Integers, Absolute Value, and Number Lines Practice Questions — ACT — Page 2

What is the sum of all integer values of $y$ that satisfy the inequality $|2y - 3| \le 7$ ?

Show answer & explanation

Answer: 12

Answer

The sum of all integer values of

y

that satisfy the inequality is 12.

Solving the inequality

|2y - 3| \le 7

requires setting up the compound inequality

-7 \le 2y - 3 \le 7

. Adding 3 to all parts gives

-4 \le 2y \le 10

, and dividing by 2 yields the interval

-2 \le y \le 5

. The integers in this closed interval are

-2, -1, 0, 1, 2, 3, 4

, and

5

. Summing these values gives 12, as the terms

-2

and

-1

cancel out with

2

and

1

.

Step-by-Step Solution

1

Set up the compound inequality

-7 \le 2y - 3 \le 7

An absolute value inequality of the form

|a| \le b

translates to

-b \le a \le b

.

2

Isolate the term containing

y

-4 \le 2y \le 10

Add 3 to all three parts of the compound inequality to eliminate the

-3

.

3

Solve for

y

-2 \le y \le 5

Divide all three parts of the inequality by 2.

4

Identify the integer solutions in the interval

-2, -1, 0, 1, 2, 3, 4, 5

The inequality includes the endpoints, so the integers satisfying the inequality are all integers from

-2

through

5

, inclusive.

5

Calculate the sum of the integers

12

Adding the integers:

(-2) + (-1) + 0 + 1 + 2 + 3 + 4 + 5 = 12

. The negative integers cancel out their corresponding positive counterparts (

-2

and

2

,

-1

and

1

).

Key Concept

Solving compound inequalities derived from absolute value inequalities and finding the sum of the integer solution set.

Estimated Time:1m 30s

Question 22Question →

On a standard number line, the coordinate of point $P$ is an integer $p$ , and the coordinate of point $Q$ is an integer $q$ . The distance between $P$ and the origin is less than 5, and the distance between $Q$ and $-3$ is exactly 4. If the product $p \cdot q$ is minimized, what is the value of $|p - q|$ ?

3

5

8

11

12

Show answer & explanation

Answer: 11

Answer

The value of

|p - q|

is 11, which corresponds to the option with a value of 11.

The distance between

P

and the origin is less than 5, so the integer coordinate

p

must satisfy

|p| < 5

, meaning

p \in \{-4, -3, -2, -1, 0, 1, 2, 3, 4\}

. The distance between

Q

and

-3

is exactly 4, so

|q - (-3)| = 4 \implies |q + 3| = 4

, which gives

q = 1

or

q = -7

. To minimize the product

p \cdot q

, we analyze the two possibilities for

q

. If

q = 1

, the minimum product is

-4

when

p = -4

. If

q = -7

, the minimum product is

-28

when

p = 4

. The absolute minimum product is

-28

, achieved when

p = 4

and

q = -7

. The value of

|p - q|

for this pair is

|4 - (-7)| = 11

.

Step-by-Step Solution

1

Determine the possible integer coordinates for point

P

.

p \in \{-4, -3, -2, -1, 0, 1, 2, 3, 4\}

The distance from

P

to the origin is less than 5, meaning

|p| < 5

. Since

p

is an integer, it can be any integer strictly between

-5

and

5

.

2

Determine the possible integer coordinates for point

Q

.

q = 1

or

q = -7

The distance from

Q

to

-3

is exactly 4, meaning

|q - (-3)| = 4 \implies |q + 3| = 4

. Solving this gives

q + 3 = 4 \implies q = 1

, or

q + 3 = -4 \implies q = -7

.

3

Evaluate products of

p

and

q

to find the pair

(p, q)

that minimizes

p \cdot q

.

p = 4

and

q = -7

, yielding the minimum product of

-28

.

If

q = 1

, the minimum product is

p \cdot 1 = -4

when

p = -4

. If

q = -7

, the product is

-7p

; to minimize this negative product, we choose the largest positive value for

p

, which is

4

, yielding

-28

. Comparing

-4

and

-28

, the absolute minimum is

-28

.

4

Compute the absolute difference

|p - q|

for the minimizing pair.

|4 - (-7)| = 11

Using the coordinates

p = 4

and

q = -7

, we find the distance between them on the number line.

Key Concept

Distance on a number line can be calculated using absolute value. Finding the minimum of a product involving signed integers requires evaluating both positive and negative cases.

Estimated Time:2m 0s

Question 23Question →

A highway has exits at mile markers $-7$ and $9$ . A new rest stop is to be built at mile marker $c$ such that its distance from the exit at mile marker $-7$ is exactly $3$ times its distance from the exit at mile marker $9$ . Which of the following is a possible value of $c$ ?

-3

1

5

8

10

Show answer & explanation

Answer: 5

Answer

The value 5 is a possible mile marker for the rest stop.

The correct answer is the value 5. The distance between the rest stop at mile marker 5 and the exit at -7 is |5 - (-7)| = 12 miles. The distance between the rest stop at mile marker 5 and the exit at 9 is |5 - 9| = 4 miles. Since 12 is exactly 3 times 4, this satisfies the given condition.

