Year : 
1991
Title : 
Mathematics (Core)
Exam : 
WASSCE/WAEC MAY/JUNE

Paper 1 | Objectives

21 - 30 of 50 Questions

# Question Ans
21.

Find the coordinates of point B

A. (0,11/2)

B. (O,2)

C. (2,0)

D. (11/2, 2)

E. (11/2, 0)

Detailed Solution

B occurs at the point y = 0.
When y = 0, we have
4x + 3(0) = 6
4x = 6
x = 1\(\frac{1}{2}\)
B = \((1\frac{1}{2}, 0)\)
22.

The first term a of an A.P is equal to twice the
common difference d. Find, in terms of d, the 5th
term of the A.P.

A. 4d

B. 5d

C. 6d

D. a + 5d

E. 2a + 4d

Detailed Solution

Tn = a + (n - 1)d; AP; a = 2d; Tn = 2d + (n - 1)d

T5 = 2d + (5 - 1)d = 2d + 4d = 6d
23.

The common ratio of a G.P. is 2. If the 5th term is
greater than the 1st term by 45, find the 5th term,

A. 3

B. 6

C. 45

D. 48

E. 90

Detailed Solution

a = 1st term ; ar4 = 5th term
∴ ar4 = a(2)4 = 16a; 16a = 45 + a, a = 3;
3(2)4 = 48
24.

Two towns, P and Q, are on (4oN 40oW) and (4oN
20oE) respectively. What is the distance between
them, along their line of latitude? (Give your
answer in teems of π and R, the radius of the earth).

A. 2πR cos4o

B. πR sin40o

C.

πR/2
cos4o

D.

πR/3
cos4o

E.

πR/6
cos4o

Detailed Solution

Angular distance = 40o + 20o = 60o
using θ/360 x 2πr for distance between p and Q
= 60/360 x 2πRcos4o = 1/3πRcos4o
25.

The graph above is a sketch of

A. y = sin 2x

B. y=sinx

C. y=cos2x

D. y=2sinx

E. y = 2cos x

D
26.

Given that sin \(\theta\) = -0.9063, where O \(\leq\) \(\theta\) \(\leq\) 270°, find \(\theta\).

A. 65o

B. 115o

C. 145o

D. 245o

E. 265o

Detailed Solution

Sin\(\theta\) = -0.9063; \(\theta\) = sin-1(0.9063)

\(\theta\) = -64.99 = -65°

\(\theta\) = 180° - (-65);

180 + 65 = 245°
27.

From the top of a cliff, the angle of depression of
a boat on the sea is 60o, if the top of the cliff is
25m above the sea level, calculate the horizontal
distance from the bottom of the cliff to the boat.

A. 50√3m

B. 25√3m

C.

25√3m/3

D.

25m/3

E.

√3m/25

Detailed Solution

x/25 = tan 30o
x =
25/√3
=
25√3/3
28.

The angle of elevation of the top of a tree 39m
away from a point on the ground is 30o. Find the height of the tree

A. 39√3m

B. 13√3m

C. 13/\(\sqrt{3}\)m

D. 13/3\(\sqrt{3}\)m

E. \(\sqrt{3}\)/13m

Detailed Solution

x/25 = tan 30o
x = 39 x 1/√3 = 13√3m
29.

In the diagram above, ATR is a tangent at the point T to the circle center O, if ∠TOB = 145°, find ∠TAO

A. 20o

B. 35o

C. 45o

D. 55o

E. 65o

Detailed Solution

From the figure, < ATO = 90°
< AOT = 180° - 145° = 35°
\(\therefore\) < TAO = 180° - (90° + 35°)
= 55° (sum of angle in a triangle)
30.

P={2, 1,3, 9, 1/2}; Q = {1,21/2,3, 7} and R = {5, 4, 21/2}. Find P∪Q∪R

A. (21/2)

B. (1, 9)

C. (1, 2, 3, 4, 5, 6, 7)

D. (5, 4, 21/2)

E. (1/2, 1, 2, 21/2, 3, 4, 5, 7, 9)

Detailed Solution

P = {2,1,3,9,1/2}; Q = {1,21/2,3,7}
R = {5,4,212}
P∪Q∪R = {1/2,1,2,21/2,3,4,5,7,9}

= {1/2,1,2,21/2,3,4,5,7,9}
21.

