If x and y are positive numbers, is (x + 1)/(y + 1) x/y

Question

If x and y are positive numbers, is  \frac{x + 1}{y + 1} > \frac{x}{y} ?

(1) x > 1
(2) x < y

Ejiroosa · Answer

x+1/y+1>x/y
cross multiply
xy +y > xy + x
subtract xy from both sides
y > x or x<y?

Ans B.

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ydmuley · Answer

If x and y are positive numbers, is  \frac{x + 1}{y + 1} > \frac{x}{y} ?

x  &   y  are POSITIVE numbers

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

Lets simplify the equation, as we are given that x & y are positive integers we can multiply the R.H.S. with the L.H.S. as there will not be any impact on the inequality SIGN of the equation.

(x + 1) * (y) > (y + 1) * (x)

xy + y > xy + x

Cancelling  xy  from both the sides.

y > x

So we want to answer, Is  y > x  ?

(1)  x > 1

This does not tell us anything about y. Hence,  Eq. (1) =====> NOT SUFFICIENT 
 
(2)  x < y

We can write it as y > x, this is what we have to prove.  Hence, Eq. (2) =====> SUFFICIENT

Hence, Answer is B

BrentGMATPrepNow · Answer

Target question :    Is (x + 1)/(y + 1) > x/y ?

Given : x and y are positive numbers

This is a great candidate for  rephrasing the target question .

We have the inequality:  (x + 1)/(y + 1) > x/y 
Since y is positive, we can multiply both sides of the inequality by y
Likewise, since y is positive, we know that y+1 is positive, which means we can multiply both sides of the inequality by (y+1)
When perform both of these multiplications we get:  (x + 1)(y) > (x)(y + 1) 
Expand to get:  xy + y > xy + x 
Subtract xy from both sides to get:  y > x 
So, we can now ask...
  REPHRASED target question :    Is y > x ?

Statement 1 : x > 1   
There's no information about y, so there's no way to determine whether or not  y > x 
Since we cannot answer the  REPHRASED target question  with certainty, statement 1 is NOT SUFFICIENT

Statement 2 : x < y  
Perfect!!! 
Since we can answer the  REPHRASED target question  with certainty, statement 2 is SUFFICIENT

Answer: B

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MarkusKarl · Answer

Might as well share that I picked C due to not Reading that x and y - positive. This would not allow crossmultiplication.

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RaguramanS · Answer

x and y are positive numbers
 x,y>0 ;
(x+1)y>(y+1)x ;
y>x

(1) x > 1
X can be a proper fraction or improper fraction and y can be proper or improper fraction
 NS 
(2) x < y 
Rephrased expression
Suff

Option B

ScottTargetTestPrep · Answer

We are given that x and y are positive numbers and need to determine whether (x + 1)/(y + 1) > x/y. Simplifying the question, we have:

Is y(x + 1) > x(y + 1)? (Note that we multiplied each side by y(y + 1), which is allowed because y > 0.)

Is yx + y > xy + x?

Is y > x?

Statement One Alone:

x > 1

Since we do not know anything about y, statement one alone is not sufficient to answer the question.

Statement Two Alone:

x < y

We see that statement two has answered the question; y is greater than x.

Answer: B

divyadamahe · Answer

(1) x <1 -- Insufficient 
(2) x< y
let's understand:

Effects of addition
if the ratio is less than 1: After adding some positive number to the numerator and denominator, the resulting ration increases
eg 2/3
adding 10 to numerator and denominator 
2+10 / 3+10 =12/13 > 2/3

if the ratio is greater than 1: After adding some positive number to the numerator and denominator, the resulting ration decreases eg. 7/5 
7+20 /5 +20 = 27/25 <75

Hence B is correct

please correct me if I am wrong

TheUltimateWinner · Answer

Here, x and y are  positive numbers  (i.e.,  x>0  and  y>0 )
If this is the case then  \frac{x}{y}  must be positive. If  \frac{x}{y}  is positive, then  \frac{x + 1}{y + 1}  must be positive too..
So, the question is:
 \frac{x + 1}{y + 1} > \frac{x}{y} ?
I can answer this question if I know value of x and y (whatever the value it is!).

In statement 2: 
Here is given that  x < y . 
I don't care what about the specific value of x and y here. I know from this statement that y is greater than x. If the statement even says that x is greater than y, i still don't care because  statement NEVER lies! . Without any calculation in statement 2, I can surely say that it gives a definite YES or a definite NO, but NOT the simultaneous YES and NO. So,  sufficient.

In statement 1: 
There is no value of y. So, insufficient.
Is my understanding wrong  Bunuel ,  IanStewart ?
Thanks__

kamal97 · Answer

It might be a silly question, but before ruling Statement 1 out, can't we use the fee values of Y (as the question stem says Y is a positive number) ?
Use few positive values of Y and plug in the question to see if we get all Yes or all No?

keshavmishra · Answer

cross multiply the given expression, 
y(x+1)>x(y+1)
xy+y>xy+x
y>x ???

a. x>1. no info on Y. Not sufficient (Cancel AD)
b. y>x. Sufficient. 
Answer B

GMATinsight · Answer

Answer: Option B

Video solution by  GMATinsight

https://www.youtube.com/watch?v=9Q9xEJikMSU

avigutman · Answer

Video solution from Quant Reasoning: Subscribe for more: https://www.youtube.com/QuantReasoning? ... irmation=1 https://www.youtube.com/watch?v=Tg3VE6uOk0k

SkDv12 · Answer

I get that we can simplify this to Y>X by multiplying.

But shouldn't the sign of inequality also be reversed according to the multiplication rule of inequality. Why it doesn't change ??

avigutman · Answer

To what rule are you referring,  SkDv12  ?

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SkDv12 · Answer

avigutman  This is my mistake. As soon as I revised, I knew my error.

"Multiplying or dividing an inequality by a negative number reverses the order of the inequality". It is only if multiplication and division is done by a  -ve number . At the time, I was remembering it as with  any number .

avigutman · Answer

Your mistake shows that you memorized the "rule" without exploring the reasoning behind it. That approach to studying will significantly limit your score potential,  SkDv12 . Attaching two images from my book.

sset92 · Answer

I think a simpler approach here is - Only in case of proper fractions (N<D) when same number is added to both N and D, the fraction gets closer to one (greater). So basically without any cross multiplication you know that you need to check if x<y. Direct answer is B.

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If x and y are positive numbers, is (x + 1)/(y + 1) > x/y

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