Step-by-Step Solution

1

Express the distance from the rest stop

c

to each exit using absolute value.

Distance to the exit at

-7

is

|c - (-7)| = |c + 7|

; distance to the exit at

9

is

|c - 9|

.

Absolute value represents the non-negative distance between two points on a number line.

2

Set up the algebraic equation using the given relationship.

|c + 7| = 3|c - 9|

The problem states that the distance to the exit at

-7

is 3 times the distance to the exit at

9

.

3

Solve the absolute value equation by separating it into two cases.

Case 1:

c + 7 = 3(c - 9)

or Case 2:

c + 7 = -3(c - 9)

An equation of the form

|x| = |y|

implies

x = y

or

x = -y

.

4

Solve Case 1 for

c

.

c + 7 = 3c - 27 \implies 2c = 34 \implies c = 17

Distribute the 3, subtract

c

from both sides, add 27 to both sides, and divide by 2.

5

Solve Case 2 for

c

.

c + 7 = -3c + 27 \implies 4c = 20 \implies c = 5

Distribute the -3, add

3c

to both sides, subtract 7 from both sides, and divide by 4.

6

Compare the solutions to the given options.

The two mathematically valid positions are

17

and

5

. Only

5

is listed among the options.

To identify the correct choice from the multiple-choice options.

Key Concept

Calculating distances and solving equations involving absolute values on a number line.

Alternative Method

Test the given answer choices directly by calculating the distances for each option. For the value 5, the distance to the exit at -7 is |5 - (-7)| = 12, and the distance to the exit at 9 is |5 - 9| = 4. Since 12 is 3 times 4, this value is correct.

Estimated Time:1m 0s

Question 24Question →

If $x$ and $y$ are integers such that $|x + 3| = 4$ and $|y - 2| = 6$ , what is the maximum possible value of $|x - y|$ ?

3

7

11

13

15

Show answer & explanation

Answer: 15

Answer

15

To find the maximum possible value of

|x - y|

, we find all possible values of

x

and

y

by solving the absolute value equations. The equation

|x + 3| = 4

yields

x = 1

or

x = -7

. The equation

|y - 2| = 6

yields

y = 8

or

y = -4

. The expression

|x - y|

represents the distance between these points on a number line. The maximum distance occurs between the two points that are furthest apart, which are

y = 8

and

x = -7

. The distance between them is

|-7 - 8| = 15

.

Step-by-Step Solution

1

Solve the absolute value equation

|x + 3| = 4

for

x

.

x + 3 = 4 \Rightarrow x = 1

or

x + 3 = -4 \Rightarrow x = -7

An absolute value equation

|u| = c

splits into two cases:

u = c

and

u = -c

.

2

Solve the absolute value equation

|y - 2| = 6

for

y

.

y - 2 = 6 \Rightarrow y = 8

or

y - 2 = -6 \Rightarrow y = -4

Similarly, split the second absolute value equation into its positive and negative cases.

3

List the possible coordinates for

x

and

y

on the number line.

x \in \{1, -7\}

and

y \in \{8, -4\}

These are the sets of values that satisfy each respective equation.

4

Find the distance

|x - y|

for all possible pairs of

(x, y)

.

For

(1, 8)

,

|1 - 8| = 7

; for

(1, -4)

,

|1 - (-4)| = 5

; for

(-7, 8)

,

|-7 - 8| = 15

; for

(-7, -4)

,

|-7 - (-4)| = 3

.

The expression

|x - y|

represents the distance between

x

and

y

on the number line. We calculate the distance for all combinations to find the maximum.

5

Identify the maximum value from the calculated distances.

The maximum value is

15

.

Comparing

7

,

5

,

15

, and

3

, the largest value is

15

.

Key Concept

Integers, Absolute Value, and Number Lines

Alternative Method

Analyze the solutions on a number line. The solutions to

|x - (-3)| = 4

are the points at a distance of

4

from

-3

, which are

-7

and

1

. The solutions to

|y - 2| = 6

are the points at a distance of

6

from

2

, which are

-4

and

8

. The maximum distance between any

x

and

y

is the distance between the leftmost point

-7

and the rightmost point

8

, which is

8 - (-7) = 15

.

Estimated Time:1m 0s

Question 25Question →

On a standard number line, point $A$ is located at $-24$ and point $B$ is located at $6$ . Point $C$ is located between point $A$ and point $B$ such that the distance from $A$ to $C$ is $\frac{2}{3}$ of the distance from $C$ to $B$ . What is the coordinate of point $C$ ?

Show answer & explanation

Answer: -12

Answer

The coordinate of point

C

is

-12

.

Because point

C

with coordinate

c

lies between

-24

and

6

, the distance from

A

to

C

is

c - (-24) = c + 24

, and the distance from

C

to

B

is

6 - c

. Setting the distance from

A

to

C

to

\frac{2}{3}

of the distance from

C

to

B

gives the equation

c + 24 = \frac{2}{3}(6 - c)

. Multiplying both sides by

3

to clear the fraction results in

3c + 72 = 12 - 2c

. Collecting like terms yields

5c = -60

, which simplifies to

c = -12

.

