M37-66

Question

|x + 5|x|| < 12 
 
What is the sum of all integers which satisfy the inequality above?

A.  -2 
B.  -1 
C.  0 
D.  1 
E.  2

Bunuel · Answer

Official Solution: |x + 5|x|| < 12 What is the sum of all integers which satisfy the inequality above? A. -2 B. -1 C. 0 D. 1 E. 2 |x + 5|x|| < 12 -12 < x + 5|x| < 12 CASE 1: x < 0 . -12 < x + 5(-x) < 12 -12 < -4x < 12 Multiply by -\frac{1}{4} (and don't forget to flip the signs since we multiply by a negative number): 3 > x > -3 Negative integers in this range (don't forget that we consider x < 0 for this case) are -1, and -2. CASE 2: x \geq 0 . -12 < x + 5x < 12 -12 < 6x < 12 Divide by 6 : -2 < x < 2 Non-negative integers in this range (don't forget that we consider x \geq 0 for this case) are 0, and 1. The sum of all integers which satisfy the inequality is therefore: -1+(-2)+0+1=-2 Answer: A

siddhantvarma · Answer

|x+5|x||<12
Can be opened to equations:
x + 5x < 12 => 6x < 12 => x < 2
x - 5x < 12 => -4x < 12 => -x < 3 => x > -3
-x -5x < 12 => -6x < 12 => -x < 2 => x > -2
-x + 5x < 12 => 4x < 12 => x < 3

From the above 4 ranges of x we can say that -2 < x < 2 so integer values are -1, 0 and 1 which give the sum as 0.

I understand that putting x = -2 satisfies the equation, but where am I wrong in my approach?

Bunuel can you help please?

Bunuel · Answer

You cannot open modulus and not pay attention to the range. When x > 0, then |x| = x, so we get: |x + 5|x|| < 12 |x + 5x| < 12 6x < 12 x < 2 When x ≤ 0, then |x| = -x, so we get: |x + 5|x|| < 12 |x - 5x| < 12 |-4x| < 12 Since x is negative or zero, then -4x is positive or 0, so we get: -4x < 12 x > -3 Combining we get: -3 < x < 2

anushree01 · Answer

Can I please get more Questions like this ? Thanks!!

Bunuel · Answer

Use filters to get inequalities and absolute value questions here: https://gmatclub.com/forum/problem-solving-ps-140/

KDM91 · Answer

I like the solution - it’s helpful.

XLmafia21 · Answer

I like the solution - it’s helpful.

PGPJr · Answer

I don’t quite agree with the solution. From Case 1 - 3>x>−3 From Case 2 - −2

Bunuel · Answer

Cases 1 and 2 are separate, and we get values of x for each case independently. Your mistake is combining the ranges. From Case 1 we get x = -2 and -1, from Case 2 we get x = 0 and 1. The sum is -2. You can also see your approach is wrong because it misses the valid root x = -2, which clearly satisfies the inequality.

RexLenny · Answer

I like the solution - it’s helpful.

kreel11 · Answer

Bunuel

my question is that when x>0 we get the rangesas -3<x<2 and x<=0 we get -2<X<3 now even if i add solutions of both cases i get -2+-1+0+1+-1+0+1+2 which equates to 0. So how are we getting sum as -2 by this method.

Bunuel · Answer

When x <= 0, we get 3 > x > -3 NOT -2 < x < 3 . Non-positive integers in this range: -2, and -1. When x > 0, we get -2 < x < 2 NOT -3 < x < 2. Positive integers in this range: 1. The sum = -2 + (-1) + 1 = -2.

M37-66

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