Mathematics (Core)
Paper 1 | Objectives | 48 Questions
WASSCE/WAEC MAY/JUNE
Year: 2000
Level: SHS
Time:
Type: Question Paper
Answers provided
FREE
No description provided
Feedbacks
This paper is yet to be rated
Paper 1 | Objectives | 48 Questions
WASSCE/WAEC MAY/JUNE
Year: 2000
Level: SHS
Time:
Type: Question Paper
Answers provided
No description provided
This paper is yet to be rated
Norway Kvaroy Arctic Women in Aquaculture Postgraduate Scholarship Program 2022 for African Women
Revision tips on test preparation for Grade A students to. Grade A students tips for any test.
The best study methods and strategies and tips for successful exam preparation for good grades
| # | Question | Ans |
|---|---|---|
| 1. |
Express \(\frac{7}{19}\) as a percentage, correct to one decimal place A. 2.7% B. 3.7% C. 27.1% D. 36.8%
Show Content
Detailed Solution\(\frac{7}{19}\) as a percentage will be\(\frac{7}{19}\times \frac{100}{1}\% = 36.8\%\) |
|
| 2. |
Express 398753 correct to three significant figures A. 398000 B. 398700 C. 398800 D. 399000
Show Content
Detailed Solution398753 \(\approxeq\) 399000 (to 3 s.f) |
|
| 3. |
Simplify \(\frac{10}{\sqrt{32}}\) A. \(\frac{5}{4}\sqrt{2}\) B. \(\frac{4}{5}\sqrt{2}\) C. \(\frac{5}{16}\sqrt{2}\) D. \(\frac{16}{5}\sqrt{2}\)
Show Content
Detailed Solution\(\frac{10}{\sqrt{32}}=\frac{10}{\sqrt{16\times 2}} = \frac{10}{4\sqrt{2}}\\\frac{10\times \sqrt{2}}{4\sqrt{2}\times\sqrt{2}}=\frac{10\sqrt{2}}{4\times 2}= \frac{5}{4}\sqrt{2}\) |
|
| 4. |
Find the missing number in the addition of the following numbers, in base seven A. 3453 B. 5556 C. 6016 D. 13453
Show Content
Detailed Solution4321\(_7\) + 1234\(_7\) = 5555\(_7\)Missing number = 12341\(_7\) - 5555\(_7\) = 3453\(_7\) |
|
| 5. |
What fraction must be subtracted from the sum of \(2\frac{1}{6}\) and \(2\frac{7}{12}\) to give \(3\frac{1}{4}\)? A. \(\frac{1}{3}\) B. \(\frac{1}{2}\) C. \(1\frac{1}{6}\) D. \(1\frac{1}{2}\)
Show Content
Detailed Solution\(2\frac{1}{6} + 2\frac{7}{12}\)= \(\frac{13}{6} + \frac{31}{12}\) = \(\frac{26 + 31}{12}\) = \(\frac{57}{12} = \frac{19}{4}\) \(\frac{19}{4} - 3\frac{1}{4}\) = \(\frac{19}{4} - \frac{13}{4}\) = \(\frac{6}{4}\) = \(1\frac{1}{2}\) |
|
| 6. |
Simplify \(\left(\frac{16}{81}\right)^{-\frac{3}{4}}\times \sqrt{\frac{100}{81}}\) A. \(\frac{80}{243}\) B. \(\frac{1}{64}\) C. \(\frac{25}{6}\) D. \(\frac{15}{4}\)
Show Content
Detailed Solution\(\left(\frac{16}{81}\right)^{-\frac{3}{4}}\times \sqrt{\frac{100}{81}}\\\frac{1}{\left(\sqrt[4]{\frac{16}{81}}\right)^3}\times \frac{10}{9}=\frac{1}{\left(\frac{2}{3}\right)^3}\times\frac{10}{9}\\ =\frac{27}{8}\times \frac{10}{9}=\frac{15}{4}\) |
|
| 7. |
Which of the following numbers is perfect cube? A. 350 B. 504 C. 950 D. 1728 |
D |
| 8. |
If \(104_x = 68\), find the value of x A. 5 B. 7 C. 8 D. 9
Show Content
Detailed Solution\(104_x = 68\\1 \times x^2 + 0 \times x + 4 \times x^0 = 68\\ x^2 = 68 - 4; x^2 = 64\\ x = \sqrt{64}=8\) |
|
| 9. |
The ages of three men are in the ratio 3:4:5. If the difference between the ages of the oldest and youngest is 18 years, find the sum of the ages of the three men A. 45 years B. 72 years C. 108 years D. 216 years
Show Content
Detailed SolutionGiven that the ages are in the ratio 3: 4: 5.Let the sum of their ages be t. \(\therefore\) The youngest age = \(\frac{3}{12} t\) The eldest age = \(\frac{5}{12} t\) \(\implies \frac{5}{12} t - \frac{3}{12} t = 18\) \(\frac{2t}{12} = 18 \implies t = \frac{18 \times 12}{2}\) t = 108 years. The sum of their ages = 108 years. |
|
| 10. |
Given that \(log_4 x = -3\), find x. A. \(\frac{1}{81}\) B. \(\frac{1}{64}\) C. 64 D. 81
Show Content
Detailed Solution\(\log_4 x = -3\)\(x = 4^{-3}\) = \(\frac{1}{64}\) |
Preview displays only 10 out of the 48 Questions