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Messages - Ezra Newton

Pages: [1]
1
EarlyFace Examination Center (EEC) / 2020 FINANCIAL ACCOUNTING ANSWERS
« on: September 22, 2020, 07:56:53 am »
l Accounting Answers
OBJ
01-10: AAADBBCACB
11-20: CAABDCCBAD
21-30: ABCCBDBCBC
31-40: CBCDCBADAB
41-50: DCABABADDA



++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++


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.

++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

(4a)
Motor Car Account

TABULATE
Dr
Jan 1 1991 cash #6,400

Cr ____

(4b)
Provision for depreciation account

TABULATE
Dr
Dec. 31 1991 balance c/d 3200

Cr
Dec. 31 1991 profit and loss #3,200



Dr
Dec 31 1992 balance c/d #4,800
_____
#4,800
_____


Cr
Jan. 1 1992 balance b/d #3,200

Dec. 31 profit and loss #1,600
___
#4,800
___



Dr
Dec. 31 1993 balance c/d #5,600
__
#5,600
___

Cr
Jan. 1 19932 balance b/d #4,800

Dec. 31 profit and loss #800
__
#5,600
___


Dr
Dec. 31 1994 balance c/d #6,000
___
#6,000
___

Cr
Jan. 1 1994 balance b/d #5,600

Dec. 31 profit and loss #400
___
#6,000
___



Dr
Dec. 31 1995 balance c/d #6,200
___
#6,200
___

Cr
Jan. 1 1995 balance b/d #6,000

Dec. 31 profit and loss #200
___
#6,200
___


(4c)
Profit and loss account

TABULATE
Dr
1991 Depreciation #3,200

1992 Depreciation #1,600

1993 Depreciation #800

1994 Depreciation #400

1995 Depreciation #200


Cr
__
__
__
__
__


(4d)
Depreciation Schedule

Year 1 2 3 4 5


Book value at beginning 6, 400 3, 200 1,600 800 400


Depreciation
3,200 1,600 800 400 200


Accumulated Depreciation 3,200 4,800 5,600 6,000 6,200


Net book value 3,200 1,600 800 400 200

++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

(5)
(i) Realization account
TABULATE

Dr
Book value of asset

Fixed asset #24,780
Stock #13,552
Debtor #30,000
Cost of liquidation #750




____
#69,082
____


Cr
Cash realized
Debtor #28,000
Stock #13,000
Fixed asset #20,000
Discount from creditor #1067
Take over by:
Bayo #800

Share of loss
Bayo=3/9 × 6215 = 2071.7
Victor=4/9 × 6215 = 2762.2
Tunde= 2/9 × 6215 = 1381.1




____
#69,082
____


(ii)
Cash account

TABULATE
Dr
Balance b/f #2,780
Debtor #28,000
Stock #13,000
Fixed asset 20000



____
#63,780
____


Cr
Cost of liquidation #750
Creditor #13,500
Capital:
Bayo #10928.3
Victor #22,237.8
Tunde #14,693.9
Loan #1670

____
#63,780
____


(iii)
Capital account

TABULATE
Dr
Particular || Bayo || Victor || Tunde

Bayo
Motor car #800
Have of loss #2,071.7
Cash #10,928.3
____
#13,800
____


Victor
Motor car ----------
Have of loss #2,762.2
Cash #22,237.8
____
#25,000
____


Tunde
Motor car ---------
Have of loss #1,381.1
Cash #14,693.9
____
#16,075
____



Cr
Particular || Bayo || Victor || Tunde


Particular
Balance b/f


Bayo
#13,800

___
#13,800
____


Victor
#25,000

___
#25,000
___


Tunde
#16,075

___
#16,075
___

++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

2
EarlyFace Examination Center (EEC) / 2020 NABTEB PYSICS ANSWERS
« on: September 22, 2020, 07:52:29 am »
2020 NABTEB Physics Answers

PHYSICS OBJ
01-10: CBBBBCCDCA
11-20: ABAADBADDC
21-30: BCAABBBBBA
31-40: BCCBBABBCD
41-50: DBDBCBCBDC


========================

THEORY

INSTRUCTIONS: You are to Answer four (4) questions in all

(1a)
(i) Acceleration=vt/t1
(ii) Retardation=0 - vt / t3 - t2
= -vt/t3 - t2
(iii) Total Distance covered=1/2(t2 - t1)(t3)(vt)

(1b)
a=v - u/t ---------------------------(i)
S=(v + u/2)t ------------------------(ii)

from equ (i)
v= u + at ------------------------(iii)
Substitute equ (iii) in equ (ii)
S=(u + at + u/2)t
=(2u + at/2)t
=(2ut + at^2/2)
=2ut/2 + at^2/2
:. S=ut + 1/2 at^2

