## ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 21 Measures of Central Tendency Chapter Test

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 21 Measures of Central Tendency Chapter Test

More Exercises

Question 1.
Arun scored 36 marks in English, 44 marks in Civics, 75 marks in Mathematics and x marks in Science. If he has scored an average of 50 marks, find x.
Solution:
Marks in English = 36
Marks in Civics = 44
Marks in Mathematics = 75
Marks in Science = x
Total marks in 4 subjects = 36 + 44 + 75 + x = 155 + x
average marks = $$\\ \frac { 155+x }{ 4 }$$
But average marks = 50 (given)
$$\\ \frac { 155+x }{ 4 }$$ = 50
⇒ 155 + x = 200
⇒ x = 200 – 155 = 45

Question 2.
The mean of 20 numbers is 18. If 3 is added to each of the first ten numbers, find the mean of new set of 20 numbers.
Solution:
Mean of 20 numbers =18
Total number = 18 × 20 = 360
By adding 3 to first 10 numbers,
The new sum will be = 360 + 3 × 10 = 360 + 30 = 390
New Mean = $$\\ \frac { 390 }{ 20 }$$ = 19.5

Question 3.
The average height of 30 students is 150 cm. It was detected later that one value of 165 cm was wrongly copied as 135 cm for computation of mean. Find the correct mean.
Solution:
In first case,
Average height of 30 students = 150 cm
Total height = 150 × 30 = 4500 cm
Difference in copying the number = 165 – 135 = 30 cm
Correct sum = 4500 + 30 = 4530 cm
Correct mean = $$\\ \frac { 4530 }{ 30 }$$ = 151 cm

Question 4.
There are 50 students in a class of which 40 are boys and the rest girls. The average weight of the students in the class is 44 kg and average weight of the girls is 40 kg. Find the average weight of boys.
Solution:
Total students of a class = 50
No. of boys = 40
No. of girls = 50 – 40 = 10
Average weight of 50 students = 44 kg
Total weight = 44 × 50 = 2200 kg
Average weight of 10 girls = 40 kg
.’. Total weight of girls = 40 × 10 = 400 kg
Then the total weight of 40 boys = 2200 – 400 = 1800kg
Average weight of boys = $$\\ \frac { 1800 }{ 40 }$$ = 45kg

Question 5.
The contents of 50 boxes of matches were counted giving the following results

Calculate the mean number of matches per box.
Solution:

Question 6.
The heights of 50 children were measured (correct to the nearest cm) giving the following results :

Solution:
Calculate the mean height for this distribution correct to one place of decimal.

Mean = $$\frac { \sum { fx } }{ \sum { f } } =\frac { 3459 }{ 50 }$$ = 69.18 = 69.2

Question 7.
Find the value of p for the following distribution whose mean is 20.6 :

Solution:

Question 8.
Find the value of p if the mean of the following distribution is 18.

Solution:

Question 9.
Find the mean age in years from the frequency distribution given below:

Solution:
Arranging the classes in proper form

Question 10.
Calculate the Arithmetic mean, correct to one decimal place, for the following frequency distribution :

Solution:

Question 11.
The mean of the following frequency distribution is 62.8. Find the value of p.

Solution:
Mean = 62.8

Hence p = 10

Question 12.
The daily expenditure of 100 families are given below. Calculate f1, and f2, if the mean daily expenditure is Rs 188.

Solution:
Mean = 188,
No. of families = 100

Question 13.
The measures of the diameter of the heads of 150 screw is given in the following table. If the mean diameter of the heads of the screws is 51.2 mm, find the values of p and q

Solution:
Mean = 51.2
No. of screws = 150

Question 14.
The median of the following numbers, arranged in ascending order is 25. Find x, 11, 13, 15, 19, x + 2, x + 4, 30, 35, 39, 46
Solution:
Here, n = 10, which is even

Question 15.
If the median of 5, 9, 11, 3, 4, x, 8 is 6, find the value of x.
Solution:
Arranging in ascending order, 3, 4, 5, x, 8, 9, 11,
Here n = 7 which is odd.
∴ Median = $$\\ \frac { n+1 }{ 2 }$$ th term = $$\\ \frac { 7+1 }{ 2 }$$ = 4th term = x
∴ but median = 6
∴ x = 6

Question 16.
Find the median of: 17, 26, 60, 45, 33, 32, 29, 34, 56 If 26 is replaced by 62, find the new median.
Solution:
Arranging the given data in ascending order
17, 26, 29, 32, 33, 34, 45, 56, 60
Here n = 9 which is odd
∴Median = $$\\ \frac { n+1 }{ 2 }$$ th term = $$\\ \frac { 9+1 }{ 2 }$$ = $$\\ \frac { 10 }{ 2 }$$ = 5th term = 33
(ii) If 26 is replaced by 62, their the order will be
17, 29, 32, 33, 34, 45, 56, 60, 62
Here 5th term is 34
∴ Median = 34

Question 17.
The marks scored by 16 students in a class test are : 3, 6, 8, 13, 15, 5, 21, 23, 17, 10, 9, 1, 20, 21, 18, 12
Find
(i) the median
(ii) lower quartile
(iii) upper quartile
Solution:
Arranging the given data in ascending order:
1, 3, 5, 6, 8, 9, 10, 12, 13, 15, 17, 18, 20, 21, 21, 23
Here n = 16 which is even.

Question 18.
Find the median and mode for the set of numbers : 2, 2, 3, 5, 5, 5, 6, 8, 9
Solution:
Here n = 9 which is odd.
∴Median = $$\\ \frac { n+1 }{ 2 }$$ th term = $$\\ \frac { 9+1 }{ 2 }$$ = $$\\ \frac { 10 }{ 2 }$$ = 5th term = 5
Here 5 occur maximum times
∴Mode = 5

Question 19.
Calculate the mean, the median and the mode of the following distribution :

Solution:

Question 20.
The daily wages of 30 employees in an establishment are distributed as follows :

Estimate the modal daily wages for this distribution by a graphical method.
Solution:

Taking daily wages on x-axis and No. of employees on the y-axis
and draw a histogram as shown. Join AB and CD intersecting each other at M.
From M draw ML perpendicular to x-axis, L is the mode
∴ Mode = Rs 23

Question 21.
Using the data given below, construct the cumulative frequency table and draw the ogive. From the ogive, estimate ;
(i) the median
(ii) the inter quartile range.

Also state the median class
Solution:

Question 22.
Draw a cumulative frequency curve for the following data :

Hence determine:
(i) the median
(ii) the pass marks if 85% of the students pass.
(iii) the marks which 45% of the students exceed.
Solution:

We hope the ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 21 Measures of Central Tendency Chapter Test help you. If you have any query regarding ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 21 Measures of Central Tendency Chapter Test, drop a comment below and we will get back to you at the earliest.

## ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 21 Measures of Central Tendency MCQS

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 21 Measures of Central Tendency MCQS

More Exercises

Choose the correct answer from the given four options (1 to 16):

Question 1.
If the classes of a frequency distribution are 1-10, 11-20, 21-30, …, 51-60, then the size of each class is
(a) 9
(b) 10
(c) 11
(d) 5.5
Solution:
In the classes 1-10, 11-20, 21-30, …, 51-60,
the size of each class is 10. (b)

Question 2.
If the classes of a frequency distribution are 1-10, 11-20, 21-30,…, 61-70, then the upper limit of the class 11-20 is
(a) 20
(b) 21
(c) 19.5
(d) 20.5
Solution:
In the classes of distribution, 1-10, 11-20, 21-30, …, 61-70,
upper limit of 11-20 is 20-5 as the classes after adjustment are
0.5-10.5, 10.5-20.5, 20.5-30.5, … (d)

Question 3.
If the class marks of a continuous frequency distribution are 22, 30, 38, 46, 54, 62, then the class corresponding to the class mark 46 is
(a) 41.5-49.5
(b) 42-50
(c) 41-49
(d) 41-50
Solution:
The class marks of distribution are 22, 30, 38, 46, 54, 62,
then classes corresponding to these class marks 46 is
46.4 – 4 = 42, 46 + 4 = 50
(Class intervals is 8 as 30 – 22 = 8, 38 – 30 = 8
i.e:, 42 – 50 (b)

Question 4.
If the mean of the following distribution is 2.6,

then the value of P is
(a) 2
(b) 3
(c) 2.6
(d) 2.8
Solution:
Mean = 2.6

Question 5.
The measure of central tendency of statistical data which takes into account all the data is
(a) mean
(b) median
(c) mode
(d) range
Solution:
A measure of central tendency of statistical data is mean. (a)

Question 6.
In a grouped frequency distribution, the mid-values of the classes are used to measure which of the following central tendency?
(a) median
(b) mode
(c) mean
(d) all of these
Solution:
In a grouped frequency distribution,
the mid-values of the classes are used to measure Mean (c)

Question 7.
In the formula: $$\overline { x } =a+\frac { \sum { { f }_{ i }{ d }_{ i } } }{ \sum { { f }_{ i } } }$$ for finding the mean of the grouped data, d’is are deviations from a (assumed mean) of
(a) lower limits of the classes
(b) upper limits of the classes
(c) mid-points of the classes
(d) frequencies of the classes
Solution:
The formula $$\overline { x } =a+\frac { \sum { { f }_{ i }{ d }_{ i } } }{ \sum { { f }_{ i } } }$$ is the finding of mean of the grouped data, d’is are mid-points of the classes

Question 8.
In the formula: $$\overline { x } =a+c\left( \frac { \sum { { f }_{ i }{ u }_{ i } } }{ \sum { { f }_{ i } } } \right)$$, for finding the mean of grouped frequency distribution, ui =
(a) $$\frac { { y }_{ i }+a }{ c }$$
(b) $$c({ y }_{ i }-a)$$
(c) $$\frac { { y }_{ i }-a }{ c }$$
(d) $$\frac { a-{ y }_{ i } }{ c }$$
Solution:
In $$\overline { x } =a+c\left( \frac { \sum { { f }_{ i }{ u }_{ i } } }{ \sum { { f }_{ i } } } \right)$$,
for finding the mean of grouped frequency, ui is $$\frac { { y }_{ i }-a }{ c }$$(c)

Question 9.
While computing mean of grouped data, we assumed that the frequencies are
(a) evenly distributed over all the classes
(b) centred at the class marks of the classes
(c) centred at the upper limits of the classes
(d) centred at the lower limits of the classes
Solution:
For computing mean of grouped data,
we assumed that frequencies are centred at class marks of the classes. (b)

Question 10.
Construction of a cumulative frequency distribution table is useful in determining the
(a) mean
(b) median
(c) mode
(d) all the three measures
Solution:
Construction of a cumulative frequency distribution table
is used for determining the median, (b)

Question 11.
The times, in seconds, taken by 150 athletes to run a 110 m hurdle race are tabulated below:

The number of athletes who completed the race in less than 14.6 seconds is
(a) 11
(b) 71
(c) 82
(d) 130
Solution:
Time taken in seconds by 150 athletes to run a 110 m hurdle race as given in the sum,
the number of athletes who completed the race in less then 14.6 second is
2 + 4 + 5 + 71 = 82 athletes. (c)

Question 12.
Consider the following frequency distribution:

The upper limit of the median class is
(a) 17
(b) 17.5
(c) 18
(d) 18.5
Solution:
From the given frequency upper limit of median class is 17.5
as total frequencies 13 + 10 + 15 + 8 + 11 = 57
$$\\ \frac { 57+1 }{ 2 }$$ = $$\\ \frac { 58 }{ 2 }$$ = 29
and 13 + 10 + 15 = 28 where class is 12-17
But actual class will be 11.5-17.5
Upper limit is 17.5 (b)

Question 13.
Daily wages of a factory workers are recorded as:

The lower limit of the modal class is
(a) Rs 137
(b) Rs 143
(c) Rs 136.5
(d) Rs 142.5
Solution:
In the daily wages of workers of a factory are 131-136, 137-142, 142-148, …
which are not a proper class
So, proper class will be 130.5-136.5, 136.5-142.5, 142.5-148.5, …
Lower limit of a model class is 136.5 as 136.5-142.5 is the modal class. (c)

Question 14.
For the following distribution:

The sum of lower limits of the median class and modal class is
(a) 15
(b) 25
(c) 30
(d) 35
Solution:
From the given distribution
Sum of frequencies = 10 + 15 + 12 + 20 + 9 = 66
and median is $$\\ \frac { 66 }{ 2 }$$ = 33
Median class will be 10-15 and modal class is 15-20
Sum of lower limits = 10 + 15 = 25 (b)

Question 15.
Consider the following data:

The difference of the upper limit of the median class and the lower limit of the modal class is
(a) 0
(b) 19
(c) 20
(d) 38
Solution:
From the given data
Total frequencies = 4 + 5 + 13 + 20 + 14 + 7 + 4 = 67
Median class $$\\ \frac { 67+1 }{ 2 }$$ = 34
which is (4 + 5 + 13 + 20) 125-145 and modal class is 125-145
Difference of upper limit of median class and the lower limit of the modal class
= 145 – 125 = 20 (c)

Question 16.
An ogive curve is used to determine
(a) range
(b) mean
(c) mode
(d) median
Solution:
An ogive curve is used to find median. (d)

We hope the ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 21 Measures of Central Tendency MCQS help you. If you have any query regarding ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 21 Measures of Central Tendency MCQS, drop a comment below and we will get back to you at the earliest.

## ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 21 Measures of Central Tendency Ex 21.6

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 21 Measures of Central Tendency Ex 21.6

More Exercises

Question 1.
The following table shows the distribution of the heights of a group of a factory workers.

(i) Determine the cumulative frequencies.
(ii) Draw the cumulative frequency curve on a graph paper.
Use 2 cm = 5 cm height on one axis and 2 cm = 10 workers on the other.
(iii) From your graph, write down the median height in cm.
Solution:
Representing the distribution in cumulative
frequency distribution :

Here, n = 83 which is even.
Now taking points (155, 6), (160, 18), (165, 36), (170, 56),
(175, 69), (180, 77) and (185, 83) on the graph.

Now join them with free hand to form the ogive
or cumulative frequency curve as shown.
Here n = 83 which is odd
Median = $$\\ \frac { n+1 }{ 2 }$$ th observation
= $$\\ \frac { 83+1 }{ 2 }$$ = 42th observation
Take a point A (42) on y-axis and from A,
draw a horizontal line parallel to x-axis meeting the curve at P.
From P draw a line perpendicular on the x-axis which meets it at Q.
∴Q is the median which is 166.5 cm. Ans.

