2024 WAEC ENGLISH: 2024 NEBTEB ACCOUNTING ANSWERS (5975)
2024/2025 WAEC GCE 2nd SERIES RUNS
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Welcome to official 2024 English WAEC answer page. We provide 2024 English WAEC Questions and Answers on Essay, Theory, OBJ midnight before the exam, this is verified & correct WAEC Eng Expo. WAEC English Questions and Answers 2024. WAEC Eng Expo for Theory & Objective (OBJ) PDF: verified & correct expo Solved Solutions, 2024 NEBTEB ACCOUNTING ANSWERS. 2024 WAEC EXAM English Questions and Answers
_2020 NABTEB financial Accounting Answers_
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TAKE NOTE: Wherever you see something like this under the credit side, just leave it blank. It means no value
*THEORY*
(1a) Software can be refered to as the programs used in computer system.
(1b)
Two types of software are
(i) Systems software.
(ii) Application software.
(1c)
Advantages of Application packages
(i) Reduction in time needed for implementation.
(ii) Reduction in systems and programming effort and cost.
(iii) Provision of expertise not normally available to the small users
(iv) Reduction in errors in design.
Disadvantages of applications Packages
(i) A package may be asked which is not completely suitable to the application.
(ii) No one on site can help when problem occur.
(iii) Inefficiency emanating from the inclusion of features not relevant to every application.
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(4 Motor Car Accoun
TABULAT
Dr
Jan 1 1991 cash #6,40
Cr ________
(4b
Provision for depreciation account
TABULAT
D
Dec. 31 1991 balance c/d 320
Dec. 31 1991 profit and loss #3,20
Dec 31 1992 balance c/d #4,80
__________
#4,80
__________
Jan. 1 1992 balance b/d #3,20
Dec. 31 profit and loss #1,6
________
#4,80
________
Dec. 31 1993 balance c/d #5,60
_____
#5,60
______
Jan. 1 19932 balance b/d #4,80
Dec. 31 profit and loss #8
_____
#5,60
______
Dec. 31 1994 balance c/d #6,000
______
#6,00
______
Jan. 1 1994 balance b/d #5,60
Dec. 31 profit and loss #4
______
#6,00
______
Dec. 31 1995 balance c/d #6,20
______
#6,20
______
Jan. 1 1995 balance b/d #6,00
Dec. 31 profit and loss #2
______
#6,20
______
(4
Profit and loss account
TABULAT
D
1991 Depreciation #3,20
1992 Depreciation #1,6
1993 Depreciation #8
1994 Depreciation #4
1995 Depreciation #2
(4
Depreciation Schedule
Year 1 2 3 4
Book value at beginning 6, 400 3, 200 1,600 800
Depreciati
3,200 1,600 800 400 20
Accumulated Depreciation 3,200 4,800 5,600 6,000 6,
Net book value 3,200 1,600 800 400 2
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(i) Realization account
TABULATE
Book value of asse
Fixed asset #24,7
Stock #13,55
Debtor #30,00
Cost of liquidation #75
#69,08
________
Cash realized
Debtor #28,00
Stock #13,00
Fixed asset #20,00
Discount from creditor #106
Take over by
Bayo #80
Share of lo
Bayo=3/9 × 6215 = 2071.
Victor=4/9 × 6215 = 2762.
Tunde= 2/9 × 6215 = 1381.
_____
#69,08
_________
(i
Cash accoun
TABULATE
D
Balance b/f #2,78
Debtor #28,00
Stock #13,00
Fixed asset 2000
______
#63,78
_________
Cost of liquidation #750
Creditor #13,50
Capital
Bayo #10928.
Victor #22,237.
Tunde #14,693.
Loan #167
_________
#63,78
_________
(ii
Capital account
TABULAT
D
Particular || Bayo || Victor || Tund
Bay
Motor car #80
Have of loss #2,071.
Cash #10,928.
_________
#13,80
_________
Vict
Motor car ---------
Have of loss #2,762.
Cash #22,237.
_________
#25,00
_________
Tu
Motor car --------
Have of loss #1,381.
Cash #14,693.
_________
#16,07
_________
Particular || Bayo || Victor || Tund
Particul
Balance b/
B
#13,80
_______
#13,80
_______
Vic
#25,00
_______
#25,00
________
Tu
#16,07
_______
#16,07
________
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Dr_0_000Cr_0_0
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Cr_0____0000r ti) _2_____127ss0:7000
Cr_2_____00280tDr 5)++00 2000on 400 5 d) ____
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Dr_0_000Cr_0_0
Dr_0_000Cr _0_0
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Cr_5_91-nde_0_82-or _0_370o erE i) _0_ 0983:0
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Dr_0_000Cr_0_0
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Cr_2_____00280tDr 5)++00 2000on 400 5 d) ____
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Cr_2_____00280tDr 5)++00 2000on 400 5 d) ____
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Dr_0_000Cr_0_
Dr_0_000Cr_0_0
Dr_0_000Cr _0_0
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Cr_2_____00280tDr 5)++00 2000on 400 5 d) ____
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Dr_0_000Cr_0_0
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Cr_2_____00280tDr 5)++00 2000on 400 5 d) ____
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Dr_0_000Cr_0_
Dr_0_000Cr_0_0
Dr_0_000Cr _0_0
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Cr_2_____00280tDr 5)++00 2000on 400 5 d) ____
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Dr_0_000Cr_0_0
Dr_0_000Cr _0_0
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Cr_0____0000r ti) _2_____127ss0:7000
Cr_2_____00280tDr 5)++00 2000on 400 5 d) ____
Cr000000000rE c) _0_000Cr_0_0
Dr_0_000Cr_0_
Dr_0_000Cr_0_0
Dr_0_000Cr _0_0
Dr0Cr0rE ) __0 E ta)++ :
__ B C D B
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