Homework SourseAmber Corporation acquired 60 percent ownership of Sparta Co
Amber Corporation acquired 60 percent ownership of Sparta Company on January 1, 20X8, at underlying book value. At that date, the fair value of the noncontrolling interest was equal to 40 percent of the book value of Sparta Company. Accumulated depreciation on Buildings and Equipment was $74,000 on the acquisition date. Trial balance data at December 31, 20X8, for Amber and Sparta are as follows:
Sparta purchased stock of Row Company on January 1, 20X8, for $34,000 and classified the investment as available-for-sale securities. The value of Row’s securities increased to $44,000 at December 31, 20X8.
Record all consolidation entries needed to prepare a three-part consolidation worksheet as of December 31, 20X8. (If no entry is required for a transaction/event, select \"No journal entry required\" in the first account field.)
Prepare a three-part consolidation worksheet for 20X8 in good form. (Values in the first two columns (the \"parent\" and \"subsidiary\" balances) that are to be deducted should be indicated with a minus sign, while all values in the \"Consolidation Entries\" columns should be entered as positive values. For accounts where multiple adjusting entries are required, combine all debit entries into one amount and enter this amount in the debit column of the worksheet. Similarly, combine all credit entries into one amount and enter this amount in the credit column of the worksheet.)
Prepare a consolidated balance sheet, income statement, and statement of comprehensive income for 20X8. (Amounts to be deducted should be indicated with a minus sign.)
| Amber Corporation acquired 60 percent ownership of Sparta Company on January 1, 20X8, at underlying book value. At that date, the fair value of the noncontrolling interest was equal to 40 percent of the book value of Sparta Company. Accumulated depreciation on Buildings and Equipment was $74,000 on the acquisition date. Trial balance data at December 31, 20X8, for Amber and Sparta are as follows: |
Solution
Ans-
mo d 4), so
(2
a;
2
b
)
2
H
, and hence
(2
a;
2
b
) +
H
=
H
. Therefore, for any
(
a;b
) +
H
2
G=H
, since
2
((
a;b
) +
H
) =
H
,
j
(
a;b
) +
H
j
2
:
We cannot say that the order of
(
a;b
) +
H
is two since it is p ossible that
1
((
a;b
) +
H
) =
H
for certain choices of
(
a;b
)
, but we can conclude that the
order of every element of
G=H
is less than or equal to two, so
G=H
has no
elements of order four. Therefore,
G=H
Z
2
Z
2
:
4.3
G=K
Let
(
a;b
)+
K
b e an element of
G=K
. Let us lo ok at integer multiples of
(
a;b
)+
K
.
2
((
a;b
) +
K
) = (2
(
a;b
)) +
K
= (2
a;
2
b
) +
K:
Is
(2
a;
2
b
) +
K
=
K
, the identity element of
G=K
? In other words, is
(2
a;
2
b
)
2
K
? In this case, it is p ossible to cho ose
(
a;b
)
so that
(2
a;
2
b
)
=
2
K
. For this to
happ en, we need to pick
b
so that
2
b
= 2
, and we need to pick
a
so that
2
a
is
not equal to 1 or 3. Thus, we can try
(
a;b
) = (0
;
1)
. Then
(2
a;
2
b
) = (0
;
2)
=
2
K
,
so in
G=K
,
2
((0
;
1) +
K
) = (2
(0
;
1)) +
K
= (0
;
2) +
K
6
=
K:
Furthermore,
(0
;
1) +
K
itself is not equal to
K
since
(0
;
1)
=
2
K
. Hence
j
(0
;
1) +
K
j
>
2
:
What could the order of
(0
;
1) +
K
b e in
G=K
? Since
j
G=K
j
= 4
, we know the
order of
(0
;
1) +
K
must divide 4. Since
j
(0
;
1) +
K
j
>
2
,
j
(0
;
1) +
K
j
= 4
:
Therefore,
G=K
has an element of order 4, so it must b e isomorphic to
Z
4
