ML Aggarwal Measures of Central Tendency Exe-21.6 Class 10 ICSE Maths Solutions

ML Aggarwal Measures of Central Tendency Exe-21.6 Class 10 ICSE Maths Solutions. We Provide Step by Step Answer of Exe-21.6 Questions for Measures of Central Tendency as council prescribe guideline for upcoming board exam. Visit official Website CISCE for detail information about ICSE Board Class-10.

ML Aggarwal Measures of Central Tendency Exe-21.6 Class 10 ICSE Maths Solutions

Board	ICSE
Subject	Maths
Class	10th
Chapter-21	Measures of Central Tendency
Writer / Book	Understanding
Topics	Solutions of Exe-21.6
Academic Session	2024-2025

Measures of Central Tendency Exe-21.6

ML Aggarwal Class 10 ICSE Maths Solutions

Question 1 : Using the data given below construct the cumulative frequency table and draw the-Ogive. From the ogive determine the median.

Answer :

Representing the given data in cumulative frequency distributions :

Taking points (10, 3), (20, 11), (30, 23), (40, 37), (50,47), (60,53), (70, 58) and (80, 60) on the graph.
Now join them in a free hand to form an ogive as shown.
Here n = 60 which is even
Median = $\frac { 1 }{ 2 } \left[ \frac { 60 }{ 2 } th\quad term+\left( \frac { 60 }{ 2 } +1 \right) th\quad term \right]$
= $\frac { 1 }{ 2 } \left[ \frac { 60 }{ 2 } +\left( \frac { 60 }{ 2 } +1 \right) th\quad term \right]$
= $\frac { 1 }{ 2 } \left[ 30th\quad term+31th\quad term \right]$
hence = 30.5 observation
Now take a point A (30.5) on y-axis and from A,
draw a line parallel to x-axis meeting the curve at P
and from P, draw a perpendicular to x-axis meeting is at Q.
∴ Q is the median which is 35.

Question 2. The weight of 50 workers is given below:

Draw an ogive of the given distribution using a graph sheet. Take 2 cm = 10 kg on one axis , and 2 cm = 5 workers along the other axis. Use a graph to estimate the following:
(i) the upper and lower quartiles.
(ii) if weighing 95 kg and above is considered overweight find the number of workers who are overweight. (2015)

Answer :

The cumulative frequency table of the given distribution table is as follows:

The ogive is as follows:

Plot the points (50, 0), (60, 4), (70, 11), (80, 22), (90, 36), (100, 42),
(110, 47), (120, 50) Join these points by using freehand drawing.
The required ogive is drawn on the graph paper.
Here n = number of workers = 50

(i) To find upper quartile:

Let A be the point on y-axis representing a frequency
$\frac { 3n }{ 4 } =\frac { 3\times 50 }{ 4 } =37.5$
Through A, draw a horizontal line to meet the ogive at B.
Through B draw a vertical line to meet the x-axis at C.
The abscissa of the point C represents 92.5 kg.
The upper quartile = 92.5 kg To find the lower quartile:
Let D be the point on y-axis representing frequency = $\frac { n }{ 4 } =\frac { 50 }{ 4 } =12.5$
Through D, draw a horizontal line to meet the ogive at E.
Through E draw a vertical line to meet the x-axis at F.
The abscissa of the point F represents 72 kg.
∴ The lower quartile = 72 kg

(ii) On the graph point, G represents 95 kg.

Through G draw a vertical line to meet the ogive at H.
Through H, draw a horizontal line to meet y-axis at 1.
The ordinate of point 1 represents 40 workers on the y-axis .
∴The number of workers who are 95 kg and above
= Total number of workers – number of workers of weight less than 95 kg
= 50 – 40 = 10

Question 3. The table shows the distribution of scores obtained by 160 shooters in a shooting competition. Use a graph sheet and draw an ogive for the distribution.
(Take 2 cm = 10 scores on the x-axis and 2 cm = 20 shooters on the y-axis)

Use your graph to estimate the following:

(i) The median.
(ii) The interquartile range.
(iii) The number of shooters who obtained a score of more than 85%.

