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−1 < a − b < 10, with b an integer such that −3 ≤ b ≤ 1. Wha

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−1 < a − b < 10, with b an integer such that −3 ≤ b ≤ 1. Wha [#permalink] New post 17 Mar 2018, 07:26
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−1 < a − b < 10, with b an integer such that −3 ≤ b ≤ 1. What most accurately describes the range of \(a^2\) ?

A. −16 < \(a^2\) < 11
B. −4 < \(a^2\) < 11
C. 0 < \(a^2\) < 16
D. 0 < \(a^2\) < 121
E. 16 < \(a^2\) < 121

Drill 2
Question: 5
Page: 497
[Reveal] Spoiler: OA

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GMAT Club Legend
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Joined: 07 Jun 2014
Posts: 4710
GRE 1: Q167 V156
WE: Business Development (Energy and Utilities)
Followers: 90

Kudos [?]: 1606 [0], given: 375

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Re: −1 < a − b < 10, with b an integer such that −3 ≤ b ≤ 1. Wha [#permalink] New post 18 Mar 2018, 07:00
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Explanation

If a range of values for a can be found, then the range of values for \(a^2\) can be found.

Start by testing the end values of b, –3 and 1.

Plug in –3 for b in the first given inequality then solve for a.

You find that –4 < a < 7. If b = 1, 0 < a <11; b could be any integer in the range –3 ≤ b ≤ 1, this means –4 < a < 11 overall.
Remember to take the last step, though!
The question is looking for the range of \(a^2\), not a; \(a^2\) is always positive (i.e., \(0 < a^2\)).
Because a < 11, \(a^2 < 121\).


This means \(0 < a^2 < 121\); the answer is choice (D).
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Re: −1 < a − b < 10, with b an integer such that −3 ≤ b ≤ 1. Wha [#permalink] New post 22 Mar 2018, 00:52
I don't understand how the answer can be D.

If b = -3 then we have -4 < a < 7.

If b = 1 then we have 0 < a < 11.

So - 4 < a < 11.

Squaring the inequality yields, 16 < a^{2} < 121.

a^{2} > 16 > 0.

sandy wrote:
Explanation

If a range of values for a can be found, then the range of values for \(a^2\) can be found.

Start by testing the end values of b, –3 and 1.

Plug in –3 for b in the first given inequality then solve for a.

You find that –4 < a < 7. If b = 1, 0 < a <11; b could be any integer in the range –3 ≤ b ≤ 1, this means –4 < a < 11 overall.
Remember to take the last step, though!
The question is looking for the range of \(a^2\), not a; \(a^2\) is always positive (i.e., \(0 < a^2\)).
Because a < 11, \(a^2 < 121\).


This means \(0 < a^2 < 121\); the answer is choice (D).
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GMAT Club Legend
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Joined: 07 Jun 2014
Posts: 4710
GRE 1: Q167 V156
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Kudos [?]: 1606 [1] , given: 375

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Re: −1 < a − b < 10, with b an integer such that −3 ≤ b ≤ 1. Wha [#permalink] New post 22 Mar 2018, 01:55
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Put a=0 and b=0.

Now \(-1< a-b <10\) is true as \(-1< 0 <10\) holds true.

Also \(-3 \leq b \leq 1\) holds true as \(-3 \leq 0 \leq 1\) is true.

So clearly \(a^2\) can have values less than 16.

Vizualize this: even though \(a>-4\) and squaring both sides yields \(a^2>16\) but a can take values such as 0 1, 2, .... so squaring would have values less than 16.
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Re: −1 < a − b < 10, with b an integer such that −3 ≤ b ≤ 1. Wha   [#permalink] 22 Mar 2018, 01:55
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