Find the coordinates of point B

A. (0,11/2)

B. (O,2)

C. (2,0)

D. (11/2, 2)

E. (11/2, 0)

Detailed Solution

B occurs at the point y = 0.
When y = 0, we have
4x + 3(0) = 6
4x = 6
x = 1\(\frac{1}{2}\)
B = \((1\frac{1}{2}, 0)\)
22.

The first term a of an A.P is equal to twice the
common difference d. Find, in terms of d, the 5th
term of the A.P.

A. 4d

B. 5d

C. 6d

D. a + 5d

E. 2a + 4d

Detailed Solution

Tn = a + (n - 1)d; AP; a = 2d; Tn = 2d + (n - 1)d

T5 = 2d + (5 - 1)d = 2d + 4d = 6d
23.

The common ratio of a G.P. is 2. If the 5th term is
greater than the 1st term by 45, find the 5th term,

A. 3

B. 6

C. 45

D. 48

E. 90

Detailed Solution

a = 1st term ; ar4 = 5th term
∴ ar4 = a(2)4 = 16a; 16a = 45 + a, a = 3;
3(2)4 = 48
24.

Two towns, P and Q, are on (4oN 40oW) and (4oN
20oE) respectively. What is the distance between
them, along their line of latitude? (Give your
answer in teems of π and R, the radius of the earth).

A. 2πR cos4o

B. πR sin40o

C.

πR/2
cos4o

D.

πR/3
cos4o

E.

πR/6
cos4o

Detailed Solution

Angular distance = 40o + 20o = 60o
using θ/360 x 2πr for distance between p and Q
= 60/360 x 2πRcos4o = 1/3πRcos4o
25.

The graph above is a sketch of

A. y = sin 2x

B. y=sinx

C. y=cos2x

D. y=2sinx

E. y = 2cos x

D
26.

Given that sin \(\theta\) = -0.9063, where O \(\leq\) \(\theta\) \(\leq\) 270°, find \(\theta\).

A. 65o

B. 115o

C. 145o

D. 245o

E. 265o

Detailed Solution

Sin\(\theta\) = -0.9063; \(\theta\) = sin-1(0.9063)

\(\theta\) = -64.99 = -65°

\(\theta\) = 180° - (-65);

180 + 65 = 245°
27.

From the top of a cliff, the angle of depression of
a boat on the sea is 60o, if the top of the cliff is
25m above the sea level, calculate the horizontal
distance from the bottom of the cliff to the boat.

A. 50√3m

B. 25√3m

C.

25√3m/3

D.

25m/3

E.

√3m/25

Detailed Solution

x/25 = tan 30o
x =
25/√3
=
25√3/3
28.

The angle of elevation of the top of a tree 39m
away from a point on the ground is 30o. Find the height of the tree

A. 39√3m

B. 13√3m

C. 13/\(\sqrt{3}\)m

D. 13/3\(\sqrt{3}\)m

E. \(\sqrt{3}\)/13m

Detailed Solution

x/25 = tan 30o
x = 39 x 1/√3 = 13√3m
29.

In the diagram above, ATR is a tangent at the point T to the circle center O, if ∠TOB = 145°, find ∠TAO

A. 20o

B. 35o

C. 45o

D. 55o

E. 65o

Detailed Solution

From the figure, < ATO = 90°
< AOT = 180° - 145° = 35°
\(\therefore\) < TAO = 180° - (90° + 35°)
= 55° (sum of angle in a triangle)
30.

P={2, 1,3, 9, 1/2}; Q = {1,21/2,3, 7} and R = {5, 4, 21/2}. Find P∪Q∪R

A. (21/2)

B. (1, 9)

C. (1, 2, 3, 4, 5, 6, 7)

D. (5, 4, 21/2)

E. (1/2, 1, 2, 21/2, 3, 4, 5, 7, 9)

Detailed Solution

P = {2,1,3,9,1/2}; Q = {1,21/2,3,7}
R = {5,4,212}
P∪Q∪R = {1/2,1,2,21/2,3,4,5,7,9}

= {1/2,1,2,21/2,3,4,5,7,9}