Test: Absolute Values/Modules - GMAT MCQ

Correct option is A
Rule is when ∣y∣<a
−a<y<a
Now when ∣x−3∣≤5, then 
−7<2x−3≤5
Adding 3 and then dividing on both sides, we get
−5+3≤ x ≤5+3
−2< x <8
Option A.

Case: -1 - x ≤ 3
In this case, we can solve for x:
-1 - x ≤ 3
Subtracting -1 from both sides:
-1 - x + 1 ≤ 3 + 1
-x ≤ 4
Multiplying both sides by -1 (remember to flip the inequality):
x ≥ -4

Case: -(-1 - x) ≤ 3
In this case, we can solve for x:
-(-1 - x) ≤ 3
Distributing the negative sign:
1 + x ≤ 3
Subtracting 1 from both sides:
x ≤ 2

Since x is a positive integer, the smallest possible value of x that satisfies the inequality is the smallest positive integer that is greater than or equal to -4 and less than or equal to 2. The only positive integer in this range is 1.

Therefore, the smallest possible value of x is 1.

The correct answer is C. 1.

We have the inequality |3 - x| < x + 5.

To simplify this inequality, we consider two cases:

Case 1: (3 - x) is positive or zero:
In this case, the absolute value |3 - x| can be simplified to (3 - x). Therefore, our inequality becomes (3 - x) < x + 5.

Expanding the inequality, we have:
3 - x < x + 5

Adding x to both sides, we get:
3 < 2x + 5

Subtracting 5 from both sides, we have:
-2 < 2x

Dividing both sides by 2, we obtain:
-1 < x

So, in this case, the statement I. x > -1 is not true. Therefore, option C: I and II only cannot be correct.

Case 2: (3 - x) is negative:
In this case, the absolute value |3 - x| can be simplified to -(3 - x), changing the direction of the inequality. Therefore, our inequality becomes -(3 - x) < x + 5.

Expanding the inequality and simplifying, we have:
-x + 3 < x + 5

Adding x to both sides, we get:
3 < 2x + 5

Subtracting 5 from both sides, we have:
-2 < 2x

Dividing both sides by 2, we obtain:
-1 < x

So, in this case, the statement I. x > -1 is not true. Therefore, option C: I and II only cannot be correct.

Case 1: x + 2 ≥ 0
In this case, the equation simplifies to |x + 2 - 2| = 2, which further simplifies to |x| = 2.
Since x is non-negative in this case, the equation becomes x = 2.
Thus, we have one root in this case.

Case 2: x + 2 < 0
In this case, the equation simplifies to |-(x + 2) - 2| = 2, which further simplifies to |-x - 4| = 2.
Removing the absolute value, we have two possibilities:

-x - 4 = 2, which gives x = -6.
-x - 4 = -2, which gives x = -2.
Thus, we have two roots in this case.
Combining the roots from both cases, we have a total of three roots: x = -6, -2, and 2.

Therefore, the correct answer is D: 3.

To find the solutions to the equation 3x = |x² - 10|, we need to consider two cases:

Case 1: x² - 10 is non-negative (x² - 10 ≥ 0):
In this case, the absolute value |x² - 10| can be simplified to (x² - 10). Therefore, our equation becomes 3x = x² - 10.

Rearranging the terms, we have:
x² - 3x - 10 = 0

Factoring the quadratic equation, we get:
(x - 5)(x + 2) = 0

Setting each factor equal to zero, we have:
x - 5 = 0 → x = 5
x + 2 = 0 → x = -2

So, in this case, the possible values of x are 5 and -2.

Case 2: x² - 10 is negative (x² - 10 < 0):
In this case, the absolute value |x² - 10| can be simplified to -(x² - 10), changing the sign of the equation. Therefore, our equation becomes 3x = -(x² - 10).

Rearranging the terms, we have:
-3x = x² - 10

Rearranging further, we get:
x² + 3x - 10 = 0

Factoring the quadratic equation, we have:
(x - 2)(x + 5) = 0

Setting each factor equal to zero, we have:
x - 2 = 0 → x = 2
x + 5 = 0 → x = -5

So, in this case, the possible values of x are 2 and -5.

Combining the solutions from both cases, we find that the possible values of x that satisfy the equation 3x = |x² - 10| are -5, -2, 2, and 5.

Therefore, the correct answer is option E: 2 and 5.

To determine the number of possible solutions to the given system of equations, let's analyze each equation individually:

50 = 3a + 2b

We are given that both a and b are integers. We need to find integer solutions for a and b that satisfy this equation.

7 > |-a|

We are given that -a lies within the absolute value function. The inequality states that 7 is greater than the absolute value of -a.

Let's consider the second equation first:

7 > |-a|

Since |-a| represents the absolute value of -a, we can rewrite the inequality as 7 > |a|.

