Mathematics 2018 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.
31. A man stands on a tree 150cm high and sees a boat at an angle of depression of 74°. Find the distance of the boat from the base of the tree.
- A. 52cm
- B. 43cm
- C. 40cm
- D. 15cm
Correct Option: Answer is B
Tan 74 = 150/x
x = 150/tan 74
= 43.01cm
32. Integrate the expression 6x\(^2\) – 2x + 1
- A. 3x\(^3\) – 2x\(^2\) + x + c
- B. 2x\(^3\) – x\(^2\) + x + c
- C. 2x\(^3\) – 3x\(^2\) + c
- D. x\(^3\) + x\(^2\) – x + c
Correct Option: Answer is B
\(\int 6x^2 – 2x + 1 = (\frac{6x^{2+1}}{2+1} – (\frac{2x^{1+1}}{1+1} + x + c\)
\(\frac{6x^3}{3} – (\frac{2x^2}{2} + x + c\)
\(2x^3 – x^2 + x + c\)
33. In how many ways can the letters LEADER be arranged?
- A. 72
- B. 144
- C. 360
- D. 720
Correct Option: Answer is C
The word LEADER has 1L 2E 1A 1D and 1R making total of 6! \(\frac{6}{1!2!1!1!1!}\) = \(\frac{6!}{2!}\)
= \(\frac{6 \times 5 \times 4 \times 3 \times 2 \times 1}{2 \times 1}\)
= 360
34.
In the figure below, /MX/ = 8cm, /XN/ = 12cm, /NZ/ = 4cm and ∠ XMN = ∠ XZY. Calculate /YM/
- A. 32cm
- B. 24 cm
- C. 16 cm
- D. 12 cm
Correct Option: Answer is C
From the figure,
∠ XMN = ∠ XZY
Angle X is common
So, ∠ XNM = ∠ XYZ
Then from the angle relationship
\(\frac{XM}{XZ}\) = \(\frac{XN}{XY}\) = \(\frac{MN}{ZY}\)
XM = 8, XZ = 12 + 4 = 16,
XN = 12, XY = 8 + YM
\(\frac{8}{16}\) = \(\frac{12}{(8 + YM) }\)
Cross multiply
8(8 + YM) = 192
64 + 8YM = 192
8YM = 128
YM = \(\frac{128}{8}\)
= 16cm
35. Express 495g as a percentage of 16.5kg
- A. 3%
- B. 3 1/3%
- C. 15%
- D. 30%
Correct Option: Answer is A
The two numbers must be expressed in the same unit. To convert 495g to kg, it will be divided by 1000
495g = \(\frac{495}{1000}\)
= 0.495kg
To express in percentage, 0.495 will be divided by 16.5 and then multiplied by 100
% will be added to the answer \(\frac{0.4950}{16.5}\) x 100
= 3%
36. Evaluate (2√3 – 4) (2√3 + 4)
- A. -4
- B. -2
- C. 2
- D. 4
Correct Option: Answer is A
2√3 – 4) ( 2√3 + 4)
= 12 + 8√3 – 8√3 – 16
= 12 – 16
= -4
37. Find the equation of the tangent at the point (2, 0) to the curve y = x² – 2x
- A. y = 2x – 4
- B. y = 2x + 4
- C. y = 2x – 2
- D. y = 2x + 2
Correct Option: Answer is A
The gradient to the curve is found by differentiating the curve equation with respect to x
So \(\frac{dy}{dx}\) 2x – 2
The gradient of the curve is the same with that of the tangent.
At point (2, 0) \(\frac{dy}{dx}\) = 2(2) – 2
= 4 – 2 = 2
The equation of the tangent is given by (y – y1) \(\frac{dy}{dx}\) (x – x1)
At point (x1, y1) = (2, 0)
y – 0 = 2(x – 2)
y = 2x – 4
38.
Use the quadratic equation curve to answer this questions
What is the minimum value of the graph?
- A. -5.3
- B. 0.5
- C. 3
- D. 8
Correct Option: Answer is A
39. Evaluate \((\frac{6}{0.32} \div \frac{2}{0.084})^{-1}\) correct to 1 decimal place.
- A. 1.3
- B. 2.5
- C. 4.6
- D. 3.2
Correct Option: Answer is D
40. Evaluate log\(_2\) 8 – log\(_3\) \(\frac{1}{9}\)
- A. -1 1\(\frac{1}{2}\)
- B. -1
- C. 1
- D. 5
Correct Option: Answer is D
log\(_2\) 8 – log\(_3\) \(\frac{1}{9}\)
= log \(_2\) 2\(^3\) – log\(_3\) 9\(^{-1}\)
= log\(_2\) 2\(^3\) – log\(_3\) 3\(^{-2}\)
Based on law of logarithm
= 3 log\(_2\) 2 – (-2 log\(_3\) 3)
But log\(_2\) 2 = 1,
log\(_3\) 3 = 1
So, = 3 + 2
= 5