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Arithmetic and Interest

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Arithmetic and Interest. Ex. 8A Q.1 (i). Calculate to the nearest cent where necessary the compound interest on (i) €600 for 2 years at 5%. Year 1 P 1 = €600 R 1 = 5% A 1 = ? I 1 = ?. Year 2 P 2 = ? R 2 = 5% A 2 = ? I 2 = ?. €630. (600)(1.05) = €630. (630)(1.05) = €661.50. - PowerPoint PPT Presentation
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ARITHMETIC AND INTEREST
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Page 1: Arithmetic and Interest

ARITHMETIC AND INTEREST

Page 2: Arithmetic and Interest

CALCULATE TO THE NEAREST CENT WHERE NECESSARY THE COMPOUND INTEREST ON (I) €600 FOR 2 YEARS AT 5%

Ex. 8A Q.1 (i)

Year 1P1 = €600R1 = 5%A1 = ?I1 = ?

(600)(1.05) = €630630 – 600 = €30

Year 2P2 = ?R2 = 5%A2 = ?I2 = ?

(630)(1.05) = €661.50661.50 – 630 = €31.50

Compound Interest = 661.50 – 600 = €61.50

€630

Page 3: Arithmetic and Interest

CALCULATE TO THE NEAREST CENT WHERE NECESSARY THE COMPOUND INTEREST ON (II) €1800 FOR 2 YEARS AT 9%

Year 1P1 = €1800R1 = 9%A1 = ?(1800)(1.09) =

€1962

Year 2P2 = ?R2 = 9%A2 = ?(1962)(1.09) =

€2138.58

Compound Interest = 2138.58 - 1800 = €338.58

€1962

Page 4: Arithmetic and Interest

A3 = ?A2 = ?

CALCULATE TO THE NEAREST CENT WHERE NECESSARY THE COMPOUND INTEREST ON (III) €3500 FOR 3 YEARS AT 7½%

Ex. 8A Q.1 (iii)

P = €3500R = 7½ %A1 = ?(3500)(1.075) = €3762.50

Compound Interest = 4348.04 - 3500 = €848.04

(3762.50)(1.075) = €4044.69

(4044.69)(1.075) = €4348.04

Page 5: Arithmetic and Interest

A3 = ?A2 = ?

CALCULATE TO THE NEAREST CENT WHERE NECESSARY THE COMPOUND INTEREST ON (III) €7800 FOR 3 YEARS AT 3½%

Ex. 8A Q.1 (iv)

P = €7800R = 3½ %A1 = ?(7800)(1.035) = €8073

Compound Interest = 8648 – 7800 = €848

(8073)(1.035) = €8355.56

(8355.56)(1.035) = €8648.00

Page 6: Arithmetic and Interest

€4600 WAS INVESTED FOR 2 YEARS AT COMPOUND INTEREST. IF THE RATE FOR THE FIRST YEAR WAS 6% AND FOR THE SECOND YEAR WAS 9%, FIND THE TOTAL INTEREST FOR THE TWO YEARS.

Ex. 8A Q.2

Year 1P1 = €4600R1 = 6%A1 = ?(4600)(1.06) =

€4876

Year 2P2 = ?R2 = 9%A2 = ?(4876)(1.09) =

€5314.84

Compound Interest = 5314.84 – 4600 = €714.84

€4876

Page 7: Arithmetic and Interest

A COMPANY BORROWED €12000 FROM A BANK AT 11% PER ANNUM, COMPOUND INTEREST. THE COMPANY REPAID €5000 AT THE END OF THE FIRST YEAR. HOW MUCH WAS OWED TO THE BANK AT THE END OF THE SECOND YEAR.

Ex. 8A Q.3

Year 1P1 = €12000R1 = 11%A1 = ?(12000)(1.11) =

€13320

Year 2P2 = ?R2 = 11%A2 = ?(8320)(1.11) =

€9235.20

ANS: Amount owed = €9235.20

13320 – 5000 = €8320

Page 8: Arithmetic and Interest

€2500 WAS INVESTED IN A BUILDING SOCIETY. IF IT AMOUNTED TO €2612.50 AFTER ONE YEAR CALCULATE THE RATE OF INTEREST?

Ex. 8A Q.4

P = €2500R = ?A = €2612.50

A/P = 2612.5/2500 = 1.045

4½%Interest rate =

Page 9: Arithmetic and Interest

A SUM OF MONEY IS INVESTED AT 7% P.A. IF IT AMOUNTS TO €6848 AFTER ONE YEAR FIND THE SUM INVESTED?

Ex. 8A Q.5

P = ?R = 7%A = €6848

€6848 is 107%

ANS: P = €6400

6848/107 = €64 is 1%

(64)(100) = €6400 is 100%

A = 107% P = 100%

Page 10: Arithmetic and Interest

Tax payable = Gross Tax – Tax credits

INCOME TAX

Page 11: Arithmetic and Interest

INCOME TAX QUESTIONS CAN BE DONE BY FILLING IN THE FOLLOWING TABLE

Gross PayTax Band 1Tax band 2Gross TaxLess Tax CreditsTax PayableNet Pay

Page 12: Arithmetic and Interest

Gross PayTax Band 1:

Tax Band 2:

Gross Tax

Less Tax Credits

Tax Payable

Net Pay


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