Step-by-Step Solution

1

Express the distances between the points on the number line using their coordinates.

The distance from

A

to

C

is

c + 24

, and the distance from

C

to

B

is

6 - c

.

Since point

C

is positioned between points

A

and

B

, the inequality

-24 < c < 6

holds. This allows the absolute value distance expressions

|c - (-24)|

and

|6 - c|

to simplify directly to positive expressions without absolute value bars.

2

Formulate an equation based on the specified ratio of distances.

c + 24 = \frac{2}{3}(6 - c)

The problem states that the distance from

A

to

C

is

\frac{2}{3}

of the distance from

C

to

B

.

3

Solve the linear equation for the coordinate

c

.

c = -12

Multiplying both sides by

3

gives

3(c + 24) = 2(6 - c)

, which expands to

3c + 72 = 12 - 2c

. Rearranging the terms by adding

2c

to both sides and subtracting

72

from both sides results in

5c = -60

. Dividing by

5

gives

c = -12

.

Key Concept

Using absolute value properties to express distances on a number line and solving partitioning coordinate problems.

Estimated Time:1m 15s

Question 26Question →

On a standard number line, point $A$ has coordinate $-15$ and point $B$ has coordinate $17$ . Point $C$ is located to the right of point $B$ such that the distance between $A$ and $C$ is exactly $3$ times the distance between $B$ and $C$ . What is the coordinate of point $C$ ?

Show answer & explanation

Answer: 33

Answer

The coordinate of point

C

is 33.

The correct coordinate is found by setting up the distance equation for point

C

(with coordinate

c > 17

) relative to

A(-15)

and

B(17)

. The distance

AC

is

c - (-15) = c + 15

, and the distance

BC

is

c - 17

. Setting

c + 15 = 3(c - 17)

and solving yields

c = 33

, which is to the right of

B

.

Step-by-Step Solution

1

Define the variable for the coordinate of point C and write the expressions for distances.

Let

c

be the coordinate of point

C

. The distance between

A

and

C

is

|c - (-15)| = |c + 15|

, and the distance between

B

and

C

is

|c - 17|

. Since point

C

is to the right of point

B

(which is at

17

), we know

c > 17

, so

|c + 15| = c + 15

and

|c - 17| = c - 17

.

To set up an algebraic equation representing the physical distance relations on the number line.

2

Set up the equation using the given relationship.

The equation is

c + 15 = 3(c - 17)

.

The problem states the distance between

A

and

C

is

3

times the distance between

B

and

C

.

3

Solve the equation for

c

.

c + 15 = 3c - 51 \implies 15 + 51 = 3c - c \implies 66 = 2c \implies c = 33

.

To find the coordinate of point

C

.

Key Concept

Calculating distances between points on a number line using absolute value and solving the resulting equations.

Alternative Method

Use geometric visualization: The distance from

A(-15)

to

B(17)

is

17 - (-15) = 32

units. Since point

C

lies to the right of

B

, the distance

AC

is the sum of

AB

and

BC

. Therefore,

AC = 32 + BC

. We are given that

AC = 3 \times BC

. Substituting this gives

32 + BC = 3 \times BC \implies 2 \times BC = 32 \implies BC = 16

. Since

C

is 16 units to the right of

B(17)

, its coordinate is

17 + 16 = 33

.

Estimated Time:1m 30s

Question 27Question →

On a number line, point $M$ has coordinate $-14$ and point $N$ has coordinate $10$ . Point $P$ is located on the number line such that the distance between $M$ and $P$ is three times the distance between $N$ and $P$ . If the coordinate of $P$ is positive, what is the sum of all possible coordinates of $P$ ?

4

8

22

26

34

Show answer & explanation

Answer: $26$

Answer

The sum of all possible coordinates of

P

is

26

.

The distance between any two points

x

and

y

on a number line is given by

|x - y|

. Thus, the distance from point

P(p)

to

M(-14)

is

|p - (-14)| = |p + 14|

, and the distance from

P(p)

to

N(10)

is

|p - 10|

. Since the distance to

M

is three times the distance to

N

, we write

|p + 14| = 3|p - 10|

. To solve this absolute value equation, we check two cases. In the first case, we have

p + 14 = 3(p - 10)

, which simplifies to

2p = 44

, or

p = 22

. In the second case, we have

p + 14 = -3(p - 10)

, which simplifies to

4p = 16

, or

p = 4

. Both coordinates are positive, satisfying the condition given in the problem. The sum of these possible coordinates is

22 + 4 = 26

.

Step-by-Step Solution

1

Set up the algebraic representation of the distances between the points on the number line using absolute value.

The distance between point

M(-14)

and point

P(p)

is

|p - (-14)| = |p + 14|

. The distance between point

N(10)

and point

P(p)

is

|p - 10|

.

The distance between two points

a

and

b

on a standard number line is always expressed as the absolute value of their difference,

|a - b|

.

