MSI FINANCE MEMO - MATERIAL FOR GRADE 12

MATHEMATICS

MATERIAL FOR GRADE 12



FINANCE


MEMORANDA
COMPILED BY EXPERTS: K. NCUBE & T. MJIKWA


1|Page

,QUESTION 1

1.1 𝑖𝑚 𝑚
(1 + ) = 1 + 𝑖𝑒
𝑚
12
𝑖 12
(1 + ) = 1,083 Substitution
12
𝑖 12 12
Simplification
= ( √1.083 − 1)
12

𝑖 12 =0,08
𝑟 = 8% Answer (3)

1.2.1 𝐼𝑛𝑡 = 𝑅8696,97 × 60 = 𝑅121 818,20 Answer (1)
11 −𝑛 11
1.2.2 9000[1−(1+ ) ] 1200
1200
400000 = 11 
1200
Substitution into
11 −𝑛 16
(1 + ) = correct formula
1200 27
16
𝑙𝑜𝑔 Use of logs
𝑛=− 27
11 
𝑙𝑜𝑔(1+ )
1200

= 57,34284187 =≈ 57 full payments Answer (4)
1.2.3 9000[1−(1+
11
)
−57
] n=57, 58
11 −58 1200
y (1 + 1200) =400000 − 11 
1200 Substitution
= 3094,83261 ≈ 3094,83 
Answer (4)
1.2.4 𝐼𝑛𝑡 = 121 818,20 − {[(9000 × 57) + 3094,83] Simplification
−400 000}
= 5723,37 is saved Answer (2)
[14]




2|Page

, QUESTION 2

2.1 𝐴 = 𝑃( 1 + 𝑖)𝑛
𝐴 ✓ making P subject
𝑃=
(1 + 𝑖)𝑛 of formula
8450
= 0,12 120 ✓
correct
(1+ 12 ) substitution
= 𝑅2560, 31 answer (3)
✓
2.2 Let the original amount be Q
𝐴 = 𝑃(1− 𝑖)𝑛
1 ✓ subst.
𝑄 = 𝑄( 1 − 0, 047)𝑛
2
1
2
= (1 − 0, 047)𝑛 logs
✓
1
log = 𝑛𝑙𝑜𝑔(1 − 0,047)
2 answer (3)
𝑛 = 14, 40 𝑦𝑒𝑎𝑟𝑠/𝑗𝑎𝑟𝑒 ✓


2.3 2.3.1 The scrap value of the tractor is ✓ scrap value (1)
R𝑥(1 − 0, 2)5

2.3.2 so Price of new tractor will be ✓ Price of new
R 𝑥(1 + 0, 18 )5 after 5 years tractor (1)


2.3.3 𝑥[(1 + 𝑖)𝑛 − 𝑖] ✓ 0,10
𝐹= 𝑖=
𝑖 12
✓
𝑛 = 60
0 ,10
8 000 [(1 + 12 ) 60 − 1]
𝐹= ✓ subst. into an
0,10
12 annuity formula
✓
= 𝑅619 496, 58 answer
NB: 𝑖 𝑎𝑛𝑑 𝑛 only get marks when used in an
(4)
annuity formula




3|Page