Data Representation and Charts Practice Questions — JAMB UTME

Question 1Question →

Match each data representation concept on the left with its corresponding mathematical formula or definition on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Sector angle of a category in a pie chart matches

\frac{\text{Category Frequency}}{\text{Total Frequency}} \times 360^\circ

; Height of a histogram bar for equal class intervals matches Frequency of the class interval; Height of a histogram bar for unequal class intervals matches

\frac{\text{Class Frequency}}{\text{Class Width}}

; Upper class boundary of an interval

a - b

matches

b + 0.5

.

Each chart concept matches its standard definition: pie chart sector angles are proportional parts of

360^\circ

; histogram heights equal class frequencies when widths are equal, but equal frequency density when widths are unequal; and class boundaries adjust discrete limits by half the gap width.

Step-by-Step Solution

1

Identify the formula for calculating pie chart sector angles.

Sector angle =

\frac{\text{Category Frequency}}{\text{Total Frequency}} \times 360^\circ

.

Pie charts distribute

360^\circ

proportionally according to the frequency of each category.

2

Determine histogram bar height representation under uniform class widths.

Bar height corresponds directly to the class frequency.

With uniform widths, the area of each rectangle is directly proportional to its height.

3

Determine histogram bar height representation under varying class widths.

Bar height corresponds to frequency density, calculated as

\frac{\text{Class Frequency}}{\text{Class Width}}

.

To maintain area proportional to frequency across varying widths, height must equal frequency divided by width.

4

Determine upper class boundary for discrete class intervals.

Upper boundary =

b + 0.5

.

Class boundaries eliminate gaps between discrete intervals by extending limits by half of the unit gap.

Key Concept

Data Representation Principles in Charts and Frequency Tables

Question 2Question →

The grouped frequency distribution table below shows the marks scored by $100$ candidates in a Mathematics screening test:

Class Interval	Frequency
$10 - 19$	15
$20 - 29$	25
$30 - 39$	40
$40 - 49$	20

Match each data representation component on the left with its correct calculated numerical value on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The sector angle matches

144^\circ

, frequency density matches

4.0

, preceding upper class boundary matches

29.5

, and median class cumulative frequency matches

80

.

The items match based on direct statistical computations: the modal sector angle is

144^\circ

, frequency density is

4.0

, upper class boundary of the preceding interval is

29.5

, and cumulative frequency at the upper boundary of the median class is

80

.

Step-by-Step Solution

1

Determine the modal class and total frequency

N

.

The highest frequency is

40

, so the modal class is

30 - 39

. Total frequency

N = 15 + 25 + 40 + 20 = 100

.

Modal class identification is essential for pie chart sector, frequency density, and class boundary calculations.

2

Calculate the sector angle for the modal class in a pie chart.

Sector Angle

= \frac{40}{100} \times 360^\circ = 144^\circ

.

The sector angle represents the class frequency as a fraction of total frequency multiplied by

360^\circ

.

3

Calculate the frequency density of the modal class.

Class boundary range for

30 - 39

is

29.5 - 39.5

, so width

= 10

. Frequency density

= \frac{40}{10} = 4.0

.

Frequency density is defined as class frequency divided by class interval width.

4

Determine the preceding upper class boundary and the cumulative frequency for the median class.

Preceding interval is

20 - 29

, so its upper boundary is

29.5

. Median is at position

50

(in interval

30 - 39

). Cumulative frequency up to boundary

39.5

is

15 + 25 + 40 = 80

.

Class boundaries are midpoints between adjacent class limits, and cumulative frequency sums all preceding frequencies up to the upper boundary.

Key Concept

Interpretation and calculation of pie chart sector angles, histogram frequency densities, real class boundaries, and cumulative frequencies from grouped data.

Question 3Question →

Match each statistical data representation term on the left with its corresponding definition or mathematical property on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Class Boundary matches with the value separating adjacent non-overlapping class intervals; Sector Angle matches with the central angle in a pie chart calculated as

\frac{\text{Frequency}}{\text{Total Frequency}} \times 360^\circ

; Frequency Density matches with the quotient of class frequency and class width; Ogive matches with a line graph produced by plotting cumulative frequencies against upper class boundaries.

Each data representation term directly corresponds to its core definition: class boundary closes gaps between discrete class intervals, sector angle measures central circle proportion in a pie chart, frequency density scales histogram height when class widths differ, and an ogive graphs cumulative frequency against upper boundaries.

Step-by-Step Solution

1

Define Class Boundary

Class boundary is the continuous point midway between adjacent class limits.

