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Absolute value in context

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5411726
A diver is at an elevation of \(-6\,\text{m}\) relative to sea level. a) Find \(\lvert-6\rvert\). b) Explain what this absolute value means in context.

Hints

- Locate \(-6\) relative to \(0\). - Absolute value records distance, not direction. - Use the original sign when describing whether the diver is above or below sea level.

Solution

1. Absolute value is distance from \(0\), so \(\lvert-6\rvert=6\). 2. The diver is \(6\,\text{m}\) from sea level; the negative sign shows that the diver is below sea level.

Answer

a) \(\lvert-6\rvert=6\) b) The diver is \(6\,\text{m}\) below sea level.
5174996
A thermometer reads \(-12\,^\circ\text{F}\) one night and \(10\,^\circ\text{F}\) the next day. Write an inequality using absolute-value notation to show which temperature is farther from \(0\,^\circ\text{F}\).

Hints

- Absolute value gives a temperature’s distance from zero. - Compare the two distances. - Use \(>\) when the first distance is greater.

Solution

1. The distances from zero are \(|-12|=12\) and \(|10|=10\). 2. Since \(12>10\), the comparison is \(|-12|>|10|\).

Answer

\(|-12|>|10|\)
5190976
A simplified latitude model uses the equator as \(0^\circ\), north latitudes as positive, and south latitudes as negative. Research station \(A\) is at \(+80^\circ\), and station \(B\) is at \(-10^\circ\). a) Which station is farther from the equator? Justify using absolute value. b) Station \(C\) is the same distance from the equator as station \(A\) but is on the opposite side. What is station \(C\)'s signed latitude?

Hints

- The equator is the zero point in this model. - Compare each station’s distance from zero, not the signed values themselves. - Points the same distance from zero on opposite sides have opposite coordinates.

Solution

1. Distance from the equator is distance from \(0\), so compare absolute values. 2. \(\lvert80\rvert=80\) and \(\lvert-10\rvert=10\). Therefore, station \(A\) is farther from the equator. 3. A point the same distance from \(0\) as \(+80\) but on the opposite side has coordinate \(-80\).

Answer

a) Station \(A\), because \(\lvert80\rvert=80>10=\lvert-10\rvert\) b) \(-80^\circ\)
5191016
An elevator uses the lobby as floor \(0\). Parking level \(P\) is floor \(-4\), and a rooftop deck is floor \(+9\). Which location is farther from the lobby, and by how many floors is its distance greater? Use absolute value to justify your answer.

Hints

- Treat the lobby as the zero point. - Find each location’s distance from zero before comparing. - Subtract the smaller distance from the larger distance to find how much farther.

Solution

1. The parking level is \(\lvert-4\rvert=4\) floors from the lobby. 2. The rooftop deck is \(\lvert9\rvert=9\) floors from the lobby. 3. Since \(9>4\), the rooftop deck is farther from the lobby. 4. Its distance is \(9-4=5\) floors greater.

Answer

The rooftop deck is farther from the lobby, by \(5\) floors.
5191026
A temperature controller records deviations from its target temperature. Sensor \(A\) reads \(-7.5^\circ\), and sensor \(B\) reads \(+6.2^\circ\), where \(0^\circ\) means exactly on target. Which sensor is farther from the target, and by how much? Justify using absolute value.

Hints

- A deviation’s sign gives direction, while its absolute value gives distance from the target. - Compare the two distances after removing the signs. - Find the difference between those distances.

Solution

1. Sensor \(A\)'s distance from the target is \(\lvert-7.5\rvert=7.5^\circ\). 2. Sensor \(B\)'s distance from the target is \(\lvert6.2\rvert=6.2^\circ\). 3. Since \(7.5>6.2\), sensor \(A\) is farther from the target. 4. The difference in distances is \(7.5-6.2=1.3^\circ\).

Answer

Sensor \(A\) is farther from the target by \(1.3^\circ\).
5411706
A budgeting app shows four account balances: \(-\$18\), \(+\$27\), \(-\$25\), and \(+\$40\). Which balances are within \(\$25\) of a zero balance? Explain using absolute values.

Hints

- Treat each balance as a signed number. - Absolute value gives its distance from a zero balance. - Include a balance whose distance is exactly \(\$25\).