2

Formulate the equation representing the relationship between the two distances.

|p + 14| = 3|p - 10|

The problem states that the distance from

M

to

P

is three times the distance from

N

to

P

.

3

Solve the absolute value equation by analyzing both positive and negative cases.

Case 1:

p + 14 = 3(p - 10) \implies p + 14 = 3p - 30 \implies 2p = 44 \implies p = 22

.
Case 2:

p + 14 = -3(p - 10) \implies p + 14 = -3p + 30 \implies 4p = 16 \implies p = 4

.

An equation of the form

|A| = |B|

is solved by evaluating the two distinct possibilities:

A = B

and

A = -B

.

4

Check if both solutions satisfy the condition of being positive, and sum them.

Both

22

and

4

are positive, so they are both valid coordinates. Their sum is

22 + 4 = 26

.

The question specifies that the coordinate of

P

must be positive, and asks for the sum of all such coordinates.

Key Concept

Representing distances on a number line using absolute value and solving absolute value equations.

Estimated Time:1m 30s

Question 28Question →

On a vertical number line representing elevation in meters, a research drone is at coordinate $d$ and a submarine is at coordinate $s$ . The coordinate of the drone is a positive integer, and the coordinate of the submarine is a negative integer. The distance between the drone and the submarine is $150$ meters. If the absolute value of the submarine's coordinate is $4$ times the coordinate of the drone, what is the coordinate of the drone?

Show answer & explanation

Answer: 30

Answer

The coordinate of the drone is 30.

The correct coordinate of 30 is found by defining the distance between the positive drone coordinate

d

and negative submarine coordinate

s

as

d - s = 150

. Since

s < 0

, its absolute value

|s|

is equal to

-s

. Substituting

-s = 4d

gives

d - (-4d) = 150

, which simplifies to

5d = 150

and yields

d = 30

.

Step-by-Step Solution

1

Set up equations based on the problem description.

d > 0

,

s < 0

, distance

|d - s| = 150

, and

|s| = 4d

.

To translate the verbal description of elevations and distances into mathematical expressions.

2

Simplify the distance and absolute value expressions using the signs of the coordinates.

Since

d

is positive and

s

is negative,

d - s = 150

. Since

s

is negative,

|s| = -s

, so

-s = 4d

or

s = -4d

.

To eliminate absolute values based on the known signs of the variables.

3

Substitute the expression for

s

into the distance equation and solve for

d

.

d - (-4d) = 150 \Rightarrow 5d = 150 \Rightarrow d = 30

.

To solve the system of linear equations to find the drone's coordinate.

Key Concept

Using absolute value to represent distance on a number line and solving equations involving sign constraints.

Question 29Question →

If $a$ and $b$ are integers such that $-5 < a < -1$ and $2 < b < 6$ , what is the difference between the maximum possible value of $|a - b|$ and the minimum possible value of $|a + b|$ ?

2

4

8

9

11

Show answer & explanation

Answer: 9

Answer

The difference between the maximum possible value of

|a - b|

and the minimum possible value of

|a + b|

is 9.

The maximum possible value of

|a - b|

is 9, which occurs when

a = -4

and

b = 5

, giving

|-4 - 5| = 9

. The minimum possible value of

|a + b|

is 0, which occurs when

a = -3

and

b = 3

(or

a = -4

and

b = 4

), giving

|-3 + 3| = 0

. The difference between these two values is

9 - 0 = 9

.

Step-by-Step Solution

1

Identify the possible integer values for

a

and

b

from the strict inequalities.

The integers satisfying

-5 < a < -1

are

a \in \{-4, -3, -2\}

. The integers satisfying

2 < b < 6

are

b \in \{3, 4, 5\}

.

Since the inequalities are strict (

<

), the endpoints

-5, -1, 2,

and

6

are excluded.

2

Determine the maximum possible value of

|a - b|

by selecting the values of

a

and

b

that maximize their distance.

Using

a = -4

and

b = 5

yields

|-4 - 5| = |-9| = 9

.

The absolute value of the difference is maximized when

a

is as small (most negative) as possible and

b

is as large (most positive) as possible.

3

Determine the minimum possible value of

|a + b|

by finding values of

a

and

b

that are closest to being additive opposites.

Using

a = -3

and

b = 3

(or

a = -4

and

b = 4

) yields

|-3 + 3| = |0| = 0

.

The absolute value of any real number is at least 0. Since we can choose integers that sum to exactly 0, the minimum possible value is 0.

4

Calculate the difference between the two extreme values found in the previous steps.

The difference is

9 - 0 = 9

.

We subtract the minimum value of the second expression from the maximum value of the first expression.

Key Concept

Absolute value represents distance from zero, and finding extreme values of absolute value expressions involving restricted integer sets requires testing boundary combinations and understanding additive inverses.

Alternative Method

Instead of checking every pair, we can analyze the extreme values of the intervals. Since

a \in [-4, -2]

and

b \in [3, 5]

, the difference

a - b

ranges from

-4 - 5 = -9

to

-2 - 3 = -5

. The absolute value

|a - b|

therefore ranges from 5 to 9, making the maximum value 9. For the sum, since the interval of

-a

is

[2, 4]

and overlaps with the interval of

b

which is

[3, 5]

, they can be equal (specifically at 3 and 4). When

-a = b

, we have

a + b = 0

, so the minimum value of

|a + b|

must be 0. Subtracting the two values gives

9 - 0 = 9

.

