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Re: Is (x + 1)/(y + 1) > x/y? (1) 0 < x < y (2) xy > 0
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24 Jun 2018, 14:33
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Bunuel wrote:
Is \(\frac{x + 1}{y + 1} > \frac{x}{y}\)?
(1) \(0 < x < y\)
(2) \(xy > 0\)
Target question:Is (x+1)/(y+1) > x/y ?
Statement 1: 0 < x < y This tells us that y is POSITIVE, which means y+1 is also POSITIVE. This means we can safely take the inequality (x+1)/(y+1) > x/y and safely multiply both sides by y When we do so, we get: (y)(x+1)/(y+1) > x We can also multiply both sides by y+1 to get: (y)(x+1) > (x)(y+1) Expand to get: xy + y > xy + x Subtract xy from both sides to get: y > x So, with the help of statement 1, our original target question Is (x+1)/(y+1) > x/y ? becomes Is y > x ? Since statement 1 tells us that y > x, the answer to the target question is a definitive YES Since we can answer the target question with certainty, statement 1 is SUFFICIENT
Statement 2: xy > 0 Let's TEST some values There are several values of x and y that satisfy statement 2 (xy > 0). Here are two: Case a: x = 1 and y = 1. In this case, (x+1)/(y+1) = (1+1)/(1+1) = 2/2 = 1, and x/y = 1/1 = 1. So, the answer to the target question is NO, (x+1)/(y+1) is NOT greater than x/y ? Case b: x = -3 and y = -2. In this case, (x+1)/(y+1) = (-3 +1)/(-2 +1) = -2/-1 = 2, and x/y = -3/-2 = 3/2. So, the answer to the target question is YES, (x+1)/(y+1) IS greater than x/y ? Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT
Answer: A
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_________________
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We'll simplify the inequalities so we understand what we need to do. This is a Precise approach.
(1) Multiplying both sides by y(y+1) (note that both are positive), gives yx+y > yx+x and cancelling out yx gives y > x. This is exactly what we're told in our statement! Sufficient.
(2) If y(y+1) is positive we can repeat the same process as above to get y > x, but since all we know is that both x,y are positive or both are negative but do not know which is larger we cannot answer. If y(y+1) is negative we'll get the expression y < x, but, for the same reason as above, we still cannot answer. Insufficient
The rephrased question now reads \(\frac{x+y}{y(y+1)} > 0\)
(1) \(0 < x < y\) Since x and y are both positive, the expression will always be positive. (Sufficient)
(2) \(xy > 0\) Here, both x and y can be positive or negative. If x and y are negative, the expression is NOT positive. We don't get a unique answer. (Insufficient)(Option A) _________________
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How can we cross multiply in this question, when we are not aware of the signs of x and y?
Warm Regards, FANJ
Hey FANewJersey, good question! In 1) we do know the sign: we are told that 0 < x < y - this means they are both bigger than 0 - they are positive! _________________
Re: Is (x + 1)/(y + 1) > x/y? (1) 0 < x < y (2) xy > 0
[#permalink]
25 Oct 2018, 04:23
Hi DavidTutorexamPAL, So basically you are reading ahead to get an idea and then applying something from the statements to firm up your stem question (is x less than y). Correct? Warmly, FANJ
Re: Is (x + 1)/(y + 1) > x/y? (1) 0 < x < y (2) xy > 0
[#permalink]
25 Oct 2018, 05:06
1
1
Bookmarks
FANewJersey wrote:
Hi DavidTutorexamPAL, So basically you are reading ahead to get an idea and then applying something from the statements to firm up your stem question (is x less than y). Correct? Warmly, FANJ
How can we cross multiply in this question, when we are not aware of the signs of x and y?
Warm Regards, FANJ
Hey FANewJersey, good question! In 1) we do know the sign: we are told that 0 < x < y - this means they are both bigger than 0 - they are positive!
You could say that, I suppose... I don't think of it as "reading ahead" since solving with statement (1) is the first thing we are supposed to try and do anyway. The question stem on its own isn't going to give us the answer.... _________________
Re: Is (x + 1)/(y + 1) > x/y? (1) 0 < x < y (2) xy > 0
[#permalink]
27 Oct 2018, 14:44
Thanks DavidTutorexamPAL, But we will have to nail down a goal, which we drive from the stem question (value vs. Yes/No, what is being asked etc. etc.). But I believe I got the idea here. Thanks
Is (x + 1)/(y + 1) > x/y? (1) 0 < x < y (2) xy > 0
[#permalink]
09 Jun 2020, 23:36
BrentGMATPrepNow wrote:
Bunuel wrote:
Is \(\frac{x + 1}{y + 1} > \frac{x}{y}\)?