Solution

1. The distances from a zero balance are \(\lvert-18\rvert=18\), \(\lvert27\rvert=27\), \(\lvert-25\rvert=25\), and \(\lvert40\rvert=40\). 2. The distances \(18\) and \(25\) are at most \(25\). 3. Therefore, the balances \(-\$18\) and \(-\$25\) are within \(\$25\) of zero.

Answer

The balances \(-\$18\) and \(-\$25\) are within \(\$25\) of zero.
5411756
An elevation \(x\), measured relative to sea level, satisfies \(\lvert x\rvert=0\). a) Find \(x\). b) Explain why there is only one possible elevation. c) Interpret the result.

Hints

- Interpret absolute value as distance from \(0\). - Consider whether a nonzero value can be zero units from \(0\). - Translate the coordinate back to elevation relative to sea level.

Solution

1. Absolute value gives distance from \(0\). 2. The only number at distance \(0\) from \(0\) is \(0\), so \(x=0\). 3. An elevation of \(0\) is exactly at sea level.

Answer

a) \(x=0\) b) Zero is the only number whose distance from \(0\) is \(0\). c) The elevation is exactly at sea level.
5411776
A temperature is \(\frac{3}{4}^\circ\text{C}\) below \(0^\circ\text{C}\), so its signed value is \(-\frac{3}{4}^\circ\text{C}\). a) Evaluate \(\left\lvert-\frac{3}{4}\right\rvert\). b) Interpret the result in this context. c) What information does the absolute value not include?

Hints

- Treat \(0^\circ\text{C}\) as the reference point. - Absolute value gives the distance from that reference point. - Separate the amount of the difference from its direction.

Solution

1. The distance of \(-\frac{3}{4}\) from \(0\) is \(\frac{3}{4}\). 2. Thus, \(\left\lvert-\frac{3}{4}\right\rvert=\frac{3}{4}\), so the temperature is \(\frac{3}{4}^\circ\text{C}\) from \(0^\circ\text{C}\). 3. The absolute value does not indicate that the temperature is below \(0^\circ\text{C}\).

Answer

a) \(\frac{3}{4}\) b) The temperature is \(\frac{3}{4}^\circ\text{C}\) from \(0^\circ\text{C}\). c) It does not include whether the temperature is above or below \(0^\circ\text{C}\).
5411786
Which numbers are less than \(1\) unit from \(0\)? \(-\frac{5}{4}, -\frac{3}{4}, \frac{1}{2}, 0.9, \frac{6}{5}\) Explain using absolute values.

Hints

- Find each number's distance from \(0\). - Compare each distance with \(1\). - Keep each original sign in the final list.

Solution

1. The absolute values are \(\frac{5}{4}\), \(\frac{3}{4}\), \(\frac{1}{2}\), \(0.9\), and \(\frac{6}{5}\). 2. The values less than \(1\) are \(\frac{3}{4}\), \(\frac{1}{2}\), and \(0.9\). 3. Therefore, the original numbers less than \(1\) unit from \(0\) are \(-\frac{3}{4}\), \(\frac{1}{2}\), and \(0.9\).

Answer

\(-\frac{3}{4}, \frac{1}{2}, 0.9\)
5411796
A pool's water level is \(2.6\,\text{cm}\) below its target level. Use a negative value for a level below the target. a) Write the signed level difference. b) Write an absolute-value equation that checks its magnitude. c) Explain why \(+2.6\) does not fit the full description.

Hints

- Use the directional word to choose the sign. - Use absolute value to check only the size of the difference. - Compare the meanings of the positive and negative values.

Solution

1. Below the target is negative, so the signed level difference is \(-2.6\,\text{cm}\). 2. The magnitude check is \(\lvert-2.6\rvert=2.6\). 3. The value \(+2.6\) has the same magnitude but represents a level above the target.

Answer

a) \(-2.6\,\text{cm}\) b) \(\lvert-2.6\rvert=2.6\) c) \(+2.6\) represents a level above the target.
5175126
Five players have these scores after one round: Player A: \(-25\) points Player B: \(18\) points Player C: \(-12\) points Player D: \(22\) points Player E: \(-3\) points a) Which player’s score has the least absolute value? b) List the players from greatest to least absolute value of their scores.

Hints

- The absolute value of a score is its distance from zero. - Find all five absolute values before ordering the players. - Ignore the signs when comparing absolute values.