Class boundaries remove gaps in discrete grouped frequency distributions.

2

Define Sector Angle formula for a pie chart

Sector Angle

= \frac{\text{Frequency}}{\text{Total Frequency}} \times 360^\circ

.

The complete circle represents total frequency, so individual sectors scale proportionally with

360^\circ

.

3

Define Frequency Density for histograms

Frequency Density

= \frac{\text{Frequency}}{\text{Class Width}}

.

Histogram area equals frequency; when widths differ, height must represent frequency per unit width.

4

Define Ogive

An Ogive is a cumulative frequency curve plotted against upper boundaries.

Each point on an ogive shows the cumulative frequency up to that class's upper boundary.

Key Concept

Data Representation Terms and Formulas

Estimated Time:1m 30s

Question 4Question →

A pie chart is used to display the distribution of $720$ candidates registered for a competitive examination across five subjects. If the sector representing Further Mathematics has a central angle of $45^\circ$ , what is the total number of candidates who registered for Further Mathematics?

Show answer & explanation

Answer: 90

Answer

The total number of candidates who registered for Further Mathematics is 90.

To calculate the number of candidates represented by a pie chart sector, multiply the total count by the ratio of the sector's central angle to 360 degrees: (45 / 360) * 720 = 90 candidates.

Step-by-Step Solution

1

Determine the fraction of the total population represented by the Further Mathematics sector.

\frac{45^\circ}{360^\circ} = \frac{1}{8}

A complete pie chart circle corresponds to an angle of 360 degrees.

2

Calculate the actual number of candidates by multiplying the fraction by the total student population.

\frac{1}{8} \times 720 = 90

The number of items in a sector is directly proportional to its central sector angle relative to 360 degrees.

Key Concept

Calculating category frequencies from pie chart sector angles

Question 5Question →

The grouped frequency table below shows the distribution of marks obtained by candidates in a Mathematics examination:

Class Interval	Frequency
10 – 19	15
20 – 29	25
30 – 39	$k$
40 – 49	20
50 �� 59	10

When this data is represented on a pie chart, the sector corresponding to the score range 30 – 39 has a central angle of $108^\circ$ . Based on a cumulative frequency curve (ogive) constructed for this distribution, what is the score corresponding to the 75th percentile ( $Q_3$ ) of the candidates?

42.0

42.5

41.5

44.5

Show answer & explanation

Answer: 42.0

Answer

42.0

The value 42.0 is obtained by first determining the missing frequency

k = 30

using the ratio

\frac{108^\circ}{360^\circ} = 0.3

, establishing

N = 100

. The 75th percentile position is at 75, which falls in the

40 – 49

class. Interpolating from the lower boundary of 39.5 yields

39.5 + \left(\frac{75 - 70}{20}\right) \times 10 = 42.0

.

Step-by-Step Solution

1

Determine the unknown frequency

k

using the pie chart sector angle.

\frac{k}{\text{Total Frequency}} = \frac{108^\circ}{360^\circ} = 0.3

Total frequency

N = 15 + 25 + k + 20 + 10 = 70 + k

.

\frac{k}{70 + k} = 0.3 \implies k = 21 + 0.3k \implies 0.7k = 21 \implies k = 30

The sector angle in a pie chart is proportional to the category frequency relative to the total frequency across

360^\circ

.

2

Calculate cumulative frequencies and locate the 75th percentile position.

Total frequency

N = 100

.
Cumulative frequencies (

cf

):
-

10 – 19

(boundary

9.5 – 19.5

):

cf = 15

-

20 – 29

(boundary

19.5 – 29.5

):

cf = 40

-

30 – 39

(boundary

29.5 – 39.5

):

cf = 70

-

40 – 49

(boundary

39.5 – 49.5

):

cf = 90

-

50 – 59

(boundary

49.5 – 59.5

):

cf = 100

75th percentile position

= 0.75 \times 100 = 75\text{th candidate}

.

The 75th percentile corresponds to the value below which 75% of the total observations lie.

3

Apply linear interpolation on the percentile class interval

40 – 49

.

The 75th score lies in class interval

40 – 49

(boundaries

39.5 – 49.5

).
- Lower class boundary

L = 39.5

- Cumulative frequency prior to class

cf_b = 70

- Frequency of percentile class

f = 20

- Class width

c = 10

Q_3 = L + \left(\frac{0.75N - cf_b}{f}\right) \times c = 39.5 + \left(\frac{75 - 70}{20}\right) \times 10 = 39.5 + 2.5 = 42.0

Linear interpolation along an ogive requires using exact class boundaries to compute specific percentile values.

