JAMB Mathematics Past Questions & Answers - Page 384

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

3¹/₂ - (2¹/₃ * 1¹/₄) + ³/₅
= ⁷/₂ - (⁷/₃ * ⁵/₄) + ³/₅
= ⁷/₂ - ³⁵/₁₂ + ³/₅
= L.C.M = 60
= (210 - 175 + 36)/60
= ⁷¹/₆₀
= 1¹¹/₆₀

\(\frac{321_4}{23_4}\\=\frac{(3\times4^{2})+(2\times4^{1})+(1\times4^{0})}{(2\times4^{0})+(3\times4^{0})}\\=\frac{3\times16+2\times4+1\times1}{2\times4+3\times1}\\=\frac{48+8+1}{8+3}\\=\frac{57}{11}=5\hspace{1mm}remainder\hspace{1mm}2\\∴r=2_{10} \\ Now\hspace{1mm}convert\hspace{1mm}2_{10} \hspace{1mm}to\hspace{1mm}base\hspace{1mm}4\\\frac{4}{2} = 2\\\frac{4}{0}=0\hspace{1mm}or\hspace{1mm}2\\∴r=2\)

If 13_m + 24_m = 41_m
1 * m¹ + 3 ⁰ + 2 * m¹ + 4 * m⁰ = 4 * m¹ + 1 * m⁰
1 * m + 3 * 1 + 2 * m + 4 * 1 = 4 * m + 1 * 1
m + 3 + 2m + 4 = 4m + 1
3m + 7 = 4m + 1
4m -3m = 7 - 1
m = 6

5000 : 5000 * 500
5000 : 2500000
5 : 2500
1 : 500