non negative. but statment mentions Y^2=-mod.
only Y=0 satisfy the equaiton so mod also 0.

so 1-2x=0, imply x=1/2.now we both values of x and y.
so option B is correct.

1) \(y=2x−1\) ==> \(y^2 =(2x−1)^2\) insufficient

2)\(y^2=−|1−2x| ==> y^2 = - 1 - 2x\) or \(y^2 = 1 + 2x\) insufficient

The two statments together:
Case 1
\((2x−1)^2= - 1 - 2x\)
\(4x^2 - 4x+1=-1-2x\)
\(4x^2-2x+2=0\)
No solution for x, discriminant (\(b^2 -4ac < 0\))


Case 2
\((2x−1)^2= 1 + 2x\)
\(4x^2 - 6x=0\)
\(x(4x-6) =0\)
\(x = 0\) or \(x =3/2\)

when \(x =0, y = -1 ==> x + y = -1\)
When \(x = 3/2, y = 2 ==> x + y = 7/2\)
We can't find a single result for the sum of x and y
Answer E

\(x + y =_________ ?\)
We are asked to find the value of \(x + y\).

Statement 1 : \(y = 2x − 1\)
--> \(y - 2x = - 1\)
There is no way this equation can be further modified to obtain the value of \(x + y\).

Not Sufficient.


Statement 2 : \(y^2 = − | 1 − 2x |\)
First, look closely at each part of this equation.

We have \(y^2\) on one side so we know that \(y^2\) on is either equal to a positive number or equal to \(0\).

We have \(− | 1 − 2x |\) on the other side. \(| 1 − 2x |\) is either equal to a positive number or to \(0\). Therefore \(− | 1 − 2x |\) will either be equal to a negative number or to \(0\).
So the only way \(y^2\) and \(− | 1 − 2x |\) can be equal to each other is if they are both equal to 0.

\(y^2 =\)0 so \(y = 0\). So you can solve the equation for x by plugging in the value of y. Once you will have done this, you will have the individual values of \(x\) and \(y\) and a fortiori the value of \(x + y\) . Note that at this stage, you don't need to calculate any further since this is a DS question.
Just for the sake of completeness, here is the end of the explanation:
\(− | 1 − 2x | = 0\) --> \(1 - 2x = 0\) --> \(x = \frac{1}{2}\)
So \(x + y = \frac{1}{2}\)

Sufficient

Answer: B

Statement 2 is sufficient.

Statement 1- We are given nothing that will help us get to the values for x and y or rather x+y.
Statement 2- we are given that square of y is equal to a negative number. This is only possible if that number is zero. Thus, y=0 and 1-2x=0 . which implies x=1/2. Thus we can get x+y = 1/2.

So, the answer is statement 2 sufficient but statement 1 is not sufficient. => option B.