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if |x+1|^2 =16 and (y-5)^2 =4, then what is the value of x/y?

1. y > x
2. y/x > 0

Let us look at the given equations

1) |x+1|^2=16, this gives 2 values of x , 3,-5 .......
2) (y-5)^2=4, this gives us only one value of y i.e 7

Let us look at the statement

A) Emmm..... y is always greater that X or in other words 7 is always greter than -5 and 3 --- not good enough to find x

B) if x is negetive then it will be the value y/x is smaller than 0, if x is +ve then is greater than 0. Hence we have only one possible value of x which wud be 3.

C) Not required since statement 2 alone is sufficient

Ans is C.
Simplifing the given data:
|x+1|^2 = 16 => |x+1| = +4 only => (x+1) = +or -4. Therefore, x =3 or -5.

(y-5)^2 = 4 => (y-5) = + or -2. Therefore, y = 7 or 3.

Statement 1 , y > x , is not sufficient since if, for example , x = -5, then y = 7 or y =3, and x/y will have 2 different values.

Statement 2 is insufficient since it merely suggests that x and y have the same sign.
combining the 2 statements implies that y=7 and x = -5 . Therefore, x/y = -5/7.

[...]
Statement 2 is insufficient since it merely suggests that x and y have the same sign.
[...]

If you establish that x and y have the same sign and you have already established that y can only take positive values (7 or 3), doesn't this lead to x being positive?

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