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Bunuel
If x and y are positive integers, is \(\sqrt{x + y + 1} > 3\sqrt{x+ 1}\)?

(1) \(y > 8x + 8\)

(2) \(x + y > 8x + 8\)

Let's simplify the given equation, \(\sqrt{x + y + 1} > 3\sqrt{x+ 1}\)

=> \((\sqrt{x + y + 1})^2 > (3\sqrt{x+ 1})^2\)
=> x+y+1 > 9(x+1)
=> x+y+1 > 9x+9
=> y > 8x+8

We can rephrase the question as Is y > 8x+8 ?

Statement 1: \(y > 8x + 8\)

Statement 1 and the simplified given statement is same. So, statement 1 is SUFFICIENT

Statement 2: \(x + y > 8x + 8\)

If x = +ve, then y is not greater than 8x + 8, the answer is NO.
If x = -ve, then y is greater than 8x+8, the answer is YES.

Therefore, statement 2 is INSUFFICIENT.

The answer is B

IMO A
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Bunuel
If x and y are positive integers, is \(\sqrt{x + y + 1} > 3\sqrt{x+ 1}\)?

(1) \(y > 8x + 8\)

(2) \(x + y > 8x + 8\)

Given: x and y are positive integers

Asked: Is \(\sqrt{x + y + 1} > 3\sqrt{x+ 1}\)?

\(\sqrt{x + y + 1} > 3\sqrt{x+ 1}\)
=> \((\sqrt{x + y + 1})^2 > (3\sqrt{x+ 1})^2\)
=> x+y+1 > 9(x+1)
=> x+y+1 > 9x+9
=> y > 8x+8

(1) \(y > 8x + 8\)
SUFFICIENT

(2) [m]x + y > 8x + 8
y>7x+8
NOT SUFFICIENT

IMO A
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anbknaga
Bunuel
If x and y are positive integers, is \(\sqrt{x + y + 1} > 3\sqrt{x+ 1}\)?

(1) \(y > 8x + 8\)

(2) \(x + y > 8x + 8\)

Let's simplify the given equation, \(\sqrt{x + y + 1} > 3\sqrt{x+ 1}\)

=> \((\sqrt{x + y + 1})^2 > (3\sqrt{x+ 1})^2\)
=> x+y+1 > 9(x+1)
=> x+y+1 > 9x+9
=> y > 8x+8

We can rephrase the question as Is y > 8x+8 ?

Statement 1: \(y > 8x + 8\)

Statement 1 and the simplified given statement is same. So, statement 1 is SUFFICIENT

Statement 2: \(x + y > 8x + 8\)

If x = +ve, then y is not greater than 8x + 8, the answer is NO.
If x = -ve, then y is greater than 8x+8, the answer is YES.

Therefore, statement 2 is INSUFFICIENT.

The answer is B

IMO A

That was a typo. thanks for notifying it to me :)
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Bunuel
If x and y are positive integers, is \(\sqrt{x + y + 1} > 3\sqrt{x+ 1}\)?

(1) \(y > 8x + 8\)

(2) \(x + y > 8x + 8\)

Let's simplify the given equation, \(\sqrt{x + y + 1} > 3\sqrt{x+ 1}\)

=> \((\sqrt{x + y + 1})^2 > (3\sqrt{x+ 1})^2\)
=> x+y+1 > 9(x+1)
=> x+y+1 > 9x+9
=> y > 8x+8

We can rephrase the question as Is y > 8x+8 ?

Statement 1: \(y > 8x + 8\)

Statement 1 and the simplified given statement is same. So, statement 1 is SUFFICIENT

Statement 2: \(x + y > 8x + 8\)

If x = +ve, then y is not greater than 8x + 8, the answer is NO.
If x = -ve, then y is greater than 8x+8, the answer is YES.


Therefore, statement 2 is INSUFFICIENT.

The answer is B

IMO A



x & y are positive integers

y>7x+8 does not mean that y is NOT >8x+8
For example If y>5 than y may or may not be >8
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Bunuel
If x and y are positive integers, is \(\sqrt{x + y + 1} > 3\sqrt{x+ 1}\)?

(1) \(y > 8x + 8\)

(2) \(x + y > 8x + 8\)

Official Explanation



On the GMAT, the radical sign denotes the positive square root of something (even though, when we solve for a variable squared, there are positive and negative possibilities), so both sides of this inequality are positive. We can therefore square it and obtain:

x + y + 1 > 9(x + 1)
y > 8x + 8

This is the question we are being asked. On to the statements, separately first.

Statement (1) tells us exactly what we want to know. It says that y > 8x + 8 is a fact, so the answer to the question of whether y > 8x + 8 will be definitively "yes." Sufficient.

Statement (2) looks similar. It simplifies to y > 7x + 8. We'll analyze by cases. Say x = 1. In that case, the statement demands that y > 7 + 8. So the statement would allow, among other possibilities, y = 16 and y = 17. These two x-y possibilities yield opposite answers to the question posed, so this statement is insufficient.

The correct answer is (A).
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Bunuel
If x and y are positive integers, is \(\sqrt{x + y + 1} > 3\sqrt{x+ 1}\)?

(1) \(y > 8x + 8\)

(2) \(x + y > 8x + 8\)


Video Explanation



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