Maths Level 4 Module 14

Mathematics NQF Level 4 *see terms and conditions Downloaded from hlayiso.com Mathematics NQF Level 4

Topic 5 Financial mathematics Downloaded from hlayiso.com Mathematics NQF Level 4

Use mathematics to plan and control financial instruments Module 14 Downloaded from hlayiso.com Mathematics NQF Level 4

Learning Activity 14.1 Downloaded from hlayiso.com Mathematics NQF Level 4

6) Simone buys furniture for R6 000. She pays a deposit of R1 000. She signs a hire purchase agreement at 15% simple interest and an insurance fee of R350. Calculate: 6.1 Her monthly instalment over 3 years. 6.2 The total amount she will pay. 6.3 The monthly interest. 6.4 The total amount of interest (including insurance). Click above to play this video. Downloaded from hlayiso.com Learning activity 14.1

Learning activity 14.3 Downloaded from hlayiso.com Mathematics NQF Level 4

Straight line & reducing balance depreciation – vid 26 Downloaded from hlayiso.com Learning activity 14.3

Summative assessment Module 14 Downloaded from hlayiso.com Mathematics NQF Level 4

1) Aaron receives a loan of R50 000. What will his loan amount to after five years if interest is calculated annually using the following methods: P = 50 000; n = 5; r = 11% Downloaded from hlayiso.com Summative assessment – Module 14

1a) 11% simple interest = P(1 − in) A  11(5)  50 000 1 +   100  = R77 500 1b) 11% compound interest = P(1 − i ) n A 5  11  = 50 000 1 +  100  = R84 252,91 Downloaded from hlayiso.com Summative assessment – Module 14

2) Peter buys a car at a cost of R130 000. After how many years will the value of the car be R50 000 if 15% depreciation is calculated using the following methods: P = 130 000; A = 50 000; r = 15% Downloaded from hlayiso.com Summative assessment – Module 14

2a) Straight-line depreciation = P(1 − in) A  15  50 000 130 000 1 − n  100  1 − 0,15n = 0,385 0,15n = 0, 615 n = 4,1 years Downloaded from hlayiso.com Summative assessment – Module 14

2b) Reducing-balance depreciation = P(1 − i ) n A n  15  = 50 000 130 000 1 −  100  (1 − 0,15) n = 0,385 0,85n = 0,385 n log 0,85 = log 0,385 log 0,385 n= log 0,85 n = 5,87 years Downloaded from hlayiso.com Summative assessment – Module 14

3) The average rate of inflation over the past five years was 6,3% per annum. If the current price of cooking oil is R26 per litre, calculate: P = R26; i = 6,3%; n = 5 3a) The expected price in five years’ time if the rate of inflation remains the same = P(1 + i ) n A 5  6,3  = 26 1 +   100  = R35,29Downloaded from hlayiso.com Summative assessment – Module 14

3b) How much 1 ℓ of cooking oil cost five years ago. = P(1 + i ) n A 5  6,3  = P 1 + 26   100  P = R19,16 Downloaded from hlayiso.com Summative assessment – Module 14

4) Zinzi invests R2 000 for five years. Calculate the final value of her investment if she invests it all at: 4a) 5% simple interest P = R2 000; i = 5%; n = 5 = P(1 + in) A  5  = 2 000 1 + (5)   100  = R2 500 Downloaded from hlayiso.com Summative assessment – Module 14

4b) 4% compound interest calculated annually P = R2 000; i = 4%; n = 5 n  r  A = P +  100  5  4  = 2 000 1 +  100  = R2 433,31 Downloaded from hlayiso.com Summative assessment – Module 14

5) How much money must John invest to have R10 000 after 10 years if he invests it at: 5a) 6% simple interest P = R10 000; i = 6%; n = 10  r  A P 1 + = ( n)   100   6  10 000 = P 1 + (10)   100  10 000 P= 1, 6 = R6 250 Downloaded from hlayiso.com Summative assessment – Module 14

5b) 5% compound interest calculated annually P = R10 000; i = 5%; n = 10 n  r  A P 1 + =  100  10  5  10 000 = P 1 +  100  10 000 P= 1, 6289 = R6 139,15 Downloaded from hlayiso.com Summative assessment – Module 14

6) The price of a house is R500 000. The expected rate of inflation over the next five years is 7%, calculated annually. Calculate the projected price of the house after five years. P = R500 000; i = 7%; n = 5 n  r A P 1 +  =  10  5  7  500 000 1 +  100  = R701 275,87 Downloaded from hlayiso.com Summative assessment – Module 14

7) Lindiwe buys a car for R150 000. She deposits R20 000 and signs a hire-purchase agreement involving equal monthly instalments over 5 years. The car insurance is R650 per month and the interest rate is 11,5%. Calculate her monthly payment. P = R130 000; i = 11,5%; n = 5 Downloaded from hlayiso.com Summative assessment – Module 14

 r  A P 1 + (n)  =  10   11,5  130 000 1 + (5)   100  = R204 750 R204 750 Monthly= payment + R650 60 = R4 062,50 Downloaded from hlayiso.com Summative assessment – Module 14

8) Mr Petersen’s gross salary was R14 500 per month and he got a bonus of R12 000. He received R30 interest on his savings account. He also paid R1 000 a month to a hospital plan on a medical scheme for himself, his wife and two children. Calculate the following: Downloaded from hlayiso.com Summative assessment – Module 14

8a) Gross income =R14500 (12) + R12000 + R30 =R186030 8b) Taxable income =R186030 - R30 =R186000 Downloaded from hlayiso.com Summative assessment – Module 14

8c) Amount of tax due Tax bracket: 0 – 189 880 Tax = 186000 × 18% = R33480 Tax rebate= 13 635 + (303 + 303 + 204 + 204) ×12 = 13 635 + 12 168 = R25 803 Tax payable = R33 480 − R25 803 = R7 677 Downloaded from hlayiso.com Summative assessment – Module 14

Mathematics NQF Level 2 Downloaded from hlayiso.com Mathematics NQF Level 4