Estimated Time:1m 30s

Question 30Question →

A scientist monitors the temperature of two research chambers. Chamber A is kept at $-5^\circ\text{C}$ and Chamber B is kept at $7^\circ\text{C}$ . The scientist sets a third chamber, Chamber C, to a temperature of $T^\circ\text{C}$ such that the distance between $T$ and the temperature of Chamber A on the Celsius scale is exactly $3$ times the distance between $T$ and the temperature of Chamber B. If the temperature of Chamber C is warmer than Chamber A but colder than Chamber B, what is the value of $T$ ?

Show answer & explanation

Answer: 4

Answer

The correct temperature value of Chamber C is 4.

The temperature of Chamber C must be

4^\circ\text{C}

because the distance from

4

to

-5

is

|4 - (-5)| = 9

and the distance from

4

to

7

is

|4 - 7| = 3

. The distance of

9

is exactly

3

times the distance of

3

. Furthermore,

4

lies between

-5

and

7

, satisfying the condition that Chamber C is warmer than Chamber A but colder than Chamber B.

Step-by-Step Solution

1

Represent the distances on the number line using absolute value expressions.

The distance to Chamber A is

|T - (-5)| = |T + 5|

and the distance to Chamber B is

|T - 7|

.

Distance between two points

x

and

y

on a number line is represented by

|x - y|

.

2

Set up the algebraic equation reflecting the relationship between the distances.

|T + 5| = 3|T - 7|

The problem states the distance to Chamber A is exactly 3 times the distance to Chamber B.

3

Solve the absolute value equation by considering both positive and negative cases.

Case 1:

T + 5 = 3(T - 7) \Rightarrow T = 13

. Case 2:

T + 5 = -3(T - 7) \Rightarrow T = 4

.

The equation

|x| = |y|

implies

x = y

or

x = -y

.

4

Verify which solution satisfies the temperature boundary condition.

Since Chamber C must be warmer than

-5^\circ\text{C}

but colder than

7^\circ\text{C}

, the only valid value is

T = 4

.

The value

T = 13

is warmer than both chambers and does not lie between them.

Key Concept

Using absolute value to represent distance on a number line and solving absolute value equations with boundary conditions.

Alternative Method

Alternatively, visualize this on a number line. The total distance between Chamber A (

-5

) and Chamber B (

7

) is

12

units. Since Chamber C lies between them and the distance from C to A is

3

times the distance from C to B, we can divide the

12

-unit interval into

3 + 1 = 4

equal parts. Each part is

12 \div 4 = 3

units. Chamber C is located

1

part away from Chamber B (towards Chamber A), which places it at

7 - 3 = 4

.

Estimated Time:1m 30s

Question 31Question →

A hiker starts at a base camp at an elevation of $120$ feet. She hikes up to an overlook and then down to a valley at an elevation of $-45$ feet. The elevation of the overlook is higher than both the base camp and the valley. On a vertical number line representing elevations, the distance between the overlook's coordinate and the valley's coordinate is exactly twice the distance between the overlook's coordinate and the base camp's coordinate. What is the elevation, in feet, of the overlook?

65

70

95

195

285

Show answer & explanation

Answer: 285

Answer

285 feet

The correct answer is

285

feet. Let

x

represent the elevation of the overlook. Because the overlook is higher than both the base camp at

120

feet and the valley at

-45

feet, we know that

x > 120

. The distance between the overlook and the valley is

x - (-45) = x + 45

, and the distance between the overlook and the base camp is

x - 120

. Setting up the relationship where the distance to the valley is twice the distance to the base camp gives the equation

x + 45 = 2(x - 120)

. Solving this equation yields

x + 45 = 2x - 240

, which simplifies to

x = 285

. This value is indeed higher than both elevations, and we can verify that the distance to the valley (

285 - (-45) = 330

) is twice the distance to the base camp (

285 - 120 = 165

).

Step-by-Step Solution

1

Define the variable and write the expressions for the distances on a number line.

Let

x

represent the elevation of the overlook in feet. The distance between the overlook and the base camp (

120

feet) is

|x - 120|

. The distance between the overlook and the valley (

-45

feet) is

|x - (-45)| = |x + 45|

.

Distance between two points

a

and

b

on a number line is defined by the absolute value of their difference,

|a - b|

.

2

Set up the equation based on the given ratio.

|x + 45| = 2|x - 120|

The problem states that the distance to the valley is exactly twice the distance to the base camp.

3

Apply the condition that the overlook is higher than both locations to simplify the absolute value expressions.

Since

x > 120

and

x > -45

, both terms inside the absolute values are positive. Therefore,

|x + 45| = x + 45

and

|x - 120| = x - 120

. The equation simplifies to:

x + 45 = 2(x - 120)

.