This inequality tells us that the distance of a from 0 must be less than 7. In other words, a can take on any integer value between -6 and 6, excluding 0.

Now let's consider the first equation:

50 = 3a + 2b

To find the possible integer solutions for a and b, we need to examine the factors of 50.

The factors of 50 are: 1, 2, 5, 10, 25, and 50.

We can see that 3a + 2b must equal one of these factors. By testing different values of a and b, we find the following integer solutions:

a = 16, b = 7 (3a + 2b = 48 + 14 = 62)
a = 14, b = 11 (3a + 2b = 42 + 22 = 64)
a = 12, b = 15 (3a + 2b = 36 + 30 = 66)
a = 10, b = 19 (3a + 2b = 30 + 38 = 68)
a = 8, b = 23 (3a + 2b = 24 + 46 = 70)
a = 6, b = 27 (3a + 2b = 18 + 54 = 72)
a = 4, b = 31 (3a + 2b = 12 + 62 = 74)

We can see that there are 7 possible integer solutions for a and b that satisfy the equation 50 = 3a + 2b.

Therefore, the correct answer is D: 7.

|ab| = ab
This equation states that the absolute value of the product ab is equal to the product ab. For this equation to hold true, it means that ab must be non-negative, or in other words, ab must be greater than or equal to zero.

|a| = -a
This equation states that the absolute value of a is equal to the negation of a. For this equation to hold true, it means that a must be negative.

Now, let's simplify the expression |b-4| + |ab-b| step by step using the given information.

|b-4|
Since there is no additional information about the value of b, we cannot determine whether b-4 is positive or negative. Therefore, we consider both cases separately.
Case 1: b-4 is positive
In this case, |b-4| simplifies to b-4.

Case 2: b-4 is negative
In this case, |b-4| simplifies to -(b-4) = 4-b.

|ab-b|
Using the information from equation 1 (|ab| = ab), we know that ab is non-negative. Thus, ab-b is also non-negative.
Since ab-b is non-negative, |ab-b| simplifies to ab-b.

Now, let's substitute the simplified expressions back into the original expression:

|b-4| + |ab-b| = (b-4) + (ab-b) = b - 4 + ab - b = ab - 4.

Therefore, the answer is D: ab - 2b + 4.

First of all |x - 3|² = (x - 3)², so we have: (x - 3)² + |x - 3| = 20.

When x < 3, x - 3 is negative, thus |x - 3| = -(x - 3). In this case we'll have (x - 3)² - (x - 3) = 20 --> x = -1 or x = 8. Discard x = 8 because it's not in the range we consider (< 3).

When x >= 3, x - 3 is non-negative, thus |x - 3| = x - 3. In this case we'll have (x - 3)² + (x - 3) = 20 --> x = -2 or x = 7. Discard x = -2 because it's not in the range we consider (>= 3).

Thus there are two solutions: x = -1 and x = 7 → the sum = 6.

Case 1: x + 5 and 2x - 5 are both non-negative or both non-positive:
In this case, the absolute values |x + 5| and |2x - 5| can be simplified to their respective expressions. Therefore, our inequality becomes x + 5 <= 2x - 5.

Rearranging the terms, we have:
x - 2x <= -5 - 5

Simplifying further, we get:
-x <= -10

Multiplying both sides by -1 (and changing the direction of the inequality), we have:
x >= 10

So, in this case, the values of x that satisfy the inequality are x >= 10.

Case 2: x + 5 and 2x - 5 have opposite signs:
In this case, the absolute values |x + 5| and |2x - 5| can be simplified by changing the signs of the expressions. Therefore, our inequality becomes -(x + 5) <= 2x - 5.

Expanding the inequality and simplifying, we have:
-x - 5 <= 2x - 5

Adding x to both sides, we get:
-5 <= 3x - 5

Adding 5 to both sides, we have:
0 <= 3x

Dividing both sides by 3, we obtain:
0 <= x

So, in this case, the values of x that satisfy the inequality are x >= 0.

Combining the solutions from both cases, we find that the values of x that satisfy the inequality |x + 5| <= |2x - 5| are x >= 10 and x >= 0.

Therefore, the correct answer is option D: x <= 0 or x >= 10.

Let's consider the first curve, |x + y| = 1. This equation represents two lines: x + y = 1 and x + y = -1. Similarly, for the second curve, |x - y| = 1, we have x - y = 1 and x - y = -1.

If we draw these lines on a graph, we can observe that they intersect at four points, forming a square with sides of length 2. The vertices of this square are (-1, -1), (-1, 1), (1, -1), and (1, 1).

Thus, the area bounded by the curves |x + y| = 1 and |x - y| = 1 is the area of this square, which is given by the formula A = side length * side length = 2 * 2 = 4.

Therefore, the correct answer is B: 4.