Key Concept

Pie chart sector angles and linear interpolation on cumulative frequency distributions

Question 6Question →

The table below displays the distribution of scores obtained by $50$ candidates in a Mathematics assessment test:

Score Interval	Frequency ( $f$ )
$10 - 19$	$8$
$20 - 29$	$12$
$30 - 39$	$18$
$40 - 49$	$7$
$50 - 59$	$5$

Match each statistical feature of the charts representing this data on the left with its corresponding numerical value on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The correct pairings are: Sector angle for the modal class matches

129.6^\circ

; Frequency density of the

30 - 39

class interval matches

1.8

; Lower class boundary of the median class matches

29.5

; Cumulative frequency up to

29.5

matches

20

.

Each chart feature correctly aligns with its mathematically derived value: the modal class sector angle is

\frac{18}{50} \times 360^\circ = 129.6^\circ

, the frequency density is

\frac{18}{10} = 1.8

, the median class lower boundary is

29.5

, and the cumulative frequency up to

29.5

is

8 + 12 = 20

.

Step-by-Step Solution

1

Determine the total frequency and locate the modal and median classes.

Total frequency

N = 8 + 12 + 18 + 7 + 5 = 50

. The modal class is

30 - 39

(highest frequency =

18

). The median position is

\frac{50}{2} = 25^{\text{th}}

, which falls within the

30 - 39

class interval since cumulative frequency reaches

38

at the end of this class.

Identifying

N

, the modal class, and the median position is required for calculating chart parameters.

2

Calculate the pie chart sector angle for the modal class (

30 - 39

).

Sector angle

= \frac{\text{Frequency}}{N} \times 360^\circ = \frac{18}{50} \times 360^\circ = 129.6^\circ

.

Pie chart sectors represent relative frequencies scaled to

360^\circ

.

3

Compute the frequency density for the histogram bar of class

30 - 39

.

Class width

= 39.5 - 29.5 = 10

. Frequency density

= \frac{\text{Frequency}}{\text{Class width}} = \frac{18}{10} = 1.8

.

Histogram height represents frequency density, defined as frequency divided by class width.

4

Identify the lower class boundary of the median class (

30 - 39

) and cumulative frequency at upper boundary

29.5

.

Lower boundary of

30 - 39

is

30 - 0.5 = 29.5

. Cumulative frequency up to

29.5

is

8 + 12 = 20

.

Class boundaries eliminate gaps between intervals for continuous plots like ogives and histograms.

Key Concept

Calculating statistical chart parameters (pie chart sector angles, histogram frequency densities, class boundaries, and cumulative frequencies) from grouped frequency distributions.

Question 7Question →

The table below shows the distribution of ages (in years) of trees in a forest reserve, presented alongside their frequency densities for a histogram representation:

Age Interval (years)	Frequency Density
$0 \le x < 10$	$3.0$
$10 \le x < 25$	$2.0$
$25 \le x < 45$	$2.25$
$45 \le x < 80$	$2.0$

If the dataset is instead displayed using a pie chart, what is the sector angle representing the age interval $25 \le x < 45$ ?

90^\circ

70^\circ

60^\circ

80^\circ

Show answer & explanation

Answer: $90^\circ$

Answer

The sector angle representing the age interval

25 \le x < 45

is

90^\circ

.

The frequency of each interval is found by multiplying its frequency density by its class width (

3.0 \times 10 = 30

,

2.0 \times 15 = 30

,

2.25 \times 20 = 45

, and

2.0 \times 35 = 75

). The total frequency is

180

. The sector angle for

25 \le x < 45

(frequency

45

) is

\frac{45}{180} \times 360^\circ = 90^\circ

.

Step-by-Step Solution

1

Calculate the class width for each interval.

Widths are:

10 - 0 = 10

,

25 - 10 = 15

,

45 - 25 = 20

, and

80 - 45 = 35

.

For histograms with unequal class widths, frequency density is defined as frequency divided by class width.

2

Calculate the frequency (

f

) for each interval using

f = \text{Frequency Density} \times \text{Class Width}

.

Interval

0 \le x < 10

:

f_1 = 3.0 \times 10 = 30

.
Interval

10 \le x < 25

:

f_2 = 2.0 \times 15 = 30

.
Interval

25 \le x < 45

:

f_3 = 2.25 \times 20 = 45

.
Interval

45 \le x < 80

:

f_4 = 2.0 \times 35 = 75

.

The frequency of a class in a histogram corresponds to the area of its bar.

