2) What amount needs to be paid back /or will we receive in each case : (a) R,500 was borrowed at \( 6 \frac{3}{8} \% \) p.a simple interest Over a period of 3 years (b) R 1200 was invested for 5 years and 3 months at a rate of \( 7 \frac{1}{4} \% \) p.a simple interest. (c) R2100 at 12\% p.a for 1 year and 2 months SI (d) R7500 at 22\% p.a SI for 219 days.

To find the amounts to be repaid or received, we'll use the simple interest formula: \[ \text{Simple Interest (SI)} = \text{Principal} \times \text{Rate} \times \text{Time} \] and \[ \text{Total Amount} = \text{Principal} + \text{SI} \] (a) For R500 borrowed at \( 6 \frac{3}{8} \% \) p.a over 3 years: Convert the rate to a decimal: \( 6.375\% = 0.06375 \). SI = \( 500 \times 0.06375 \times 3 = R95.625 \). Total Amount = \( 500 + 95.625 = R595.625 \). (b) For R1200 invested for 5 years and 3 months (5.25 years) at \( 7 \frac{1}{4} \% \): Convert the rate to a decimal: \( 7.25\% = 0.0725 \). SI = \( 1200 \times 0.0725 \times 5.25 = R914.25 \). Total Amount = \( 1200 + 914.25 = R2114.25 \). (c) For R2100 at 12% p.a for 1 year and 2 months (1.1667 years): SI = \( 2100 \times 0.12 \times 1.1667 = R292.08 \). Total Amount = \( 2100 + 292.08 = R2392.08 \). (d) For R7500 at 22% p.a for 219 days (approximately 0.6 years): SI = \( 7500 \times 0.22 \times 0.6 = R990 \). Total Amount = \( 7500 + 990 = R8490 \). So the amounts to be repaid or received are: (a) R595.63 (b) R2114.25 (c) R2392.08 (d) R8490