Paper 1  Objectives  50 Questions
WASSCE/WAEC MAY/JUNE
Year: 2019
Level: SHS
Time:
Type: Question Paper
Source: Nigeria
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#  Question  Ans 

1. 
Express, correct to three significant figures, 0.003597. A. 0.359 B. 0.004 C. 0.00360 D. 0.00359
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Detailed Solution0,00 3597 = 0.00360 to 3 s.f 

2. 
Evaluate: (0.064)  \(\frac{1}{3}\) A. \(\frac{5}{2}\) B. \(\frac{2}{5}\) C. \(\frac{2}{5}\) D. \(\frac{5}{2}\)
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Detailed Solution(0.064)\(^{ \frac{1}{3}}\)= (\(\frac{64}{1000}\))\(^{\frac{1}{3}}\) = 3\(\sqrt{\frac{1000}{64}}\) = \(\frac{10}{4}\) = \(\frac{5}{2}\) 

3. 
Solve: \(\frac{y + 1}{2}  \frac{2y  1}{3}\) = 4 A. y = 19 B. y = 19 C. y = 29 D. y = 29
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Detailed Solution\(\frac{y + 1}{2}  \frac{2y  1}{3}\) = \(\frac{4}{1}\) \(\frac{3(y + 1)  2(2y  1)}{6} = \frac{4}{1}\) 3y + 3  4y + 2 = 24  y + 5 = 24  y = 24  5 = 19 y =  19 

4. 
Simplify, correct to three significant figures, (27.63)\(^2\)  (12.37)\(^2\) A. 614 B. 612 C. 611 D. 610
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Detailed Solution(27.63)\(^2\)  (12.37)\(^2\)= (27.63 + 12.37)(27.63  12.37) = 40 x 15.26 = 610 

5. 
If 7 + y = 4 (mod 8), find the least value of y, 10 \(\leq y \leq 30\) A. 11 B. 13 C. 19 D. 21
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Detailed Solution7 + y = 4 (mod 8)y = 4  7 (mod 8) y = 3 + 8 (mod 8) y = 5 + 8 (mod 8) y = 13 

6. 
If T = {prime numbers} and M = {odd numbers} are subsets of \(\mu\) = {x : 0 < x < 10} and x is an integer, find (T\(^{\prime}\) \(\mu\) M\(^{\prime}\)). A. {4, 6, 8, 10} B. {1 C. {1, 2, 4, 6, 8, 10} D. {1, 2, 3, 5, 7, 8, 9}
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Detailed SolutionT = {2, 3, 5, 7}M = {1, 3, 5, 7, 9} \(\mu\) = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} T\(^{\prime}\) = = {1, 4, 6, 8, 9, 10} M\(^{\prime}\) = {2, 4, 6, 8, 10} (T\(^{\prime}\) \(\cap\) M\(^{\prime}\)) = {4, 6, 8, 10} 

7. 
Evaluate; \(\frac{\log_3 9  \log_2 8}{\log_3 9}\) A. \(\frac{1}{3}\) B. \(\frac{1}{2}\) C. \(\frac{1}{3}\) D. \(\frac{1}{2}\)
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Detailed Solution\(\frac{\log_3 9  \log_2 8}{\log_3 9}\)= \(\frac{\log_3 3^2  \log_2 2^3}{\log_3 3^2}\) = \(\frac{2 3}{2}\) = \(\frac{1}{2}\) 

8. 
If 23\(_y\) = 1111\(_{\text{two}}\), find the value of y A. 4 B. 5 C. 6 D. 7
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Detailed Solution23\(_y\) = 1111\(_{\text{two}}\),2 x y\(^1\) + 3 x y\(^0\) = 1 x 2\(^3\) + 1 x 2\(^1\) + 1 x 2\(^o\) 2y + 3 = 8 + 4 + 2 + 1 2y + 3 = 15 \(\frac{2y}{2}\) \(\frac{12}{2}\) y = 6 

9. 
If 6, P, and 14 are consecutive terms in an Arithmetic Progression (AP), find the value of P. A. 9 B. 10 C. 6 D. 8
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Detailed Solution6, p, 1414  p = p  6 14 + 6 = p  6 14 + 6 = p + p \(\frac{2p}{2}\) = \(\frac{20}{2}\) p = 10 

10. 
Evaluate: 2\(\sqrt{28}  3\sqrt{50} + \sqrt{72}\) A. 4\(\sqrt{7}  21 \sqrt{2}\) B. 4\(\sqrt{7}  11 \sqrt{2}\) C. 4\(\sqrt{7}  9 \sqrt{2}\) D. 4\(\sqrt{7} + \sqrt{2}\)
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Detailed Solution2\(\sqrt{28}  3\sqrt{50} + \sqrt{22}\)4\(\sqrt{7}  15\sqrt{2} + 6\sqrt{2}\) 6\(\sqrt{7}  9\sqrt{2}\) 
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