(1) \(0 < x < y\)
(2) \(xy > 0\)
Target question:Is (x+1)/(y+1) > x/y ?
Statement 1: 0 < x < y This tells us that y is POSITIVE, which means y+1 is also POSITIVE. This means we can safely take the inequality (x+1)/(y+1) > x/y and safely multiply both sides by y When we do so, we get: (y)(x+1)/(y+1) > x We can also multiply both sides by y+1 to get: (y)(x+1) > (x)(y+1) Expand to get: xy + y > xy + x Subtract xy from both sides to get: y > x So, with the help of statement 1, our original target question Is (x+1)/(y+1) > x/y ? becomes Is y > x ? Since statement 1 tells us that y > x, the answer to the target question is a definitive YES Since we can answer the target question with certainty, statement 1 is SUFFICIENT
Statement 2: xy > 0 Let's TEST some values There are several values of x and y that satisfy statement 2 (xy > 0). Here are two: Case a: x = 1 and y = 1. In this case, (x+1)/(y+1) = (1+1)/(1+1) = 2/2 = 1, and x/y = 1/1 = 1. So, the answer to the target question is NO, (x+1)/(y+1) is NOT greater than x/y ? Case b: x = -3 and y = -2. In this case, (x+1)/(y+1) = (-3 +1)/(-2 +1) = -2/-1 = 2, and x/y = -3/-2 = 3/2. So, the answer to the target question is YES, (x+1)/(y+1) IS greater than x/y ? Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT
Answer: A
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Hi BrentGMATPrepNow Why do we need to make sure that x and y are positive before we simplify the inequality? Couldn't we simplify the inequality first and then change the truth of the inequality if we find out that x or y are negative?
i.e. can we not just simplify the question stem to say: "is y>x?" before being given the information in (1)?
Re: Is (x + 1)/(y + 1) > x/y? (1) 0 < x < y (2) xy > 0
[#permalink]
10 Jun 2020, 04:35
1
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Expert Reply
silverprince wrote:
Hi BrentGMATPrepNow Why do we need to make sure that x and y are positive before we simplify the inequality? Couldn't we simplify the inequality first and then change the truth of the inequality if we find out that x or y are negative?
i.e. can we not just simplify the question stem to say: "is y>x?" before being given the information in (1)?
Re: Is (x + 1)/(y + 1) > x/y? (1) 0 < x < y (2) xy > 0
[#permalink]
10 Jun 2020, 12:41
BrentGMATPrepNow wrote:
silverprince wrote:
Hi BrentGMATPrepNow Why do we need to make sure that x and y are positive before we simplify the inequality? Couldn't we simplify the inequality first and then change the truth of the inequality if we find out that x or y are negative?
i.e. can we not just simplify the question stem to say: "is y>x?" before being given the information in (1)?
Sure, so here's how I approached solving the problem.
The question initially asks: is \(\frac{x+1}{y+1}>\frac{x}{y}\)?
I eliminated the fraction by initially multiplying both sides first by \(y\) and then by \((y+1)\).
This results in: is \(y(x+1) > x(y+1)\)?
Finally expanding the brackets gives: is \(xy+y > xy+x\)?
\(xy\) cancels out on both sides of the inequality and the question stem can be rephrased as: is \(y>x\)?
Now when we look at the data provided, we're searching specifically for evidence that proves or disproves that y is greater than x.
(1) clearly shows that y is greater than x --> SUFFICIENT
(2) shows that either x and y are both positive or both negative. Since the combinations x=1;y=2 and x=-1 y=-2 satisfy this condition but yield different results, then INSUFFICIENT
Would you agree with the initial approach of simplifying the inequality given in the question stem?
Sure, so here's how I approached solving the problem.
The question initially asks: is \(\frac{x+1}{y+1}>\frac{x}{y}\)?
I eliminated the fraction by initially multiplying both sides first by \(y\) and then by \((y+1)\).
This results in: is \(y(x+1) > x(y+1)\)?
Finally expanding the brackets gives: is \(xy+y > xy+x\)?
\(xy\) cancels out on both sides of the inequality and the question stem can be rephrased as: is \(y>x\)?
Now when we look at the data provided, we're searching specifically for evidence that proves or disproves that y is greater than x.
(1) clearly shows that y is greater than x --> SUFFICIENT
(2) shows that either x and y are both positive or both negative. Since the combinations x=1;y=2 and x=-1 y=-2 satisfy this condition but yield different results, then INSUFFICIENT
Would you agree with the initial approach of simplifying the inequality given in the question stem?
Thanks
That approach seems valid for this particular question. I have a feeling it could cause issues with other questions (but I can't think of an example at the moment) Also, the hard part is remembering that the rephrased target question may or may not be accurate. _________________
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