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