Exam (elaborations)
Quantitative Modelling I(DSC1520 Exam pack 2024)
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Quantitative Modelling I(DSC1520 Exam pack 2024). Questions and accurate answers.
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DSC1520 EXAM PACK2023
QUESTIONS AND
ANSWERS
For assistance contact
Email:gabrielmusyoka940@gmail.com
, Page 1 of 167
Study Unit 1 : Mathematical preliminaries
Chapter 1 : Sections 1.1 – 1.6
1. Basics
• Numbers: different type of numbers – Natural, Real, etc.
Also called constants
• Basic operations
o + (add); 2 + 3 = 5
o – (subtract); 3 – 2 = 1
za
ls S
o x (multiply); also •; 3 x 2 = 3•2 = 6
ria L
o.
to RIA
.c
o ÷ (division) also / or fraction ( 1 = 1 divide by 2);
2
rtu O
6
6 ÷ 3 = 6/3 = =2
.g T
3
w U
Remember:
w T
1 x anything = anything
w .R.
1x8=8
G
0 x anything = 0
0x4=0
1 + anything = one more than anything
1 + 345 = 346
0 + anything = anything
0 + 34 = 34
anything ÷ 0 = not allowed
12 ÷ 0 = not allowed
0 ÷ anything = 0
0÷7=0
, Page 2 of 167
• Brackets ( ) : group operations together
(3 + 4) – 3
=7–3
=4
• Order of operation: BODMAS
Brackets; Of; Divide; Multiply; Add; Subtract
40 – 4 x (5 + 8) + 20
= 40 – 4 x (13) + 20
= 40 – 52 + 20
za
ls S
=8
ria L
o.
to RIA
.c
• Variables: used for unknown or generalisation of things: place
rtu O
holder: use alphabetic characters for example X or A or Y. Can
.g T
take on different values
w U
w T
3x + 2y +7g + x
w .R.
G
3x is known as a term with coefficient 3 and variable x
Remember : the last term x has a coefficient value of 1 in front of
it namely 1x
o Operations on variables or unknown:
+ and – : only if same variable, then + or – coefficients and
variable stays the same
3x + 4x + 3 = (3 + 4)x + 3 = 7x + 3
5x – x – 6 = (5 – 1)x – 6 = 4x –6
, Page 3 of 167
x and ÷ : only if same variable, then x and ÷ coefficient
and unknowns
3a x 4a = (3x4) (a x a) = 12a 2
2 x2 xx
2x ÷ x = (2÷1) = 2 = 2x
x x
• Laws of operations
o Commutative law : order
a+b=b+a 3+4=4+3=7
axb=bxa 3 × 4 = 4 × 3 = 12
za
≠ b–a ≠
ls S
a–b 4–3=1 3 – 4 = –1
ria L
o.
≠b ÷ a ≠
to RIA
a÷b 4÷2=2 2 ÷ 4 = 0,5
.c
rtu O
o Associative law: ( )
.g T
w U
(a + b) + c = a + (b + c)
w T
(3 + 4) + 2 = 7 + 2 = 9
w .R.
3 + (4 + 2) = 3 + 6 = 9
G
(a x b) x c = a x (b x c)
(3 × 4) ×2 = 12 × 2 = 24
3 × (4 × 2) = 3 × 8 = 24
(a – b) – c ≠ a – (b – c)
(3 – 4) – 2 = –1 – 2 = –3
3 – (4 – 2) = 3 – 2 = 1
(a ÷ b) ÷ c ≠ a ÷ (b ÷ c)
(12 ÷ 2) ÷ 2 = 6 ÷ 2 = 3
12 ÷ (2 ÷ 2) = 12 ÷ 1 = 12
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