# COVER PAGE COMMUNICATION

```SCHOOL OF COMMECE
BACHELOR OF BUSINESS MANAGEMENT AND ENTREPRENUERSHIP (BBME)
OR
BACHELOR OF APPLIED ACCOUNTING AUDITING AND INFORMATION
SYSTEM
SUBJECT:
I.B.M 114
SEMESTER:
ONE
ASSGNMENT:
ONE
STUDENT NO:
KA1227
PRESENTED TO:
MUSSI
DUE DATE OF SUBMISSION:
9th September 2019
Question one:
(a) i. to calculate the value of first machine 1 January 2017.
Solution
Cost of first machine
=
k 350,000.00
Scrap value
=
k 25,000.00
Useful life
=
10 years
Depreciation charge =
cost of first machine – scrap value
Useful life
= k350,000.00- k25,000.00
10 years
= k 325,000.00
10
= k 32,500.00/year
Value of first machine
= depreciation charge * time used
Total depreciation
=
k 32,500.00 * 7
=
k 227,500.00
Value of machine on I January 2017 =
=
k 350 000.00 – 227,500.00
k122,500.00
Question one: ii.
To calculate profit or loss on the sale of first machine
Solution
amount sold for – value of machine
Profit or loss =
k 150, 000.00 – k 122,500.00
=
=
k27,500.00
There was a profit of 27,500.00 on the sale of the first machine
Question one: b.
To calculate the value, the new machine on 1 January 2021
Solution
cost of second machine (B)= k 400,000.00
deprecation rate (i)
=
20%
useful life
=
4 years
future value =
B(1-i)n
=
k 400,000.00(1- 0.2)4
=
k 400,000.00(0.8)4
=
k 400,000.00 (0.4096)
Value of second machine on 1 January 1017 was
=
k163,840.00
Question two: (b) to find time within which Sandra’s deposit will earn an interest of k 50,000.00
at 4.5% per annum simple interest.
Solution
Simple interest =
P*R*T
100
Where P (principal amount) =
k 220,000.00
R (interest rate)
=
4.5%
T (Period of investment)
=
?
Simple interest
=
k 50,000.00
K 50,000.00
=
220,000.00*4.5*T
100
K 50,000.00
=
990,000.00T
100
K 50,000.00
=
K 9900
9900T
9900
Sandra’s deposit will take 5 years to yield a simple interest of k 50, 000.00
Question two ii. Option 1
To calculate interest earned if the amount is deposited for four years at 4.5% per
annum compound interest directly credited to her account at the end of each year.
Solution
compound interest = p(1 + i)n
where p (principal amount) =
n (time of investment)
k 220,000.00
=
4 years
I (interest rate)
=
4.5%
Compound interest
=
k 220,000.00(1 + 4.5/100)4
=
k 220,000.00(1 + 0.045)4
=
k 220,000.00(1.045)4
=
k220,000.00(1.1925)
Compound interest will be k 262,354.09
To calculate option (b)
To calculate compound interest if the amount is deposited for four years at
4%per compound interest compounded every six months.
Solution
Compound interest
p (1 + i)n
=
Where p (principal amount) =
k 220,000.00
I (interest rate as proportion) =
4%/2
N (period of investment)
=
2*4=8
Compound interest
=
k 220,000.00(1 + 2%/100)8
=
k 220,000.00(1+0.02)8
=
k 220,000.00(1.02)8
=
k 220,000.00(0.1716)
=
k 257, 765.06
The best option will = compound interest under option 1 – option 2
As such the best option would be option one as it’s giving a higher interest .
Question four: (a)
To calculate length and width of Asha’s garden
Solution
Area = length * width
Area = 60ft2
Let width = x
Length =
60ft2=
x+4
x*(x+4)
X2 + 4x – 60 ft2 = 0
X2 + 10x – 6x – 60ft2 = 0
X (x + 10) – 6 (x – 10) =0
(X + 10) (x – 6) = 0
X = -10
X= 6
Width of the garden = 6ft
Length =
=
6ft + 4ft
10ft
Question four (b):
To find price an adults meal and a child’s meal.
Solution
Childs price
=
x
Sue and mike plus their two children
2y + 2x = k 55
Ahmed and his three children
Y + 3x = k48
In equation 1.
When x = 0
2y + 2(0) = k 55
2y = k 55
Y = 27.5
And in equation 1
When y = 0
2(0) + 2x = k55
2x = k55
X=27.5
Therefore, when x = 0, y is 27.5 and when y = 0, x = 27.5
In equation two
Y + 3x = k48
When y= 0
0 +3x = k48
3x=k48
X=16
And when x =0
Y + 3 (0) = k48
Y = k48
Therefore:
2y + 2x = k55
x
y
0
27.5
27.5
0
Y + 3x =k48
X
y
0
48
16
0
Question three:
Solution
Book value of machine =
Scrap value
=
k 1,000,000.00
150,000.00
Useful life
=
5 year
book value of machine – scrap value
Net cash outflows =
=
k 1000,000.00 – k 150,000.00
Yea
r
Cash in
flow
Cash out
flow
Net cash
flow
0
-
K
(850,000.00
)
1
K
200,000.0
0
K
350,000.0
0
K
(850,000.00
)
K
200,000.00
2
Discoun
t factor
(18%)
1.00
Present
value
Discoun
t factor
(24%)
1.00
Present
value
0.847
K
169,400.00
0.806
0.718
K
251,300.00
0.650
K
161,20
0
K
217,50
0
K
350,000.00
(850,000.00
)
3
K
350,00.00
K
350,000.00
0.609
K
213,150.00
0.524
4
K
350,000.0
0
K
350,000.0
0
K
350,000.00
0.516
K
180,600.00
0.423
K
350,000.00
0.437
K
152,950
0.341
5
(850,0
00.00)
K
183,40
0.00
K148,0
50.00
119.35
0
Total present value 1 =
=
Total present value 2 =
(k 850,000.00) + 967400
k 117 400
(k 850,000.00) + k 839, 500.00
(k 10, 500)
Internal rate of return =
NPVi1 – NPVi2
NPV1 – NPV2
=
k 117,400(18%) – (k 10,500*24)
117,400 – (10,500)
=
K 2, 113,200.00 – (k252,000.00)
K 117, 400 – (k10,500)
=
K 2, 365,700.00
K 127, 900.00
=
Discount factor=
20.07%
1
(1+i)n
Internal rate return falls between 18% and 24% as such the project is viable.
Question two:
to calculate the net amount paid by Sandra to nur and udoka in January.
Solution
Rate/hour =
10\$
Normal hours=
150 hours
Statutory deductions= 10%
Other deductions=
Nurs pay
5%
= normal hours × rate /hour+ ( over time × 1.5)
=
150 × 10\$ (12 × 1.5)
=
1500\$ + 180 \$
=
1680 \$
Stator deduction
Other deductions
= (1689\$ - 1400 \$) ×10%÷100
=
280 ×0.1
=
28 \$
=
280 \$- 28 \$
=
252\$
=
252\$ × 5%÷100
=
252 × 0.05
=
12.6\$
Total deduction =
Total nurs pay
Udoka pay
280\$ - ( 28\$ + 12.6 )
=
239.4
=
1400 \$ + 239.4
=
1639.4\$
=
135 hours × 10 \$/hour
=
1350
Total money payed to nur and udoka by Sandra
=
1350\$ + 1639.4\$
=
2989.4\$
```