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# −1 < a − b < 10, with b an integer such that −3 ≤ b ≤ 1. What most acc

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Math Expert
Joined: 02 Sep 2009
Posts: 52430
−1 < a − b < 10, with b an integer such that −3 ≤ b ≤ 1. What most acc  [#permalink]

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13 Jun 2018, 10:26
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Difficulty:

65% (hard)

Question Stats:

53% (02:06) correct 47% (01:42) wrong based on 104 sessions

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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

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Joined: 27 Oct 2016
Posts: 8
Re: −1 < a − b < 10, with b an integer such that −3 ≤ b ≤ 1. What most acc  [#permalink]

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13 Jun 2018, 11:03
IMO D
For b=-3 ; -1<a+3<10
-4<a<7 , so a^2 =0,1,4,9,16,25,36
For b=-2 ; -1<a+2<10
-3<a<8 , so a^2 =0,1,4,9,16,25,36,49
b=-1; -1<a+1<10
-1<a<9 , so a^2 =0,1,4,9,16 ,25,36,49,64
b=0 ; -1<a<10
so a^2 =0,1,4,9,16,25,36,49,64,81
b=1; -1<a-1<10
0<a<11 so a^2 =0,1,4,9,16,25,36,49,64,81,100
Option D has possible range values for a^2
Manager
Joined: 07 Feb 2017
Posts: 188
Re: −1 < a − b < 10, with b an integer such that −3 ≤ b ≤ 1. What most acc  [#permalink]

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13 Jun 2018, 13:35
Test the end cases
b = -3
a - b = -1 = a+3; a=-4
a - b = 10 = a+3; a=7
b = 1
a - b = -1 = a-1; a= 0
a - b = 10 = a-1; a=11

0 < a^2 < 121
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Posts: 2830
Re: −1 < a − b < 10, with b an integer such that −3 ≤ b ≤ 1. What most acc  [#permalink]

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17 Jun 2018, 18:48
Bunuel wrote:
−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

Since b is an integer between -3 and 1, inclusive, b could be -3, -2, -1, 0 or 1.We only need to test the two extreme values of b, that is, b = -3 and b = 1:

If b = -3, then we have

-1 < a - (-3) < 10

-1 < a + 3 < 10

-4 < a < 7

0 ≤ a^2 < 49 (note: since a could be 0, the minimum value of a^2 is 0, not 16.)

If b = 1, then we have

-1 < a - 1 < 10

0 < a < 11

0 < a^2 < 121

So we see that a^2 can be almost as large as 121 and almost as small as 0 (a^2 can’t be exactly 0 since when b = 1, a can’t be exactly 0).

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Posts: 91
Location: New Zealand
−1 < a − b < 10, with b an integer such that −3 ≤ b ≤ 1. What most acc  [#permalink]

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23 Jul 2018, 10:52
2
Bunuel wrote:
−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

To find range of an inequality, we need to take the extreme values of the equation −3 ≤ b ≤ 1; which are -3 & 1

Case 1; b = -3

-1 < a - (-3) < 10 which gives

-4 < a < 7 (Squaring the inequality)

16 < a^2 < 49

Case 2; b = 1

-1 < a - (1) < 10 which gives

0 < a < 11 (Squaring the inequality)

0 < a^2 < 121

So, we need to take the extreme values to find the range and can conclude that range = 0 < a^2 < 121 Option D is the right answer.

Kudos Appreciated
−1 < a − b < 10, with b an integer such that −3 ≤ b ≤ 1. What most acc &nbs [#permalink] 23 Jul 2018, 10:52
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