Year : 
2014
Title : 
Mathematics (Core)
Exam : 
WASSCE/WAEC MAY/JUNE

Paper 1 | Objectives

11 - 20 of 49 Questions

# Question Ans
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 = \(\sqrt{\frac{2v^2Eg}{m}}\)

Detailed Solution

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
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

Detailed Solution

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

y \(\alpha z\)

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

Detailed Solution

(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
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)

A
15.

\(\begin{array}{c|c}height & 2 & 3 & 4 & 5 & 6 \\ \hline frequency & 2 & 4 & 5 & 3 & 1
\end{array}\)
The table shows the distribution of the height of plants in a nursery. Calculate the mean height of the plants.

A. 3.8

B. 3.0

C. 2.8

D. 2.3

Detailed Solution

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

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

= \(\frac{57}{15}\)

= 3.8
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

Detailed Solution

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
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

Detailed Solution

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
18.

Which of the following is a measure of dispersion?

A. range

B. percentile

C. median

D. quartile

A
19.

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

A
20.

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

Detailed Solution

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

subst
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 = \(\sqrt{\frac{2v^2Eg}{m}}\)

Detailed Solution

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
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

Detailed Solution

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

y \(\alpha z\)

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

Detailed Solution

(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
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)

A
15.

\(\begin{array}{c|c}height & 2 & 3 & 4 & 5 & 6 \\ \hline frequency & 2 & 4 & 5 & 3 & 1
\end{array}\)
The table shows the distribution of the height of plants in a nursery. Calculate the mean height of the plants.

A. 3.8

B. 3.0

C. 2.8

D. 2.3

Detailed Solution

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

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

= \(\frac{57}{15}\)

= 3.8
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

Detailed Solution

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
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

Detailed Solution

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
18.

Which of the following is a measure of dispersion?

A. range

B. percentile

C. median

D. quartile

A
19.

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

A
20.

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

Detailed Solution

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

subst