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Bunuel
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If 0 > y > –1, then which of the following inequalities must be true? 24 Feb 2022, 23:15
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56% (02:20) correct 44% (02:12) wrong based on 468 sessions

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If 0 > y > –1, then which of the following inequalities must be true?

I. \(y^4 > y^3\)

II. \(y^5 > y^3 + y^2\)

III. \(y^5 – y^6 > y^3 – y^4\)

(A) I only
(B) II only
(C) I and II only
(D) I and III only
(E) I, II, and III



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D
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false9
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If 0 > y > –1, then which of the following inequalities must be true? 24 Feb 2022, 23:58
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I think D.

Y is negative. Y lies between -1 to 0.

I. \(Y^4\) is positive and \(Y^3\) is negative so I is correct.

II. \(Y^2\) is positive and will be greater than \(Y^3\) as Y lies between -1 to 0. So expression \(Y^3\)+\(Y^2\) is positive, while \(Y^5\) is a negative quantity. So this inequality is incorrect.

III. In LHS we have \(Y^5\) - \(Y^6\) , which is, \(Y^5\)\((1-Y)\)


Similarly RHS can be written as \(Y^3\)\((1-Y)\)

Now \((1-Y)\) is common in both sides and it is positive, which ultimately leaves us with \(Y^5\)\(>\)\(Y^3\) , which is correct, so this inequality is correct.

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If 0 > y > –1, then which of the following inequalities must be true? 25 Jun 2025, 05:29
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What did I do wrong here?

Condition B:

\(y^{5} > y^{3} + y^{2}\)
\(y^{5} > y^{2} (y + 1)\)

\(\frac{y^{5}}{y^{2}} < (y + 1)\) (flipped the sign as y is negative)
\(y^{3} < (y + 1)\) ; which is a true statement since y+1 would always be positive
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If 0 > y > –1, then which of the following inequalities must be true? 26 Jun 2025, 10:00
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you dont flip the inequality sign since y^2 is always positive
ZIX
What did I do wrong here?

Condition B:

\(y^{5} > y^{3} + y^{2}\)
\(y^{5} > y^{2} (y + 1)\)

\(\frac{y^{5}}{y^{2}} < (y + 1)\) (flipped the sign as y is negative)
\(y^{3} < (y + 1)\) ; which is a true statement since y+1 would always be positive
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If 0 > y > –1, then which of the following inequalities must be true? 26 Jun 2025, 23:49
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If 0 > y > –1, then which of the following inequalities must be true?

I. \(y^4 > 0 > y^3\) : MUST BE TRUE

II. \(y^5 > y^3 + y^2\) : 0 > y^5 > y^3; But y^2 >0: MAY OR MAY NOT BE TRUE

III. \(y^5 – y^6 > y^3 – y^4\) : y^4 - y^6 = y^4 (1 - y^2) > y^3 - y^5 = y^3(1-y^2) : MUST BE TRUE

(A) I only
(B) II only
(C) I and II only
(D) I and III only
(E) I, II, and III

IMO D
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If 0 > y > –1, then which of the following inequalities must be true? 28 Jun 2025, 01:53
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Y^5 > Y^3

Can you explain why is this assumed correct by default.
I can understand it to be correct if Y > 1 but here -1<Y<0

Also, can the above expression be written as Y^2 >1
If yes, the this statement is definitely false

false9
I think D.

Y is negative. Y lies between -1 to 0.

I. \(Y^4\) is positive and \(Y^3\) is negative so I is correct.

II. \(Y^2\) is positive and will be greater than \(Y^3\) as Y lies between -1 to 0. So expression \(Y^3\)+\(Y^2\) is positive, while \(Y^5\) is a negative quantity. So this inequality is incorrect.

III. In LHS we have \(Y^5\) - \(Y^6\) , which is, \(Y^5\)\((1-Y)\)


Similarly RHS can be written as \(Y^3\)\((1-Y)\)

Now \((1-Y)\) is common in both sides and it is positive, which ultimately leaves us with \(Y^5\)\(>\)\(Y^3\) , which is correct, so this inequality is correct.

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If 0 > y > –1, then which of the following inequalities must be true? 28 Jun 2025, 02:58
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RBgmatPre
Y^5 > Y^3

Can you explain why is this assumed correct by default.
I can understand it to be correct if Y > 1 but here -1<Y<0

Also, can the above expression be written as Y^2 >1
If yes, the this statement is definitely false

false9
I think D.

Y is negative. Y lies between -1 to 0.

I. \(Y^4\) is positive and \(Y^3\) is negative so I is correct.

II. \(Y^2\) is positive and will be greater than \(Y^3\) as Y lies between -1 to 0. So expression \(Y^3\)+\(Y^2\) is positive, while \(Y^5\) is a negative quantity. So this inequality is incorrect.

III. In LHS we have \(Y^5\) - \(Y^6\) , which is, \(Y^5\)\((1-Y)\)


Similarly RHS can be written as \(Y^3\)\((1-Y)\)

Now \((1-Y)\) is common in both sides and it is positive, which ultimately leaves us with \(Y^5\)\(>\)\(Y^3\) , which is correct, so this inequality is correct.

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Since -1 < y < 0, then y^5 > y^3 because raising a negative number between -1 and 0 to a higher odd power makes it closer to zero (less negative). So y^5 is greater than y^3 on the number line, even though both are negative:

--(-1)---(y^3)---(y^5)---0---

Also, yes, you could divide y^5 > y^3 by y^3, but since y^3 is negative, you need to flip the sign. That gives y^2 < 1, which holds true for all -1 < y < 0.
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