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1. Grade 7 – Mathematics
Quarter I
MULTIPLICATION AND
DIVISION OF INTEGERS

2. • apply the rules in multiplying and
dividing integers; and
• solve problems involving
multiplication and division of
integers.

3. Perform the indicated operation.
6
20
12
56
151. 3 x 5 =
2. 2 x 3 =
3. 5 x 4 =
4. 6 x 2 =
5. 8 x 7 =
2
3
8
6
46. 8 ÷ 2 =
7. 10 ÷ 5 =
8. 18 ÷ 6 =
9. 24 ÷ 3 =
10. 30 ÷ 5 =

4. MULTIPLICATION AND DIVISION OF INTEGERS
SYMBOLS
MULTIPLICATION
(2)(4) parenthesis
dot operator/bullet
operator
𝟒
𝟐
fraction bar
÷
division
symbol
DIVISION
2 x 4 multiplication
symbol
2 · 4

5. MULTIPLICATION AND DIVISION OF INTEGERS
RULE:
SAME SIGNS = POSITIVE
DIFFERENT SIGNS = NEGATIVE
(+)(+) = (+)
(-)(-) = (+)
(+)(-) = (-)
(-)(+) = (-)
(+) ÷ (+) = (+)
(-) ÷ (-) = (+)
(+) ÷ (-) = (-)
(-) ÷ (+) = (-)

6. MULTIPLICATION OF INTEGERS
1. ( 2 )( 4 ) = 8
positive positive positive
2. ( -3 )( 5 ) = -15
negative positive negative
3. ( 8 )(-7 ) = -56
positive negative negative
4. ( -9 )(-5 ) = 45
negative negative positive
5. ( 3 )(-7 )(2) = -42
1
odd number
negative
6. (-3)(-4)(1)(-2)= -24
1
odd number
negative2 3
7. (-3)(5)(-1)(2) = 30
1
even number
positive2

7. DIVISION OF INTEGERS
1. 25 ÷ (-5) = -5
positive negative negative
2. -42 ÷ (-7) = 6
negative negative positive
3. 16 ÷ 2 = 8
positive positive positive
4. -20 ÷ 4 = -5
negative positive negative
5. -30 ÷ 5 = -6
negative
6. -90 ÷ (-15) = 6
positive
7. 120 ÷ (-10)= -12
negative
8. [(-3)(4)] ÷ (-6) = 2
positive
-12
negativenegative

8. Problem Solving.
For each late arrival in class, a student gets
2 demerits. Maria arrives late in class for the fifth
time since the school year started. How many
demerits has she received?
(-2) (5) = -10 demerits
Based on the given condition, if a student
gets 14 demerits at the end of the year, how
many times has she been late for school?
(-14) ÷ (-2) = 7 times

9. Think about this.
If a and b are integers, what
conditions would make a x b > 0 ?
If a and b are integers, what
conditions would make a x b < 0 ?
Answer: a and b must be integers with the
same signs (both positive or both negative)
Answer: a and b must be integers with different
signs (one positive and one negative).

10. Test Yourself.
2. (4)(-3)(-2) =
1. (20)(-4) =
4. [(3)(-6)] ÷ (-2) =
3. (-9)(-2)(2)(-3) =
5. [(9)(4)] ÷ [(3)(-2)] =

11. Answer Key
1. -80
2. 24
3. -108
4. 9
5. -6