If y≠0 and y≠-1, is x/y x/(y+1)?

Question

If y≠0 and y≠-1, is x/y > x/(y+1)? (1) x≠0 (2) x > y question21.JPG OPEN DISCUSSION OF THIS QUESTION IS HERE: if-y-0-and-y-1-is-x-y-x-y-109984.html

sreehari · Answer

C

x=3, y=2 => x/y=1.5, x/(y+1)=1 => x/y is greater
x=3, y=-2 => x/y=-1.5, x/(y+1)=-3 => x/y is greater
x=-2,y=-3 => x/y=-0.67, x/(y+1)=-1 => x/y is greater

akkifreaky · Answer

When x=-2,y=-3, then x/y=0.67 and x/y+1 = 1 and hence x/y+1 > x/y. So condition fails here and the ans shud be E.

subhashghosh · Answer

x/y - x/(y+1)

= (xy + x - xy)/y(y+1)

= x/y(y+1)

(1)

x != 0, so let x = -1, y = 1

So, x/y(y+1) = -1/2 < 0

x = 1, y = 1

So, x/y(y+1) = 1/2 > 0

Insufficient

(2)

x > y

let x = 0, y = any value < 0 (except -1)

Both the values are equal

x = 1, y = -2

So, x/y(y+1) = 1/(-2 * -1) = 1/2

x = -1, y = -2

So, x/y(y+1) = -1/(-2 * -1) = -1/2

Insufficient

(1) + (2)

Still Insufficient as per above cases

Answer - E

amit2k9 · Answer

a differs for y = +|- values.

b becomes equal for x = 0.

a+b for x = 3|-2 and y = 2|-3 the values are different.

Hence E.

Bunuel · Answer

If y≠0 and y≠-1, is x/y > x/(y+1)? Is \frac{x}{y}>\frac{x}{y+1} ? --> is \frac{x}{y}-\frac{x}{y+1}>0 ? --> is \frac{xy+x-xy}{y(y+1)}>0 ? --> is \frac{x}{y(y+1)}>0 ? (1) x≠0. Not sufficient. (2) x > y. Not sufficient. (1)+(2) Still not sufficient, for example: if x>0>(y=-1/2) answer is NO but if x>y>0 answer is YES. Answer: E. OPEN DISCUSSION OF THIS QUESTION IS HERE: if-y-0-and-y-1-is-x-y-x-y-109984.html

If y≠0 and y≠-1, is x/y > x/(y+1)?

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