12 multiple-choice questions, progressively harder.
Order these from least to greatest by their actual (signed) values: −8-8−8, ∣−3∣\lvert -3 \rvert∣−3∣, ∣5∣\lvert 5 \rvert∣5∣, −∣2∣-\lvert 2 \rvert−∣2∣.
Solution
Correct answer: D
First simplify each value: ∣−3∣=3\lvert -3 \rvert = 3∣−3∣=3, ∣5∣=5\lvert 5 \rvert = 5∣5∣=5, and −∣2∣=−2-\lvert 2 \rvert = -2−∣2∣=−2 (the minus is outside).
−8, −2, 3, 5-8,\; -2,\; 3,\; 5−8,−2,3,5
Ordered from least to greatest, that is −8<−2<3<5-8 < -2 < 3 < 5−8<−2<3<5, which matches −8, −∣2∣, ∣−3∣, ∣5∣-8,\; -\lvert 2 \rvert,\; \lvert -3 \rvert,\; \lvert 5 \rvert−8,−∣2∣,∣−3∣,∣5∣.
Evaluate ∣5−14∣\lvert 5 - 14 \rvert∣5−14∣.
Correct answer: A
The bars group the subtraction, so work inside first: 5−14=−95 - 14 = -95−14=−9. Then take the absolute value of that result.
∣5−14∣=∣−9∣=9\lvert 5 - 14 \rvert = \lvert -9 \rvert = 9∣5−14∣=∣−9∣=9
How many integers sit less than 333 units from zero?
Correct answer: C
Start at zero and stay strictly nearer than three units, in either direction. List what you land on: −2,−1,0,1,2-2, -1, 0, 1, 2−2,−1,0,1,2.
∣−2∣,∣−1∣,∣0∣,∣1∣,∣2∣ are all <3\lvert -2 \rvert, \lvert -1 \rvert, \lvert 0 \rvert, \lvert 1 \rvert, \lvert 2 \rvert \text{ are all } < 3∣−2∣,∣−1∣,∣0∣,∣1∣,∣2∣ are all <3
That is five integers. (Note ∣−3∣=3\lvert -3 \rvert = 3∣−3∣=3 is not less than 333.)
Evaluate −∣−11∣-\lvert -11 \rvert−∣−11∣.
Correct answer: B
Work the bars first: ∣−11∣=11\lvert -11 \rvert = 11∣−11∣=11. The minus sign sits outside the bars, so it applies to that result.
−∣−11∣=−(11)=−11-\lvert -11 \rvert = -(11) = -11−∣−11∣=−(11)=−11
The bars measure only what stands between them, so the outer minus takes the opposite of the distance they returned.
Evaluate ∣7−7∣−∣0−4∣\lvert 7 - 7 \rvert - \lvert 0 - 4 \rvert∣7−7∣−∣0−4∣.
Evaluate each group separately. The first is ∣7−7∣=∣0∣=0\lvert 7 - 7 \rvert = \lvert 0 \rvert = 0∣7−7∣=∣0∣=0, and the second is ∣0−4∣=∣−4∣=4\lvert 0 - 4 \rvert = \lvert -4 \rvert = 4∣0−4∣=∣−4∣=4.
0−4=−40 - 4 = -40−4=−4
Subtracting the second result from the first gives −4-4−4.
What is the distance between −15-15−15 and 666 on the number line?
Distance is the absolute value of the difference.
∣−15−6∣=∣−21∣=21\lvert -15 - 6 \rvert = \lvert -21 \rvert = 21∣−15−6∣=∣−21∣=21
From −15-15−15 it is fifteen one-unit steps to zero and six more to 666, which is twenty-one steps.
A diver is at −40-40−40 meters and a second diver is at −25-25−25 meters. How far apart are they?
How far apart they are is the distance between the two depths, which is the absolute value of their difference.
∣−40−(−25)∣=∣−40+25∣=∣−15∣=15\lvert -40 - (-25) \rvert = \lvert -40 + 25 \rvert = \lvert -15 \rvert = 15∣−40−(−25)∣=∣−40+25∣=∣−15∣=15
The divers are 151515 meters apart.
Evaluate ∣3−3∣+∣−7∣\lvert 3 - 3 \rvert + \lvert -7 \rvert∣3−3∣+∣−7∣.
Evaluate each group separately. The first is ∣3−3∣=∣0∣=0\lvert 3 - 3 \rvert = \lvert 0 \rvert = 0∣3−3∣=∣0∣=0, and the second is ∣−7∣=7\lvert -7 \rvert = 7∣−7∣=7.
0+7=70 + 7 = 70+7=7
Which list is ordered from least to greatest by absolute value: −7-7−7, 444, −1-1−1, 666?
Take the absolute value of each: ∣−7∣=7\lvert -7 \rvert = 7∣−7∣=7, ∣4∣=4\lvert 4 \rvert = 4∣4∣=4, ∣−1∣=1\lvert -1 \rvert = 1∣−1∣=1, ∣6∣=6\lvert 6 \rvert = 6∣6∣=6.
1<4<6<71 < 4 < 6 < 71<4<6<7
Ordering the original numbers by those distances gives −1, 4, 6, −7-1,\; 4,\; 6,\; -7−1,4,6,−7.
Evaluate ∣2×(1−6)∣\lvert 2 \times (1 - 6) \rvert∣2×(1−6)∣.
Work from the innermost grouping outward. Inside the parentheses, 1−6=−51 - 6 = -51−6=−5, then multiply by 222 to get 2×(−5)=−102 \times (-5) = -102×(−5)=−10. Finally the bars measure its distance from zero.
∣2×(1−6)∣=∣−10∣=10\lvert 2 \times (1 - 6) \rvert = \lvert -10 \rvert = 10∣2×(1−6)∣=∣−10∣=10
A submarine sits at −90-90−90 meters and a fish swims at −35-35−35 meters. How far apart are they?
∣−90−(−35)∣=∣−90+35∣=∣−55∣=55\lvert -90 - (-35) \rvert = \lvert -90 + 35 \rvert = \lvert -55 \rvert = 55∣−90−(−35)∣=∣−90+35∣=∣−55∣=55
They are 555555 meters apart.
For which value of xxx is ∣x∣=−x\lvert x \rvert = -x∣x∣=−x true?
The rule ∣x∣=−x\lvert x \rvert = -x∣x∣=−x holds when xxx is negative (or zero), since then the distance from zero is the positive opposite of xxx.
∣−6∣=6=−(−6)\lvert -6 \rvert = 6 = -(-6)∣−6∣=6=−(−6)
For a positive xxx it fails, because ∣x∣=x\lvert x \rvert = x∣x∣=x, not −x-x−x. So it is true for x=−6x = -6x=−6.
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