Multiplying and Dividing Integers and Rational Numbers: Grade 7 Math Worksheet Answer Key – PDF Download

Problem 1: Repeated Addition

1. Four jumps of \(-3\): \(4 \times (-3) = -3 + (-3) + (-3) + (-3) = -12\).
2. \(5 \times (-2) = (-2) + (-2) + (-2) + (-2) + (-2) = -10\).

Answers: \(-12\), \(-10\).

Problem 2: Jumps of Negative Three

1. There are 5 jumps, and each jump is \(-3\). \(5 \times (-3) = (-3) + (-3) + (-3) + (-3) + (-3) = -15\).
2. \(3 \times (-5) = (-5) + (-5) + (-5) = -15\). Both products are \(-15\): the order of the factors does not change the product.

Answers: \(-15\); \(-15\).

Problem 3: Sign Rules for Products

1. Different signs: \(7 \cdot 6 = 42\), negative. \(-42\).
2. Same signs: \(5 \cdot 8 = 40\), positive. \(40\).
3. Different signs: \(-36\).
4. Same signs: \(36\).
5. Zero times any number is \(0\).

Answers: \(-42\), 40, \(-36\), 36, 0.

Problem 4: Count the Negative Factors

1. Three negative factors, so the product is negative. \(2 \times 3 \times 4 = 24\). \(-24\)
2. Four negative factors, so the product is positive. \(1\)
3. Two negative factors, so the product is positive. \(5 \times 2 \times 3 = 30\). \(30\)
4. Five negative factors, so the product is negative. \(2 \times 2 \times 2 \times 2 \times 2 = 32\). \(-32\)

Problem 5: A Pattern of Products

The products go up by \(4\) each time the factor goes down by \(1\).

1. \((-4)(-1) = 4\) and \((-4)(-2) = 8\).
2. Each product increases by \(4\). The pattern continues into positive values, so a negative times a negative is positive.

Problem 6: Find the Missing Number

1. \(56 : (-7) = -8\). Check: \(-7 \times (-8) = 56\). \(-8\)
2. \(9 \times (-4) = -36\). Check: \(-36 : (-4) = 9\). \(-36\)
3. \(-45 : 9 = -5\). Check: \(-45 : (-5) = 9\). \(-5\)
4. \(-54 : (-6) = 9\). Check: \(-6 \times 9 = -54\). \(9\)

Problem 7: Dividing Integers

1. Different signs: \(48 : 6 = 8\), so \(-8\).
2. Same signs: \(45 : 9 = 5\), so \(5\).
3. Different signs: \(-8\).
4. Zero divided by a nonzero number is \(0\).
5. Same signs: \(90 : 15 = 6\), so \(6\).

Answers: \(-8\), 5, \(-8\), 0, 6.

Problem 8: Tape Diagram for Division

1. \(-56 : 8 = -7\), so each part is \(-7\).
2. \(8 \times (-7) = -56\). It checks.
3. \(-56 : (-7) = 8\). It tells how many parts of size \(-7\) fit in \(-56\): 8 parts.

Problem 9: Division as Equal Parts

1. \(-24 : 6 = -4\), so each part is \(-4\).
2. \(6 \times (-4) = -24\). It checks.
3. \(-35 : 5 = -7\) because \(5 \times (-7) = -35\).

Answers: \(-4\), checks, \(-7\).

Problem 10: Division by Zero

1. \(0 : 8 = 0\) because \(8 \cdot 0 = 0\).
2. \(8 : 0 = r\) would mean \(0 \cdot r = 8\). Zero times any number is \(0\), so no number works. The quotient is undefined.
3. The rule holds only when the divisor is nonzero. Dividing by \(0\) is never allowed, so \(0 : 0\) is also undefined.

