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#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here. # −1 < a − b < 10, with b an integer such that −3 ≤ b ≤ 1. Wha  Question banks Downloads My Bookmarks Reviews Important topics
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GRE Prep Club Legend  Joined: 07 Jun 2014
Posts: 4809
GRE 1: Q167 V156 WE: Business Development (Energy and Utilities)
Followers: 123

Kudos [?]: 1983 , given: 397

−1 < a − b < 10, with b an integer such that −3 ≤ b ≤ 1. Wha [#permalink]
Expert's post 00:00

Question Stats: 0% (00:00) correct 100% (02:10) wrong based on 5 sessions
−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

_________________

Sandy
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GRE Prep Club Legend  Joined: 07 Jun 2014
Posts: 4809
GRE 1: Q167 V156 WE: Business Development (Energy and Utilities)
Followers: 123

Kudos [?]: 1983 , given: 397

Re: −1 < a − b < 10, with b an integer such that −3 ≤ b ≤ 1. Wha [#permalink]
Expert's post
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).
_________________

Sandy
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Manager Joined: 15 Feb 2018
Posts: 53
Followers: 1

Kudos [?]: 18 , given: 33

Re: −1 < a − b < 10, with b an integer such that −3 ≤ b ≤ 1. Wha [#permalink]
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). GRE Prep Club Legend  Joined: 07 Jun 2014
Posts: 4809
GRE 1: Q167 V156 WE: Business Development (Energy and Utilities)
Followers: 123

Kudos [?]: 1983  , given: 397

Re: −1 < a − b < 10, with b an integer such that −3 ≤ b ≤ 1. Wha [#permalink]
1
KUDOS
Expert's post
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.
_________________

Sandy
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Try our free Online GRE Test 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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