# Question: Is 2x less than y+1?

## Comment on Is 2x less than y+1?

### In stmnt 1, even 2x can be y

In stmnt 1, even 2x can be y<2x<y+1 like logic given for stmnt 2. Please explain In your suggested inequality (y<2x<y+1), y is less than 2x. This is not true.
In statement 1, we first concluded that 2x < y (after we multiplied both sides by 2). So y is definitely greater than 2x.
Since y < y+1, our only conclusion can be 2x < y < y + 1

### the condition y<y+1 is valid

the condition y<y+1 is valid only if Y is positive right ? ### The inequality y<y+1 is valid

The inequality y<y+1 is valid for ALL y-values.
When we add 1 to any value, that value becomes greater.

If you`re not convinced, try testing some y-values.

Try y = 0
We get: 0 < 0 + 1
Simplify: 0 < 1 (works)

Try y = 5
We get: 5 < 5 + 1
Simplify: 5 < 6 (works)

Try y = -3
We get: -3 < -3 + 1
Simplify: -3 < -2 (works)

And so on.

### I still dont understand why y

I still dont understand why y + 2 in statement 2 can go on the either side. If the question is asking if 2x < y + 1, and you know 2x < y + 2 to be true, how can it be that y + 1 go either side? ### All we know for certain is

All we know for certain is that y+1 < y+2
Statement two tells us (indirectly) that 2x < y+2
So where does y+1 fit here? We know that it must go to the left of y+2, but where?

Let's test some values...

CASE A: x = 2.75 and y = 10
This satisfies statement 2 (2x < y+2)
With these x- and y-values, y+1 fits BETWEEN 2x and y+2
We get: 2x < y+1 < y+2
So, we can conclude that 2x < y+1

CASE B: x = 2.75 and y = 4
This satisfies statement 2 (2x < y+2)
With these x- and y-values, y+1 fits to the LEFT of 2x
We get: y+1 < 2x < y+2
So, we can conclude that y+1 < 2x

Since the two CASES provide different answers to the target question, statement 2 is not sufficient. ### When to rephrase the Target

When to rephrase the Target Question . Like Here the target question can be rephrased as 2x-y<1, And that led me to the wrong conclusion. Please answer? ### There's nothing wrong with

There's nothing wrong with rephrasing the target question as "Is 2x-y < 1?" (as you have done)

For statement 1, we get: 2x < y
Subtract y from both sides to get: 2x - y < 0
This means 2x-y is NEGATIVE, which means 2x-y is definitely less than 1 (using your rephrased target question)
So, statement 1 is sufficient

For statement 2, we get: 2x < y + 2
Subtract y from both sides to get: 2x - y < 2
So, it COULD be the case that 2x - y = 1.5, in which case 2x-y > 1 (using your rephrased target question)
Or it COULD be the case that 2x - y = 0, in which case 2x-y < 1 (using your rephrased target question)
Since we cannot answer the rephrased target question with certainty, statement 2 is NOT sufficient

### Hey Guys,

Hey Guys,
When I rephrased, I got "Is x <= Y/2 and statement 1 addressed that easily and statement 2 was ambiguous. ### I like the fact that you're

I like the fact that you're looking for ways to rephrase the target question, but I'm not sure how you rephrased "Is 2x < y + 1?" to get "Is x ≤ y/2?"

Can you explain?

Cheers,
Brent

### Perhaps it was just an

Perhaps it was just an intuition that without the +1, the question would be Is 2x < y which would imply is x < y/2. However, adding 1 to y implies that the question is now asking if x <= y/2.
Did I confuse you?...lol ### I was a bit confused :-)

I was a bit confused :-)

Your rephrased target question worked THIS TIME, but I'd probably avoid such approximations in the future; they could lead to the wrong answer.

### Can you please verify my

Can you please verify my approach to this DS question?
(or suggest any alternative?)

Is x < 5?
1) x^2 > 5
2) x^2 + x <5

my approach:
1)x > 5^(1/2) and -5^(1/2),
x = 6 and -6, so insufficient

2) x < x^2 + x <5,
so, sufficient 