december 2010 fa2a

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STRICTLY CONFIDENTIAL

THE PUBLIC ACCOUNTANTS EXAMINATION

COUNCIL OF MALAWI

2010 EXAMINATIONS

CERTIFICATE IN FINANCIAL ACCOUNTING

PROGRAMME

PAPER FA 2: PRACTICAL MATHEMATICS & COMPUTING

(DECEMBER 2010)

TIME ALLOWED: 3 HOURS

SUGGESTED SOLUTIONS

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1

SECTION A

Answer ALL questions in this section

1. (a) (i) 10½ ÷ 2⅓

=

=

= 4½

(ii) (½)2 ×

¼ ×

¼ ×

¼

= ¼

×

½

=

OR

(½)2 ×

¼ ×

=

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(b)

1.6853 =

1.11 =

.11 =

11% = i

OR

1.6853 = 5log

2. (a)

3 log(62 + 82)

3 log(36 + 64)

3 × 2

= 6

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(b)( x + 1) = 5 – x

=

×

= 3

OR

( x + 1) = 5 – x

x=3

3. (a) Complete the following table:

base 2 base 8 base 10

1110111 167 119

100011 43 35

1100110 146 102

10111

+ 0110110111_11011 2

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(b) 16m

5m 5

24m 8

Area of the office = area of the carpet

Area of the rectangle5 × 16 = 80m2

Area of triangle = 1 × 8 × 5 = 20m

2

Total area = 80 + 20 = 100m

2

OR

Area of carpet = Area of trapezium

=

=(16 + 24) × 5

= 100m2

Cost of carpet

100m2

× 575.60= K57,560

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SECTION B

Answer THREE Questions ONLY from this section

4. (a)

Age f Midpoint

X

f x

) <18 0 9 0

18 – 20 50 19 950

20 – 22 65 21 1365

22 – 24 58 23 1334

24 – 26 27 25 675

Total 200 (1) 4324

Mean age 4324200= 21.62 years

= 22 years

(b)

A

<30m

B

30 – 60m

C

>60m

Total

M 524 455 221 1200

W 413 263 124 800

Total 937 718 345 2000

(i) P (commutes >hr or is a man)

= P(c) + P(m) – P(c and m)

=

=

= 0.662

(ii) P(W and A)

= 413_

2000

= 0.2065

(iii) P(B/W) = 263

800= 0.32875

= 0.33

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5. (a) Solve the following simultaneous equation:

5 x – 3 y = 26 x + 7 y = 13

30 x – 18 y = 1230 x + 35 y = 65

– 53 y = – 53

y = – 53

– 53 y = 1

5 x – 3 × 1 = 25 x = 5

x = 55

x = 1

(b) Electricity

Fixed charge 182 × 800 = 145,600Variable charge 15 × 15000 = 225,000

370,600

Heating oil

Fixed K8000 × 6 = 48,000

Variable charge K80 × 8000 = 640,000 688,000

Total K1,058,600

6. (a) Advantages of using a computer in an office are:

- More efficient and effective control procedures in production sales costs, budgets.

- Improved accounting procedures.- Improved cashflow controls.

-

Systems integration in that output of one part of business may becomeinput for another part and hence avoiding delays and duplication.

- Easy flow of information.

- Improved customer relations.- Improves employees’ morale.- Managers concentrate on strategic issues.

- Information is readily available and up to date.- Improves decision making.

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(b) (i) Operating Systems

Operating system is a suite of programs which take control over theoperation of the computer to the extent of being able to allow a

number of programs to run without human intervention. The more

advanced operating systems allow a number of application programsto run concurrently and in sequence without the intervention of theuser.The purpose of the operating system is to contro l the way the

software uses the hardware. The purpose of this control is to makethe computer operate in the way intended by the user and in a

systematic, reliable and efficient manner.

(ii) Functions of an operating system

Operating system has the following functions:

(1)

The scheduling and loading of programs: in order to providea continuous job processing sequence or to provideappropriate response to events.

(2) Control over hardware recourses: e.g. control over the

selection and operation of devices used for input, output orstorage.

(3) Handling errors when they occur and using correctiveroutines where possible.

(4)

Passing on control from one program to another under asystem of priority when more than one application program

occupies main storage.

(5) Protecting hardware, software and data from improper use.

(6) Communication with computer user or operator by means of

terminals or consoles through the use of monitor commandsand responses. The operator or the user may also be able to

communicate with the operating system by means ofcommand language.

(7) Furnishing a complete record of what has happened duringoperation. Some of the details of this log may be stored for

accounting purposes.

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. (a)

(b) (i) Mr Banda’s share = = K16,000

Total profit × = 56,000

Mr Akonda’s share = ×

= K8000

Mr Chayamba’s share = 8000 x 4

= K32,000

(ii) The difference between Mr Chayamba’s and Mr Akonda’s share

= K32,000 – K8,000

= K24,000

(iii) The ratio of Akonda’s : Chayamba’s

16000 : 32000

1:2

(c) (i) 4 hrs 15 minutes = 4.25 hrs

Speed 2614.25

= 61.41km/hr

(ii) Speed in miles/hr

= 61.41 x 0.62

= 38m/hr

E N D

december 2010 fa2a

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