Question 2.
Using the data given below construct the cumulative frequency table and draw the-Ogive. From the ogive determine the median.

Solution:
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 = $$\frac { 1 }{ 2 } \left[ \frac { 60 }{ 2 } th\quad term+\left( \frac { 60 }{ 2 } +1 \right) th\quad term \right]$$
= $$\frac { 1 }{ 2 } \left[ \frac { 60 }{ 2 } +\left( \frac { 60 }{ 2 } +1 \right) th\quad term \right]$$
= $$\frac { 1 }{ 2 } \left[ 30th\quad term+31th\quad term \right]$$
= 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 3.
Use graph paper for this question.
The following table shows the weights in gm of a sample of 100 potatoes taken from a large consignment:

(i) Calculate the cumulative frequencies.
(ii) Draw the cumulative frequency curve and from it determine the median weight of the potatoes. (1996)
Solution:
Representing the given data in cumulative frequency table :

Now plot the points (60, 8), (70, 18), (80, 30), (90, 46), (100, 64),
(110, 78), (120, 90), (130, 100) on the graph and join them
in a free hand to form an ogive as shown

Here n =100 which is even.
Median = $$\frac { 1 }{ 2 } \left[ \frac { n }{ 2 } th\quad term+\left( \frac { n }{ 2 } +1 \right) th\quad term \right]$$
= $$\frac { 1 }{ 2 } \left[ \frac { 100 }{ 2 } +\left( \frac { 100 }{ 2 } +1 \right) th\quad term \right]$$
= $$\frac { 1 }{ 2 } \left[ 50th\quad term+51th\quad term \right]$$
= 50.5
Now take a point A (50.5) on they-axis and from A
draw a line parallel to x-axis meeting the curve at R
From P, draw a perpendicular on x-axis meeting it at Q.
Q is the median which is = 93 gm.

Question 4.
Attempt this question on graph paper.

(i) Construct the ‘less than’ cumulative frequency curve for the above data, using 2 cm = 10 years, on one axis and 2 cm = 10 casualties on the other.
(ii) From your graph determine (1) the median and (2) the upper quartile
Solution:
Representing the given data in less than cumulative frequency.

Now plot the points (15, 6), (25, 16), (35, 31), (45, 44), (55, 68), (65,76)
and (75, 83) on the graph and join these points in free hand
to form a cumulative frequency curve (ogive) as shown.
Here n = 83, which is odd.

(i) Median = $$\frac { n+1 }{ 2 } =\frac { 83+1 }{ 2 } +\frac { 84 }{ 2 } =42$$
Now we take point A (42) 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 it at Q.
Q is the median which is = 43
(ii) Upper quartile = $$\frac { 3(n+1) }{ 4 } =\frac { 3\times (83+1) }{ 4 } =\frac { 252 }{ 4 } =63$$
Take a point B 63 on y-axis and from B,
draw a parallel line to x-axis meeting the curve at L.
From L, draw a perpendicular to x-axis meeting it at M which is 52.
∴ Upper quartile = 52 years

Question 5.
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)
Solution:
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 6.
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%.
Solution:

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

Question 7.
The daily wages of 80 workers in a project are given below

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

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 8.
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.
Solution:

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

Question 9.
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.
Solution:

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 10.
Using a graph paper, draw an ogive for the following distribution which shows a record of the weight in kilograms of 200 students

Use your ogive to estimate the following:
(i) The percentage of students weighing 55 kg or more.
(ii) The weight above which the heaviest 30% of the students fall,
(iii) The number of students who are :
1. under-weight and
2. over-weight, if 55.70 kg is considered as standard weight.
Solution:

Plot the points (45, 5), (50, 22), (55, 44), (60, 89), (65, 140),
(70, 171), (75, 191) and (80, 200) on the graph
and join them in free hand to get an ogive as shown From the graph,
number of students weighing 55 kg or more = 200 – 44 = 156
Percentage = $$\frac { 156 }{ 200 } \times 100$$ = 78%
(ii) 30% of 200 = $$\frac { 200\times 30 }{ 100 }$$ = 60
∴ Heaviest 60 students in weight = 9 + 20 + 31 = 60
(From the graph, the required weight is 65 kg or more but less than 80 kg)
(iii) Total number of students who are
1. under weight = 47 and
2. over weight = 152
(∴ Standard weight is 55.70 kg)

Question 11.
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 :
Solution:
We represent the given data in the
cumulative frequency table as given below:

Question 12.
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)
Solution:
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 13.
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.
Solution:
The cumulative frequency table may be prepared as follows:

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

Question 14.
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).
Solution:
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.

We hope the ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 21 Measures of Central Tendency Ex 21.6 help you. If you have any query regarding ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 21 Measures of Central Tendency Ex 21.6, drop a comment below and we will get back to you at the earliest.

## ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 21 Measures of Central Tendency Ex 21.5

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 21 Measures of Central Tendency Ex 21.5

More Exercises

Question 1.
Draw an ogive for the following frequency distribution:

Solution:

Plot the points (160, 8), (170, 11), (180, 15), (190, 25) and (200, 27)
on the graph and join them with the free hand. We get an ogive as shown:

Question 2.
Draw an ogive for the following data:

Solution:

Plot the points (10.5, 3), (20.5, 8), (30.5, 16), (40.5, 23),
(50.5, 29), (60.5, 31) on the graph and join them with a free hand,
we get an ogive as shown:

Question 3.
Draw a cumulative frequency curve for the following data:

Solution:

Plot the points (29, 1), (34, 3), (39, 8), (44, 14), (49, 18),
(54, 21) and (59, 23) on the graph and join them
with a free hand to get an ogive as shown:

We hope the ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 21 Measures of Central Tendency Ex 21.5 help you. If you have any query regarding ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 21 Measures of Central Tendency Ex 21.5, drop a comment below and we will get back to you at the earliest.

## ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 21 Measures of Central Tendency Ex 21.4

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 21 Measures of Central Tendency Ex 21.4

More Exercises

Question 1.
Draw a histogram for the following frequency distribution and find the mode from the graph :

Solution:

Mode = 14

Question 2.
Find the modal height of the following distribution by drawing a histogram :

Solution:

Now present the Height on x-axis and No. of students (frequency) on the y-axis
and draw a histogram as shown. In the histogram join AB and CD intersecting at M.
From M, draw MN to the x-axis. N shows the mode.
Hence mode = 174 cm

Question 3.
A Mathematics aptitude test of 50 students was recorded as follows :

Draw a histogram for the above data using a graph paper and locate the mode. (2011)
Solution:

Hence, the required mode is 82.5.