Solution

1. The absolute values of the scores are \(25,18,12,22,3\) for Players A through E. 2. The least absolute value is \(3\), so Player E answers part a). 3. From greatest to least absolute value, the order is Player A, Player D, Player B, Player C, Player E.

Answer

a) Player E b) Player A, Player D, Player B, Player C, Player E
5331446
Several points are marked on the coordinate plane. Write each coordinate pair. Which point or points are farthest from the x-axis?
Figure for problem 533144

Hints

- The vertical distance to the x-axis depends on the y-coordinate. - Distance is nonnegative, even for a point below the x-axis. - Compare the absolute values of the y-coordinates.

Solution

1. The coordinates are \(P(5, 1)\), \(Q(-4, -3)\), \(R(1, 4)\), \(S(-3, 1)\), and \(T(0, -4)\). 2. A point’s distance from the x-axis is the absolute value of its y-coordinate. 3. The distances are \(1, 3, 4, 1,\) and \(4\) units, respectively. 4. Points \(R\) and \(T\) are each \(4\) units from the x-axis, so they are tied for the greatest distance.

Answer

The coordinates are \(P(5, 1)\), \(Q(-4, -3)\), \(R(1, 4)\), \(S(-3, 1)\), and \(T(0, -4)\). Points \(R\) and \(T\) are farthest from the x-axis, at a distance of \(4\) units.
5411696
Use the number line to answer the questions. a) Which marked temperature is closest to \(0^\circ\text{C}\)? b) Which marked temperature is farthest from \(0^\circ\text{C}\)? c) Write the absolute value used for each choice.
Figure for problem 541169

Hints

- Read each marked temperature from the number line. - Absolute value gives distance from \(0\). - Compare those distances to identify the closest and farthest readings.

Solution

1. The graph shows \(A=-2.5\), \(B=1.75\), \(C=-0.75\), and \(D=3.25\). 2. Their absolute values are \(\lvert-2.5\rvert=2.5\), \(\lvert1.75\rvert=1.75\), \(\lvert-0.75\rvert=0.75\), and \(\lvert3.25\rvert=3.25\). 3. The least absolute value is \(0.75\), so \(C\), at \(-0.75^\circ\text{C}\), is closest to \(0^\circ\text{C}\). 4. The greatest absolute value is \(3.25\), so \(D\), at \(3.25^\circ\text{C}\), is farthest from \(0^\circ\text{C}\).

Answer

a) \(C\), at \(-0.75^\circ\text{C}\) b) \(D\), at \(3.25^\circ\text{C}\) c) \(\lvert-0.75\rvert=0.75\) and \(\lvert3.25\rvert=3.25\)
5411716
Four elevator positions relative to the lobby are \(-6\), \(+2\), \(-9\), and \(+4\) floors. Order the signed positions from nearest to farthest from the lobby. Keep each original sign in your list.

Hints

- Interpret each signed floor as a position above or below the lobby. - Compare distances from floor \(0\), not ordinary signed order. - Restore each original sign after ordering the distances.

Solution

1. The distances from the lobby are \(\lvert-6\rvert=6\), \(\lvert2\rvert=2\), \(\lvert-9\rvert=9\), and \(\lvert4\rvert=4\). 2. Ordering the distances gives \(2<4<6<9\). 3. Matching those distances to the original signed positions gives \(+2, +4, -6, -9\).

Answer

\(+2, +4, -6, -9\)
5411736
An east-west trail uses the trailhead as position \(0\), with east positive. A hiker is \(3.2\,\text{mi}\) from the trailhead. a) Give both possible signed positions. b) Write an absolute-value equation satisfied by both positions. c) Explain why the equation does not identify a unique position.

Hints

- Mark the trailhead as \(0\) and consider both directions. - Use one variable for the signed position. - Identify the directional information that absolute value does not preserve.

Solution

1. The hiker could be \(3.2\,\text{mi}\) east of the trailhead, at \(+3.2\), or \(3.2\,\text{mi}\) west, at \(-3.2\). 2. Both positions satisfy \(\lvert x\rvert=3.2\). 3. Absolute value gives distance from \(0\) but does not identify the side of \(0\).

