Is x less than y?

Is \(x<y\)? --> is \(x-y<0\)?

(1) x-y+1<0 --> \(x-y<-1\). Sufficient.
(2) x-y-1<0 --> \(x-y<1\). Not sufficient.

Answer: A.
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yes A it is

1.) x-y<-1 This is satisfied when x<y

2.) x-y<+1 Not sufficient

example x=1.5 and y=1; 1.5-1 < 1
x = - 2 and y= -1; -2 - (-1) < 1

A it is.

That's a very easy trap to fall for, I to fell for it.
Added the two inequalities and got x-y <0 and condluded both are necessary
But if you look at x-y +1 < 0 this means that x+y < -1
which means that X-Y < 0 as -1 < 0 , so we can get the answer from one only.

Now looking at 2) X-Y -1< 0 this means that x-y < 1 which doesn,t clearly mean that x<y as something less than 1 can be positive and negative both.

So A is the answer.
Damn i need more practice !

(1) \(x-y+1<0\) implies \(x-y<-1<0\), which means \(x-y<0\) or \(x<y\).
Hence sufficient.

(2) Take \(x = y = 0\) for which obviously x is not less than y, but (2) holds. For x = 0 and y = 1, (2) still holds, but now x < y.
Not sufficient.

Answer A.
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Given x>y

Then we need to find x-y > 0 ?

Stat 1: x-y-1>0
=> x-y > 1...Sufficient.

Stat 2: x-y + 1 > 0.
=> x-y > -1... Insufficient ( Since we need to seek x-y > 0 )

Thus A is correct answer.

The way you explained makes a lot more sense than the official explanation of Magoosh.

I can now see that statement 1 is sufficient by itself.
Thanks!

Bumping for DS butler Projec
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Target question: Is x < y?

Statement 1: x-y+1<0
Add y to both sides to get x+1 < y
Since x < x + 1, we can write: x < x+1 < y
From this, we can definitely see that x < y
So, statement 1 is SUFFICIENT


Statement 2: x - y - 1 < 0
Add y to both sides to get x - 1 < y
Since x - 1 < x,
We can see that y and x are both greater than x - 1, but we can't tell whether x < y
So, statement 2 is NOT SUFFICIENT

Answer: A

Cheers,
Brent
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Now we will solve this DS question using the Variable Approach

You can visit http://www.mathrevolution.com to learn and master this Approach along with other tips to quickly solve DS questions.

The first step of the Variable Approach: Modify the original condition and the question to suit the type of information given in the conditions. This is a priority and the most important step. The first step alone gives us a 30% chance of answering the question correctly.

=> We have to find whether x is less than y?

We have to apply CMT 1 as we are looking for a yes or no answer.

=> Is x < y => Is x - y < 0

Lets look at Condition (1) : It tells us that x - y + 1 < 0

=> Rearranging gives: x - y < - 1 that tells us x - y is less than '0'

Since the answer is yes, the condition is sufficient by CMT 1 which means that the answer must be in terms of a unique yes or no.

Lets look at Condition (2) : It tells us that x - y - 1 < 0

=> Rearranging gives: x - y < 1

=> If x = 2 and y = 1.5 then x - y = 2 - 1.5 = 0.5 < 1 but 0.5 is not less than 0 [NO]

=> But if x = - 2 and y = 1.5 then x - y = -2.5 - 1.5 = -4 < 1 and it is less than 0 [YES]

So we will not get a unique value. So, the condition is not sufficient by CMT 1 which means that the answer must be in terms of a unique yes or no.

Condition (1) ALONE is sufficient.

So, A is the correct answer.

Answer: A

In about 30% of cases, just knowing the first step of the Variable Approach will get you to the correct answer.
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Bunuel what if for ST 1 i have 4-6 < 1 then y could be -6 and x could be 4

then x is more than y

In your example, y is 6 not -6.
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MathRevolution What does CMT mean ? i think all DS questions require yes/no answer

Is x less than y?

x < y?

(1) x - y + 1 < 0
x + 1 < y

SUFFICIENT.

(2) x - y - 1 < 0

x - 1 < y

We can't conclude if x < y. INSUFFICIENT.

Answer is A.
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Bunuel not primarily for this question but for such questions that require us to plug in values for x/y to arrive to a ">" or "<" relation between them, can we do it by graphical method to find out the regions for this inequality to hold and thus avoid such number plug ins that may end up in us missing a test value.

I can plot the lines but I am not able to define which regions signify if X or greater or y. Kindly advise.