Question 4.
Draw a histogram and estimate the mode for the following frequency distribution :

Solution:

Representing classes on x-axis and frequency on the y-axis,
we draw a histogram as shown.
In the histogram, join AB and CD intersecting at M.
From M, draw ML perpendicular to the x-axis. L shows the mode
Hence Mode = 23

Question 5.
IQ of 50 students was recorded as follows

Draw a histogram for the above data and estimate the mode.
Solution:

Representing the IQ scores on x-axis and number of students on the y-axis,
we draw a histogram as shown. Join AB and CD intersecting each other at M.
From M draw ML ⊥ x-axis. L is the mode which is 107

Question 6.
Use a graph paper for this question. The daily pocket expenses of 200 students in a school are given below:

Draw a histogram representing the above distribution and estimate the mode from the graph.
Solution:

Mark the upper comers of the highest rectangle
and the corners of the adjacent rectangles as A, B, C, D as shown.
Join AC and BD to intersect at P. Draw PM ⊥ x-axis.
Then the abscissa of M is 21, which is the required mode.
Hence, mode = 21

Question 7.
Draw a histogram for the following distribution :

Hence estimate the modal weight.
Solution:
We write the given distribution in the continuous form :

Representing the weight (in kg) on x-axis
and No. of students on y-axis. We draw a histogram as shown.
Now join AB and CD intersecting each other at M.
From M, draw ML perpendicular to x-axis.
L is the mode which is 51.5 kg

Question 8.
Find the mode of the following distribution by drawing a histogram

Also state the modal class.
Solution:

Representing class on x-axis and frequency on the y-axis,
we draw a histogram as shown.
Join AB and CD intersecting each other at M.
From M, draw ML perpendicular to the x-axis.
L shows the mode which is 30.5 and class is 27-33.

We hope the ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 21 Measures of Central Tendency Ex 21.4 help you. If you have any query regarding ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 21 Measures of Central Tendency Ex 21.4, drop a comment below and we will get back to you at the earliest.

## ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 21 Measures of Central Tendency Ex 21.3

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 21 Measures of Central Tendency Ex 21.3

More Exercises

Question 1.
Find the mode of the following sets of numbers ;
(i) 3, 2, 0, 1, 2, 3, 5, 3
(ii) 5, 7, 6, 8, 9, 0, 6, 8, 1, 8
(iii) 9, 0, 2, 8, 5, 3, 5, 4, 1, 5, 2, 7
Solution:
(i) ∵ The number 3 occurs maximum times
Mode = 3
(ii) ∵ The number 8 occurs maximum times
Mode = 8
(iii) ∵ The number 5, occurs maximum times
Mode = 5

Question 2.
Calculate the mean, the median and the mode of the numbers : 3, 2, 6, 3, 3, 1, 1, 2
Solution:
Arranging in ascending order 1, 1, 2, 2, 3, 3, 3, 6
(i) Mean = $$\frac { \sum { { x }_{ i } } }{ n }$$
= $$\frac { 1+1+2+2+3+3+3+6 }{ 8 }$$
= $$\\ \frac { 21 }{ 8 }$$
= 2.625
(ii) Here n = 8 which is even

Question 3.
Find the mean, median and mode of the following distribution : 8, 10, 7, 6, 10, 11, 6, 13, 10
Solution:
Mean = $$\\ \frac { 8+10+7+6+10+11+6+13+10 }{ 2 }$$
= $$\\ \frac { 81 }{ 9 }$$ = 9
Given nos. in ascending order are as follows:
6, 6, 7, 8, 10, 10, 10, 11, 13
Median = $$\\ \frac { n+1 }{ 2 }$$ th term = $$\\ \frac { 9+1 }{ 2 }$$ = 5th term = 10
Mode = 10 (having highest frequency 3 times)

Question 4.
Calculate the mean, the median and the mode of the following numbers : 3, 1, 5, 6, 3, 4, 5, 3, 7, 2
Solution:
Arranging in ascending order 1, 2, 3, 3, 3, 4, 5, 5, 6, 7
(i) Mean = $$\frac { \sum { { x }_{ i } } }{ n }$$
= $$\frac { 1+2+3+3+3+4+5+5+6+7 }{ 8 }$$
= $$\\ \frac { 39 }{ 10 }$$
= 3.9

Question 5.
The marks of 10 students of a class in an examination arranged in ascending order are as follows: 13, 35, 43, 46, x, x +4, 55, 61,71, 80
If the median marks is 48, find the value of x. Hence, find the mode of the given data. (2017)
Solution:
Given marks are 13, 35, 43, 46, x, x + 4, 55, 61, 71, 80
n = 10 (even), median = 48

Question 6.
A boy scored the following marks in various class tests during a term each test being marked out of 20: 15, 17, 16, 7, 10, 12, 14, 16, 19, 12, 16
(i) What are his modal marks ?
(ii) What are his median marks ?
(iii) What are his mean marks ?
Solution:
Arranging in ascending order 7, 10, 12, 12, 14, 15, 16, 16, 16, 17, 19
(ii) Here n = 11 which is odd

Question 7.
Find the mean, median and mode of the following marks obtained by 16 students in a class test marked out of 10 marks : 0, 0, 2, 2, 3, 3, 3, 4, 5, 5, 5, 5, 6, 6, 7, 8
Solution:
Here, n = 16

Question 8.
Find the mode and median of the following frequency distribution :

Solution:

Question 9.
The marks obtained by 30 students in a class assessment of 5 marks is given below:

Calculate the mean, median and mode of the above distribution.
Solution:

Question 10.
The distribution given below shows the marks obtained by 25 students in an aptitude test. Find the mean, median and mode of the distribution.

Solution:

Question 11.
At a shooting competition, the scores of a competitor were as given below :

(i) What was his modal score ?
(ii) What was his median score ?
(iii) What was his total score ?
(iv) What was his mean ?
Solution:
Writing the given distribution in cumulative frequency distribution:

Question 12.
(i) Using step-deviation method, calculate the mean marks of the following distribution.
(ii) State the modal class.

Solution:

Question 13.
The following table gives the weekly wages (in Rs.) of workers in a factory :

Calculate:
(i) The mean.
(ii) the modal class
(iii) the number of workers getting weekly wages below Rs. 80.
(iv) the number of workers getting Rs. 65 or more but less than Rs. 85 as weekly wages.
Solution:
Representing the given distribution in cumulative frequency distribution

We hope the ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 21 Measures of Central Tendency Ex 21.3 help you. If you have any query regarding ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 21 Measures of Central Tendency Ex 21.3, drop a comment below and we will get back to you at the earliest.

## ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 21 Measures of Central Tendency Ex 21.2

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 21 Measures of Central Tendency Ex 21.2

More Exercises

Question 1.
A student scored the following marks in 11 questions of a question paper : 3, 4, 7, 2, 5, 6, 1, 8, 2, 5, 7 Find the median marks.
Solution:
Arranging in the ascending order, 1, 2, 2, 3, 4, 5, 5, 6, 7, 7, 8
Here, n = 11 i.e. odd,
The middle term = $$\\ \frac { n+1 }{ 2 }$$ = $$\\ \frac { 11+1 }{ 2 }$$ = $$\\ \frac { 12 }{ 2 }$$ = 6th term
Median = 5

Question 2.
(a) Find the median of the following set of numbers : 9, 0, 2, 8, 5, 3, 5, 4, 1, 5, 2, 7 (1990)
(b)For the following set of numbers, find the median: 10, 75, 3, 81, 17, 27, 4, 48, 12, 47, 9, 15.
Solution:
(a) Arranging in ascending order :
0, 1, 2, 2, 3, 4, 5, 5, 5, 7, 8, 9
Here, n = 12 which is even

Question 3.
Calculate the mean and the median of the numbers : 2, 1, 0, 3, 1, 2, 3, 4, 3, 5
Solution:
Writing in ascending order 0, 1, 1, 2, 2, 3, 3, 3, 4, 5
Here, n = 10 which is even

Question 4.
The median of the observations 11, 12, 14, (x – 2), (x + 4), (x + 9), 32, 38, 47 arranged in ascending order is 24. Find the value of x and hence find the mean.
Solution:
Observation are :
11, 12, 14, (x – 2), (x + 4), (x + 9), 32, 38, 47
n = 9
Median = $$\\ \frac { 9+1 }{ 2 }$$ th term
i.e, 5th term = x + 4

Question 5.
The mean of the numbers 1, 7, 5, 3, 4, 4, is m. The numbers 3, 2, 4, 2, 3, 3, p have mean m – 1 and median q. Find
(i) p
(ii) q
(iii) the mean of p and q.
Solution:
(i) Mean of 1, 7, 5, 3, 4, 4 is m.
Here n = 6

Question 6.
Find the median for the following distribution:

Solution:
Writing the distribution in cumulative frequency table:

Question 7.
Find the median for the following distribution.

