2014 - WAEC Mathematics Past Questions and Answers - page 2

E = (\frac{m}{2g})(v² - u²)

multiply both sides by 2g

2Eg = 2g ((\frac{M}{2g} (V^2 - U^2))

2Eg = m(V² - U²)

2Eg - mV² - mU²

mU² = mV² - 2Eg

divide both sides by m

(\frac{mU^2}{m} = \frac{mV^2 - 2Eg}{m})

U² = (\frac{mV^2 - 2Eg}{m})

= (\frac{mV^2}{m} - \frac{2Eg}{m})

U² = V² - (\frac{2Eg}{m})

U = (\sqrt{V^2 - \frac{2Eg}{m}})

x (\alpha \frac{1}{y})

y (\alpha z)

the relationship = x (\alph \frac{1}{z})

(2, -3)(2, 5)

gradient = (\frac{y_ = y_1}{x_2 = x_1} = \frac{5 - (-3)}{2 - 2})

= (\frac{5 + 3}{0})

= (\frac{8}{0})

= undefined

(\begin{array}{c|c} height(x) & frequency(f) & fx \ \hline 2 & 2 & 4\ 3 & 4 & 12\ 4 & 5 & 20 \ 5 & 3 & 15\ 6 & 1 & 6\end{array})

mean(\bar{x} = \frac{\sum fx}{\sum f})

= (\frac{57}{15})

= 3.8

diameter = 12cm

radius = (\frac{12}{2})cm = 6cm

area of sector = 66cm²

area of sector = area of cone

66 = (\pi rl)

66 = (\frac{22}{7} \times r \times 6)

divide both sides by 6

11 = (\frac{22}{7}r)

22r = 11 x 7

r = (\frac{77}{22})

= 3.5cm

let the chord be AB = 7cm

radius OA = 3.7cm distance of the = OM using Pythagoras theorem

OA² = AM² + OM²

3.7² = 3.5² + OM²

13.69 = 12.25 + OM²

13.69 - 12.25 = OM²

1.44 = OM²

OM = (\sqrt{1.44})

OM = 1.2cm

distance of chord = 1.2cm

let the number of N50 notes = x

let the number of N20 notes = y

from the first statement

x + y = 13.....(1)

from the second statement

50x + 20y = 530 ......(2)

dividing through by 10, we have

5x + 2y = 53 ....(20

solving simultaneously by eliminating

x + y = 12....(1)

5x + 2y = 53.....(2)

multiply eqn (1) by 2

multiply eqn (2) by 1

2x + 2y = 26

-5x + 2y = 53

-3x = -27

x = 9

substitute x = 9 into equation (1)

x + y = 13

y = 13 - x

= 13 - 9 = 4

y = 4

the number of N50 notes = 9