3

Compute the total frequency (

N

).

N = 30 + 30 + 45 + 75 = 180

.

The total frequency represents the entire dataset needed for pie chart sector calculations.

4

Calculate the sector angle (

\,\theta\,

) for the interval

25 \le x < 45

.

\theta = \frac{f_3}{N} \times 360^\circ = \frac{45}{180} \times 360^\circ = 90^\circ

.

The sector angle in a pie chart is proportional to the relative frequency of the class out of

360^\circ

.

Key Concept

Conversion between Histogram Frequency Density and Pie Chart Sector Angles

Estimated Time:2m 0s

Question 8Question →

A continuous grouped frequency distribution consists of four class intervals: $10 - 14$ , $15 - 24$ , $25 - 29$ , and $30 - 44$ . The total frequency of the distribution is $160$ , and the frequency of the class interval $25 - 29$ is $22$ .

In a histogram representing this data, the height of the rectangle for the interval $15 - 24$ corresponds to a frequency density of $6$ . In a pie chart representing the same distribution, the sector angle for the class interval $30 - 44$ is $108^\circ$ .

What is the frequency density of the class interval $10 - 14$ ?

Show answer & explanation

Answer: 6

Answer

The frequency density of the class interval

10 - 14

is

6

.

By converting the pie chart sector angle of

108^\circ

into a frequency of

48

out of

160

, and using the frequency density of

6

with class width

10

to find a frequency of

60

for

15 - 24

, the remaining frequency for

10 - 14

is

30

. Dividing this by the true class width of

5

(from boundaries

9.5

to

14.5

) gives a frequency density of

6

.

Step-by-Step Solution

1

Calculate the frequency of the class interval

30 - 44

from the pie chart sector angle.

Frequency

f_{30-44} = \frac{108^\circ}{360^\circ} \times 160 = 48

.

The sector angle in a pie chart is directly proportional to the frequency relative to the

360^\circ

total.

2

Determine the class width and frequency of the class interval

15 - 24

.

Class boundaries are

14.5

and

24.5

, so width

w = 10

. Frequency

f_{15-24} = 6 \times 10 = 60

.

Frequency density is defined as frequency divided by class width, so frequency equals frequency density multiplied by class width.

3

Determine the frequency of the class interval

10 - 14

.

Frequency

f_{10-14} = 160 - (60 + 22 + 48) = 30

.

The sum of all class frequencies must equal the total frequency of

160

.

4

Calculate the class width and frequency density of

10 - 14

.

Class width

w_{10-14} = 14.5 - 9.5 = 5

. Frequency density

= \frac{30}{5} = 6

.

Dividing the frequency of the class (

30

) by its exact class boundary width (

5

) yields the frequency density.

Key Concept

Integration of Frequency Density and Pie Chart Sector Angles

Question 9Question →

A pie chart illustrates the annual allocation of funds for an agricultural research station across four sectors: Crop Research, Livestock, Irrigation, and Equipment Maintenance. The sector angles for Crop Research, Livestock, and Irrigation are $135^\circ$ , $90^\circ$ , and $60^\circ$ respectively. If the total annual budget is $\text{N}7,200,000$ , what is the amount allocated to Equipment Maintenance in Naira?

Show answer & explanation

Answer: 1500000

Answer

The amount allocated to Equipment Maintenance is 1,500,000 Naira.

A complete pie chart has a total angle of

360^\circ

, representing the full amount of

\text{N}7,200,000

. The angle corresponding to Equipment Maintenance is

360^\circ - (135^\circ + 90^\circ + 60^\circ) = 75^\circ

. The dollar/naira allocation is given by

\frac{75^\circ}{360^\circ} \times 7,200,000 = 1,500,000

.

Step-by-Step Solution

1

Calculate the sum of the given sector angles

Sum of known angles = 285 degrees

Determining the total angular measure already accounted for by the three known sectors.

2

Subtract the sum of known angles from 360 degrees

Sector angle for Equipment Maintenance = 75 degrees

The sum of all sector angles in a pie chart is always 360 degrees.

3

Multiply the fraction of the pie chart by the total funds

1,500,000 Naira

Converting the sector angle representation to its corresponding quantitative value.

Key Concept

Pie chart sector angle computation and value conversion

Question 10Question →

A pie chart illustrates the distribution of undergraduate students enrolled across four faculties at a university: Arts, Science, Law, and Medicine. The central angles for the sectors representing Arts, Science, and Law are $120^\circ$ , $90^\circ$ , and $75^\circ$ , respectively. If $300$ students are enrolled in the Faculty of Medicine, what is the total number of students enrolled in the university?