Solution:
Writing the distribution in cumulative frequency table :

Question 8.
Marks obtained by 70 students are given below :

Calculate the median marks.
Solution:
Arranging the variates in ascending order and in c.f. table.

Question 9.
Calculate the mean and the median for the following distribution :

Solution:
Writing the distribution in c.f. table :

Question 10.
The daily wages (in rupees) of 19 workers are
41, 21, 38, 27, 31, 45, 23, 26, 29, 30, 28, 25, 35, 42, 47, 53, 29, 31, 35.
Find
(i) the median
(ii) lower quartile
(iii) upper quartile range,
(iv) interquartile range.
Solution:
Arranging the observations in ascending order
21, 23, 25, 26, 27, 28, 29, 29, 30, 31, 31, 35, 35, 38, 41, 42, 45, 47, 53
Here n = 19 which is odd.
(i) Median = $$\\ \frac { n+1 }{ 2 }$$ th term = $$\\ \frac { 19+1 }{ 2 }$$ = $$\\ \frac { 20 }{ 2 }$$ = 10th term = 31

Question 11.
From the following frequency distribution, find :
(i) the median
(ii) lower quartile
(iii) upper quartile
(iv) inter quartile range

Solution:
Writing frequency distribution in c.f. table :

Question 12.
For the following frequency distribution, find :
(i) the median
(ii) lower quartile
(iii) upper quartile

Solution:
Writing the distribution in cumulative frequency (c.f.) table :

We hope the ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 21 Measures of Central Tendency Ex 21.2 help you. If you have any query regarding ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 21 Measures of Central Tendency Ex 21.2, drop a comment below and we will get back to you at the earliest.

## ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 21 Measures of Central Tendency Ex 21.1

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 21 Measures of Central Tendency Ex 21.1

More Exercises

Question 1.
(a) Calculate the arithmetic mean of 5.7, 6.6, 7.2, 9.3, 6.2.
(b) The weights (in kg) of 8 new born babies are 3, 3.2, 3.4, 3.5, 4, 3.6, 4.1, 3.2. Find the mean weight of the babies.
Solution:
(a) Sum of 5 observations = 5.7 + 6.6 + 7.2 + 9.3 + 6.2 = 35.0
∴ Mean = $$\\ \frac { 35.0 }{ 5 }$$ = 7
(b) Weights of 8 babies (in kg) are 3, 3.2, 3.4, 3.5, 4, 3.6, 4.1, 3.2
∴ Total weights of 8 babies
= 3 + 3.2 + 3.4 + 3.5 + 4 + 3.6 + 4.1 + 3.2 = 28.0 kg
Mean weight = $$\frac { \sum { { x }_{ i } } }{ n }$$
= $$\\ \frac { 28.0 }{ 8 }$$ (Here n = 8)
= 3.5 kg

Question 2.
The marks obtained by 15 students in a class test are 12, 14, 07, 09, 23, 11, 08, 13, 11, 19, 16, 24, 17, 03, 20 find
(i) the mean of their marks.
(ii) the mean of their marks when the marks of each student are increased by 4.
(iii) the mean of their marks when 2 marks are deducted from the marks of each student.
(iv) the mean of their marks when the marks of each student are doubled.
Solution:
Sum of marks of 15 students.
= 12 + 14 + 07 + 09 + 23 + 11 + 08 + 13 + 11 + 19 + 16 + 24 + 17 + 03 + 20
= 207
(i) Mean = $$\\ \frac { 207 }{ 15 }$$
= 13.8

Question 3.
(a) The mean of the numbers 6, y, 7, x, 14 is 8. Express y in terms of x.
(b) The mean of 9 variates is 11. If eight of them are 7, 12, 9, 14, 21, 3, 8 and 15 find the 9th variate.
Solution:
(a) Sum of numbers = 6 + y + 7 + x + 14
= 27 + x + y …(i)
But mean of 5 numbers = 8
∴ Sum = 8 × 5 = 40 …(ii)
From (i) and (ii)
27 + x + y = 40
⇒ x + y = 40 – 27 = 13
∴ y = 13 – x
(b) Mean of 9 variates = 11
∴ Total sum =11 × 9 = 99
But sum of 8 of these variates
= 7 + 12 + 9 + 14 + 21 + 3 + 8 + 15 = 89
∴ 9th variate = 99 – 89 = 10

Question 4.
(a) The mean age of 33 students of a class is 13 years. If one girl leaves the class, the mean becomes $$12 \frac { 15 }{ 16 }$$ years. What is the age of the girl ?
(b) In a class test, the mean of marks scored by a class of 40 students was calculated as 18.2. Later on, it was detected that marks of one student was wrongly copied as 21 instead of 29. Find the correct mean.
Solution:
(a) Mean age of 33 students = 13 years
Total age = 13 × 33 = 429 years
After leaving one girl, the mean of 32

Question 5.
Find the mean of 25 given numbers when the mean of 10 of them is 13 and the mean of the remaining numbers is 18.
Solution:
Mean of 10 numbers = 13
Sum = 13 × 10 = 130
and mean of remaining 15 numbers = 18
Sum = 18 × 15 = 270
Total sum of 25 numbers = 130 + 270 = 400
Mean of 25 numbers = $$\\ \frac { 400 }{ 25 }$$ = 16

Question 6.
Find the mean of the following distribution:

Solution:

Question 7.
The contents of 100 match boxes were checked to determine the number of matches they contained

(i) Calculate, correct to one decimal place, the mean number of matches per box.
(ii) Determine how many extra matches would have to be added to the total contents of the 100 boxes to; bring the mean upto exactly 39 matches. (1997)
Solution:

Question 8.
Calculate the mean for the following distribution :
>
Solution:

Question 9.
Six coins were tossed 1000 times, and at each toss the number of heads were counted and the results were recorded as under :

Calculate the mean for this distribution.
Solution:

Question 10.
Find the mean for the following distribution

Solution:

Question 11.

(i) Calculate the mean wage correct to the nearest rupee (1995)
(ii) If the number of workers in each category is doubled, what would be the new mean wage ?
Solution:

Question 12.
If the mean of the following distribution is 7.5, find the missing frequency ” f “.

Solution:

Question 13.
Find the value of the missing variate for the following distribution whose mean is 10

Solution:
Let missing variate be x, then

Question 14.
Marks obtained by 40 students in a short assessment are given below, where a and b are two missing data.

If the mean of the distribution is 7.2, find a and b.
Solution:

Question 15.
Find the mean of the following distribution

Solution:

Question 16.
Calculate the mean of the following distribution:

Solution:
Consider the following distribution :

Question 17.
Calculate the mean of the following distribution using step deviation method:

Solution:

Question 18.
Find the mean of the following frequency distribution:

Solution:

Question 19.
The following table gives the daily wages of workers in a factory:

Calculate their mean by short cut method.
Solution:

Question 20.
Calculate the mean of the distribution given below using the short cut method.