(1ci)
(view image above)
(i) x=utAB + 1/2 at^2AB
and
x=(40 + 80/2)tAB
=(120/2)tAB
x=60tAB
but a=40 - 80 / tAB
a= -40/tAB
:. x=80tAB - 1/2 x 40/tAB x (tAB)^2
x=80tAB - 20tAB
=60tAB

FOR BC
100=(40 + 0/2)tBC
100=20tBC
:. tBC=100/20
tBC=5m/s

100 + x = (80 + 0/2)(tAB + tBC)
100 + x =40(tAB + 5)
100 + x = 40(x/60 + 5/1)
100 + x = 40x/60 + 200/1
collect like terms
x - 40x/60 = 200 - 100
60x - 40x/60 = 100
Cros multiply
60x - 40x = 60 x 100
20x = 6000
x=6000/20
x=300m

(1cii)
tAB=x/60
tAB=300/60
tAB=5sec

(1ciii)
tAC=tAB + tBC
tAC=5 + 5
tAC=10sec

=========================

(2a)
Boiling point is the temperature at which the vapour pressure is equal to the atmospheric pressure.

Procedures to determine boiling point
(i) Take about 25ml of distilled water in a boiling tube and add 2 - 3 small pieces pumice stone.
(ii) Close the mouth of the boiling tube with a rubber cork that has two bores and clamp it with the stand.
(iii) Introduce a thermometer(temperature range -10 to 110C) in one bore of the cork of the boiling tube, keep the tulb if the thermometer about 3-5cm above the surface of the water.
(iv) Then introduce one end of a delivery tube in the second bore of the cork.
(v) Place a 250ml beaker below the second end of the delivery tube to collect a condensed water.
(vi) Heat the boiling tube gently, preferably by rotating the flame.
(vii) Note the temperature(t1) when the water starts boiling.
(viii) Continue to heat the water till the temperature becomes constant and the water remains boiling. Note the constant temperature (t2).
(ix) Record the observation.

(2c)
Mc=300g =0.3kg
tc=950C
tw=25C
t=100C
Mw=250g =0.25kg
Heat loss by copper = heat gained by water
McCc(Tc - t) = MwCw(t - tw) + ml
0.3 x 4 x 10^2(950 - 100) = 0.25 x 4.2 x 10^3 (100 -25) + 2.26 x 10^6m
102000=78750 + 2.26 x 10^6m
2.26 x 10^6m=102000 - 78750
2.26 x 10^6m=23250
Divide both sides by 2.26 x 10^6
m=23250/2.26 x 10^6
m=0.0103kg
m=10.3g

Effect of Heat
(i) change in state.
(ii) Increase in temperature.
(iii) Chemical action
(iv) Change in physical properties

==========================

(3ai)
Resonance means the production of forced vibration of maximum amplitude of a vibrating body.

(3aii)
Two identical turning forks and sounding boxes are placed next to one another. Striking one turning fork will cause the other to resonate at the same frequency.
when a weight is attached to one turning fork. They are no longer identical. Thus, one will not cause the other to resonate, when two different turning forks are struck at the same time, the interference of their particles produce beats.

(3aiii)
The sound is louder when a struck turning fork is held against a table because the surface of the table is set into vibration. Thus, the result in more air molecules vibrating which moves the sound louder.

(3bi)
(View image above)

(3bii)
(View image above)


==========================

(4a)
(View image above)
==========================
(4b)
(i) Eddy current:- Reduced by leminating the core
(ii) Hysteresis loss:- Reduced by using a low loss nickel iron alloys in the core.
(iii) I^2R or Heat loss:- Reduced by using thick wires.

(4ci)
Vp=4400v
Ps=600w
Vs=220v

k=Ip/Is
=Vs/Vp
=Ns/Np
:. Ns/Np=220/4400
=1:20

(4cii)
95%=Ps/Pp x 100%
95%=60/Ip x 4400 x 100%
Ip=60 x 100/95 x 4400
Ip=0.014A.

==========================

COMPLETED

3
Ask For Help and Get Answers / Re: Account
« on: September 05, 2020, 10:23:46 am »
*3a*

Fixed capital is conventionally defined as the stock of tangible, durable fixed assets owned or used by resident enterprises for more than one year. This includes plant, machinery, vehicles and equipment, installations and physical infrastructures, the value of land improvements, and buildings.

4
EarlyFace Examination Center (EEC) / Re: C R S
« on: September 02, 2020, 10:04:32 am »
Pls more

Pages: [1]


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