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March 26, 2023
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PROBLEM 6.3
Time diagram (
Bid A):
i = 9%
W69,000
PV
–
OA = R =
? 3,000
3,000 3,000
3,000
69,000 3,000
3,000 3,000
3,000
0
n = 9
Present value
of initial cost
12,000 X W5.75
= W69,000 (i
ncurred today)
……………
W
69,000
Present value
of maintena
nce cost (years 1
–
4)
12,000 X W.25 =
W3,000
R (PVF
–
OA
4, 9%
) = W3,000
(3.23972)
………………………
Present value
of resurfacing
Present value
of maintena
nce cost (years 6
–
9)
R (PVF
–
OA
9
–
5, 9%
) = W3,000 (5
.99525
–
3.88965)
…….
PROBLEM 6.3 (C
ontinued)
Time diagram
(Bid B):
i = 9%
W126,000
PV
–
OA = R =
? 1,080
1,080 1,080
1,080
1,080 1,080
1,080 1
,080 1,080
0
Present value
of initial cost
12,000 X W10.50
= W126,000 (inc
urred today)
…….
W126,000.00
Present value
of maintena
nce cost
12,000 X W.09 =
W1,080
R (PV
–
OA
9, 9%
) = W1,080 (5.99525
)
……………………
PROBLEM 6.4
Lump sum al
ternative: Prese
nt Value = $500,000
X (1
–
.46) = $270,000
.
Annuity alte
rnative: Payme
nts = $36,000 X (1
–
.25) =
$27,000.
PROBLEM 6.5
(a)
The present va
lue of
€
55,000 cash
paid today is
€
55,0
00.
(b)
Time diagram
:
i = 2
1
/
2
% per qua
rter
PV
–
OA =
€
62,357
(c)
Time diagram
:
i = 2
1
/
2
% per qua
rter
€
18,000
PV
–
AD =
PROBLEM 6.5
(Continued)
Formula:
PV
–
AD = R (PV
F
–
AD
n, i
)
(d)
Time diagram
:
i = 2
1
/
2
% per qua
rter
PV
–
OA =
R =
?
€
1,500
€
1,500
€
1,500
€
1,500
PV
–
OA =
R =
?
€
4,000
€
4,000
€
4
,000
0
1
11
12 13
14
36 37
PROBLEM 6.5 (C
ontinue
d)
Present values:
(a)
€
55,000.
Time diagram
:
i = 12%
PV
–
OA = ? R =
(
€
39,000)
(
€
39,000)
€
1
8,000
€
18,0
00
€
68,000
€
68,00
0
€
68,000
€
68,000
€
38,000
€
38,000
€
38,000
€
12,000)
€
12,000)
€
12,000)
Formulas:
PV
–
OA = R (PVF
–
OA
n, i
)
PV
–
OA = R (PVF
–
OA
n, i
)
PV
–
OA = R (PVF
–
OA
n, i
)
PV
–
OA =R (PVF
–
OA
n, i
)
PV
–
OA
= (
€
39
,
0
0
0
)
(
PV
F
–
OA
5,
1
2
%
)
PV
–
OA =
€
18,000 (PVF
–
OA
10
–
5,
12%
)
PV
–
OA =
€
68,000 (PVF
–
OA
30
–
10, 12%
)
PV
–
OA =
€
38,000 (PVF
–
OA
40
–
30, 12%
)
PV
–
OA = (
€
39,000)(3.60478)
PV
–
OA
=
€
1
8,
0
00
(5
.
65
0
22
–
3.
6
04
7
8)
PV
–
OA =
€
68,000 (8.05518
–
5.65022)
PV
–
OA =
€
38,000 (8.24378
–
8.05518)
PV
–
OA = (
€
140,586)
PV
–
OA
=
€
18,000 (2.04544)
PV
–
OA =
€
68,000 (2.40496)
PV
–
OA =
€
38,000 (.18860)
PV
–
OA =
€
36,818
PV
–
OA =
€
163,537
PV
–
OA =
€
7,167
Present value of future ne
t cash inflows:
€
(140,586)
36,818
Copyright © 2018 Wiley
Kieso,
IFRS,
3/
e, Solutions Manual
(For Instructor Use Only)
6-
47
PROBLEM 6.7
(a)
Time diagram
(alternative o
ne):
i = ?
PV
–
OA =
$600,000
R =
$80,000
$80,000
$80,000 $80,000
$80,000
7.50 is
the present value of
an annuity
factor
of $1
for 12
years
discounte
d at an interest rate of
approximately 8%.
Time diagram
(alternative two):
i = ?
