JAMB Mathematics Past Questions & Answers - Page 319

Volume of cube = L³

3³ = 27cm³

volume of rectangular tank = L x B X h

= 3 x 4 x 5

= 60cm³

volume of H₂O the tank can now hold

= volume of rectangular tank - volume of cube

= 60 - 27

= 33cm³

From Pythagoras theorem

|OA|2 = |AN|2 + |ON|2

72 = |AN|2 + (5)2

49 = |AN|2 + 25

|AN|2 = 49 - 25 = 24

|AN| = \(\sqrt {24}\)

= \(\sqrt {4 \times 6}\)

= 2√6 cm

|AN| = |NB| (A line drawn from the centre of a circle to a chord, divides the chord into two equal parts)

|AN| + |NB| = |AB|

2√6 + 2√6 = |AB|

|AB| = 4√6 cm

Interior angle = (n - 2)180

but, n = 12

= (12 -2)180

= 10 x 180

= 1800

let each interior angle = x

x = \(\frac{(n - 2)180}{n}\)

x = \(\frac{1800}{12}\)

= 150^o

N = [2 3]

N^-1 = \(\frac{adj N}{|N|}\)

adj N = \(\begin{vmatrix} 4 & -3 \\ -1 & 2 \end{vmatrix}\)

|N| = (2 x4) - (1 x 3)

= 8 - 3

=5

N^-1 = \(\frac {1}{5}\) \(\begin{vmatrix} 4 & -3 \\ -1 & 2 \end{vmatrix}\)

\(\begin{vmatrix} 4 & 2 & -1 \\ 2 & 3 & -1 \\ -1 & 1 & 3 \end{vmatrix}\)

4 \(\begin{vmatrix} 3 & -1 \\ 1 & 3 \end{vmatrix}\) -2 \(\begin{vmatrix} 2 & -1 \\ -1 & 3\end{vmatrix}\) -1 \(\begin{vmatrix} 2 & 3 \\ -1 & 1 \end{vmatrix}\)

4[(3 x 3) - (-1 x 1)] -2 [(2x 3) - (-1 x -1)] -1 [(2 x 1) - (-1 x 3)]

= 4[9 + 1] -2 [6 - 1] -1 [2 + 3]

= 4(10) - 2(5) - 1(5)

= 40 - 10 - 5

= 25