For any expression

y > 0

,

|y| = y

. Establishing the location of

x

relative to

120

and

-45

allows us to remove the absolute value bars.

4

Solve the simplified linear equation for

x

.

x + 45 = 2x - 240 \Rightarrow x = 285

Distribute the

2

on the right side and isolate

x

by subtracting

x

and adding

240

to both sides.

Key Concept

Representing distances between points on a number line using absolute value equations.

Alternative Method

Instead of setting up equations, you can test the options starting from the constraint that the elevation must be greater than

120

feet. This eliminates

65

,

70

, and

95

. Test the remaining options: for

195

, the distance to

120

is

75

and the distance to

-45

is

240

(not twice

75

). For

285

, the distance to

120

is

165

and the distance to

-45

is

330

, which is exactly twice

165

.

Estimated Time:1m 30s

Question 32Question →

If $k$ and $m$ are integers such that $|k - 3| = 8$ and $|m + 2| = 5$ , where $k < 0$ and $m < 0$ , what is the distance on the number line between $k$ and $m$ ?

2

4

8

12

18

Show answer & explanation

Answer: $2$

Answer

The distance on the number line between

k

and

m

is

2

.

To find

k

, solve

|k - 3| = 8

, which gives

k - 3 = 8 \implies k = 11

or

k - 3 = -8 \implies k = -5

. Since

k < 0

,

k = -5

. To find

m

, solve

|m + 2| = 5

, which gives

m + 2 = 5 \implies m = 3

or

m + 2 = -5 \implies m = -7

. Since

m < 0

,

m = -7

. The distance on the number line between

-5

and

-7

is

|-5 - (-7)| = 2

.

Step-by-Step Solution

1

Solve the absolute value equation for

k

using the given constraint

k < 0

.

k - 3 = 8 \implies k = 11

or

k - 3 = -8 \implies k = -5

. Since

k < 0

,

k = -5

.

An absolute value equation

|X| = c

splits into

X = c

and

X = -c

.

2

Solve the absolute value equation for

m

using the given constraint

m < 0

.

m + 2 = 5 \implies m = 3

or

m + 2 = -5 \implies m = -7

. Since

m < 0

,

m = -7

.

An absolute value equation

|X| = c

splits into

X = c

and

X = -c

.

3

Calculate the distance between

k = -5

and

m = -7

on the number line.

\text{Distance} = |k - m| = |-5 - (-7)| = |-5 + 7| = |2| = 2.

The distance between two points on a number line is given by the absolute value of their difference.

Key Concept

Evaluating absolute value equations and finding distance between two integers on a number line.

Estimated Time:1m 15s

Question 33Question →

If $a$ and $b$ are nonzero integers such that $|a| < |b|$ and $a \cdot b < 0$ , which of the following expressions must have the greatest value?

|a - b|

|a + b|

a - b

|a| - |b|

|b| - |a|

Show answer & explanation

Answer: $|a - b|$

Answer

The expression

|a - b|

must have the greatest value because

a

and

b

have opposite signs, making

|a - b| = |a| + |b|

.

Because

a

and

b

have opposite signs (

a \cdot b < 0

), the distance between them on the number line is the sum of their individual distances from zero,

|a| + |b|

. The expression

|a - b|

measures this exact distance, which is guaranteed to be greater than any net difference or signed value.

Step-by-Step Solution

1

Analyze the given conditions

Since

a \cdot b < 0

,

a

and

b

have opposite signs (one positive, one negative). Since

|a| < |b|

, the absolute value of

b

is strictly greater than the absolute value of

a

.

Determining the signs and relative magnitudes of

a

and

b

is necessary to evaluate absolute value expressions.

2

Evaluate

|a - b|

in terms of magnitudes

When two numbers have opposite signs, subtracting them adds their magnitudes:

|a - b| = |a| + |b|

.

If

a > 0

and

b < 0

,

|a - b| = a - (b) = |a| + |b|

. If

a < 0

and

b > 0

,

|a - b| = |- |a| - |b|| = |a| + |b|

.

3

Compare

|a - b|

to the other expressions

|a - b| = |a| + |b|

is strictly greater than

|a + b| = |b| - |a|

,

|b| - |a|

,

|a| - |b|

, and

a - b

.

Since both

a

and

b

are non-zero,

|a| + |b| > |b| - |a| > 0 > |a| - |b|

.

Key Concept

Absolute value distance and sign properties on a number line

Question 34Question →

On a standard number line, point $R$ has coordinate $-18$ and point $T$ has coordinate $22$ . Point $S$ lies on the line segment $RT$ such that the distance from $R$ to $S$ is $\frac{3}{5}$ of the distance from $R$ to $T$ . What is the coordinate of point $S$ ?

Show answer & explanation

Answer: 6

Answer

6

The distance between

R(-18)

and

T(22)

is calculated using absolute value:

|22 - (-18)| = 40

. Taking

\frac{3}{5}

of 40 yields 24 units. Adding 24 to the starting coordinate

-18

gives the coordinate of point

S

as 6.

Step-by-Step Solution

1

Calculate the total distance between points

R

and

T

Distance

= |22 - (-18)| = |40| = 40

units

The distance between two points on a number line is given by the absolute difference of their coordinates.

