If x and y are positive numbers, is x

Question

If x and y are positive numbers, is  x<y ?

(1)  \frac{(x+1)}{(y+1)}>\frac{x}{y}

(2)  \frac{(x-1)}{(y-1)} < \frac{(x+1)}{(y+1)}

chetan2u · Accepted Answer

TIPS for these question :- The below observations will be true when both x and y are greater than a. 1) x and y are positive numbers and x\frac{x}{y} b) If you subtract same positive number from x and y, the fraction \frac{x-a}{y-a}<\frac{x}{y} 2) x and y are positive numbers and x>y or fraction \frac{x}{y}>1 a) If you ADD same positive number to x and y, the fraction \frac{x+a}{y+a}<\frac{x}{y} b) If you subtract same positive number from x and y, the fraction \frac{x-a}{y-a}>\frac{x}{y} let's see the choices 1. \frac{(x+1)}{(y+1)}>\frac{x}{y} TIP 1a, x

Theprince · Answer

Hello, EgmatQuantExpert I have a doubt regarding 2nd statement. question has mentioned x and y are positive numbers so if we consider 2 posibilities 1. x=2 ; y=3 it follows the inequality (2-1)/(3-1) < (2+1)/(3+1) -->1/2<3/4 gives us x2/-0.5<4/1.5 it follows inequality but we have x > y shoudn't answer be A? am i making any silly mistake? Do we need to take only integers into account. if yes, how do we decide as it's not mentioned in the question? Hello chetan2u, could you please help regarding my doubt?

mysterymanrog · Answer

I believe this question is best resolved by solving the inequalities. 
As we know that x and y are both positive, there is no need for us to worry about sign change. This implies that we can distribute the variables, and divide and multiply by them as we wish. 
Doing this for both 1 and 2, the end result is that y>x, a definite yes answer.

Theprince · Answer

mysterymanrog thanks for the help.but i still have a doubt. I think when you solved the inequalities you assumed that y-1 can't be negative this means that y is greater than 1 when you cross multiply we don't look at the individual value but value of the denominator. I think we can solve the 2nd statement by using 2 conditions 1.y>1 As you said this will give us x y is still positive but when you take this condition we get x >y I feel ans can be D when we have intergers only.

mysterymanrog · Answer

You raise an excellent point. https://www.wolframalpha.com/input?i2d= ... 9%2C1.5%5D Is the wolframalpha solution to the inequality, when y=0.5 Clearly the range given is (x>0.5) which implies that x>y. We have the opposite case (y>x) if y is some number >1. From statement B we can only derive that y!=1, letting us know that 01, which gives two conflicting answers. egmat could you shed some light on this please?

gmatophobia · Answer

Hello chetan2u - I have a general question about the property itself In the highlighted sentence (2a), for the property to hold true, shouldn't the value of a be less than the value of x? If a \geq x , then the numerator becomes non-positive and we can have \frac{(x-a) }{ (y - a) }< \frac{x }{ y} Example : x = 3; y = 2; a = 1 \frac{x}{y} < \frac{(x-a)}{(y-a)} \frac{3}{2} < \frac{2}{1} ⇒ This is in line with the property. x = 3; y = 2; a = 3 \frac{x}{y} > \frac{(x-a)}{(y-a)} \frac{3}{2} > \frac{0}{-1} \frac{3}{2} > 0 x = 3; y = 2; a = 4 \frac{x}{y} > \frac{(x-a)}{(y-a)} \frac{3}{2} > \frac{-1}{-2} \frac{3}{2} > \frac{1}{2} Please let me know if I am missing anything.

chetan2u · Answer

Hi gmatophobia

You are absolutely correct with your observation. 
If a is between x and y, (x-a)/(y-a) will always be negative and less than x/y. And a greater than both x and y will also have opposite outcome.

So, the general observations made above were for both x and y greater than a. I’ll put the caveat in the original post. 
Thanks 😊

siddhantvarma · Answer

KarishmaB  Is there a way to algebraically solve statement 2 to derive that x<y?

siddhantvarma · Answer

Statement (2) says: (x-1)/(y-1) < (x+1)/(y+1)

Case 1: x < y
Let x = 2, y = 3
(x-1)/(y-1) = 1/2 = 0.5
(x+1)/(y+1) = 3/4 = 0.75 
 (x-1)/(y-1) < (x+1)/(y+1) holds true

Case 2: x > y
Let x = 0.8, y = 0.5
(x-1)/(y-1) = -0.2/-0.5 = 2/5 = 0.4
(x+1)/(y+1) = 1.8/1.5 = 18/15 = 6/5 = 1.2
 (x-1)/(y-1) < (x+1)/(y+1) holds true

If the inequality holds true for both scenarios, how can we say that x < y from Statement (2) alone?

Bunuel   chetan2u

KarishmaB · Answer

Statement 2 is not sufficient alone. The answer needs to be (A), not (D).
When you take y-1 into account, you are including negative values because y can be a proper fraction (lies between 0 and 1).
Put x = 2/3 and y = 1/2 in statement 2 and it holds. But here x > y.

Also to solve it algebraically, start with statement II and take 2 cases:

y > 1
Cross multiply, inequality sign stays the same, you get x < y

and
0 < y < 1
Cross multiply, inequality sign flips and you get x > y

siddhantvarma

KarishmaB Is there a way to algebraically solve statement 2 to derive that x<y?

If x and y are positive numbers, is x < y?

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