Real Numbers, Number Line, and Absolute Value Practice Questions — GRE General Test — Page 2

Question 21Question →

If $x$ and $y$ are real numbers such that $|x - 3| \le 2$ and $|y + 1| \le 4$ , which of the following could be the value of $|x - y|$ ? Select all such values.

Select all that apply

-2

0

5

10

12

Show answer & explanation

Answer: $0$ ; $5$ ; $10$

Answer

The possible values of

|x - y|

are 0, 5, and 10.

Solving the inequalities yields

1 \le x \le 5

and

-5 \le y \le 3

. The expression

|x - y|

represents the distance between

x

and

y

on the number line. Because the intervals overlap between 1 and 3,

x

and

y

can be equal, making the minimum distance 0. The maximum distance occurs at the extreme points

x = 5

and

y = -5

, giving a distance of

|5 - (-5)| = 10

. Therefore, any value from 0 to 10 inclusive is possible, making 0, 5, and 10 valid values.

Step-by-Step Solution

1

Solve the absolute value inequality for

x

.

-2 \le x - 3 \le 2 \implies 1 \le x \le 5

Unwrapping

|x - 3| \le 2

gives the bounds for

x

on the real number line.

2

Solve the absolute value inequality for

y

.

-4 \le y + 1 \le 4 \implies -5 \le y \le 3

Unwrapping

|y + 1| \le 4

gives the bounds for

y

on the real number line.

3

Determine the minimum and maximum possible values for

|x - y|

.

Minimum value is 0 (since the intervals

[1, 5]

and

[-5, 3]

overlap at

[1, 3]

). Maximum value is

|5 - (-5)| = 10

. Thus,

0 \le |x - y| \le 10

.

The absolute value

|x - y|

represents the distance between

x

and

y

on the number line, which can take any real value from 0 to 10.

4

Evaluate the given choices against the range

[0, 10]

.

The values 0, 5, and 10 fall within

[0, 10]

, whereas

-2

is impossible for absolute values and

12

exceeds the maximum bound.

Determines which specific options are valid outcomes for

|x - y|

.

Key Concept

Absolute value as distance on the real number line and range of differences between bounded real variables

Estimated Time:1m 30s

Question 22Question →

On the real number line, the distance between $x$ and $3$ is equal to twice the distance between $x$ and $-9$ . What is the sum of all possible real values of $x$ ?

-26

-21

-14

-5

22

Show answer & explanation

Answer: $-26$

Answer

The sum of all possible real values of

x

is

-26

.

The distance between

x

and

a

on the real number line is expressed as

|x - a|

. Thus, the condition translates to

|x - 3| = 2|x - (-9)|

, which simplifies to

|x - 3| = 2|x + 9|

. Setting up the two algebraic cases gives

x - 3 = 2(x + 9)

, leading to

x = -21

, and

x - 3 = -2(x + 9)

, leading to

x = -5

. Adding these two solutions yields

(-21) + (-5) = -26

.

Step-by-Step Solution

1

Translate the geometric statement on the number line into an absolute value equation.

The distance between

x

and

3

is

|x - 3|

, and the distance between

x

and

-9

is

|x - (-9)| = |x + 9|

. Therefore,

|x - 3| = 2|x + 9|

.

Distance between two points

a

and

b

on the number line is given by

|a - b|

.

2

Solve Case 1 where the expressions inside the absolute values have the same sign.

x - 3 = 2(x + 9) \implies x - 3 = 2x + 18 \implies x = -21

.

Removing absolute values with identical signs gives a linear equation in

x

.

3

Solve Case 2 where the expressions inside the absolute values have opposite signs.

x - 3 = -2(x + 9) \implies x - 3 = -2x - 18 \implies 3x = -15 \implies x = -5

.

Removing absolute values with opposite signs accounts for the second possible geometric location.

4

Sum the valid real solutions.

(-21) + (-5) = -26

.

The question asks for the sum of all possible real values of

x

.

Key Concept

Absolute Value as Distance on the Real Number Line

Question 23Question →

On a number line, point $A$ is located at $10$ and point $B$ is located at $2$ . If point $P$ , with coordinate $x > 0$ , is three times as far from point $A$ as it is from point $B$ , what is the value of $x$ ?

Show answer & explanation

Answer: 4

Answer

4

The distance from

P(x)

to

A(10)

is

|x - 10|

and to

B(2)

is

|x - 2|

. Setting

|x - 10| = 3|x - 2|

gives two equations:

x - 10 = 3x - 6

, which yields

x = -2

, and

x - 10 = -3x + 6

, which yields

x = 4

. Since

x > 0

, the correct value is 4.

