WAEC Mathematics Past Questions & Answers - Page 7

Given that radius = 3cm.

volume of sphere = $\frac{4}{3}\times\pi\times r^3$

= $\frac{4}{3}\times\pi\times 3^3$

= $\frac{4}{3}\times\pi\times 27$

= $4\times\pi\times9$

= $36\pi cm^3$

radius = 8cm , height = 14cm  and $\pi = \frac{22}{7}$

total surface area of a solid cylinder =$2πrh+2πr^2$ = 2πr( h + r )

 $2 \times \frac{22}{7} \times 8( 8 + 14)$

 $2 \times \frac{22}{7} \times 8 \times 22$

$\frac{7744}{7}$

= $1,106.29cm^2$

%error=5%, measured height = 5.98m

let the actual height = y 

error=x - 5.98 (since 'y' is more than 5.98)

%error = $\frac{error}{actual height}\times 100%$

5% =  $\frac{y - 5.98}{y}\times 100%$

$\frac{5}{100} = \frac{y - 5.98}{y}$

5y = 100(y - 5.98)

5y = 100y - 598

5y - 100y = - 598

-95y = - 598

y = $\frac{-598}{-95}$

y = 6.29m( to 2 d.p).

Find the roots of the equations: $3m^2 - 2m - 65 = 0$

= $m^2 - 15m + 13m - 65 = 0$

= 3m(m - 5) + 13( m - 5) = 0

( m - 5)(3m + 13) = 0 

m-5 = 0 or 3m + 13 = 0

therefore, m = 5 or $\frac{-13}{3}$

therefore the roots of the quadratic equation = ( 5, $\frac{-13}{3})$

Given: $log_a 3$ = m and $log_a 5$ = p
$log_a 75$ = $log_a (3 × 25)$
= $log_a (3 × 5^2)$
= $log_a 3 + log_a 5^2$
= $log_a 3 + 2log_a 5$
Since $log_a 3$ = m and $log_a 5$ = p
∴ $log_a 75$ = m + 2p