Year : 
2010
Title : 
Mathematics (Core)
Exam : 
WASSCE/WAEC MAY/JUNE

Paper 1 | Objectives

1 - 10 of 49 Questions

# Question Ans
1.

Simplify 0.000215 x 0.000028 and express your answer in standard form

A. 6.03 x 109

B. 6.02 x 109

C. 6.03 x 10-9

D. 6.02 x 10-9

Detailed Solution

0.000215215 x 0.000028

= 215 x 10-4 x 28 x 10-4

= 215 x 28 x 10-6-6

= 6020 x 10-12

= 6.020 x 103 x 10-12

= 6.02 x 103 - 12

= 6.02 x 10-9
2.

Factorize: x + y - ax = ay

A. (x - y)(1 - a)

B. (x + y)(1 + a)

C. (x + y)(1 - a)

D. (x - y)(1 + a)

Detailed Solution

x + y - ax = ay

= (x + y) -a(x + y) = (x + y)(1 - a)
3.

A car uses one litre of petrol for every 14km. If one of petrol cost N63.00, how far can the car go with N900.00 worth of petrol?

A. 420km

B. 405km

C. 210km

D. 200km

Detailed Solution

1 litre = N63.00

xlitres = N900.00

x x N63 = N900 x 1 litre

x = \(\frac{900}{63} \times litre\)

x = \(\frac{100}{7}\) litres

Also, 1 litre = 14km

\(\frac{100}{7}\) = y

y x 1 litre = \(\frac{100litre}{7}\) x 14km

y = 20km
4.

Correct 0.002473 to 3 significant figure

A. 0.002

B. 0.0024

C. 0.00247

D. 0.0025

C

5.

Simplify 1\(\frac{1}{2} + 2\frac{1}{3} \times \frac{3}{4} - \frac{1}{2}\)

A. -2\(\frac{1}{3}\)

B. -2\(\frac{1}{4}\)

C. 2\(\frac{1}{8}\)

D. 2\(\frac{3}{4}\)

Detailed Solution

1\(\frac{1}{2} + 2\frac{1}{3} \times \frac{3}{4} - \frac{1}{2}\)

\(\frac{3}{2} + \frac{7}{3} \times \frac{3}{4} - \frac{1}{2} = \frac{3}{2} + \frac{7}{4} - \frac{1}{2}\)

= \(\frac{6 + 7 - 2}{4} = \frac{11}{4}\)

= 2\(\frac{3}{4}\)
6.

The sum of 2 consecutive whole numbers is \(\frac{5}{6}\) of their product, find the numbers

A. 3, 4

B. 1, 2

C. 2, 3

D. 0, 1

Detailed Solution

Let the no. be x and x + 1

x + (x + 1) = \(\frac{5}{6}\) of x(x + 1)

2x + 1 = \(\frac{5}{6}\) x(x + 1)

6(2x + 1) = 5x2 + 5x

12x + 6 = 5x2 + 5x

5x2 + 5x - 12x - 6 = 0

5x2 - 7x - 6 = 0

5x2 - 10x + 3x - 6 = 0

5x(x - 2) + 3(x - 2) = 0

(5x + 3)(x - 2) = 0

(5x + 3)(x - 2) = 0

(5x + 3) = 0

x - 2 = 0

for (5x + 3) = 0

5x = -3
7.

A casting is made up of Copper and Zinc. If 65% of the casting is Zinc and there are 147g of Copper. What is the mass of the casting?

A. 320g

B. 420g

C. 520g

D. 620g

Detailed Solution

%Zinc + %Copper = 100%

65% + %Copper = 100%

%Copper = 100% - 65% = 35%

%Copper = \(\frac{\text{Mass of % Copper}}{\text{Mass of Casting}}\) x 100%

Mass of Casting = \(\frac{147}{35}\) x 100%

= 420g
8.

Given that P = {x : 1 \(\leq x \leq 6\)}, and Q = {x : 2 \(\leq x \leq 10\)} where x is an integer. Find n(P \(\cap\) Q)

A. 5

B. 6

C. 8

D. 10

Detailed Solution

P = {1, 2, 3, 4, 5, 6}; Q = {3, 4, 5, 6, 7, 8, 9}

P \(\cap\) Q) = {2, 3, 4, 5, 6}

n(P \(\cap\) Q) = 5
9.

The sum of 6 and one-third of x is one more than twice x, find x

A. x = 7

B. x = 5

C. x = 3

D. x = 2

Detailed Solution

6 = \(\frac{1}{3}x\) = 1 + 2x

6 = 1 = \(\frac{2x}{1} - \frac{x}{3}\)

5 = \(\frac{6x - x}{3} = \frac{5x}{3}\)

5 = \(\frac{5x}{3}\)

3 x 5 = 5x

15 = 5x

x = \(\frac{15}{5}\)

= 3
10.

Given that = {x: -2 < x \(\leq\) 9}, where x is an integer what is n(T)?