Solution:

Question 21.
A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of days a students was absent.

Solution:

Question 22.
The mean of the following distribution is 23.4. Find the value of p.

Solution:

Question 23.
The following distribution shows the daily pocket allowance fo children of a locality. The mean pocket allowance is Rs. 18. Find the value of f

Solution:
Mean = Rs. 18

Question 24.
The mean of the following distribution is 50 and the sum of all the frequencies is 120. Find the values of p and q.

Solution:
Mean = 50, Total number of frequency = 120

Question 25.
The mean of the following frequency distribution is 57.6 and the sum of all the frequencies is 50. Find the values of p and q.

Solution:
Mean = 57.6
and sum of all frequencies = 50

Question 26.
The following table gives the life time in days of 100 electricity tubes of a certain make :

Find the mean life time of electricity tubes.
Solution:

Question 27.
Using the information given in the adjoining histogram, calculate the mean correct to one decimal place.
Solution:
From the histogram given, we represent the information in the following table :

We hope the ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 21 Measures of Central Tendency Ex 21.1 help you. If you have any query regarding ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 21 Measures of Central Tendency Ex 21.1, drop a comment below and we will get back to you at the earliest.

## ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 20 Heights and Distances Chapter Test

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 20 Heights and Distances Chapter Test

More Exercises

Question 1.
The angle of elevation of the top of a tower from a point A (on the ground) is 30°. On walking 50 m towards the tower, the angle of elevation is found to be 60°. Calculate
(i) the height of the tower (correct to one decimal place).
(ii) the distance of the tower from A.
Solution:
Let TR be the tower and A is a point on the ground
and angle of elevation of the top of tower = 30°
AB = 50 m
and from B, the angle of elevation is 60°
Let TR = h and AR = x
BR = x – 50

Question 2.
An aeroplane 3000 m high passes vertically above another aeroplane at an instant when the angles of elevation of the two aeroplanes from the same point on the ground are 60° and 45° respectively. Find the vertical distance between the two planes.
Solution:
Let A and B are two aeroplanes
and P is a point on the ground such that
angles of elevations from A and B are 60° and 45° respectively.
AC = 3000 m
Let AB = x
∴ BC = 3000 – x
Let PC = y

Question 3.
A 7m long flagstaff is fixed on the top of a tower. From a point on the ground, the angles of elevation of the top and bottom of the flagstaff are 45° and 36° respectively. Find the height of the tower correct to one place of demical.
Solution:
Let TR be the tower and PT is the flag on it such that PT = 7m
Let TR = h and AR = x
Angles of elevation from P and T are 45° and 36° respectively.
Now in right ∆PAR

Question 4.
A boy 1.6 m tall is 20 m away from a tower and observes that the angle of elevation of the top of the tower is 60°. Find the height of the tower.
Solution:
Let AB be the boy and TR be the tower
∴ AB = 1.6 m
Let TR = h
from A, show AE || BR
∴ ER = AB = 1.6 m
TE = h – 1.6
AE = BR = 20 m

h = 36.24
∴ Height of tower = 36.24 m

Question 5.
A boy 1.54 m tall can just see the sun over a wall 3.64 m high which is 2.1 m away from him. Find the angle of elevation of the sun.
Solution:
Let AB be the boy and CD be the wall which is at a distance of 2.1 m

Question 6.
In the adjoining figure, the angle of elevation of the top P of a vertical tower from a point X is 60° ; at a point Y, 40 m vertically above X, the angle of elevation is 45°. Find
(i) the height of the tower PQ
(ii) the distance XQ

Solution:
Let PQ be the tower and let PQ = h
and XQ = YR = y
XY = 40 m
∴ PR = h – 40

Question 7.
An aeroplane is flying horizontally 1 km above the ground is observed at an elevation of 60°. After 10 seconds, its elevation is observed to be 30°. Find the speed of the aeroplane in km/hr.
Solution:
A and D are the two positions of the aeroplane ;
AB is the height and P is the point
∴ AB = 1 km,
Let AD = x and PB = y
and angles of elevation from A and D at point P are 60° and 30° respectively.

Question 8.
A man on the deck of a ship is 16 m above the water level. He observes that the angle of elevation of the top of a cliff is 45° and the angle of depression of the base is 30°. Calculate the distance of the cliff from the ship and the height of the cliff.
Solution:
Let A is the man on the deck of a ship B and CE is the cliff.
AB = 16 m and angle of elevation from the top of the cliff in 45°
and the angle of depression at the base of the cliff is 30°.
Let CE = h, AD = x, then
CD = h – 16, AD = BE = x

Question 9.
There is a small island in between a river 100 metres wide. A tall tree stands on the island. P and Q are points directly opposite to each other on the two banks and in the line with the tree. If the angles of elevation of the top of the tree from P and Q are 30° and 45° respectively, find the height of the tree.
Solution:
The width of the river (PQ) = 100 m.
B is the island and AB is the tree on it.

Question 10.
A man standing on the deck of the ship which is 20 m above the sea-level, observes the angle of elevation of a bird as 30° and the angle of depression of its reflection in the sea as 60°. Find the height of the bird
Solution:
Let P is the man standing on the deck of a ship
which is 20 m above the sea level and B is the bird.
Now angle of elevation of the bird from P = 30°
and angle of depression from P to the shadow of the bird in the sea

We hope the ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 20 Heights and Distances Chapter Test help you. If you have any query regarding ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 20 Heights and Distances Chapter Test, drop a comment below and we will get back to you at the earliest.

## ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 20 Heights and Distances MCQS

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 20 Heights and Distances MCQS

More Exercises

Choose the correct answer from the given four options (1 to 9):

Question 1.
In the given figure, the length of BC is
(a) 2 √3 cm
(b) 3 √3 km
(c) 4 √3 cm
(d) 3 cm

Solution:
In the given figure, $$\\ \frac { BC }{ AC }$$ = sin 30°
⇒ $$\\ \frac { BC }{ 6 }$$ = $$\\ \frac { 1 }{ 2 }$$
⇒ BC = $$\\ \frac { 6 }{ 2 }$$ = 3cm (d)

Question 2.
In the given figure, if the angle of elevation is 60° and the distance AB = 10 √3 m, then the height of the tower is
(a) 20 √3 cm
(b) 10 m
(c) 30 m
(d) 30 √3 m

Solution:
In the given figure,
∠A = 60°, AB = 10 √3 m
Let BC = h
tan 60° = $$\frac { h }{ 10\sqrt { 3 } }$$
⇒ $$\sqrt { 3 } =\frac { h }{ 10\sqrt { 3 } }$$
⇒ h = 10 √3 × √3 = 10 × 3 = 30 m (c)

Question 3.
If a kite is flying at a height of 40 √3 metres from the level-ground, attached to a string inclined at 60° to the horizontal, then the length of the string is
(a) 80 m
(b) 60 √3 m
(c) 80 √3 m
(d) 120 m
Solution:
Let K is kite
Height of KT = 40 √3 m
Angle of elevation of string at the ground = 60°
Let length of string AK = x m

Question 4.
The top of a broken tree has its top touching the ground (shown in the given figure) at a distance of 10 m from the bottom. If the angle made by the broken part with ground is 30°, then the length of the broken part is
(a) 10 √3 m
(b) $$\frac { 20 }{ \sqrt { 3 } }$$
(c) 20 m
(d) 20 √3 m

Solution:
From the figure, AC is the height of tree and from B, it was broken
AB = A’C
Angle of elevation = 30°
A’C = 10 m
Let AC = hm’
and A’B = x m
BC = h – x m
$$cos\theta =\frac { A’C }{ A’B }$$
$$cos{ 30 }^{ o }=\frac { 10 }{ x }$$

Question 5.
If the angle of depression of an object from a 75 m high tower is 30°, then the distance of the object from the tower is
(a) 25 √3 m
(b) 50√ 3 m
(c) 75 √3 m
(d) 150 m
Solution:
Height tower AB = 75 m
C is an object on the ground and angle of depression from A is 30°.

