Mathematics 2020 Past Questions | WAEC
Study the following Mathematics past questions and answers for JAMB, WAEC NECO and Post-JAMB. Get prepared with official past questions and answers for upcoming examinations.
- Evaluate and correct to two decimal places, 75.0785 – 34.624 + 9.83
- A. 30.60
- B. 50.29
- C. 50.28
- D. 30.62
Correct Option: Answer is C
Solution & Explanation:
75.0785
– 34.624
40.4545
+ 9.83
49.2845 = 50.28 to 2 decimal places
2. If x = {x : x < 7} and Y = {y : y is a factor of 24} are subsets of μ= {1, 2, 3…10} find X ∩ Y
- A. {2, 3, 4, 6}
- B. {1, 2, 3, 4, 6}
- C. {2, 3, 4, 6, 8}
- D. {1, 2, 3, 4, 6, 8}
Correct Option: Answer is B
Solution & Explanation:
μ = {1, 2, 3, 4…,10}
X = {1, 2, 3, 4, 5, 6}
Y = {1, 2, 3, 4, 6, 8}
Then,
- x ∩y = {1, 2, 3, 4, 6}
3. Simplify [(\(\frac{16}{9}\))\(^{\frac{-3}{2}}\) x 16\(^{\frac{-3}{2}}\)]\(^{\frac{1}{3}}\)
- A. \(\frac{3}{4}\)
- B. \(\frac{9}{16}\)
- C. \(\frac{3}{8}\)
- D. \(\frac{1}{4}\)
Correct Option: Answer is C
Solution & Explanation
[(\(\frac{16}{9}\))\(^{\frac{-3}{2}}\) x 16\(^{\frac{-3}{2}}\)]\(^{\frac{1}{3}}\)
= [(\(\frac{9}{16}\))]\(^{\frac{3}{2}}\) x [(\(\frac{1}{16}\))\(^{\frac{3}{4}}\)]\(^{\frac{1}{3}}\)
= [(\(\sqrt{\frac{9}{10}}\))\(^3\) x (4\(\sqrt{\frac{1}{16}})^3\)]\(^{\frac{1}{3}}\)
= [(\(\frac{3}{4})^3 \times (\frac{1}{2})^3\)]\(^\frac{1}{3}\)
(\(\frac{27}{64} \times \frac{1}{8}\))\(^\frac{1}{3}\) = \({3}\sqrt{\frac{27}{64} \times \frac{1}{8}}\)
= \(\frac{3}{4} \times \frac{1}{2}\) = \(\frac{3}{8}\)
- A. log10\(^{90}\)
- B. log10\(^{19}\)
- C. log10\(^{9}\)
- D. log10\(^{6}\)
Correct Option: Answer is A
Solution & Explanation1 + 2log\(_{10}^3\)
= log\(_{10}^{10} + log_{10}^{3^2}\)
= log\(_{10}^{10} + log_{10}^{9}\)
= log\(_{10}^{10 \times 90}\) = log\(_{10}^{90}\)
5. If 101two+ 12y = 3.3five. Find the value of y
- A. 8
- B. 7
- C. 6
- D. 5
Correct Option: Answer is C
Solution & Explanation:
101two + 12y = 3.3five
= 1(2)0 + 0(2)1 + 1(2)2 + 1(y)0 + 2(y)1 = 3(5)0 + 3(5)1
= 1 + 0 + 4 + 1 + 2y = 3 + 15
6 + 2y =18: 18-6 = 2y
12 = 2y: y =6
6. An amount of N550,000.00 was realized when a principal, x was saved at 2% simple interest for 5 years. Find the value of x
- A. N470,000.00
- B. N480,000.00
- C. N490,000.00
- D. N500,000.00
Correct Option: Answer is D
Solution & Explanation:
S.I = PTR/100 = X x 5 x 2/100 = 10x/100
A = P + I, where A= Amount, P = Principle, I = Simple Interest
Therefore,
550,000 = x + 0.01x
= 550,000/1.1x = 500,000
7. Given that \(\frac{\sqrt{3} + \sqrt{5}}{\sqrt{5}}\)
= x + y\(\sqrt{15}\), find the value of (x + y)
- A. 1\(\frac{3}{5}\)
- B. 1\(\frac{2}{5}\)
- C. 1\(\frac{1}{5}\)
- D. \(\frac{1}{5}\)
Correct Option: Answer is C
\(\frac{\sqrt{3} + \sqrt{5}}{\sqrt{5}}\) = x + y\(\sqrt{15}\)
cross multiply to have: \(\sqrt{3}\) + \(\sqrt{5}\) = x\(\sqrt{5}\) + 5y\(\sqrt{3}\)
Collect like roots : x\(\sqrt{5}\) = \(\sqrt{5}\) → x = 1
5y\(\sqrt{3}\) = \(\sqrt{3}\) → y = \(\frac{1}{5}\)
∴ ( x + y ) = 1 + \(\frac{1}{5}\)
= 1\(\frac{1}{5}\)
8. If x = 3 and y = -1, evaluate 2(x\(^2\) – y\(^3\))
- A. 24
- B. 22
