Mathematics (Core)
Paper 1 | Objectives | 49 Questions
WASSCE/WAEC MAY/JUNE
Year: 2012
Level: SHS
Time:
Type: Question Paper
Answers provided
FREE
No description provided
Feedbacks
This paper is yet to be rated
Paper 1 | Objectives | 49 Questions
WASSCE/WAEC MAY/JUNE
Year: 2012
Level: SHS
Time:
Type: Question Paper
Answers provided
No description provided
This paper is yet to be rated
There's no secret to passing WAEC or BECE apart from the merit of hard work and adequate preparation
Apply tips for stress-free eating before and during exams and what to avoid for good grades.
A guide to passing your exams without studying hard, how to pass any test and top your class
| # | Question | Ans |
|---|---|---|
| 1. |
Express 302.10495 correct to five significant figures A. 302.10 B. 302.11 C. 302.105 D. 302.1049 |
A |
| 2. |
Simplify; \(\frac{3\sqrt{5} \times 4\sqrt{6}}{2 \sqrt{2} \times 3\sqrt{2}}\) A. \(\sqrt{2}\) B. \(\sqrt{5}\) C. 2\(\sqrt{2}\) D. 2\(\sqrt{5}\)
Show Content
Detailed Solution\(\frac{3\sqrt{5} \times 4\sqrt{6}}{2 \sqrt{2} \times 3\sqrt{2}}\)= \(\frac{\sqrt{5} \times 2\sqrt{6}}{\sqrt{2} \times \sqrt{3}}\) = \(\frac{\sqrt{5} \times 2 \sqrt{6}}{\sqrt{6}}\) = 2\(\sqrt{5}\) |
|
| 3. |
In 1995, the enrollment of two schools X and Y were 1,050 and 1,190 respectively. Find the ration of the enrollments of X and Y A. 50:11 B. 15:17 C. 13:55 D. 12:11
Show Content
Detailed SolutionDivide cash enrollment by their common factors then the answer of each division gives the required ratio. i.e. x = \(\frac{1050}{70}\) = 15and y = \(\frac{1190}{70}\) = 17 ratio of enrollment of x to y = 15:17 |
|
| 4. |
Convert 3510 to number in base 2 A. 1011 B. 10011 C. 100011 D. 11001 |
C |
| 5. |
The nth term of a sequence is Tn = 5 + (n - 1)2. Evaluate T4 - T6 A. 30 B. 16 C. -16 D. -30
Show Content
Detailed SolutionT4 = 5 + (4 - 1)2; where n = 4= 5 + (3)2 = 5 + 9 = 14 T = 5 + (6 - 1)2 where n = 6 = 5 + (5)2 = 5 + 25 = 30 T4 + T6 = 14 - 30 = -16 |
|
| 6. |
Mr. Manu travelled from Accra to Pamfokromb a distance of 720km in 8 hours. What will be his speed in m/s? A. 25m/s B. 150m/s C. 250m/s D. 500m/s
Show Content
Detailed Solution\(\frac{720 \times 1000}{8 \times 60 \times 60}\) = \(\frac{20 \times 10}{8}\)= \(\frac{200}{8}\) = 25 m/s |
|
| 7. |
If N2,500.00 amounted to N3,50.00 in 4 years at simple interest, find the rate at which the interest was charged A. 5% B. 7\(\frac{1}{2}\)% C. 8% D. 10%
Show Content
Detailed SolutionA = 3,500, P = N2,500A = P + I But I = N3,500 - N2,500 I = N1,000 I = \(\frac{PRT}{100}\) N1,000 = \(\frac{2.500 \times R \times 4}{100}\) 1000 = 100R R = \(\frac{1000}{100}\) = 10% |
|
| 8. |
Solve for x in the equation; \(\frac{1}{x} + \frac{2}{3x} = \frac{1}{3}\) A. 5 B. 4 C. 3 D. 1
Show Content
Detailed Solution\(\frac{1}{8} + \frac{2}{3x} = \frac{1}{3}\)= \(\frac{1}{2}\) \(\frac{5}{3x} = \frac{1}{3}\) 3x = 15 x = \(\frac{15}{3}\) = 5 |
|
| 9. |
Simplify: \(\frac{54k^2 - 6}{3k + 1}\) A. 6(1 - 3k2) B. 6(3k2 - 1) C. 6(3k - 1) D. 6(1 - 3k)
Show Content
Detailed Solution\(\frac{54k^2 - 6}{3k + 1} = \frac{6(9k^2 - 1)}{3k + 1}\)= \(\frac{6(3k + 1) - (3k - 1)}{3k + 1}\) = 6(3k - 1) |
|
| 10. |
Make p the subject of the relation: q = \(\frac{3p}{r} + \frac{s}{2}\) A. p = \(\frac{2q - rs}{6}\) B. p = 2qr - sr - 3 C. p = \(\frac{2qr - s}{6}\) D. p = \(\frac{2qr - rs}{6}\)
Show Content
Detailed Solutionq = \(\frac{3p}{r} + \frac{s}{2}\)q = \(\frac{6p + rs}{2r}\) 6p + rs = 2qr 6p = 2qr - rs p = \(\frac{2qr - rs}{6}\) |
Preview displays only 10 out of the 49 Questions