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19 Feb 2026, 20:00
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
49% (01:44) correct
51%
(02:03)
wrong
based on 158
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The rate at which a certain investment appreciates is R. If R is a function of the federal interest rate i, such that R = xi^3 + yi^2 + z, where x, y, and z are constants and i > 0. Is there a value of i for which R is 0?
(1) x > y > z
(2) x > 0
(1) x > y > z
(2) x > 0
20 Feb 2026, 05:23
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This is a beautifully constructed Data Sufficiency question — it looks intimidating because of the polynomial, but it's actually testing a clean concept once you strip it down.
Key concept being tested: Behavior of polynomials / whether a cubic function has a real root, combined with DS sufficiency logic
Restating the question: R = xi^3 + yi^2 + z, where i > 0. Can R equal 0 for some positive value of i?
Before touching the statements, think about what determines whether this polynomial can hit zero. It's a function of i, with coefficients x, y, z. The question is essentially: does the curve cross the x-axis for some i > 0?
Statement (1): x > y > z
This tells us the relative sizes of x, y, and z, but says nothing about their signs. For example:
- x=3, y=2, z=1 → R = 3i^3 + 2i^2 + 1. Since all terms are positive for i > 0, R > 0 always. No root.
- x=1, y=0, z=−2 → This satisfies x > y > z (1 > 0 > −2), and here R = i^3 − 2 = 0 when i = cube root of 2. Root exists.
Two different answers → Statement (1) is NOT sufficient.
Statement (2): x > 0
Again, tells us one coefficient is positive but nothing else. For example:
- x=1, y=1, z=1 → R = i^3 + i^2 + 1 > 0 for all positive i. No root.
- x=1, y=0, z=−5 → R = i^3 − 5 = 0 when i = cube root of 5. Root exists.
Two different answers → Statement (2) is NOT sufficient.
Combined: Even knowing x > y > z AND x > 0, we still can't determine the sign of z or whether R crosses zero. The examples from Statement (1) above (where x=3, y=2, z=1 gives no root and x=1, y=0, z=−2 gives a root) both satisfy x > 0 as well.
Combined statements are NOT sufficient.
Answer: E
Common trap: Students see a cubic polynomial and assume it must cross zero somewhere — because odd-degree polynomials always have at least one real root over all reals. But the domain here is restricted to i > 0, so you can't apply that general rule. The function might never reach zero on the positive side.
Takeaway: In Data Sufficiency, always check whether the stated domain restriction changes the behavior of a function before assuming general mathematical rules apply.
Key concept being tested: Behavior of polynomials / whether a cubic function has a real root, combined with DS sufficiency logic
Restating the question: R = xi^3 + yi^2 + z, where i > 0. Can R equal 0 for some positive value of i?
Before touching the statements, think about what determines whether this polynomial can hit zero. It's a function of i, with coefficients x, y, z. The question is essentially: does the curve cross the x-axis for some i > 0?
Statement (1): x > y > z
This tells us the relative sizes of x, y, and z, but says nothing about their signs. For example:
- x=3, y=2, z=1 → R = 3i^3 + 2i^2 + 1. Since all terms are positive for i > 0, R > 0 always. No root.
- x=1, y=0, z=−2 → This satisfies x > y > z (1 > 0 > −2), and here R = i^3 − 2 = 0 when i = cube root of 2. Root exists.
Two different answers → Statement (1) is NOT sufficient.
Statement (2): x > 0
Again, tells us one coefficient is positive but nothing else. For example:
- x=1, y=1, z=1 → R = i^3 + i^2 + 1 > 0 for all positive i. No root.
- x=1, y=0, z=−5 → R = i^3 − 5 = 0 when i = cube root of 5. Root exists.
Two different answers → Statement (2) is NOT sufficient.
Combined: Even knowing x > y > z AND x > 0, we still can't determine the sign of z or whether R crosses zero. The examples from Statement (1) above (where x=3, y=2, z=1 gives no root and x=1, y=0, z=−2 gives a root) both satisfy x > 0 as well.
Combined statements are NOT sufficient.
Answer: E
Common trap: Students see a cubic polynomial and assume it must cross zero somewhere — because odd-degree polynomials always have at least one real root over all reals. But the domain here is restricted to i > 0, so you can't apply that general rule. The function might never reach zero on the positive side.
Takeaway: In Data Sufficiency, always check whether the stated domain restriction changes the behavior of a function before assuming general mathematical rules apply.
21 Feb 2026, 11:15
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ExpertsGlobal5
Explanation:
R = xi3 + yi2 + z
The values of x, y, and z are unknown.
i > 0
We need to find whether there exists a value of i for which R is 0, which can also be expressed as:
We need to find whether there exists a value of i for which xi3 + yi2 + z is 0.
Statement (1)
x > y > z
Possibility 1: If x = 1, y = 0, z = -1 and i = 1 , then R = xi3 + yi2 + z = (1)(1)3 + (0)(1)2 – 1 = 1 – 1 = 0. In this case, there exists a value of i for which R is 0.
Possibility 2: If x = 3, y = 2, and z = 1, then R = xi3 + yi2 + z = (3)i3 + (2)i2 + 1. In this case, R will NOT be 0 for any value of i > 0.
It is NOT possible to determine with certainty whether there exists a value of i for which R is 0. Hence, Statement (1) is insufficient.
Statement (2)
Similar reasoning as above can be applied using the same values for x, y, and z to show that it is NOT possible to determine with certainty whether there exists a value of i for which R is 0. Hence, Statement (2) is insufficient.
As Statement (1) alone as well as Statement (2) alone is insufficient to answer the question, we need to now combine the two statements.
Statement (1) and Statement (2) combined
Similar reasoning as above can be applied using the same values for x, y, and z to show that it is NOT possible to determine with certainty whether there exists a value of i for which R is 0. Hence, Statement (1) and Statement (2) combined are insufficient.
E is the correct answer choice.
