Paper 1  Objectives  49 Questions
WASSCE/WAEC MAY/JUNE
Year: 2021
Level: SHS
Time:
Type: Question Paper
Answers provided
No description provided
This paper is yet to be rated
Effective ways of coping with exam stress and test anxiety which is supported by Science
The goal of mocks tests is to create a benchmarking tool to help students assess their performances.
Scholarship in Norway universities are open for application in 2022 also for developing countries.
#  Question  Ans 

1. 
Correct 0.007985 to three significant figures. A. 0.0109 B. 0.0800 C. 0.00799 D. 0.008
Show Content
Detailed Solution


2. 
Simplify: (11\(_{two}\))\(^2\) A. 1001\(_2\) B. 1101\(_2\) C. 101\(_2\) D. 10001\(_2\)
Show Content
Detailed Solution(11\(_{two}\))\(^2\) = (11\(_2\) \(\times\) (11\(_2\))= \({1 \times 2^1 + 1 \times 2^0} \times ({1 \times 2^1 + 1 \times 2^0})\) = \({1 \times 2 + 1 \times 1} \times ({1 \times 2 + 1 \times 1})\) = \({2 + 1} \times ({2 + 1})\) = 3 \(\times\) 3 = 9\(_{10}\) or 1001\(_2\) from 

3. 
Solve: 2\(^{√2x + 1}\) = 32 A. 13 B. 24 C. 12 D. 11
Show Content
Detailed Solution2\(^{√2x + 1}\) = 322\(^{√2x + 1}\) = 2\(^5\) √2x + 1 = 5 square both sides 2x + 1 = 5\(^2\) 2x + 1 = 25 2x = 25  1 2x = 24 x = \(\frac{24}{2}\) x = 12 

4. 
If log\(_{10}\) 2 = m and log\(_{10}\) 3 = n, find log\(_{10}\) 24 in terms of m and n. A. 3m + n B. m + 3n C. 4mn D. 3mn
Show Content
Detailed Solutionlog\(_{10}\) 24 = log\(_{10}\) 8 \(\times\) log\(_{10}\) 3where log\(_{10}\) 8 = 3 log\(_{10}\) 2 = 3 \(\times\) m and log\(_{10}\) 3 = n : log\(_{10}\) 24 = 3m + n 

5. 
Find the 5th term of the sequence 2,5,10,17....? A. 22 B. 24 C. 36 D. 26
Show Content
Detailed SolutionSimply add odd number starting from '3' to the next number2 2 + 3 = 5 5 + 5 = 10 10 + 7 = 17 17 + 9 = 26 The fifth term = 26 

6. 
If p = {3<x<1} and Q = {1<x<3}, where x is a real number, find P n Q. A. 0 B. 3, 2, 1, 0 and 1 C. 2, 1 and 0 D. 1, 0 and 1
Show Content
Detailed Solutionp = {3<x<1} = {2,1 and 0}Q = {1<x<3} = {0,1 and 2} P n Q = {0} or {1<x<1} 

7. 
Factorize 6pq3rs3ps+6qr A. 3(r p)(2q + s) B. 3(p + r)( 2q  2q  s) C. 3(2q  s)(p + r) D. 3(r  p)(s  2q)
Show Content
Detailed Solution6pq3rs3ps+6qr = 3 (2pq  rs  ps + 2qr)= 3 ({2pq + 2qr} {ps  rs}) = 3 (2q{ p + r} s{p + r}) = 3 ({2q  s}{p + r}) 

8. 
What number should 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. 1\(\frac{1}{2}\) C. 1\(\frac{1}{6}\) D. \(\frac{1}{2}\)
Show Content
Detailed SolutionThe sum of 2 \(\frac{1}{6}\) and 2\(\frac{7}{12}\)= \(\frac{13}{6}\) + \(\frac{31}{12}\) = \(\frac{13 \times 2 + 31}{12}\) = \(\frac{26 + 31}{12}\) = \(\frac{57}{12}\) What should be subtracted from \(\frac{57}{12}\) to give 3\(\frac{1}{4}\) \(\frac{57}{12}\)  y = 3\(\frac{1}{4}\) : y = \(\frac{57}{12}\)  3\(\frac{1}{4}\) = \(\frac{57}{12}\)  \(\frac{13}{4}\) y = \(\frac{57  3 \times 13}{12}\) = \(\frac{57  39}{12}\) y = \(\frac{18}{12}\) y = \(\frac{3}{2}\) or 1\(\frac{1}{2}\) 

9. 
Mensah is 5 years old and joyce is thrice as old as mensah. In how many years will joyce be twice as old as Mensah? A. 3 years B. 10 years C. 5 years D. 15 years
Show Content
Detailed SolutionMensah’s age is 5. Thus,Joyce’s age is 15 (5*3=15) The difference between their ages is 10 (15–5=10) As we ought to find how many years Joyce’s age will be twice of Mensah’s age, we should write down the following : 15+X=2*(5+X) 15+X=10+2X lets add (10X) to both sides of the equation and 15+X10X = 10+2X10X 5=X —> X=5 After 5 years Joyce’s age will be 20 (15+5=20) After 5 years Mensah’s age will be 10 (5+5=10) After 5 years Joyce will be twice as old as Mensah (10*2=20) 

10. 
If 16 * 2\(^{(x + 1)}\) = 4\(^x\) * 8\(^{(1  x)}\), find the value of x. A. 4 B. 4 C. 1 D. 1
Show Content
Detailed Solution16 * 2\(^{(x + 1)}\) = 4\(^x\) * 8\(^{(1  x)}\)= 2\(^4\) * 2\(^{(x + 1)}\) = 2\(^{2x}\) * 2\(^{3(1  x)}\) > 4 + x + 1 = 2x + 3  3x collect like terms > x  2x + 3x = 3  1  4 > 2x = 2 > x = 1 
Preview displays only 10 out of the 49 Questions