index

Pre-Algebra Workbook 1: Integers & Absolute Value

· 1min
Problem 01
Integer Addition & Subtraction

Evaluate:

−7−12-7 - 12

Choices
A −5-5
B 55
C 1919
D −19-19
Strategy

Rewrite subtraction as “adding the opposite”:

−7−12=−7+(−12).-7 - 12 = -7 + (-12).

Then add the two negative numbers by adding their absolute values and keeping the negative sign.

Solution

Rewrite the subtraction as addition of the opposite:

−7−12=−7+(−12)-7 - 12 = -7 + (-12)

Add the integers:

−7+(−12)=−19-7 + (-12) = -19

So the correct choice is D. −19-19

Problem 02
Integer Addition & Subtraction

Evaluate:

15+(−9)15 + (-9)

Choices
A 2424
B −6-6
C −24-24
D 66
Strategy

You’re adding a positive and a negative number. Compare their absolute values, subtract the smaller from the larger, and keep the sign of the number with the larger absolute value.

Solution

We are adding integers with different signs. Subtract the absolute values and keep the sign of the number with the larger absolute value.

15+(−9)=15−9=615 + (-9) = 15 - 9 = 6

Since 1515 has the larger absolute value and is positive, the result is positive. So the correct choice is D. 66

Problem 03
Integer Multiplication & Division

Evaluate:

−4⋅6-4 \cdot 6

Choices
A 2424
B −24-24
C −10-10
D 1010
Strategy

Use the sign rule for multiplication:

  • Negative ×\times positive = negative.

Multiply the absolute values, then apply the correct sign.

Solution

Use the sign rule for multiplication: negative times positive is negative.

Multiply the absolute values, then attach the sign:

−4⋅6=−(4⋅6)=−24-4 \cdot 6 = -(4 \cdot 6) = -24

So the correct choice is B. −24-24

Problem 04
Integer Multiplication & Division

Evaluate:

−48÷(−6)-48 \div (-6)

Choices
A −8-8
B −42-42
C 88
D 4242
Strategy

Use the sign rule for division:

  • Negative ÷\div negative = positive.

Then divide the absolute values.

Solution

Use the sign rule for division: negative divided by negative is positive.

Divide the absolute values:

−48÷(−6)=48÷6=8-48 \div (-6) = 48 \div 6 = 8

So the correct choice is C. 88

Problem 05
Integer Addition & Subtraction

Evaluate:

−2−(−5)−3-2 - (-5) - 3

Choices
A 00
B −10-10
C 1010
D 44
Strategy

First rewrite subtraction as addition of the opposite, then work left to right:

−2−(−5)−3=−2+5−3.-2 - (-5) - 3 = -2 + 5 - 3.

Solution

First rewrite subtraction as addition of the opposite:

−2−(−5)−3=−2+5−3-2 - (-5) - 3 = -2 + 5 - 3

Then simplify from left to right:

−2+5=3,3−3=0-2 + 5 = 3,\quad 3 - 3 = 0

So −2−(−5)−3=0-2 - (-5) - 3 = 0 and the correct choice is A. 00

Problem 06
Absolute Value

Evaluate:

∣−3∣+∣−7∣|-3| + |-7|

Choices
A 44
B 1010
C −10-10
D 1414
Strategy

Use the definition of absolute value: it’s the distance from 00 always nonnegative. Replace each absolute value with a positive number, then add.

Solution

Use the definition of absolute value: it is the distance from 00 so it is always nonnegative.

∣−3∣=3,∣−7∣=7|-3| = 3,\quad |-7| = 7

Add the results:

3+7=103 + 7 = 10

So the correct choice is B. 1010

Problem 07
Absolute Value

Evaluate:

−∣7−3−2∣−∣−5∣-\bigl|7 - 3 - 2\bigr| - |-5|

Choices
A −7-7
B 77
C −3-3
D 33
Strategy

First simplify inside each absolute value, then take the absolute value, then apply the outside signs. Finally combine the results.

Solution

Simplify inside each absolute value, then apply the outside signs.

7−3−2=4−2=2,∣7−3−2∣=∣2∣=2,∣−5∣=57 - 3 - 2 = 4 - 2 = 2,\quad \bigl|7 - 3 - 2\bigr| = |2| = 2,\quad |-5| = 5

Now substitute into the original expression:

−∣7−3−2∣−∣−5∣=−2−5=−7-\bigl|7 - 3 - 2\bigr| - |-5| = -2 - 5 = -7

So the correct choice is A. −7-7

Problem 08
Absolute Value

Evaluate:

∣−5−8∣|-5 - 8|

Choices
A −13-13
B 33
C 1313
D −3-3
Strategy

First combine the integers inside the absolute value to get a single number, then take its absolute value.

Solution

Combine the integers inside the absolute value, then take the absolute value.

−5−8=−13,∣−13∣=13-5 - 8 = -13,\quad |-13| = 13

So the correct choice is C. 1313

Problem 09
Integer Word Problems

The temperature at sunrise is 8∘C8^\circ\text{C} By midday, it drops by 15∘C15^\circ\text{C} and by evening it rises by 4∘C4^\circ\text{C} What is the temperature in the evening?

Choices
A −3-3°C
B 33°C
C 1111°C
D −11-11°C
Strategy

Represent each change as adding an integer: a drop is adding a negative number, a rise is adding a positive number. Start from 88 and apply the changes:

8+(change1)+(change2).8 + (\text{change}_1) + (\text{change}_2).

Solution

Interpret each temperature change as adding an integer.

Start at 8∘C8^\circ\text{C} then subtract 1515 and add 44

8+(−15)+4=8−15+4=−7+4=−38 + (-15) + 4 = 8 - 15 + 4 = -7 + 4 = -3

So the evening temperature is −3∘C-3^\circ\text{C} The correct choice is A.

Problem 10
Integer Word Problems

A bank account starts with a balance of 120120 A withdrawal of 7575 is made. Then a deposit of 4040 is made. Finally, another withdrawal of 2525 is made.

What is the final balance?

Choices
A 2020
B 4040
C 6060
D 8080
Strategy

Write each transaction as an integer (withdrawal = negative, deposit = positive) and add them with the starting balance:

120+(change1)+(change2)+(change3).120 + (\text{change}_1) + (\text{change}_2) + (\text{change}_3).

Solution

Write each transaction as an integer and add them to the starting balance.

120−75+40−25120 - 75 + 40 - 25

Now simplify step by step:

120−75=45,45+40=85,85−25=60120 - 75 = 45,\quad 45 + 40 = 85,\quad 85 - 25 = 60

So the final balance is 6060 The correct choice is C. 6060