Mathematics 2016 Jamb Examination

Mathematics 2016 Jamb Examination
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Question 41

 If Q is a factor of 18 and T is prime numbers between 2 and 18. What is \( Q \cap T \)?
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First, list the elements of set Q (factors of 18): \( Q = \{1, 2, 3, 6, 9, 18\} \). Next, list set T (prime numbers between 2 and 18, inclusive of 2): \( T = \{2, 3, 5, 7, 11, 13, 17\} \). The intersection \( Q \cap T \) consists of elements common to both sets: \[ Q \cap T = \{2, 3\} \]
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Question 42


From the cyclic quadrilateral above, find ∠TSV
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< TVS = 180o - (80 + 20) = 180 - 100 = 80o
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Question 43

Find the mean of 10, 8, 5, 11, 12, 9, 6, 3, 15, and 23.
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The mean is calculated by finding the sum of all values and dividing by the count (n = 10): \[ \text{Sum} = 10 + 8 + 5 + 11 + 12 + 9 + 6 + 3 + 15 + 23 = 102 \] \[ \text{Mean} = \dfrac{\sum x}{n} = \dfrac{102}{10} = 10.2 \]
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Question 44

the cyclic quadrilateral above . Find < PRO
In the cyclic quadrilateral above . Find ∠PRO
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PRO = 180° - (110 + 50)°
= 180 - 160 = 20°
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Question 45

OGIVE is constructed using
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An Ogive, also known as a cumulative frequency polygon, is a graph that represents the cumulative frequencies of a dataset. It is constructed by plotting cumulative frequencies against the upper class boundaries of the data intervals.
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Question 46

Rationalize \( \dfrac{\sqrt{6} - \sqrt{4}}{\sqrt{6} + \sqrt{4}} \)
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To rationalize the expression, multiply the numerator and denominator by the conjugate of the denominator, which is \( \sqrt{6} - \sqrt{4} \). Since \( \sqrt{4} = 2 \), we simplify it to: \[ \dfrac{\sqrt{6} - 2}{\sqrt{6} + 2} \times \dfrac{\sqrt{6} - 2}{\sqrt{6} - 2} = \dfrac{(\sqrt{6})^2 - 2(2)\sqrt{6} + 2^2}{(\sqrt{6})^2 - 2^2} \] \[ = \dfrac{6 - 4\sqrt{6} + 4}{6 - 4} = \dfrac{10 - 4\sqrt{6}}{2} = 5 - 2\sqrt{6} \]
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Question 47

The bar chart above shows the marks obtained by students in a mathematics test. How many students in all took the test?
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3 + 2 + 5 + 5 + + 5 + 10 + 4 + 3 + 1 + 2 = 40
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Question 48


Determine the mean score of the student that took the mathematics test?
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To calculate the mean:
\[ \begin{aligned}
\text{Mean} &= \frac{(3 \times 0) + (1 \times 2) + (5 \times 2) + (5 \times 3) + (5 \times 4)}{40} \\
&\quad + \frac{(10 \times 5) + (4 \times 6) + (3 \times 7) + (1 \times 8) + (2 \times 9)}{40} \\
&= \frac{168}{40} \\
&= 4.2
\end{aligned}\]
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Question 49


From the diagram above. Find the fraction of the shaded position?
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\(\theta\) = 180o -(90 + 60) \(\theta\)
= 180o - 150o = 30o
Fraction of shaded position
= \(\dfrac{30}{360}\) + \(\dfrac{30}{360}\)
= \(\dfrac{1}{12}\) + \(\dfrac{1}{12}\) = \(\dfrac{1}{6}\)
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Question 50

From the diagram above, find the required roots/region of x.
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The graph represents a quadratic function (parabola) that opens upwards and intersects the x-axis at \( x = -1 \) and \( x = 4 \). The shaded region is the part of the curve below the x-axis, which corresponds to the inequality where the function is less than or equal to zero. \[ \text{The region is } -1 \le x \le 4 \]
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Mathematics 2016 Jamb Examination
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