Answer :

Plot the points (10, 9), (20, 22), (30, 42), (40, 68), (50, 98),
(60, 120), (70, 135), (80, 145), (90, 153), (100, 160)
on the graph and join them with free hand to get an ogive as shown:

Here n = 160
$\frac { n }{ 2 } =\frac { 160 }{ 2 } =80$

(i)Median :

Take a point 80 on 7-axis and through it,
draw a line parallel to x-axis-which meets the curve at A.
Through A, draw a perpendicular on x-axis which meet it at B.
B Is median which is 44.

(ii) Interquartile range (Q1)

$\frac { n }{ 4 } =\frac { 160 }{ 4 } =40$
From a point 40 ony-axis, draw a line parallel to x-axis
which meet the curve at C and from C draw a line perpendicular to it
which meet it at D. which is 31.
The interquartile range is 31.

(iii) Number of shooter who get move than 85%.

Scores : From 85 on x-axis, draw a perpendicular to it meeting the curve at P.
From P, draw a line parallel to x-axis meeting y-axis at Q.
Q is the required point which is 89.
Number of shooter getting more than 85% scores = 160 – 149 = 11.

Measures of Central Tendency Exe-21.6

ML Aggarwal Class 10 ICSE Maths Solutions

Page 518

Question 4. The daily wages of 80 workers in a project are given below:

Use a graph paper to draw an ogive for the above distribution. ( a scale of 2 cm = Rs 50 on x- axis and 2 cm = 10 workers on y-axis). your ogive to estimate:

(i) the median wage of the workers.
(ii) the lower quartile wage of the workers.
(iii) the number of workers who earn more than Rs 625 daily. (2017)

Answer :

Number of workers = 80
(i) Median = ${ \left( \frac { n }{ 2 } \right) }^{ th }$ term = 40 th term
Through mark 40 on the y-axis, draw a horizontal line
which meets the curve at point A.

Through point A, on the curve draw a vertical line which meets the x-axis at point B.
The value of point B on the x-axis is the median, which is 604.
(ii) Lower Quartile (Q1) = ${ \left( \frac { 80 }{ 4 } \right) }^{ th }$ term = 20 th term = 550
(iii) Through mark 625 on x-axis, draw a vertical line which meets the graph at point C.
Then through point C, draw a horizontal line which meets the y-axis at the mark of 50.
Thus, number of workers that earn more Rs 625 daily = 80 – 50 = 30

Question 5. Marks obtained by 200 students in an examination are given below :

Draw an ogive for the given distribution taking 2 cm = 10 marks on one axis and 2 cm = 20 students on the other axis. Using the graph, determine

(i) The median marks.
(ii) The number of students who failed if minimum marks required to pass is 40.
(iii) If scoring 85 and more marks is considered as grade one, find the number of students who secured grade one in the examination.

Answer :

(i) Median is 57.
(ii) 44 students failed.
(iii) No. of students who secured grade one = 200 – 188 = 12.

Question 6. The monthly income of a group of 320 employees in a company is given below :

Draw an ogive of the given distribution on a graph sheet taking 2 cm = Rs. 1000 on one axis and 2 cm = 50 employees on the other axis. From the graph determine

(i) the median wage.
(ii) the number of employees whose income is below Rs. 8500.
(iii) If the salary of a senior employee is above Rs. 11500, find the number of senior employees in the company.
(iv) the upper quartile.

