Mathematics 2008 Jamb Examination

Mathematics 2008 Jamb Examination
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2. Career
4. Job
1...678

Question 71

Find the median of 4, 1, 4, 1, 0, 4, 4, 2 and 0
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Order the values:
0, 0, 1, 1, 2, 4, 4, 4, 4
The middle (5th) term is:\[ 2 \]
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Question 72

If x > 0, find the range of the numbers:
x − 3, 3x + 2, x − 1, 4x, 2x − 1, x − 2, 2x − 2, 3x, 3x + 1
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Find maximum and minimum expressions.
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

Find the mean deviation of 2, 4, 5, and 9
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Mean deviation is the average of the absolute deviations from the mean.
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

In how many ways can the letters of the word ACCEPTANCE be arranged?
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The word ACCEPTANCE has 10 letters in total.
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

Find the number of ways of selecting 6 out of 10 subjects for an examination
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Since order does not matter, this is a combination problem.
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 probability of picking a letter T from the word OBSTRUCTION is
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Count total letters in OBSTRUCTION: 11.
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

The result of rolling a fair die 150 times is summarised in the table below:

Number       1     2   3   4    5     6
Frequency  12  18   x  30  2x   45

What is the probability of obtaining a 5?
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The total number of rolls is 150.
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

angle diagram
In the diagram, find the size of the angle marked a°.
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Angle at centre = 2 × angle at circumference\[ 2 \times s = 280^{\circ}\] \[ s = \dfrac{280^{\circ}}{2} = 140^{\circ}\] \[ \angle O = 360^{\circ} - 280^{\circ} = 80^{\circ}\] Sum of angles in a quadrilateral\[ 60^{\circ} + 80^{\circ} + 140^{\circ} + a = 360^{\circ}\] \[ 280^{\circ} + a = 360^{\circ}\] \[ a = 360^{\circ} - 280^{\circ} = 80^{\circ}\]
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Use the information below to answer question:
Trading account for the year ended 31st December 2009

₦               ₦              
Opening Stock 32,000 Sales 48,000
Purchases 40,000 Less Return 2,000
Carriage inwards 1,000
41,000
Less Return 2,000 39,000
Cost of goods available ??
Less closing stock 9,000
Cost of goods sold ??

Calculate the cost of goods sold




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Mathematics 2008 Jamb Examination
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