Question 6.
A ladder 14 m long rests against a wall. If the foot of the ladder is 7 m from the wall, then the angle of elevation is
(a) 15°
(b) 30°
(c) 45°
(d) 60°
Solution:
Length of a ladder AB = 14 m

Question 7.
If a pole 6 m high casts shadow 2 √3 m long on the ground, then the sun’s elevation is
(a) 60°
(b) 45°
(c) 30°
(d) 90°
Solution:
Height of pole AB = 6 m
and its shadow BC = 2√3 m

Question 8.
If the length of the shadow of a tower is √3 times that of its height, then the angle of elevation of the sun is
(a) 15°
(b) 30°
(c) 45°
(d) 60°
Solution:
Let height of a tower AB = h m
Then its shadow BC = √3 hm

Question 9.
In ∆ABC, ∠A = 30° and ∠B = 90°. If AC = 8 cm, then its area is
(a) 16 √3 cm²
(b) 16 m²
(c) 8 √3 cm²
(d) 6 √3 cm²
Solution:
In ∆ABC, ∠A = 30°, ∠B = 90°
AC = 8 cm

We hope the ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 20 Heights and Distances MCQS help you. If you have any query regarding ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 20 Heights and Distances MCQS, drop a comment below and we will get back to you at the earliest.

## ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 20 Heights and Distances Ex 20

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 20 Heights and Distances Ex 20

More Exercises

Question 1.
An electric pole is 10 metres high. If its shadow is 10√3 metres in length, find the elevation of the sun.
Solution:
Let AB be the pole and

Question 2.
The angle of elevation of the top of a tower from a point on the ground and at a distance of 150 m from its foot is 30°. Find the height of the tower correct to one place of decimal
Solution:
Let BC be the tower and
A is the point on the ground such that
∠A= 30° and AC = 150 m

Question 3.
A ladder is placed against a wall such that it just reaches the top of the wall. The foot of the ladder is 1.5 metres away from the wall and the ladder is inclined at an angle of 60° with the ground. Find the height of the wall.
Solution:
Let AB be the wall and AC be the ladder
whose foot C is 1.5 m away from B
Let AB = x m and angle of inclination is 60°

Question 4.
What is the angle of elevation of the sun when the length of the shadow of a vertical pole is equal to its height.
Solution:
Let AB be the pole and CB be its shadow
and θ is the angle of elevation of the sun.
Let AB = x m, then BC = x m

Question 5.
A river is 60 m wide. A tree of unknown height is on one bank. The angle of elevation of the top of the tree from the point exactly opposite to the foot of the tree on the other bank is 30°. Find the height of the tree.
Solution:
Let AB be the tree and BC is the width of the river
and C is the point exactly opposite to B on the other bank
and angle of elevation is 30°.

Question 6.
From a point P on level ground, the angle of elevation of the top of a tower is 30°. If the tower is 100 m high, how far is P from the foot of the tower ?
Solution:
Let AB be the tower and P is at a distance of x m from B, the foot of the tower.
While the height of the tower AB = 100 m
and angle of elevation = 30°

Question 7.
From the top of a cliff 92 m high, the angle of depression of a buoy is 20°. Calculate to the nearest metre, the distance of the buoy from the foot of the cliff. (2005)
Solution:
Let AB be cliff whose height is 92 m
and C is buoy making depression angle of 20°.

Question 8.
A boy is flying a kite with a string of length 100 m. If the string is tight and the angle of elevation of the kite is 26°32′, find the height of the kite correct to one decimal place, (ignore the height of the boy).
Solution:
Let AB be the height of the kite A and AC is the string
and angle of elevation of the kite is 26°32′

Question 9.
An electric pole is 10 m high A steel wire tied to the top of the pole is affixed at a point on the ground to keep the pole upright. If the wire makes an angle of 45° with the horizontal through the foot of the pole, find the length of the wire.
Solution:
Let AB be the pole and AC be the wire
which makes an angle of 45° with the ground.
Height of the pole AB = 10 m
and let the length of wire AC = x m

Question 10.
A bridge across a river makes an angle of 45° with the river bank. If the length of the bridge across the river is 200 metres, what is the breadth of the river.

Solution:
Let AB be the width of river = xm
Length of the bridge AC = 200 m
and angle with the river bank = 45°
sin θ = $$\\ \frac { AB }{ AC }$$
⇒ sin 45° = $$\\ \frac { x }{ 200 }$$

Question 11.
A vertical tower is 20 m high. A man standing at some distance from the tower knows that the cosine of the angle of elevation of the top of the tower is 0.53. How far is he standing from the foot of the tower ? (2001)
Solution:
Let AB be the tower and
let a man C stands at a distance from the foot of the tower = x m
and cos θ = 0.53

Question 12.
The upper part of a tree broken by wind, falls to the ground without being detached. The top of the broken part touches the ground at an angle of 38°30′ at a point 6 m from the foot of the tree. Calculate.
(i) the height at which the tree is broken.
(ii) the original height of the tree correct to two decimal places.
Solution:
Let TR be the total height of the tree
and TP is the broken part which touches the ground
at the distance of 6 m from the foot of the tree
making an angle of 38°30′ with the ground.
Let PR = x and TR = x + y
PQ = PT = y
In right ∆PQR

Height of the tree = 4.7724 + 7.6665 = 12.4389 = 12.44 m
and height of the tree at which it is broken = 4.77 m

Question 13.
An observer 1.5 m tall is 20.5 metres away from a tower 22 metres high. Determine the angle of elevation of the top of the tower from the eye of the observer.
Solution:
In the figure, AB is tower and CD is an observer.
θ is the angle of observation from

Question 14.
From a boat 300 metres away from a vertical cliff, the angles of elevation of the top and the foot of a vertical concrete pillar at the edge of the cliff are 55°40′ and 54°20′ respectively. Find the height of the pillar correct to the nearest metre.

Solution:
Let CB be the cliff and AC be the pillar
and D be the boat which is 300 m away from
the foot of the cliff i.e. BD = 300 m.
Angles of elevation of the top and foot of the pillar
are 55°40′ and 54°20′ respectively.
Let CB = x and AC = y
In right ∆CBD,

Question 15.
From a point P on the ground, the angle of elevation of the top of a 10 m tall building and a helicopter hovering over the top of the building are 30° and 60° respectively. Find the height of the helicopter above the ground.
Solution:
let AB be the building and H is the helicopter hovering over it.
P is a point on the ground,
the angle of elevation of the top of building and helicopter are 30° and 60°

Question 16.
An aeroplane when flying at a height of 3125 m from the ground passes vertically below another plane at an instant when the angles of elevation of the two planes from the same point on the ground are 30° and 60° respectively. Find the distance between the two planes at the instant.
Solution:
Let the distance between the two planes = h m
Given that, AD = 3125 m and ∠ACB = 60° and ∠ACD = 30°

Question 17.
A person standing on the bank of a river observes that the angle subtended by a tree on the opposite bank is 60° ; when he retires 20 m from the bank, he finds the angle to be 30°. Find the height of the tree and the breadth of the river. .
Solution:
Let TR be the tree and PR be the width of the river.

