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10 Jun 2021, 02:30
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C
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Difficulty:
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Question Stats:
82% (00:55) correct
18%
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wrong
based on 567
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Which of the following describes all possible solutions to the inequality |a + 4| < 7?
(A) a < 3
(B) a > -11
(C) 3 > a > -11
(D) -11 > a > 3
(E) a > 11 or a < -11
(A) a < 3
(B) a > -11
(C) 3 > a > -11
(D) -11 > a > 3
(E) a > 11 or a < -11
10 Jun 2021, 02:41
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Bunuel
|a + 4| < 7
-7 < a + 4 < 7
Subtracting 4 throughout:
-11 < a < 3
3 > a > -11, (C) IMO!
Updated on: 18 Jul 2025, 04:25
Originally posted by BrushMyQuant on 21 May 2022, 11:55.
Last edited by BrushMyQuant on 18 Jul 2025, 04:25, edited 1 time in total.
Last edited by BrushMyQuant on 18 Jul 2025, 04:25, edited 1 time in total.
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|a + 4| < 7
We will have to take two cases
Case 1: Whatever is inside the modulus is >= 0
=> a+4 >= 0 => a >= -4
=> |a+4| = a+4 (as |X| = X when X >= 0)
=> a+4 < 7
=> a < 7-4
=> a < 3
But condition was a >=-4
=> So solution will be the intersection of two of them (As shown in the image below)
=> -4 <= a < 3
Line image.JPG [ 25.48 KiB | Viewed 4743 times ]
Case 2: Whatever is inside the modulus is < 0
=> a+4 < 0 => a < -4
=> |a+4 | = -(a+4 ) (as |X| = -X when X < 0)
=> -(a+4 ) < 7
=> -a - 4 < 7
=> -a < 7 + 4
=> -a < 11
=> a > -11 (multiplying both sides with negative sign reverses the inequality sign)
But the condition was a < -4
=> So solution will be the intersection of two of them (As shown in the image below)
=> -11 < a < -4
temp image.JPG [ 22.8 KiB | Viewed 4500 times ]
=> -11 < a < -4 and -4 <= a < 3
=> -11 < a < 3
Or, 3 > a > -11
So, Answer will be C
Hope it helps!
Playlist on Solved problems on Absolute Values here
Watch the following video to learn how to Solve Inequality + Absolute value Problems
We will have to take two cases
Case 1: Whatever is inside the modulus is >= 0
=> a+4 >= 0 => a >= -4
=> |a+4| = a+4 (as |X| = X when X >= 0)
=> a+4 < 7
=> a < 7-4
=> a < 3
But condition was a >=-4
=> So solution will be the intersection of two of them (As shown in the image below)
=> -4 <= a < 3
Attachment:
Line image.JPG [ 25.48 KiB | Viewed 4743 times ]
Case 2: Whatever is inside the modulus is < 0
=> a+4 < 0 => a < -4
=> |a+4 | = -(a+4 ) (as |X| = -X when X < 0)
=> -(a+4 ) < 7
=> -a - 4 < 7
=> -a < 7 + 4
=> -a < 11
=> a > -11 (multiplying both sides with negative sign reverses the inequality sign)
But the condition was a < -4
=> So solution will be the intersection of two of them (As shown in the image below)
=> -11 < a < -4
Attachment:
temp image.JPG [ 22.8 KiB | Viewed 4500 times ]
=> -11 < a < -4 and -4 <= a < 3
=> -11 < a < 3
Or, 3 > a > -11
So, Answer will be C
Hope it helps!
Playlist on Solved problems on Absolute Values here
Watch the following video to learn how to Solve Inequality + Absolute value Problems