PV = $600
,000
FV = $1,900,000
PROBLEM 6.7
(Continued)
Future value
approach
Present value
approach
FV = PV (FV
F
n, i
)
PV = FV (PVF
n, i
)
$1,900,000 = $60
0,000 (FVF
12, i
)
$600,000 = $1,90
0,000 (PVF
12, i
)
(b)
Time diagram
:
i = ?
($824,150
–
$200,000)
PV
–
OA = R =
$624,150 $76,952
$76,952 $76,952
$76,952
PROBLEM 6.7 (C
ontinued)
Formulas:
PV
–
OA = R (PV
F
–
OA
n, i
)
(c)
Time diagram
:
i = 5% per
six months
(10% ÷
5)
PV = ?
PV
–
OA = R =
?
$32,000 $3
2,000
$32,000
$32,000 $32
,000 ($800,000 X
8% X 6/12)
PROBLEM 6.7
(Continued)
(d)
Time diagram
(future value
of $200,000 dep
osit)
i = 2
1/2
% per quarte
r
(10% ÷
4)
PV =
$200,000
FV = ?
Amount to wh
ich quarterly de
posits m
ust grow:
$1,300,000
–
$537,
012 = $762,988.
Time diagram
(future value
of quarterly
deposits)
PROBLEM 6.7 (C
ontinued)
Formulas:
FV
–
OA =
R (FVF
–
OA
n, i
)
PROBLEM 6.8
Vendor A:
£18,000
Annual Payme
nt
X 6.14457
(PV of ordina
ry annuity 10%
, 10 periods)
£110,602
+ 55,000
down payment
+ 10,000
maintenance c
ontract
£175,602
total cost f
rom Vendor A
Vendor B:
£9,500
semiannual
payment
X 18.01704
(PV of annuity
due *5%,
**
40 periods)
£171,162
*
(10 ÷
2) **(20 period
s x 2)
Vendor C:
£1,000
X 3.79079
(PV of ordina
ry annuity of 5
periods, 10%)
£ 3,791
PV of first 5 ye
ars of maintena
nce
£2,000
[PV of ordina
ry annuity 15
per., 10% (7.60608
)
–
X 3.81529
PV of ordi
nary annuity 5 pe
r., 10% (3.79079)
]
£3,000
[(PV of ordina
ry annuity
20 per., 10% (8
.51356)
–
X .90748
PV of ordinary
annuity 15 per
., 10% (7.60608)]
£ 2,722
PV of last 5
years of maintena
nce
Total cost
of press a
nd maintenance Ve
ndor C:
£150,000.00
cash purchase
price
3,791
maintenance y
ears 1
–
5
7,631
maintenance y
ears 6
–
15
2,722
£164,144
PROBLEM 6.9
(a)
Time diagram
for the first te
n payments:
i = 10%
PV
–
AD = ?
R =
$800,000 $800,00
0 $800,000
$800,000 $8
00,000 $800,000 $
800,000
Formula for
the first ten pay
ments:
PV
–
AD = R (PV
F
–
AD
n, i
)
Formula for
the last ten pay
ments:
PV
–
AD = R (PV
F
–
AD
n, i
)
Note: The p
resent value of
an
annuity due
is used h
ere, not
the
present val
ue of an ordinary annuity
, although it
may be used
.
PROBLEM 6.9
(Continued)
The total cos
t for leasing t
he facilities is:
$5,407,216 + $1
,042,360 = $6,449
,576.
Formulas for
the last ten
payments:
(i)
Present value
of the last ten
payments:
PV
–
A = R (PVF
–
AD
n, i
)
PROBLEM 6.9 (C
ontinued)
(ii)
Present value of the last ten payments at the beginni
ng of current
year:
PV = FV (PVF
n, i
)
(b)
Time diagram
:
i = 11%
PV
–
OA = ?
R =
$15,000
$15,000 $15,000
$15,000 $15,000
$15,000 $15,000
PROBLEM 6.9
(Continued)
Formula:
PV
–
OA = R (PV
F
–
OA
n, i
)
(c)
Time diagram
:
Amount paid
=
$792,000
I
f
th
e
c
o
m
p
a
n
y
de
ci
de
s no
t
t
o
t
a
k
e
th
e
c
a
s
h
di
sc
ou
n
t
,
t
h
e
n
th
e
c
o
m
p
a
n
y
can
use
the
$792
,000
fo
r
an
ad
ditional
20
days.
The
implied
interest
rate for post
poning the
payment can be
calculated as
follows:
(i)
Im
plied
interest
for
the
period
from
the
end
o
f
discount
period
to
the due date:
PROBLEM 6.9 (C
ontinued)
(ii)
Convert the
implied interest
rate to annua
l basis:
Daily interest
= 0.010101/20 =
0.000505