2

Determine the distance from

R

to

S

Distance

= \frac{3}{5} \times 40 = 24

units

Point

S

is located at a distance equal to

\frac{3}{5}

of the total segment length starting from

R

.

3

Calculate the coordinate of point

S

Coordinate of

S = -18 + 24 = 6

Since

S

lies between

R(-18)

and

T(22)

, moving from

R

toward

T

corresponds to increasing the coordinate value by 24.

Key Concept

Calculating distance between signed integers using absolute value and finding a point dividing a number line segment in a given ratio

Question 35Question →

On a standard number line, point $A$ has coordinate $-28$ and point $B$ has coordinate $12$ . Point $C$ is the midpoint of segment $AB$ . Point $D$ is positioned on the line such that point $B$ is the midpoint of segment $CD$ . What is the distance between point $C$ and point $D$ on the number line?

Show answer & explanation

Answer: 40

Answer

The distance between point

C

and point

D

on the number line is 40.

The midpoint

C

of segment

AB

is found by averaging the coordinates:

\frac{-28 + 12}{2} = -8

. Because point

B(12)

is the midpoint of segment

CD

, set up the equation

\frac{-8 + D}{2} = 12

, yielding

D = 32

. The distance between point

C(-8)

and point

D(32)

is the absolute value of their difference:

|32 - (-8)| = 40

.

Step-by-Step Solution

1

Calculate the coordinate of point

C

, which is the midpoint of segment

AB

.

The coordinate of point

C

is

-8

.

The midpoint of two points on a number line is given by their average:

\frac{-28 + 12}{2} = -8

.

2

Determine the coordinate of point

D

using the midpoint relationship for segment

CD

.

The coordinate of point

D

is

32

.

Since point

B(12)

is the midpoint of segment

CD

,

\frac{C + D}{2} = 12 \implies \frac{-8 + D}{2} = 12

, which solves to

D = 32

.

3

Find the distance between point

C

and point

D

.

The distance is

40

.

Distance on a number line is the absolute value of the difference between coordinates:

|32 - (-8)| = |40| = 40

.

Key Concept

Midpoint and Absolute Value Distance on a Number Line

Alternative Method

Alternatively, note that the distance of segment

AB

is

|12 - (-28)| = 40

. Since

C

is the midpoint of

AB

, the length of segment

CB

is

\frac{40}{2} = 20

. Since

B

is the midpoint of segment

CD

, the length of segment

BD

must equal the length of segment

CB

, which is

20

. Therefore, the total distance from

C

to

D

is

CB + BD = 20 + 20 = 40

.

Estimated Time:1m 15s

Question 36Question →

Let $a$ and $b$ be integers. On a standard number line, $a = -14$ , and the distance between $a$ and $b$ is $23$ . If $b > 0$ , what is the value of $|2b - |a||$ ?

Show answer & explanation

Answer: 4

Answer

The value of

|2b - |a||

is 4.

The distance between

a = -14

and

b

on the number line is

|b - (-14)| = |b + 14| = 23

. Solving for a positive value of

b

gives

b + 14 = 23

, so

b = 9

. The absolute value of

a

is

|-14| = 14

. Substituting these values into

|2b - |a||

yields

|2(9) - 14| = |18 - 14| = 4

.

Step-by-Step Solution

1

Determine the value of integer

b

using the distance relationship on the number line.

The distance formula

|b - (-14)| = 23

simplifies to

|b + 14| = 23

. Since

b > 0

,

b = 9

.

The distance between two points

x

and

y

on a number line is given by

|x - y|

.

2

Calculate the absolute value of

a

.

|a| = |-14| = 14

.

The absolute value of a negative number is its positive magnitude.

3

Substitute

b = 9

and

|a| = 14

into the expression

|2b - |a||

.

|2(9) - 14| = |18 - 14| = 4

.

Perform operations inside the outer absolute value first, then take the absolute value of the result.

Key Concept

Absolute Value as Distance on a Number Line

Estimated Time:1m 0s

Question 37Question →

What is the sum of all integer solutions to the equation $|2x - 8| = 14$ ?

-3

8

11

14

22

Show answer & explanation

Answer: 8

Answer

The sum of all integer solutions is 8.

To solve

|2x - 8| = 14

, consider both possible cases:

2x - 8 = 14

and

2x - 8 = -14

. The first equation gives

2x = 22

, so

x = 11

. The second equation gives

2x = -6

, so

x = -3

. Adding these two solutions yields

11 + (-3) = 8

.

Step-by-Step Solution

1

Set up the two linear equations represented by the absolute value equation.

2x - 8 = 14

or

2x - 8 = -14

The absolute value expression

|x| = k

for

k > 0

splits into

x = k

and

x = -k

.

2

Solve the first equation for

x

.

2x = 22 \implies x = 11

Add 8 to both sides and divide by 2.

3

Solve the second equation for

x

.

2x = -6 \implies x = -3

Add 8 to both sides and divide by 2.

4

Calculate the sum of both integer solutions.

