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# If -1 < a < 0, q = a - 1, r = a^2, and s = a^3, then which of the foll

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If -1 < a < 0, q = a - 1, r = a^2, and s = a^3, then which of the foll  [#permalink]

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28 Oct 2018, 23:43
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If $$-1 < a < 0$$, $$q = a - 1$$, $$r = a^2$$, and $$s = a^3$$, then which of the following is true?

A. $$q < r < s$$

B. $$q < s < r$$

C. $$r < q < s$$

D. $$s < q < r$$

E. $$s < r < q$$

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Re: If -1 < a < 0, q = a - 1, r = a^2, and s = a^3, then which of the foll  [#permalink]

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29 Oct 2018, 01:07
1
From the question a is negative.

Then q will also be negative.
r will be positive.
and s will be negative.
Since s is cube of a then s will be lesser than q.

$$s<q<r$$

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Re: If -1 < a < 0, q = a - 1, r = a^2, and s = a^3, then which of the foll  [#permalink]

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29 Oct 2018, 07:28
Taking a= -0.5
Then Q= - 1.5
R = 0.25
S= -0.125

Q< A < S < R

B, methinks.
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Re: If -1 < a < 0, q = a - 1, r = a^2, and s = a^3, then which of the foll  [#permalink]

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13 Nov 2018, 16:48
Bunuel wrote:
If $$-1 < a < 0$$, $$q = a - 1$$, $$r = a^2$$, and $$s = a^3$$, then which of the following is true?

A. $$q < r < s$$

B. $$q < s < r$$

C. $$r < q < s$$

D. $$s < q < r$$

E. $$s < r < q$$

a is a negative fraction.

$$a^2 > a^3.$$

q = a -1. As a is negative q must be negative.

2 negative we have . but a^3 yields much smaller than q.

Thus , $$a^2>a^3>q.$$
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Re: If -1 < a < 0, q = a - 1, r = a^2, and s = a^3, then which of the foll  [#permalink]

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27 Jan 2019, 19:46
Bunuel wrote:
If $$-1 < a < 0$$, $$q = a - 1$$, $$r = a^2$$, and $$s = a^3$$, then which of the following is true?

A. $$q < r < s$$

B. $$q < s < r$$

C. $$r < q < s$$

D. $$s < q < r$$

E. $$s < r < q$$

We can choose -½ as the value of a.

Thus, r = ¼, s = -1/8, and q = -3/2.

We see that q < s < r.

Alternate Solution:

Subtracting 1 from each side of the inequality -1 < a < 0, we find -2 < a - 1 < -1, and since q = a - 1, we have -2 < q < -1.

Since a is between -1 and 0, the square of a will be between 0 and 1; therefore 0 < r < 1.

Finally, the third power of any negative number is also negative, and we cannot obtain a number less than -1 by taking the third power of any number between -1 and 0; therefore, -1 < s < 0.

Using the three inequalities obtained above, we find q < s < r.

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Re: If -1 < a < 0, q = a - 1, r = a^2, and s = a^3, then which of the foll  [#permalink]

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27 Jan 2019, 20:04
Bunuel wrote:
If $$-1 < a < 0$$, $$q = a - 1$$, $$r = a^2$$, and $$s = a^3$$, then which of the following is true?
A. $$q < r < s$$

B. $$q < s < r$$

C. $$r < q < s$$

D. $$s < q < r$$

E. $$s < r < q$$

take a value here a = -0.5

r will always be the greatest, Since it is a square, remove A, C and E

Now out of B and D

a^3 > a-1, for the given range $$-1 < a < 0$$

- 0.125 > -1.5

B
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Re: If -1 < a < 0, q = a - 1, r = a^2, and s = a^3, then which of the foll   [#permalink] 27 Jan 2019, 20:04
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