Show answer & explanation

Answer: 1440

Answer

The total number of students enrolled in the university is 1440.

The total sum of central angles in any pie chart is

360^\circ

. Subtracting the given angles for Arts (

120^\circ

), Science (

90^\circ

), and Law (

75^\circ

) from

360^\circ

gives the sector angle for Medicine as

75^\circ

. Since

75^\circ

represents

300

students, each degree represents

\frac{300}{75} = 4

students. Multiplying

4

students per degree by the total

360^\circ

yields

1440

total students in the university.

Step-by-Step Solution

1

Determine the sector angle for Medicine.

Sector angle for Medicine =

75^\circ

The sum of central angles in a pie chart is

360^\circ

. Subtracting

120^\circ + 90^\circ + 75^\circ = 285^\circ

from

360^\circ

gives

75^\circ

.

2

Formulate the proportion relating sector angle to frequency.

\frac{75^\circ}{360^\circ} \times N = 300

, where

N

represents total students.

The fractional portion of the angle (

75^\circ

out of

360^\circ

) equals the fractional portion of the total student count (

300

out of

N

).

3

Calculate the total student population

N

.

N = \frac{300 \times 360}{75} = 1440

Dividing

300

by

75

yields

4

students per degree. Multiplying

4

by

360

gives

1440

students.

Key Concept

Pie Chart Sector Angle and Total Population Calculation

Question 11Question →

Match each statistical chart term or parameter in Column A with its corresponding definition or formula in Column B.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Class boundary matches the exact continuous limit of a class interval; Frequency density matches the quotient of class frequency and class width; Cumulative frequency matches the running total of frequencies plotted on an ogive; Pie chart sector angle matches the central angle formula involving multiplication by 360 degrees.

Each chart parameter in Column A directly corresponds to its standard mathematical definition, formula, or geometric representation in Column B.

Step-by-Step Solution

1

Identify the statistical definition for continuous class intervals in histograms.

Class boundaries remove gaps between non-overlapping class limits.

Histograms require continuous real class boundaries on the horizontal axis.

2

Recall the formula for histogram bar height when class intervals vary.

Frequency density =

\frac{\text{Frequency}}{\text{Class Width}}

.

Bar area must remain proportional to frequency.

3

Determine the parameter used to plot an ogive curve.

Cumulative frequency tracks accumulated totals across upper class boundaries.

An ogive represents cumulative distribution.

4

Identify the angular calculation for circular charts.

Sector angle =

\frac{\text{Class Frequency}}{\text{Total Frequency}} \times 360^\circ

.

A complete pie chart represents 360 degrees.

Key Concept

Data Representation and Chart Properties

Question 12Question →

The frequency distribution table below displays the examination marks of a group of candidates:

Mark Class	Frequency ( $f$ )
$10 - 19$	$5$
$20 - 29$	$9$
$30 - 39$	$16$
$40 - 49$	$12$
$50 - 59$	$8$

If an ogive (cumulative frequency curve) is constructed to represent this dataset, which of the following represents the correct coordinate pair for the point corresponding to the modal class?

(39.5, 30)

(29.5, 30)

(34.5, 30)

(39.5, 16)

Show answer & explanation

Answer: $(39.5, 30)$

Answer

The correct coordinate pair is

(39.5, 30)

.

The modal class is

30 - 39

because it has the highest frequency of

16

. The upper class boundary for this interval is

39.5

. Summing frequencies up to this class gives

5 + 9 + 16 = 30

. Therefore, the plotted point on the ogive must be

(39.5, 30)

.

Step-by-Step Solution

1

Identify the modal class interval from the frequency table.

The class with the highest frequency (

16

) is

30 - 39

.

The modal class is defined as the interval with the maximum frequency.

2

Determine the upper class boundary of the modal class.

Upper class boundary

= 39 + 0.5 = 39.5

.

Cumulative frequency (ogive) curves are plotted using the upper class boundaries on the horizontal axis.

3

Calculate the cumulative frequency up to and including the modal class.

Cumulative frequency

= 5 + 9 + 16 = 30

.

An ogive plots accumulated frequencies up to each boundary.

4

Form the coordinate pair

(x, y) = (\text{Upper Class Boundary}, \text{Cumulative Frequency})

.

(39.5, 30)

.

Points on an ogive take the form

(\text{Upper Boundary}, \text{Cumulative Frequency})

.

Key Concept

Plotting Cumulative Frequency Curves (Ogives)

Data Representation and Charts

Answer

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