11 + (-3) = 8

The problem asks for the sum of all integer solutions.

Key Concept

Solving Absolute Value Equations with Integers

Estimated Time:1m 0s

Question 38Question →

On a vertical number line representing elevation relative to sea level ( $0$ meters), Submersible $X$ is located at a coordinate of $-45$ meters. Submersible $Y$ is located at a depth such that the distance between Submersible $X$ and Submersible $Y$ is $30$ meters. If Submersible $Y$ is closer to sea level than Submersible $X$ , what is the value of $|2Y - X|$ ?

15

60

75

105

135

Show answer & explanation

Answer: $15$

Answer

The value of

|2Y - X|

is

15

.

Submersible

X

is at

-45

. Being

30

meters away means Submersible

Y

is at either

-15

or

-75

. Since Submersible

Y

is closer to sea level (

0

), its coordinate is

-15

. Substituting

X = -45

and

Y = -15

into

|2Y - X|

gives

|2(-15) - (-45)| = |-30 + 45| = |15| = 15

.

Step-by-Step Solution

1

Determine the possible coordinates for Submersible

Y

.

Since

X = -45

and the distance between

X

and

Y

is

30

,

Y = -45 + 30 = -15

or

Y = -45 - 30 = -75

.

Distance on a number line from coordinate

x

is given by

x \pm d

.

2

Select the correct coordinate for

Y

based on the given constraint.

Y = -15

, because

|-15| = 15 < 45 = |-45|

, meaning

Y

is closer to sea level (

0

).

The problem states that Submersible

Y

is closer to sea level than Submersible

X

.

3

Substitute

X = -45

and

Y = -15

into the expression

|2Y - X|

and evaluate.

|2(-15) - (-45)| = |-30 + 45| = |15| = 15

.

Multiplying

2

by

-15

yields

-30

, and subtracting

-45

is equivalent to adding

45

.

Key Concept

Distance on a Number Line and Absolute Value Evaluation

Estimated Time:1m 15s

Question 39Question →

On a standard number line, point $P$ is located at $-17$ and point $Q$ is located at $31$ . Point $R$ lies between $P$ and $Q$ such that the ratio of the distance between $P$ and $R$ to the distance between $R$ and $Q$ is $3:5$ . What is the value of $|R - (-5)|$ ?

Show answer & explanation

Answer: 6

Answer

The value of

|R - (-5)|

is

6

.

The total distance between point

P (-17)

and point

Q (31)

is

31 - (-17) = 48

. Since point

R

divides segment

PQ

in a

3:5

ratio, the distance from

P

to

R

is

\frac{3}{8} \times 48 = 18

. Adding this distance to

-17

gives

R = 1

. Substituting

R = 1

into

|R - (-5)|

yields

|1 - (-5)| = |6| = 6

.

Step-by-Step Solution

1

Calculate the total distance between points

P

and

Q

Distance

PQ = 31 - (-17) = 48

The distance between two points on a number line is found by taking the absolute difference of their coordinates.

2

Determine the coordinate of point

R

Coordinate of

R = 1

The ratio of

PR

to

RQ

is

3:5

, meaning

PR

is

\frac{3}{3+5} = \frac{3}{8}

of the total distance

PQ

. Adding

\frac{3}{8} \times 48 = 18

to the starting coordinate

-17

gives

R = 1

.

3

Evaluate the absolute value expression

|R - (-5)|

|1 - (-5)| = |6| = 6

Substitute

R = 1

into the target expression and simplify the double negative before taking the absolute value.

Key Concept

Calculating distance on a number line, segment ratio partitioning, and absolute value evaluation.

Estimated Time:1m 30s

Question 40Question →

On a standard number line, point $m$ is located at coordinate $-9$ , and point $n$ is located to the right of point $m$ . If the distance between point $m$ and point $n$ is given by the expression $|3k - 6|$ where $k = -4$ , what is the coordinate of point $n$ ?

-27

-15

-3

9

27

Show answer & explanation

Answer: $9$

Answer

The coordinate of point

n

is

9

.

First, evaluate the distance expression

|3k - 6|

when

k = -4

:

3(-4) - 6 = -18

, and

|-18| = 18

. Since point

n

is located to the right of point

m

(which is at

-9

), add the distance of

18

to

-9

:

-9 + 18 = 9

. Thus, the coordinate of point

n

is

9

.

Step-by-Step Solution

1

Evaluate the expression inside the absolute value for

k = -4

3(-4) - 6 = -12 - 6 = -18

Substitute

k = -4

into

3k - 6

following the order of operations.

2

Calculate the absolute value to find the distance

|-18| = 18

Distance is non-negative, so taking the absolute value yields a distance of 18 units.

3

Find the position of point

n

relative to point

m

n = -9 + 18 = 9

Since point

n

lies to the right of point

m = -9

on the number line, add the distance of 18 units to

-9

.

Key Concept

Distance on a Number Line using Absolute Value

Estimated Time:1m 0s

Integers, Absolute Value, and Number Lines

Answer

Step-by-Step Solution

Key Concept

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Common Mistakes

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Common Mistakes

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