## 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