Is x less than y? (1) x - y + 1 < 0 (2) x - y - 1 < 0

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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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

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!

Now we will solve this DS question using the Variable Approach

You can visit https://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.

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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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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x = - 2 and y= -1; -2 - (-1) < 1 but will not this results to (-2+1<-1 ?)

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