Answer :

Now plot the points (7000,20), (8000,65), (9000,130),
(10000,225), (11000,285), (12000,315) and(13000, 320)
on the graph and join them in order with a free hand
to get an ogive as shown in the figure

(i) Total number of employees = 320

$\frac { N }{ 2 } =\frac { 320 }{ 2 } =160$
From 160 on the y-axis, draw a line parallel to x-axis meeting the curve at P.
From P. draw a perpendicular on x-axis meeting it at M, M is the median which is 9300

(ii)

From 8500 on the x-axis, draw a perpendicular which meets the curve at Q.
From Q, draw a line parallel to x-axis meeting y-axis at N. Which is 98

(iii)

From 11500 on the x-axis, draw a line perpendicular to x-axis meeting the curve at R.
From R, draw a line parallel to x-axis meeting y-axis at L. Which is 300
No. of employees getting more than Rs. 11500 = 320

(iv) Upper quartile (Q1)

$\frac { 3N }{ 4 } =\frac { 320\times 3 }{ 4 } =240$
From 240 on y-axis, draw a line perpendicular on the x-axis which meets the curve at S.
From S, draw a perpendicular on x-axis meeting it at T. Which is 10250.
Hence Q3 = 10250

Question 7. Use Graph paper for this question.

A survey regarding height (in cm) of 60 boys belonging to Class 10 of a school was conducted. The following data was recorded :

Taking 2 cm = height of 10 cm along one axis and 2 cm = 10 boys along the other axis draw and give of the above distribution. Use the graph to estimate the following:
(i) the median
(ii) lower Quartile
(iii) if above 158 cm is considered as the tall boys of the class. Find the number of boys in the class who are tall.

Answer :

Given data was recorded as :

Plot the points (140,4), (145,12), (150,32), (155,46), (160,53), (165,59) and (170,60). Join them free hand to get the required ogive.
Now, from the graph, we obtain :

(i) Median height (in cm) = 149.5 cm
(ii) Lower Quartile = 146 cm ‘
(iii) Number of boys who are tall e., height above 158 cm = 60 – 51 = 9.

Measures of Central Tendency Exe-21.6

ML Aggarwal Class 10 ICSE Maths Solutions

Page 519

Question 8. The marks obtained by 100 students in a Mathematics test are given below :

Draw an ogive on a graph sheet and from it determine the :

(i) median
(ii) lower quartile
(iii) number of students who obtained more than 85% marks in the test.
(iv) number of students who did not pass in the test if the pass percentage was 35. We represent the given data in cumulative frequency table as given below :

Answer :

Plot the points (10, 3), (20, 10), (30, 22), (40, 39), (50, 62), (60, 76), (70, 85), (80, 91), (90, 96) and (100, 100) on the graph.

Join the points with the free hand. We get an ogive as shown:

(i) Here n = 100

Median = n/2

= 50^thterm

Mark a point A (50) on Y axis. Draw a line parallel to X axis from A.

Let it meet the curve at B. From B draw a perpendicular which meets X axis at C.

The point C is 45.

(ii) Lower quartile = n/4

= 100/4

= 25^thterm

Mark a point P (25) on Y axis. Draw a line parallel to X axis from that point.

Let it meet the curve at Q. From that point draw a perpendicular which meets X axis at R.

The point R is 32.

(iii) no. of students who obtained more than 85% = 100 – 94 = 6 [from graph]

(iv) No of students who failed if 35% is the pass percentage = 25 [from graph]

Question 9. The marks obtained by 120 students in a Mathematics test are-given below:

Draw an ogive for the given distribution on a graph sheet. Use a suitable scale for ogive to estimate the following:

(i) the median
(ii) the lower quartile
{iii) the number of students who obtained more than 75% marks in the test.
(iv) the number of students who did not pass in the test if the pass percentage was 40. (2002)

Answer :

We represent the given data in cumulative frequency table as given below :

Now we plot the points (10, 5), (20, 14), (30, 30), (40, 52), (50, 78), (60, 96), (70, 107),
(80, 113), (90, 117) and (100, 120) on the graph
and join the points in a free hand to form an ogive as shown.
Here n = 120 which is an even number

(i)

Median = $\frac { 1 }{ 2 } \left[ \frac { 120 }{ 2 } +\left( \frac { 120 }{ 2 } +1 \right) \right]$ = $\frac { 1 }{ 2 } \left( 60+61 \right)$ = 60.5
Now take a point A (60.5) on y-axis and from A
draw a line parallel to x- axis meeting the curve in P
and from P, draw a perpendicular to x-axis meeting it at Q.
∴ Q is the median which is 43.00 (approx.)

