Mathematics 2008 Jamb Examination
Question 71
0, 0, 1, 1, 2, 4, 4, 4, 4
The middle (5th) term is:\[ 2 \]
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Question 72
x − 3, 3x + 2, x − 1, 4x, 2x − 1, x − 2, 2x − 2, 3x, 3x + 1
Minimum term: \(x-3\).
Maximum term: \(3x+2\).
Range:\[ (3x+2) - (x-3) = 3x + 2 - x + 3 = 2x + 5 \]
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Question 73
The mean of the numbers is:\[ \dfrac{2 + 4 + 5 + 9}{4} = 5 \]Find the absolute deviations:
\( |2 - 5| = 3 \), \( |4 - 5| = 1 \), \( |5 - 5| = 0 \), \( |9 - 5| = 4 \)
Sum of deviations:\[ 3 + 1 + 0 + 4 = 8 \]Mean deviation:\[ \dfrac{8}{4} = 2 \]The mean deviation shows how far each value is from the mean on average.
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Question 74
Repeated letters:
A appears 2 times, C appears 2 times, and E appears 3 times.
The number of distinct permutations of a word with repeated letters is:\[ \dfrac{10!}{2! \cdot 2! \cdot 3!} \]This accounts for over-counting caused by identical letters appearing in multiple positions.
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Question 75
Use the formula:\[ ^{10}C_6 = \dfrac{10!}{6!4!} \]Simplifying:\[ ^{10}C_6 = 210 \]This gives the number of possible groups of 6 subjects from 10 available subjects.
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Question 76
The letter T appears 2 times in the word.
Probability is therefore:\[ \dfrac{2}{11} \]This expresses the ratio of favourable outcomes (selecting T) to the total possible selections.
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Question 77
Number    1   2   3   4  5   6
Frequency  12  18  x  30  2x  45
What is the probability of obtaining a 5?
Sum all frequencies:\[ 12 + 18 + x + 30 + 2x + 45 = 150 \]Combine like terms:\[ 105 + 3x = 150 \]Solve for x:\[ 3x = 45 \Rightarrow x = 15 \]Frequency of rolling a 5 is \(2x = 30\).
Probability:\[ \dfrac{30}{150} = \dfrac{1}{5} \]This gives the likelihood of landing on 5 when rolling the die.
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Question 78

In the diagram, find the size of the angle marked a°.
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