Basic Numeracy
BNU1501
Department of Decision Sciences
Exam May/June 2020
, Question 1
The expression
35 ÷ 5 + 2 − ((3 × 4 − 2) − (18 − 3 × 4))
can be rewritten as
[1] 9 − 6 + 42.
[2] 9 − 10 + 6.
[3] 5 − 6 + 42.
[4] 5 − 10 + 6.
Answer:
Remember to use BODMAS:
35 ÷ 5 + 2 − ((3 × 4 − 2) − (18 − 3 × 4))
7 + 2 − ((12 − 2) − (18 − 12))
7 + 2 − ((10) − (6))
7 + 2 − (4) = 5
9 − 10 + 6 = 5
Question 2
Subtracting the sum of 2𝑥 + 𝑥 3 and 𝑥 + 7 from −8 − 5𝑥 3 + 4𝑥 gives
[1] −15 − 6𝑥 3 + 𝑥
[2] 3𝑥 + 𝑥 3 + 7
[3] −𝑥 + 6𝑥 3 + 15
[4] 7𝑥 − 4𝑥 3 + 15
Answer:
−8 − 5𝑥 3 + 4𝑥 − (2𝑥 + 𝑥 3 + 𝑥 + 7)
−8 − 5𝑥 3 + 4𝑥 − (𝑥 3 + 3𝑥 + 7)
−8 − 5𝑥 3 + 4𝑥 − 𝑥 3 − 3𝑥 − 7
BNU1501
Department of Decision Sciences
Exam May/June 2020
, Question 1
The expression
35 ÷ 5 + 2 − ((3 × 4 − 2) − (18 − 3 × 4))
can be rewritten as
[1] 9 − 6 + 42.
[2] 9 − 10 + 6.
[3] 5 − 6 + 42.
[4] 5 − 10 + 6.
Answer:
Remember to use BODMAS:
35 ÷ 5 + 2 − ((3 × 4 − 2) − (18 − 3 × 4))
7 + 2 − ((12 − 2) − (18 − 12))
7 + 2 − ((10) − (6))
7 + 2 − (4) = 5
9 − 10 + 6 = 5
Question 2
Subtracting the sum of 2𝑥 + 𝑥 3 and 𝑥 + 7 from −8 − 5𝑥 3 + 4𝑥 gives
[1] −15 − 6𝑥 3 + 𝑥
[2] 3𝑥 + 𝑥 3 + 7
[3] −𝑥 + 6𝑥 3 + 15
[4] 7𝑥 − 4𝑥 3 + 15
Answer:
−8 − 5𝑥 3 + 4𝑥 − (2𝑥 + 𝑥 3 + 𝑥 + 7)
−8 − 5𝑥 3 + 4𝑥 − (𝑥 3 + 3𝑥 + 7)
−8 − 5𝑥 3 + 4𝑥 − 𝑥 3 − 3𝑥 − 7
