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

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

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New post 13 Jun 2018, 11:26
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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

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

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New post 13 Jun 2018, 12:03
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
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Re: −1 < a − b < 10, with b an integer such that −3 ≤ b ≤ 1. What most acc  [#permalink]

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New post 13 Jun 2018, 14:35
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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Re: −1 < a − b < 10, with b an integer such that −3 ≤ b ≤ 1. What most acc  [#permalink]

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New post 17 Jun 2018, 19:48
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).

Answer: D
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−1 < a − b < 10, with b an integer such that −3 ≤ b ≤ 1. What most acc  [#permalink]

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New post 23 Jul 2018, 11:52
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 :cool:
−1 < a − b < 10, with b an integer such that −3 ≤ b ≤ 1. What most acc &nbs [#permalink] 23 Jul 2018, 11:52
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