N1.50
N7.50
N15.00
N18.00
Correct answer is A
\(I = N4.50, P = N150,T=2\frac{1}{2}\hspace{1mm}years\\I=\frac{P\times T\times R}{100}\\4.50=\frac{150 \times 2\frac{1}{2} \times R}{100}\\\frac{4.50}{1}=\frac{150 \times 5 \times R}{100\times 2}\\4.50\times 4 = 15R\\R=\frac{4.50\times5}{15}\\R = \frac{6}{5}\\Again\hspace{1mm}I\hspace{1mm}=\frac{P\times T \times R}{100}\\=\frac{250\times 1 \times 6}{100\times 2\times 5}\\=\frac{3}{2}=N1.50\)
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