Timed Quant Section (15 Questions, Hard) #9 (Question 8)

[Archi]

I believe the answer to this question should be D. While I reviewed Greg’s solution, I think we should also consider negative values for 𝑥 and 𝑦. This is because under the radical, the expression involves (𝑥 + 𝑦)^2, which is always non-negative regardless of whether 𝑥 or 𝑦 are negative. If we account for the possibility of negative values, there are multiple ways to achieve a sum of 4 other than 𝑥 = 3 and 𝑦 = 1. For example, 𝑥 = 5 and 𝑦 = − 1 also work. I tested these values in the radical and confirmed they do not result in a negative value. Additionally, what about the case 𝑥 = 4 and 𝑦 = 0 (Question doesn’t mention it can’t be 0) ? If we include this possibility as well, then the answer becomes D.

I’d appreciate your insights on why these cases are not being considered in the solution provided.

Thank you!

[Leaderboard]

@Archi

For example, 𝑥 = 5 and 𝑦 = − 1 also work

Hint: x < y

[Archi]

aah…I misread it as y>x. Thanks for pointing that out but we can use negatives right like (8, -4).Since Greg, mentioned in the solution video that negatives aren’t allowed so I wanted to understand why is that so because we are always getting a positive expression inside the root

[Leaderboard]

@Archi

we can use negatives right like (8, -4).

Even then, if x = 8 and y = -4, x > y.

[Archi]

yeah but then x would be greater than 3 (quantity B) right? So, answer would be D?

[Leaderboard]

It doesn’t meet the “x > y” condition?

[Archi]

makes sense. I got unnecessarily confused. Thank you for your help.