Problem 11: Decimal Products and Quotients

1. Different signs, so negative. \(0.6 \times 0.9 = 0.54\). \(-0.54\)
2. Same signs, so positive. \(1.5 \times 4 = 6\). \(6\)
3. Same signs, so positive. \(72 : 8 = 9\). \(9\)
4. Different signs, so negative. \(45 : 9 = 5\). \(-5\)

Problem 12: Signs in Fractions

1. \(-\frac{12}{4} = -3\), \(\frac{-12}{4} = -3\), \(\frac{12}{-4} = -3\). All three are equal.
2. \(-\frac{-10}{2} = -(-5) = 5\) and \(\frac{10}{2} = 5\).
3. \(\frac{20}{-5} = \frac{-20}{5} = -\frac{20}{5} = -4\).

Answers: \(-3\) (all three), 5 and 5, \(-4\).

Problem 13: Fraction Products and Quotients

1. Same signs, so positive. \(\frac{5}{6} \times \frac{3}{10} = \frac{15}{60} = \frac{1}{4}\). \(\frac{1}{4}\)
2. Different signs, so negative. \(\frac{2}{3} \times \frac{9}{4} = \frac{18}{12} = \frac{3}{2}\). \(-\frac{3}{2}\)
3. Same signs, so positive. \(\frac{3}{4} \times \frac{8}{3} = \frac{24}{12} = 2\). \(2\)

Problem 14: Powers and Order of Operations

1. \((-2)(-2)(-2) = 4 \times (-2) = -8\). \(-8\)
2. \((-3)^{2} = (-3)(-3) = 9\). \(-(3^{2}) = -(9) = -9\). In the first, the base is \(-3\); in the second, only 3 is squared and the negative sign is applied afterward. \(9\) and \(-9\)
3. Parentheses: \(4 – 9 = -5\). Multiply and divide left to right: \(-3(-5) = 15\) and \(-20 : 5 = -4\). Add: \(15 + (-4) = 11\). \(11\)

Problem 15: Falling Temperatures

1. \(5 \times (-3) = -15\). The total change is \(-15^{\circ}\text{F}\).
2. \(4 + (-15) = -11\). The temperature at 7 p.m. is \(-11^{\circ}\text{F}\).

Problem 16: Submarine Dive

1. The depth changes by \(-15\) feet each minute: \(7 \times (-15) = -105\) feet.
2. \(-40 + (-105) = -145\), so the depth is \(-145\) feet.
3. The change needed is \(-190 – (-40) = -150\) feet. \(-150 : (-15) = 10\), so it takes 10 minutes.

Problem 17: Weekly Fees

1. \(6 \times (-15) = -90\), a change of \(-\$90\).
2. \(120 + (-90) = 30\), so \(\$30\).
3. After \(8\) weeks: \(120 – 120 = 0\). After \(9\) weeks: \(120 – 135 = -15\). The balance first goes below \(\$0\) after 9 weeks.

Problem 18: Mean Temperature Change

(a) The sum is \((-6) + (-9) + (-12) + (-3) + (-10) = -40\). Divide by 5: \(-40 : 5 = -8\). The mean change is \(-8^{\circ}\text{F}\) per day.

(b) Five changes of \(-8\): \(5 \times (-8) = -40\). \(20 + (-40) = -20\), so the final temperature is \(-20^{\circ}\text{F}\).

(c) Five equal changes of the mean add up to the sum of the five actual changes, \(-40\), so both ways reach \(-20^{\circ}\text{F}\).

\(-8^{\circ}\text{F}\) per day; \(-20^{\circ}\text{F}\).

Problem 19: Stock Price Tracker

(a) \(5 \times (-4) + 5 \times 2 = -20 + 10 = -10\), so the total change is \(-\$10\).

(b) \(\$50 + (-\$10) = \$40\).

(c) \(-10 : 10 = -1\), so the price changes by \(-\$1\) per day on average.

Total change -$10; price $40; mean change -$1 per day.

Problem 20: Multi-Step Products and Quotients

1. \((-6)(-4) = 24\). Then \(24 : (-8) = -3\).
2. Way 1: \(-4(3) = -12\). Way 2: \(-4 \cdot 5 + (-4)(-2) = -20 + 8 = -12\). Both give \(-12\).
3. \(\frac{-3 \cdot 8}{4 \cdot 9} = \frac{-24}{36} = -\frac{2}{3}\).

Answers: \(-3\), \(-12\), \(-\frac{2}{3}\).

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