Step-by-Step Solution

1

Formulate the distance relationship using absolute value notation.

|x - 10| = 3|x - 2|

The distance between two points

u

and

v

on the real number line is given by

|u - v|

.

2

Split the absolute value equation into two linear equations representing possible cases.

x - 10 = 3(x - 2)

or

x - 10 = -3(x - 2)

The equality

|a| = |b|

implies

a = b

or

a = -b

.

3

Solve each case algebraically.

Case 1 gives

x - 10 = 3x - 6 \implies 2x = -4 \implies x = -2

. Case 2 gives

x - 10 = -3x + 6 \implies 4x = 16 \implies x = 4

.

Standard linear equation solving.

4

Apply the given domain condition

x > 0

.

x = 4

The solution

x = -2

is negative and therefore violates the condition

x > 0

.

Key Concept

Distance on a number line represented by absolute value equations

Question 24Question →

On the real number line, the distance between a real number $k$ and $-3$ is strictly less than $7$ , and the distance between $k$ and $5$ is at least $4$ . Which of the following inequalities represents the complete set of all possible values of $k$ ?

-10 < k \le 1

-10 \le k \le 1

-10 < k < 1

-4 < k \le 1

1 \le k < 4

Show answer & explanation

Answer: $-10 < k \le 1$

Answer

-10 < k \le 1

The correct inequality

-10 < k \le 1

properly combines the strict bound from the distance to

-3

(which gives

-10 < k < 4

) with the non-strict bound from the distance to

5

(which gives

k \le 1

or

k \ge 9

). Taking their intersection gives

-10 < k \le 1

.

Step-by-Step Solution

1

Express the first condition using absolute value notation and solve for

k

.

|k - (-3)| < 7 \implies |k + 3| < 7 \implies -7 < k + 3 < 7 \implies -10 < k < 4

.

Distance on a number line between

x

and

y

is given by

|x - y|

.

2

Express the second condition using absolute value notation and solve for

k

.

|k - 5| \ge 4 \implies k - 5 \le -4

or

k - 5 \ge 4 \implies k \le 1

or

k \ge 9

.

The phrase 'at least 4' means greater than or equal to 4.

3

Find the overlap (intersection) of the two solution sets.

(-10 < k < 4) \cap (k \le 1 \text{ or } k \ge 9) = -10 < k \le 1

.

Since

k

must be less than

4

, the region

k \ge 9

contains no valid solutions, leaving only

-10 < k \le 1

.

Key Concept

Absolute value as distance on the number line and solving compound absolute value inequalities.

Estimated Time:1m 30s

Question 25Question →

If $a$ and $b$ are real numbers such that $|a - 3| = 5$ and $|2b + 1| = 9$ , what is the minimum possible value of $|a - b|$ ?

1

2

3

4

6

Show answer & explanation

Answer: 3

Answer

The minimum possible value of

|a - b|

is

3

.

Solving

|a - 3| = 5

gives two possible values for

a

:

a = 8

and

a = -2

. Solving

|2b + 1| = 9

gives two possible values for

b

:

b = 4

and

b = -5

. Evaluating the distance

|a - b|

for all four pairs

(a,b)

gives

|8 - 4| = 4

,

|8 - (-5)| = 13

,

|-2 - 4| = 6

, and

|-2 - (-5)| = 3

. The minimum possible value is

3

.

Step-by-Step Solution

1

Solve the absolute value equation

|a - 3| = 5

for all possible values of

a

.

a - 3 = 5 \implies a = 8

or

a - 3 = -5 \implies a = -2

. Thus,

a \in \{-2, 8\}

.

An absolute value equation

|x| = k

splits into two linear equations:

x = k

and

x = -k

.

2

Solve the absolute value equation

|2b + 1| = 9

for all possible values of

b

.

2b + 1 = 9 \implies 2b = 8 \implies b = 4

or

2b + 1 = -9 \implies 2b = -10 \implies b = -5

. Thus,

b \in \{-5, 4\}

.

An absolute value equation

|2b + 1| = 9

has two cases:

2b + 1 = 9

and

2b + 1 = -9

.

3

Calculate

|a - b|

for all four possible pairs of

(a, b)

.

For

(8, 4): |8 - 4| = 4

.
For

(8, -5): |8 - (-5)| = 13

.
For

(-2, 4): |-2 - 4| = 6

.
For

(-2, -5): |-2 - (-5)| = 3

.

To find the minimum possible value of

|a - b|

, every valid combination of

a

and

b

must be tested.

4

Identify the minimum value among the calculated absolute differences.

The minimum calculated value is

3

.

Comparing

4, 13, 6,

and

3

yields

3

as the smallest value.

Key Concept

Solving absolute value equations and finding distances between points on the real number line

Estimated Time:1m 30s

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