- C. 20
- D. 16
Correct Option: Answer is D
Solution & Explanation
2(x2−y2)
= 2(x + y)(x – y)
= 2(3 + (-1))(3 – (-1))
= 2(2)(4) = 16
9. Solve 3x – 2y = 10 and x + 3y = 7 simultaneously
- A. x = -4 and y = 1
- B. x = -1 and y = -4
- C. x = 1 and y = 4
- D. x = 4 and y = 1
Correct Option: Answer is A
Solution & Explanation:
3x – 2y = 10 – – x 3
x + 3y = 7 —x 2
9x – 6y = 30
2x + 6y = 14
11x + 0 = 44
x = 4
From x + 3y = 7
3y = 7 – 4; 3y = 3
y = 1
10. The implication x →y is equivalent to
- A. ~ y → ~ x
- B. y → ~ x
- C. ~ x → ~ y
- D. y → x
Correct Option: Answer is A
If x : y : z = 3 : 3 : 4, evaluate \(\frac{9x + 3y}{6x – 2y}\)
\(\frac{x}{y}\) = \(\frac{2}{3}\) and \(\frac{y}{z}\) = \(\frac{3}{4}\)
Thus; x = \(\frac{2}{3}T_1\) and z = \(\frac{3}{5}T_1\)
y = \(\frac{3}{7}T_2\) and z = \(\frac{4}{7}T_2\)
Using y = y
\(\frac{3}{5}T_1\) = \(\frac{3}{7}T_2\); \(\frac{T_1}{T_2}\) = \(\frac{3}{7}\) x \(\frac{5}{3}\)
\(\frac{T_1}{T_2}\) = \(\frac{15}{21}\)
\(T_1\) = 15 and \(T_2\) = 21
Therefore;
x = \(\frac{2}{5}\) x 15 = 6
y = \(\frac{3}{5}\) x 15 = 9
y = \(\frac{3}{7}\) x 21 = 9 (again)
z = \(\frac{4}{7}\) x 21 = 12
Hence;
\(\frac{9x + 3y}{6z – 2y}\) = \(\frac{9(6) + 3(9)}{6(12) – 2(9)}\)
\(\frac{54 + 27}{72 – 18}\) = \(\frac{81}{54}\) = \(\frac{3}{2}\)
= 1\(\frac{1}{2}\)
11. The first term of a geometric progression (G.P) is 3 and the 5th term is 48. Find the common ratio.
- A. 2
- B. 4
- C. 8
- D. 16
Correct Option: Answer is A
Solution & Explanation:
First a = 3,
fifth ar4 = 48,
Common ratio r = ?
Therefore, 3 =3r4 =48
r4 =16, r = 4 \(\sqrt{16}\) = 2
12. Solve \(\frac{1}{3}\)(5 – 3x) < \(\frac{2}{5}\)(3 – x)
- A. x > \(\frac{7}{22}\)
- B. x < \(\frac{7}{22}\)
- C. x > \(\frac{-7}{27}\)
- D. x < \(\frac{-7}{27}\)
Correct Option: Answer is D
Solution & Explanation:
\(\frac{1}{3}\)(5 – 3x) < \(\frac{2}{5}\)(3 – 7x)
5(5 – 3x) < 6(3 – 7x)
25 – 15x < 18 – 42x
– 15x + 42x < 18 – 25
\(\frac{27x}{27}\) < \(\frac{-7}{27}\)
x < \(\frac{-7}{27}\)
13. Make m the subject of the relation k = \(\frac{m – y}{m + 1}\)
- A.
m = \(\frac{y + k^2}{k^2 + 1}\) - B.
m = \(\frac{y + k^2}{1 – k^2}\) - C.
m = \(\frac{y – k^2}{k^2 + 1}\) - D.
m = \(\frac{y – k^2}{1 – k^2}\)
Correct Option: Answer is B
Solution & Explanation:k = \(\frac{m – y}{m + 1}\)
k\(^2\) = \(\frac{m – y}{m + 1}\)
k\(^2\)m + k\(^2\) = m – y
k\(^2\) + y = m – k\(^2\)m
\(\frac{k^2 + y}{1 – k^2}\) = m\(\frac{(1 – k^2)}{1 – k^2}\)
m = \(\frac{y + k^2}{1 – k^2}\)
14. Find the quadratic equation whose roots are \(\frac{1}{2}\) and -\(\frac{1}{3}\)
- A. 3x\(^2\) + x + 1 = 0
- B. 6x\(^2\) + x – 1 = 0
- C. 3x\(^2\) + x – 1 = 0
- D. 6x\(^2\) – x – 1 = 0
Correct Option: Answer is D
Solution & Explanation:
1/2 and -1/3
(2x – 1) = 0 and (3x + 1) = 0
(2x – 1) (3x + 1) = 0
6x2 – x – 1 = 0
15. Given that x is directly proportional to y and inversely proportional to Z, x = 15 when y = 10 and Z = 4, find the equation connecting x, y and z
- A. x = \(\frac{6y}{z}\)
- B. x = \(\frac{12y}{z}\)
- C. x = \(\frac{3y}{z}\)
- D. x = \(\frac{3y}{2z}\)
Correct Option: Answer is A