A. 9

B. 10

C. 11

D. 12

Detailed Solution

T = {-1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9}

n(T) = 11
1.

Simplify 0.000215 x 0.000028 and express your answer in standard form

A. 6.03 x 109

B. 6.02 x 109

C. 6.03 x 10-9

D. 6.02 x 10-9

Detailed Solution

0.000215215 x 0.000028

= 215 x 10-4 x 28 x 10-4

= 215 x 28 x 10-6-6

= 6020 x 10-12

= 6.020 x 103 x 10-12

= 6.02 x 103 - 12

= 6.02 x 10-9
2.

Factorize: x + y - ax = ay

A. (x - y)(1 - a)

B. (x + y)(1 + a)

C. (x + y)(1 - a)

D. (x - y)(1 + a)

Detailed Solution

x + y - ax = ay

= (x + y) -a(x + y) = (x + y)(1 - a)
3.

A car uses one litre of petrol for every 14km. If one of petrol cost N63.00, how far can the car go with N900.00 worth of petrol?

A. 420km

B. 405km

C. 210km

D. 200km

Detailed Solution

1 litre = N63.00

xlitres = N900.00

x x N63 = N900 x 1 litre

x = \(\frac{900}{63} \times litre\)

x = \(\frac{100}{7}\) litres

Also, 1 litre = 14km

\(\frac{100}{7}\) = y

y x 1 litre = \(\frac{100litre}{7}\) x 14km

y = 20km
4.

Correct 0.002473 to 3 significant figure

A. 0.002

B. 0.0024

C. 0.00247

D. 0.0025

C

5.

Simplify 1\(\frac{1}{2} + 2\frac{1}{3} \times \frac{3}{4} - \frac{1}{2}\)

A. -2\(\frac{1}{3}\)

B. -2\(\frac{1}{4}\)

C. 2\(\frac{1}{8}\)

D. 2\(\frac{3}{4}\)

Detailed Solution

1\(\frac{1}{2} + 2\frac{1}{3} \times \frac{3}{4} - \frac{1}{2}\)

\(\frac{3}{2} + \frac{7}{3} \times \frac{3}{4} - \frac{1}{2} = \frac{3}{2} + \frac{7}{4} - \frac{1}{2}\)

= \(\frac{6 + 7 - 2}{4} = \frac{11}{4}\)

= 2\(\frac{3}{4}\)
6.

The sum of 2 consecutive whole numbers is \(\frac{5}{6}\) of their product, find the numbers

A. 3, 4

B. 1, 2

C. 2, 3

D. 0, 1

Detailed Solution

Let the no. be x and x + 1

x + (x + 1) = \(\frac{5}{6}\) of x(x + 1)

2x + 1 = \(\frac{5}{6}\) x(x + 1)

6(2x + 1) = 5x2 + 5x

12x + 6 = 5x2 + 5x

5x2 + 5x - 12x - 6 = 0

5x2 - 7x - 6 = 0

5x2 - 10x + 3x - 6 = 0

5x(x - 2) + 3(x - 2) = 0

(5x + 3)(x - 2) = 0

(5x + 3)(x - 2) = 0

(5x + 3) = 0

x - 2 = 0

for (5x + 3) = 0

5x = -3
7.

A casting is made up of Copper and Zinc. If 65% of the casting is Zinc and there are 147g of Copper. What is the mass of the casting?

A. 320g

B. 420g

C. 520g

D. 620g

Detailed Solution

%Zinc + %Copper = 100%

65% + %Copper = 100%

%Copper = 100% - 65% = 35%

%Copper = \(\frac{\text{Mass of % Copper}}{\text{Mass of Casting}}\) x 100%

Mass of Casting = \(\frac{147}{35}\) x 100%

= 420g
8.

Given that P = {x : 1 \(\leq x \leq 6\)}, and Q = {x : 2 \(\leq x \leq 10\)} where x is an integer. Find n(P \(\cap\) Q)

A. 5

B. 6

C. 8

D. 10

Detailed Solution

P = {1, 2, 3, 4, 5, 6}; Q = {3, 4, 5, 6, 7, 8, 9}

P \(\cap\) Q) = {2, 3, 4, 5, 6}

n(P \(\cap\) Q) = 5
9.

The sum of 6 and one-third of x is one more than twice x, find x

A. x = 7

B. x = 5

C. x = 3

D. x = 2

Detailed Solution

6 = \(\frac{1}{3}x\) = 1 + 2x

6 = 1 = \(\frac{2x}{1} - \frac{x}{3}\)

5 = \(\frac{6x - x}{3} = \frac{5x}{3}\)

5 = \(\frac{5x}{3}\)

3 x 5 = 5x

15 = 5x

x = \(\frac{15}{5}\)

= 3
10.

Given that = {x: -2 < x \(\leq\) 9}, where x is an integer what is n(T)?

A. 9

B. 10

C. 11

D. 12

Detailed Solution

T = {-1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9}

n(T) = 11