2014 - WAEC Mathematics Past Questions and Answers - page 2
11
Make u the subject of formula, E = \(\frac{m}{2g}\)(v2 - u2)
A
u = \(\sqrt{v^2 - \frac{2Eg}{m}}\)
B
u = \(\sqrt{\frac{v^2}{m} - \frac{2Eg}{4}}\)
C
u = \(\sqrt{v- \frac{2Eg}{m}}\)
D
u = \(\sqrt{\frac{2v^2Eg}{m}}\)
correct option: a
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 = \(\sqrt{V^2 - \frac{2Eg}{m}}\)
Users' Answers & Commentsmultiply 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 = \(\sqrt{V^2 - \frac{2Eg}{m}}\)
12
F x varies inversely as y and y varies directly as Z, what is the relationship between x and z?
A
x \(\alpha\) z
B
x \(\alpha \frac{1}{z}\)
C
x \(\alpha z^2\)
D
x \(\alpha \frac{1}{z^2}\0
correct option: b
x \(\alpha \frac{1}{y}\)
y \(\alpha z\)
the relationship = x \(\alph \frac{1}{z}\)
Users' Answers & Commentsy \(\alpha z\)
the relationship = x \(\alph \frac{1}{z}\)
13
Find the gradient of the line joining the points (2, -3) and 2, 5)
A
9
B
1
C
2
D
undefined
correct option: d
(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
Users' Answers & Commentsgradient = \(\frac{y_ = y_1}{x_2 = x_1} = \frac{5 - (-3)}{2 - 2}\)
= \(\frac{5 + 3}{0}\)
= \(\frac{8}{0}\)
= undefined
14
If (x - a) is a factor pf bx - ax + x2, find the other factor.
A
(x + b)
B
(x - b)
C
(a + b)
D
(a - b)
correct option: a
Users' Answers & Comments15
\(\begin{array}{c|c}height & 2 & 3 & 4 & 5 & 6 \ \hline frequency & 2 & 4 & 5 & 3 & 1
\end{array}\)
The table shows the distribution of the height of plants in a nursery. Calculate the mean height of the plants.
\end{array}\)
The table shows the distribution of the height of plants in a nursery. Calculate the mean height of the plants.
A
3.8
B
3.0
C
2.8
D
2.3
correct option: a
\(\begin{array}{c|c} height(x) & frequency(f) & fx \ \hline 2 & 2 & 4\ 3 & 4 & 12\ 4 & 5 & 20 \ 5 & 3 & 15\ 6 & 1 & 6\end{array}\)
mean\(\bar{x} = \frac{\sum fx}{\sum f}\)
= \(\frac{57}{15}\)
= 3.8
Users' Answers & Commentsmean\(\bar{x} = \frac{\sum fx}{\sum f}\)
= \(\frac{57}{15}\)
= 3.8
16
The area of a sector a circle with diameter 12cm is 66cm2. If the sector is folded to form a cone, calculate the radius of the base of the cone [Take \(\pi = \frac{22}{7}\)]
A
3.0cm
B
3.5cm
C
7.0cm
D
7.5cm
correct option: b
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
Users' Answers & Commentsradius = \(\frac{12}{2}\)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
17
A chord, 7cm long, is drawn in a circle with radius 3.7cm. Calculate the distance of the chord from the centre of the circle
A
0.7cm
B
1.2cm
C
2.0cm
D
2.5cm
correct option: b
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
Users' Answers & Commentsradius 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
18
Which of the following is a measure of dispersion?
A
range
B
percentile
C
median
D
quartile
correct option: a
Users' Answers & Comments19
A ship sails x km due east to a point E and continues x km due north to F. Find the bearing the bearing of f from the starting point.
A
045o
B
090o
C
135o
D
225o
correct option: a
Users' Answers & Comments20
A box contains 13 currency notes, all of which are either N50 or N20 notes. The total value of the currency notes is N530. How many N50 notes are in the box?
A
4
B
6
C
8
D
9
correct option: d
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
Users' Answers & Commentslet 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