1994 - JAMB Mathematics Past Questions and Answers - page 3

The inverse of 2 is 6 since 2 x 6 = 12; under mod 10 12 = 2 which is also the value required

\(\begin{pmatrix} 1 & 1 \ 3 & y \end{pmatrix}\)\(\begin{pmatrix} x \ 1 \end{pmatrix}\) = \(\begin{pmatrix} 4 \ 1\end{pmatrix}\) = x + 1 = 4 x = 4 - 1 = 3 3x + y =1 3(3) = y = 1 = 9 + y = 1 y = 1 - 9 = -8

4 x 2 = -8 upward arrows = +ve 2 x 5 x 3 = 30 0 x 6 = \(\frac{0}{22}\) downward arrows = -ve -1(1 x 5x - 1) = 5 - (2 x 6 x 2) = -24 = -(4 x 0 x 1) = \(\frac{0}{-19}\) therefore 22 - 19 = 3

The sum of angles of a quadrilateral is 360^o

∴ (3y - x - z)^o + 3x^o + (2z - 2y - x)^o + p^o = 360^o

Where P is the fourth angle

3y - x - z + 3x + 2z - 2y - x + p = 360^o

p = 360 - (x + y + z)

∴ p = (360 - x - y - z)^o

Internal dimension are 50cm, 36cm and 20cm

internal volume = 50 x 36 x 20cm³

1000 x 36cm³

= 36000cm³

External dimension are 54cm x 40cm x 22cm

= 2160cm² x 22cm = 47520cm³

Volume of wood = Ext. volume - Int. volume

= 47,520cm³ - 36,000cm³

= 11,520cm³

Perimeter = OP + OQ + PQ = 8 + 8 + PQ length PQ = \(\frac{\theta}{360 \times 2\pi r}\) = \(\frac{45}{360}\) x 2 x \(\pi\) x 8 = 2\(\pi\) Perimeter of sector 2r + L Where l = length of arc and r = radius ∴ P = 2(8) + 2\(\pi\) = 16 + 2\(\pi\)

Since XY is a fixed line and

XPY = 90^o P is on one side of XY

P₁P₂P₃......Pⁿ are all possible cases where

XPY = 90^o the only possible tendency is a semicircle because angles in semicircle equals 90^o

\(\frac{p + q}{2}\) = 4 p + q = 8 ....(i) \(\frac{p - 2}{2}\) = q p - 2q 2 ....(ii) q = 2, p = 6 3q = 6

RQP = 90^o

PQS = 30^o but

RQS = 90^o = \(\frac{RS}{10}\)

RS = 10 tan 60^o

RS = 10 tan 60^o

RS = 10√3m