Answer

a) \(+3.2\,\text{mi}\) and \(-3.2\,\text{mi}\) b) \(\lvert x\rvert=3.2\) c) The equation gives the distance from the trailhead but not the direction.
5411746
On an east-west trail, east is positive. Two hikers are at \(+0.7\,\text{km}\) and \(-0.7\,\text{km}\) relative to the trailhead. a) Compare the signed positions using \(<\), \(>\), or \(=\). b) Compare their absolute values. c) Explain what the two comparisons say about the hikers.

Hints

- Use left-to-right number-line order for the signed comparison. - Use distance from \(0\) for the absolute-value comparison. - Interpret sign and magnitude as different features of position.

Solution

1. As signed positions, \(-0.7<+0.7\). 2. Their distances from the trailhead are equal: \(\lvert-0.7\rvert=\lvert+0.7\rvert=0.7\). 3. The hikers are on opposite sides of the trailhead but equally far from it.

Answer

a) \(-0.7<+0.7\) b) \(\lvert-0.7\rvert=\lvert+0.7\rvert\) c) They are on opposite sides of the trailhead and the same distance from it.
5411766
An elevation \(m\), measured relative to sea level, has magnitude \(12\) and is less than \(0\). a) Write an absolute-value equation for \(m\). b) Find \(m\). c) Interpret the sign of \(m\).

Hints

- Begin with the two numbers that are \(12\) units from \(0\). - Use the inequality to select one of them. - Translate the selected sign into elevation relative to sea level.

Solution

1. A magnitude of \(12\) gives \(\lvert m\rvert=12\). 2. The two numbers with that magnitude are \(12\) and \(-12\). 3. Since \(m<0\), \(m=-12\), which represents an elevation \(12\) units below sea level.

Answer

a) \(\lvert m\rvert=12\) b) \(m=-12\) c) The elevation is \(12\) units below sea level.
5411806
For the signed position \(-9\), both \(\lvert-9\rvert\) and \(-(-9)\) equal \(9\). Explain why the expressions have the same value but represent different ideas.

Hints

- State the definition of each notation separately. - Distinguish a distance from a new signed position. - Explain the shared result without claiming the operations always mean the same thing.

Solution

1. The expression \(\lvert-9\rvert\) gives the distance of \(-9\) from \(0\), which is \(9\). 2. The expression \(-(-9)\) gives the opposite of \(-9\), which is the signed number \(9\). 3. One operation finds magnitude, while the other reverses direction.

Answer

Both expressions equal \(9\), but \(\lvert-9\rvert\) means distance from \(0\), while \(-(-9)\) means the opposite of \(-9\).
5411816
Two elevations relative to sea level are \(-10\,\text{m}\) and \(+3\,\text{m}\). a) Compare the elevations using \(<\), \(>\), or \(=\). b) Compare their absolute values. c) Explain why the lower elevation is farther from sea level.

Hints

- First compare the signed values on a number line. - Then find each value's distance from \(0\). - Interpret signed order and absolute value as answers to different questions.

Solution

1. As signed elevations, \(-10<+3\), so \(-10\,\text{m}\) is lower. 2. Their distances from sea level are \(\lvert-10\rvert=10\) and \(\lvert+3\rvert=3\), so \(\lvert-10\rvert>\lvert+3\rvert\). 3. Signed order describes vertical position, while absolute value describes distance from sea level.

Answer

a) \(-10<+3\) b) \(\lvert-10\rvert>\lvert+3\rvert\) c) The elevation \(-10\,\text{m}\) is \(10\,\text{m}\) from sea level, while \(+3\,\text{m}\) is \(3\,\text{m}\) from sea level.
5411826
An integer \(e\) satisfies \(2<\lvert e\rvert<4\). a) Find all possible values of \(e\) and explain how you know your list is complete. b) Interpret the values as positions on opposite sides of \(0\).

Hints

- Identify the integer magnitudes strictly between the two bounds. - Check both endpoints carefully because the inequalities are strict. - For each nonzero magnitude, consider positions on both sides of \(0\).

Solution

1. The only integer magnitude strictly between \(2\) and \(4\) is \(3\). 2. The integers with magnitude \(3\) are \(-3\) and \(3\), so these are all possible values. 3. They represent positions \(3\) units from \(0\) on opposite sides.

Answer

a) \(e=-3\) or \(e=3\); \(3\) is the only integer magnitude strictly between \(2\) and \(4\). b) The positions are \(3\) units to the left and right of \(0\).

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