(ii)

Lower quartile = $\frac { n }{ 4 } =\frac { 120 }{ 4 } =30$
Now take a point B (30) on y-axis and from B,
draw a line parallel to x-axis meeting the curve in L
and from L draw a perpendicular to x-axis meeting it at M.
M is the lower quartile which is 30.

(iii)

Take a point C (75) on the x-axis
and from C draw a line perpendicular to it meeting the curve at R.
From R, draw a line parallel to x-axis meeting y-axis at S.
∴S shows 110 students getting below 75%
and 120 – 110 = 10 students getting more than 75% marks.

(iv)

Pass percentage is 40%
Now take a point D (40) on x-axis and from D
draw a line perpendicular to x-axis meeting the curve at E
and from E, draw a line parallel to x-axis meeting the y-axis at F.
∴ F shows 52
∴ No of students who could not get 40% and failed in the examination are 52.

Question 10. The following distribution represents the height of 160 students of a school.

Draw an ogive for the given distribution taking 2 cm = 5 cm of height on one axis and 2 cm = 20 students on the other axis. Using the graph, determine :

(i)The median height.
(ii)The inter quartile range.
(iii) The number of students whose height is above 172 cm.

Answer :

The cumulative frequency table may be prepared as follows:

Plot the points (145, 12), (150, 32), (155, 62), (160, 100), (165, 124), (170, 140), (175, 152), and (180, 160) on the graph.

Join the points with the free hand. We get an ogive as shown:

Now, we take height along x-axis and number of students along the y-axis.
Now, plot the point (140, 0), (145, 12), (150, 32), (155, 62), (160, 100), (165, 124),
(170, 140), (175, 152) and (180, 160). Join these points by a free hand curve to get the ogive.

(i)

Here N = 160 => $\\ \frac { N }{ 2 }$ = 80
On the graph paper take a point A on the y- axis representing 80.
A draw horizontal line meeting the ogive at B.
From B, draw BC ⊥ x-axis, meeting the x-axis at C. The abscissa of C is 157.5
So, median = 157.5 cm

(ii)

Proceeding in the same way as we have done in above,
we have, Q1 = 152 and Q3 = 164 So, interquartile range = Q3 – Q1 = 164 – 152 = 12 cm

(iii)

From the ogive, we see that the number of students whose height is less than 172 is 145.
No. of students whose height is above 172 cm = 160 – 145 = 15

(ML Aggarwal Measures of Central Tendency Exe-21.6 Class 10 ICSE)

Question 11. 100 pupils in a school have heights as tabulated below :

Draw the ogive for the above data and from it determine the median (use graph paper).

Answer :

Representing the given data in cumulative frequency table (in continuous distribution):

∴ Here n = 100 which is an even number
∴ Median = $\frac { 1 }{ 2 } \left[ \frac { n }{ 2 } +\left( \frac { n }{ 2 } +1 \right) \right] =\frac { 1 }{ 2 } \left[ \frac { 100 }{ 2 } +\left( \frac { 100 }{ 2 } +1 \right) \right] =\frac { 1 }{ 2 } (50+51)=\frac { 101 }{ 2 } =50.5$
Now plot points (130.5, 12), (140.5, 28), (150.5, 58), (160.5, 78), (170.5, 92)
and (180.5, 100) on the graph and join them in free hand to form an ogive as shown.
Now take a point A (50-5) on y-axis and from A
draw a line parallel to x-axis meeting the curve at P
and from P, draw a line perpendicular to x-axis meeting it at Q
.’. Q (147.5) is the median

–: End of ICSE Measures of Central Tendency Exe-21.6 Class 10 ML Aggarwal Maths Solutions :–

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