Mathematics 2019 Past Questions | JAMB
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.
- Make q the subject of the formula in the equation \(\frac{mn}{a^2} – \frac{pq}{b^2} = 1\)
- A. \(q = \frac{b^2(mn – a^2)}{a^2 p}\)
- B. \(q = \frac{m^2 n – a^2}{p^2}\)
- C. \(q = \frac{mn – 2b^2}{a^2}\)
- D. \(q = \frac{b^2 (n^2 – ma^2)}{n}\)
Correct Option: Answer is A
Solution:
\(\frac{mn}{a^2} – \frac{pq}{b^2} = 1\)
\(\frac{mn}{a^2} – 1 = \frac{pq}{b^2}\)
\(\frac{mn – a^2}{a^2} = \frac{pq}{b^2}\)
\(pq = \frac{b^2 (mn – a^2)}{a^2}\)
\(q = \frac{b^2(mn – a^2)}{a^2 p}\)
2. The angle of elevation of the top of a tree from a point on the ground 60m away from the foot of the tree is 78°. Find the height of the tree correct to the nearest whole number.
- A. 148m
- B. 382m
- C. 282m
- D. 248m
Correct Option: Answer is C
Solution:
\(\tan 78 = \frac{h}{60}\)
\(h = 60 \tan 78\)
\(h = 60 \times 4.705 = 282.27m\)
\(\approxeq\) 282m to the nearest whole number.
3. A binary operation ⊗⊗ is defined by m⊗n=mn+m−nm⊗n=mn+m−n on the set of real numbers, for all m, n ∈∈ R. Find the value of 3 ⊗⊗ (2 ⊗⊗ 4).
- A. 6
- B. 25
- C. 15
- D. 18
Correct Option: Answer is C
Solution:
m⊗n=mn+m−n
3 (2 4)
2 4 = 2(4) + 2 – 4 = 6
3 otime 6 = 3(6) + 3 – 6 = 15
Age in years | 7 | 8 | 9 | 10 | 11 |
No of pupils | 4 | 13 | 30 | 44 | 9 |
- A. 48.6°
- B. 56.3°
- C. 46.8°
- D. 13°
Correct Option: Answer is C
Solution:
Total number of pupils : 4 + 13 + 30 + 44 + 9 = 100
The number of 8 – year olds = 13
The angle represented by the 8-year olds on the pie chart = 13/100×360°
= 46.8°
5. In a class of 50 students, 40 students offered Physics and 30 offered Biology. How many offered both Physics and Biology?
- A. 42
- B. 20
- C. 70
- D. 54
Correct Option: Answer is B
Solution:
n(Total) = 50
n(Physics) = 40
n(Biology) = 30
Let n(Physics and Biology) = x
n(Physics only) = 40 -x
n(Biology only) = 30 – x
40 – x + 30 – x + x = 50
70 – x = 50
x = 20
- A. \(-5 – 2\sqrt{6}\)
- B. \(-5 + 3\sqrt{2}\)
- C. \(5 – 2\sqrt{3}\)
- D. \(5 + 2\sqrt{6}\)
Correct Option: Answer is A
Solution:
\(\frac{\sqrt{2} + \sqrt{3}}{\sqrt{2} – \sqrt{3}}\)
= \((\frac{\sqrt{2} + \sqrt{3}}{\sqrt{2} – \sqrt{3}})(\frac{\sqrt{2} + \sqrt{3}}{\sqrt{2} + \sqrt{3}})\)
= \(\frac{2 + \sqrt{6} + \sqrt{6} + 3}{2 – \sqrt{6} + \sqrt{6} – 3}\)
= \(\frac{5 + 2\sqrt{6}}{-1}\)
= \(- 5 – 2\sqrt{6}\)
7.
Find the length of the chord |AB| in the diagram shown above.
- A. 4.2 cm
- B. 4.3 cm
- C. 3.2 cm
- D. 3.4 cm
Correct Option: Answer is D
Solution:Length of chord = \(2r \sin (\frac{\theta}{2})\)
= \(2(3) \sin (\frac{68}{2})\)
= \(6 \sin 34\)
= \(6 \times 0.559\)
= 3.354 cm \(\approxeq\) 3.4 cm
8. Given
find p.
- A. 48°
- B. 58°
- C. 32°
- D. 52°
Correct Option: Answer is C
Solution:
sinθ=cos(90−θ)
sinθ=cos(90−58)
= cos32
9. \(\frac{\frac{2}{3} \div \frac{4}{5}}{\frac{1}{4} + \frac{3}{5} – \frac{1}{3}}\)
- A. \(\frac{31}{50}\)
- B. \(\frac{20}{31}\)
- C. \(\frac{31}{20}\)
- D. \(\frac{50}{31}\)
Correct Option: Answer is D
Solution:
\(\frac{\frac{2}{3} \div \frac{4}{5}}{\frac{1}{4} + \frac{3}{5} – \frac{1}{3}}\)
\(\frac{2}{3} \div \frac{4}{5} = \frac{2}{3} \times \frac{5}{4}\)
= \(\frac{5}{6}\)
\(\frac{1}{4} + \frac{3}{5} – \frac{1}{3} = \frac{15 + 36 – 20}{60}\)
= \(\frac{31}{60}\)
\(\therefore \frac{\frac{2}{3} \div \frac{4}{5}}{\frac{1}{4} + \frac{3}{5} – \frac{1}{3}} = \frac{5}{6} \div \frac{31}{60}\)
= \(\frac{5}{6} \times \frac{60}{31}\)
= \(\frac{50}{31}\)
10. If \(6x^3 + 2x^2 – 5x + 1\) divides \(x^2 – x – 1\), find the remainder.
- A. 9x + 9
- A. 2x + 6
- B. 6x + 8
- C. 5x – 3