Year : 
1988
Title : 
Mathematics (Core)
Exam : 
WASSCE/WAEC MAY/JUNE

Paper 1 | Objectives

11 - 20 of 41 Questions

# Question Ans
11.

What is the difference in longitude between P (lat. 50°N. long. 50°W) and Q (lat.50°N, long. 150°W)?

A. 300o

B. 200o

C. 130o

D. 100o

E. 30o

Detailed Solution

Let \(\theta\) be the angular difference between P (50°N. 50°W), and Q (50°N, 150°W),

\(\theta\) = 150° - 50° =100°
12.

The number 186,470 was corrected to 186,000. Which of the following describe the degree of approximation made?
I. to the nearest hundred II. to the nearest thousand III. to 3 significant figures IV. to 4 significant figures

A. I & III only

B. I & IV only

C. II & III only

D. II & IV only

E. All of the above

C

13.

Points P and Q respectively 24m north and 7m east point R. Calculate |PQ| in meters

A. 20

B. 24

C. 25

D. 31

E. 84

Detailed Solution

PQ\(^2\) = 24\(^2\) + 7\(^2\)
PQ\(^2\) = 625
PQ = 25
14.

Points P and Q respectively 24m north and 7m east point R. What is the bearing of Q from P to the nearest whole degree?

A. 16o

B. 17o

C. 73o

D. 106o

E. 164o

Detailed Solution

Using pythagoras' theorem:
Hyp\(^2\) = Adj\(^2\) + Opp\(^2\) → 24\(^2\) + 7\(^2\)
Hyp = √625 → 25
\(\tan P = \frac{7}{24}\)
\(\tan P = 0.2917\)
\(P = \tan^{-1} (0.2917)\)
= 16.26°
Bearing of Q from P = 180° - 16.26°
= 163.74° \(\approxeq\) 164°
15.

If sinθ = 3/5 find tanθ for 0 < θ < 90o

A. 4\5

B. 3\4

C. 5\8

D. 1\2

E. 3\8

Detailed Solution

sinθ = 3/5
tanθ = 3/4
16.

Simplify (271/3)2

A. 41/2

B. 6

C. 9

D. 18

E. 81

Detailed Solution

(271/3)2
= (33 x 1/3)2 = 32 = 9
17.

A student told to draw the graph of y = x2 + 4x - 6. He is then told to draw a liner graph on the same axis such that the intersection of the two graphs will give the solutions to the equation x2 + 4x - 7 = 0, What is the equation of the linear graph he needs to draw?

A. x = 1

B. x = -1

C. y = 1

D. y = -1

E. x + y = 1

Detailed Solution

The linear equation will be the difference between the equation.

y=x2 + 4x - 6 and x2 + 4x - 7 = 0

y=x2 + 4x - 6, ( 0 = x2 + 4x - 7)

y = -(-1) y = 1
18.

Calculate the value of x in the diagram above

A. 18o

B. 30o

C. 40o

D. 60o

E. 15°

Detailed Solution

vertically opposite angles are equal, hence
3x + 6x + 3x = 180°
12x = 180°
x = 15°
19.

Find the quadratic equation whose roots are x = -2 or x = 7

A. x2 + 2x - 7 = 0

B. x2 - 2x + 7 = 0

C. x2 + 5 +14 = 0

D. x2 - 5x - 14 = 0

E. x2 + 5x - 14 = 0

Detailed Solution

x + 2 and x - 7 are factors
(x+2)(x+7) = 0

x(x-7) + 2(x-7) = 0

x2 - 7x + 2x - 14 = 0

x2 - 5x - 14 = 0
20.

A sales girl gave a change of N1.15 to a customer instead of N1.25. Calculate her percentage error

A. 10%

B. 7%

C. 8.0%

D. 2.4%

E. 10%

Detailed Solution

Error = 1.25 - 1.15 = 0.10

0.10/1.25
* 100 = 8.0%

11.

What is the difference in longitude between P (lat. 50°N. long. 50°W) and Q (lat.50°N, long. 150°W)?

A. 300o

B. 200o

C. 130o

D. 100o

E. 30o

Detailed Solution

Let \(\theta\) be the angular difference between P (50°N. 50°W), and Q (50°N, 150°W),

\(\theta\) = 150° - 50° =100°
12.

The number 186,470 was corrected to 186,000. Which of the following describe the degree of approximation made?
I. to the nearest hundred II. to the nearest thousand III. to 3 significant figures IV. to 4 significant figures

A. I & III only

B. I & IV only

C. II & III only

D. II & IV only

E. All of the above

C

13.

Points P and Q respectively 24m north and 7m east point R. Calculate |PQ| in meters

A. 20

B. 24

C. 25

D. 31

E. 84

Detailed Solution

PQ\(^2\) = 24\(^2\) + 7\(^2\)
PQ\(^2\) = 625
PQ = 25
14.

Points P and Q respectively 24m north and 7m east point R. What is the bearing of Q from P to the nearest whole degree?

A. 16o

B. 17o

C. 73o

D. 106o

E. 164o

Detailed Solution

Using pythagoras' theorem:
Hyp\(^2\) = Adj\(^2\) + Opp\(^2\) → 24\(^2\) + 7\(^2\)
Hyp = √625 → 25
\(\tan P = \frac{7}{24}\)
\(\tan P = 0.2917\)
\(P = \tan^{-1} (0.2917)\)
= 16.26°
Bearing of Q from P = 180° - 16.26°
= 163.74° \(\approxeq\) 164°
15.

If sinθ = 3/5 find tanθ for 0 < θ < 90o

A. 4\5

B. 3\4

C. 5\8

D. 1\2

E. 3\8

Detailed Solution

sinθ = 3/5
tanθ = 3/4
16.

Simplify (271/3)2

A. 41/2

B. 6

C. 9

D. 18

E. 81

Detailed Solution

(271/3)2
= (33 x 1/3)2 = 32 = 9
17.

A student told to draw the graph of y = x2 + 4x - 6. He is then told to draw a liner graph on the same axis such that the intersection of the two graphs will give the solutions to the equation x2 + 4x - 7 = 0, What is the equation of the linear graph he needs to draw?

A. x = 1

B. x = -1

C. y = 1

D. y = -1

E. x + y = 1

Detailed Solution

The linear equation will be the difference between the equation.

y=x2 + 4x - 6 and x2 + 4x - 7 = 0

y=x2 + 4x - 6, ( 0 = x2 + 4x - 7)

y = -(-1) y = 1
18.

Calculate the value of x in the diagram above

A. 18o

B. 30o

C. 40o

D. 60o

E. 15°

Detailed Solution

vertically opposite angles are equal, hence
3x + 6x + 3x = 180°
12x = 180°
x = 15°
19.

Find the quadratic equation whose roots are x = -2 or x = 7

A. x2 + 2x - 7 = 0

B. x2 - 2x + 7 = 0

C. x2 + 5 +14 = 0

D. x2 - 5x - 14 = 0

E. x2 + 5x - 14 = 0

Detailed Solution

x + 2 and x - 7 are factors
(x+2)(x+7) = 0

x(x-7) + 2(x-7) = 0

x2 - 7x + 2x - 14 = 0

x2 - 5x - 14 = 0
20.

A sales girl gave a change of N1.15 to a customer instead of N1.25. Calculate her percentage error

A. 10%

B. 7%

C. 8.0%

D. 2.4%

E. 10%

Detailed Solution

Error = 1.25 - 1.15 = 0.10

0.10/1.25
* 100 = 8.0%