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JAMB UTME - Mathematics - 2004

Question 1 Report

The shadow of a pole 5√3m high is 5m. Find the angle of elevation of the sun.

45o

75o

30o

60o

Answer Details

The height of the pole is 5√3m and its shadow on the ground is 5m. Let x be the angle of elevation of the sun. From the geometry of the situation, we can see that tan(x) = opposite/adjacent = (5√3)/5 = √3. Taking the inverse tangent of both sides, we get: x = tan^(-1)(√3) ≈ 60° Therefore, the angle of elevation of the sun is 60°. So the answer is (D) 60°.

Read lesson note on Trigonometry (JAMB)

Trigonometry

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Question 5 Report

The shaded region in the Venn diagram above is

P $^{\circ}$ $\cap$ (Q $\cap$ R)

P $\cap$ Q

P $^{\circ}$ $\cup$ (Q $\cap$ R)

P $^{\circ}$ $\cup$ (Q $\cup$ R)

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Question 13 Report

The graph above shows the cumulative frequency curve of the distribution of marks in a class test. What percentage of the students scored more than 20 marks?

68%

28%

17%

8%

Answer Details

\begin{matrix} class-marks(x) & (freq. (f) & cum. freq. \\ 9 & 3.....3 \\ 14 & 6....3 + 6 = 9 \\ 19 & 9.....9 + 9 = 18 \\ 24 & 3.....18 + 3 = 21 \\ 29 & 2.....21 + 2 = 23 \\ 34 & 2.....23 + 2 = 25 \\ 25 \end{matrix}

% of students scoring more than 20 marks

= $\frac{7}{25}$ x 100% = 28%

Read lesson note on Percentages (WAEC)

Percentages

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Question 25 Report

Evaluate \( \frac{\frac{1}{10}\times\frac{2}{3}+\frac{1}{4}}{\frac{\frac{1}{2}}{\frac{3}{5}}-\frac{1}{4}}} \)

\frac{7}{12}

\frac{19}{35}

\frac{2}{25}

\frac{19}{60}

Answer Details

Explanation:

\frac{\frac{1}{10} \times \frac{2}{3} + \frac{1}{4}}{\frac{\frac{1}{2}}{\frac{3}{5}} - \frac{1}{4}} N u m e r a t o r \frac{1}{10} \times \frac{2}{3} + \frac{1}{4} = \frac{1}{5} + \frac{1}{4} = \frac{4 + 15}{60} = \frac{19}{60} d e n o m i n a t o r = \frac{\frac{1}{2}}{\frac{3}{5}} - \frac{1}{4} = \frac{1}{2} \times \frac{5}{3} - \frac{1}{4} = \frac{5}{6} - \frac{1}{4} = \frac{10 - 3}{12} = \frac{7}{12} \frac{N u m e r a t o r}{d e n o m i n a t o r} = \frac{\frac{19}{60}}{\frac{7}{12}} = \frac{19}{60} \times \frac{12}{7} = \frac{19}{35}

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Question 26 Report

PQRSTV is a regular polygon of side 7 cm inscribed in a circle. Find the circumference of the circle PQRSTV.

56 cm

22 cm

42 cm

44 cm

Answer Details

Each interior Angle of the polygon =	(n-2)	180
	_n

=	(6-2)	180
	_6

= 4 x 30
= 120

^{\circ}

i.e ∠PQR = 120

^{\circ}

∴ RQD = 120/2 = 60

^{\circ}

also
∴ Δ QOR is equilateral Δ ∴ QO = 7cm
∴ PO = QO = RO ...... = 7cm radius of the circumference of the circle = 2πr
= 2 x 22/7 x 7/1
= 44 cm

Read lesson note on Circles (WAEC)

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Question 35 Report

Evaluate \( \int_{1}^{3}(X^2 - 1)\,dx \)

\frac{2}{3}

- \frac{2}{3}

- 6 \frac{2}{3}

6 \frac{2}{3}

Answer Details

\int_{1}^{3} (x^{2} - 1) d x = {[\frac{1}{3} x^{2} - x]}_{1}^{2} = (9 - 3) - (\frac{1}{3} - 1) = 6 - (- \frac{2}{3}) = 6 + \frac{2}{3} = 6 \frac{2}{3}

Read lesson note on Integration (JAMB)

Read lesson note on Trigonometry (JAMB)

Integration

Trigonometry

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Question 36 Report

The pie chart above shows the distribution of the crops harvested from a farmland in a year. If 3000 tonnes of millet is harvested, what amount of beans is harvested

6000 tonnes

1500 tonnes

1200 tonnes

9000 tonnes

Answer Details

Sectorial angle of beans = 360 - (150 + 90 + 60)
= 360 -300
= 60

^{\circ}

if 150

^{\circ}

represents 3000 tonnes
1

^{\circ}

3000/150
60

^{\circ}

will represents (3000/150) * (60/1)
= 1200 tonnes

Read lesson note on Sets (JAMB)

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