Mathematics 2004 Jamb Examination

Mathematics 2004 Jamb Examination
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Question 41

I. Rectangular bars of equal width 
II. The height of each rectangular bar is proportional to the frequency of the 3 corresponding class interval.
III. Rectangular bars have common sides with no gaps in between.

A histogram is described by

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A histogram is described by:

\[ \boxed{\text{C. I, II and III}}\]  

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Question 42

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

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\[ \begin{array}{|c|c|c|}
\hline
\text{Class mark } (x) & \text{Cumulative Frequency} & \text{Frequency} \\
\hline
9 & 1 & 1 \\
14 & 6 & 5 \\
19 & 15 & 9 \\
24 & 17 & 2 \\
29 & 24 & 7 \\
34 & 25 & 1 \\
\hline
\end{array}\]

Total number of students = 25
Students who scored more than 20 marks:\[ 2\ (\text{from }24) + 7\ (\text{from }29) + 1\ (\text{from }34) = 10\]

\[ \text{Percentage} = \dfrac{10}{25} \times 100 = \boxed{40\%}\]  

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Question 43

The mean age of a group of students is 15 years. When the age of a teacher, 45 years old, is added to the ages of the students, the mean of their ages becomes 18 years. Find the number of students in the group.

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Let the number of students be \( n \). Since the mean age of the students is 15: \[ \text{Total age of students} = 15n \] After adding the teacher (age 45), the new total age is: \[ 15n + 45 \] and the total number of people becomes: \[ n + 1 \] The new mean is 18: \[ \frac{15n + 45}{n + 1} = 18 \] Multiply both sides: \[ 15n + 45 = 18(n + 1) = 18n + 18 \] Rearrange: \[ 15n + 45 = 18n + 18 \Rightarrow 45 - 18 = 18n - 15n \Rightarrow 27 = 3n \] Solve: \[ n = \frac{27}{3} = \boxed{9} \]
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Question 44

The weights of 10 pupils in a class are 15kg, 16kg, 17kg, 18kg, 16kg, 17kg, 17kg, 17kg, 18kg and 16kg.
What is the range of this distribution?

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Range = Max value - Min value = 18 - 15 = 3

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Question 45

Find the mean deviation of 1, 2, 3 and 4

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Given values: 1, 2, 3, 4 First, calculate the mean: \[ \text{Mean} = \frac{1 + 2 + 3 + 4}{4} = \frac{10}{4} = 2.5 \] Next, find the absolute deviations from the mean: \[ |1 - 2.5| = 1.5, \quad |2 - 2.5| = 0.5, \quad |3 - 2.5| = 0.5, \quad |4 - 2.5| = 1.5 \] Calculate the Mean Deviation: \[ \text{Mean Deviation} = \frac{1.5 + 0.5 + 0.5 + 1.5}{4} = \frac{4}{4} = \boxed{1.0} \]
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Question 46

In how many ways can 2 students be selected from a group of 5 students in a debating competition?

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Number of ways to choose 2 from 5\[ {}^5C_2 = \dfrac{5!}{2!(5 - 2)!} = \dfrac{5 \times 4}{2 \times 1} = \boxed{10}\]  

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Question 47

A committee of six is to be formed by a state governor from nine state commissioners and three members of the state house of assembly. In how many ways can the members of the committee be chosen so as to include one member of the house of assembly?

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Choose 5 commissioners from 9: \[ {}^{9}C_{5} = \frac{9!}{5!(9-5)!} = \frac{9 \times 8 \times 7 \times 6 \times 5!}{5! \times 4!} = 126 \] Total combinations: \[ \text{Total combinations} = 3 \times 126 = \boxed{378} \]
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Question 48

Some white balls were put in a basket containing twelve red balls and sixteen black balls. If the probability of picking a white ball from the basket is 3/7, how many white balls were introduced?

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Let the number of white balls be x.

\[ \text{Total balls} = 12 + 16 + x = 28 + x\]\[ \text{Given: } \dfrac{x}{28 + x} = \dfrac{3}{7}\] ​​​​​​​Cross-multiply:\[7x = 3(28 + x) \Rightarrow 7x = 84 + 3x\]\[\Rightarrow 4x = 84 \Rightarrow x = \boxed{21}\]  

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Question 49

Some white balls were put in a basket containing twelve red balls and sixteen black balls. If the probability of picking a white ball from the basket is 3/7, how many white balls were introduced?

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Let the number of white balls be \( x \). Total balls: \[ \text{Total balls} = 12 + 16 + x = 28 + x \] Given the probability of picking a white ball is \( \frac{3}{7} \): \[ \frac{x}{28 + x} = \frac{3}{7} \] Cross-multiply: \[ 7x = 3(28 + x) \] \[ 7x = 84 + 3x \] Solve for \( x \): \[ 4x = 84 \Rightarrow x = \frac{84}{4} \] \[ \boxed{x = 21} \]
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Question 50

An unbiased die is rolled 100 times and the outcome is tabulated as follows:

\[
\begin{array}{|c|c|c|c|c|c|c|}
\hline
\textbf{Outcome} & 1 & 2 & 3 & 4 & 5 & 6 \\
\hline
\textbf{Frequency} & 13 & 15 & 22 & 14 & 20 & 16 \\
\hline
\end{array}
\]

What is the probability of obtaining 5?

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Number of times 5 was obtained = 20
P(obtaining 5) = 20/100
= 1/5

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