Question 18.
The shadow of a vertical tower on a level ground increases by 10 m when the altitude of the sun changes from 45° to 30°. Find the height of the tower, correct to two decimal places. (2006)
Solution:
In the figure, AB is the tower,
BD and BC are the shadow of the tower in two situations.
Let BD = x m and AB = h m
In ∆ABD,

Question 19.
From the top of a hill, the angles of depression of two consecutive kilometer stones, due east are found to be 30° and 45° respectively. Find the distance of two stones from the foot of the hill.
Solution:
Let A and B be the position of two consecutive kilometre stones.
Then AB = 1 km = 1000m
Let the dIstance BC = x m
∴ Distance AC = (1000 + x) m

Question 20.
A man observes the angles of elevation of the top of a building to be 30°. He walks towards it in a horizontal line through its base. On covering 60 m the angle of elevation changes to 60°. Find the height of the building correct to the nearest me he.
Solution:
Given that
AB is a building CD = 60 m

Question 21.
At a point on level ground, the angle,of elevation of a vertical lower is found to be such that its tangent is $$\\ \frac { 5 }{ 12 }$$. On walking 192 m towards the tower,the tangent of the angle is found to be $$\\ \frac { 3 }{ 4 }$$. Find the height of the tower. (1990)
Solution:
Let TR be the tower and P is the point on the
ground such that tan θ = $$\\ \frac { 5 }{ 12 }$$

Question 22.
In the figure, not drawn to scale, TF is a tower. The elevation of T from A is x° where tan x = $$\\ \frac { 2 }{ 5 }$$ and AF = 200 m. The elevation of T from B, where AB = 80 m, is y°. Calculate :
(i) The height of the tower TF.
(ii) The angle y, correct to the nearest degree. (1997)

Solution:
Let height of the tower TF = x
tan x = $$\\ \frac { 2 }{ 5 }$$, AF = 200 m, AB = 80 m
(i) In right ∆ATF,

Question 23.
From the top of a church spire 96 m high, the angles of depression of two vehicles on a road, at the same level as the base of the spire and on the same side of it are x° and y°, where tan x° = $$\\ \frac { 1 }{ 4 }$$ and tan y° = $$\\ \frac { 1 }{ 7 }$$. Calculate the distance between the vehicles. (1994)
Solution:
Height of the church CH.
Let A and B are two vehicles which make the angle of depression
from C are x° and y° respectively.

Question 24.
In the adjoining figure, not drawn to the scale, AB is a tower and two objects C and D are located on the ground, on the same side of AB. When observed from the top A of the tower, their angles of depression are 45° and 60°. Find the distance between the two objects. If the height of the tower is 300 m. Give your answer to the nearest metre. (1998)

Solution:
Let CB = x and
DB = y
AB = 300 m

Question 25.
The horizontal distance between two towers is 140 m. The angle of elevation of the top of the first tower when seen from the top of the second tower is 30°. If the height of the second tower is 60 m, find the height of the first tower.
Solution:
Let the height of first tower TR = x
height of second tower PQ = 60 m
Distance between the two towers QR = 140 m

Question 26.
As observed from the top of a 80 m tall lighthouse, the angles of depression of two ships on the same side of the ,lighthouse in horizontal line with its base are 30° and 40° respectively. Find the distance between the two ships. Give your answer correct to the nearest metre.
Solution:
Let AB be the lighthouse and C and D be the two ships.

Question 27.
The angle of elevation of a pillar from a point A on the ground is 45° and from a point B diametrically opposite to A and on the other side of the pillar is 60°. Find the height of the pillar, given that the distance between A and B is 15 m.
Solution:
Let CD be the pillar and let CD = x
Angles of elevation of points A and B are 45° and 60° respectively.

Question 28.
From two points A and B on the same side of a building, the angles of elevation of the top of the building are 30° and 60° respectively. If the height of the building is 10 m, find the distance between A and B correct to two decimal places
Solution:
In ∆DBC, tan 60° = $$\\ \frac { 10 }{ BC }$$
⇒ √3= $$\\ \frac { 10 }{ BC }$$
⇒ BC = $$\frac { 10 }{ \sqrt { 3 } }$$
∆DBC ,tan 30° = $$\\ \frac { 10 }{ BC+AB }$$

Question 29.
(i) The angles of depression of two ships A and B as observed from the top of a light house 60 m high are 60° and 45° respectively. If the; two ships are on the opposite sides of the light house, find the distance between the two ships. Give your answer correct to the nearest whole number. (2017)
(ii) An aeroplane at an altitude of 250 m observes the angle of depression of two boats on the opposite banks of a river to be 45° and 60° respectively. Find the width of the river. Write the answer correct to the nearest whole number. (2014)
Solution:
(i) Let AD be the height of the lighthouse CD = 60 m
Let AD = x m, BD = y m

Question 30.
From a tower 126 m high, the angles of depression of two rocks which are in a horizontal line through the base of the tower are 16° and 12°20′ Find the distance between the rocks if they are on
(i) the same side of the tower
(ii) the opposite sides of the tower.
Solution:
Let CD be the tower and CD = 126 m
Let A and B be the two rocks on the same line
and angles of depression are 16° and 12°20′ respectively,

Question 31.
A man 1.8 m high stands at a distance of 3.6 m from a lamp post and casts a shadow of 5.4 m on the ground. Find the height of the lamp post.
Solution:
AB is the lamp post CD is the height of man.
BD is the distance of man from the foot of the lamp
and FD is the shadow of man.
CE || DB.

Question 32.
The angles of depression of the top and the bottom of an 8 m tall building from the top of a multi-storeyed building are 30° and 45° respectively. Find the height of tire multi-storeyed building and the distance between the two buildings, correct to two decimal places.
Solution:
Let AB be the CD be the building
The angles of depression in from A, to C
and D are 30° and 45° respectively
∠ACE = 30° and ∠ADB = 45°
CD = 8 m

Question 33.
A pole of height 5 m is fixed on the top of a tower. The angle of elevation of the top of the pole as observed from a point A on the ground is 60° and the angle of depression of the point A from the top of the tower is 45°. Find the height of the tower. (Take √3 = 1.732).
Solution:
Let QR be the tower and PQ be the pole on it
Angle of elevation from P to a point A is ∠PAR = 60°
and angle of depression from Q to A = 45°
∠QAR = 45° (alternate angle)
PQ = 5 m,

Question 34.
A vertical pole and a vertical tower are on the same level ground. From the top of the pole the angle of elevation of the top of the tower is 60° and the angle of depression of the foot of the tower is 30°. Find the height of the tower if the height of the pole is 20 m.
Solution:
Let TR is tower and
PL is the pole on the same level, ground PL = 20m
From P, draw PQ || LR
then ∠ TPQ = 60° and ∠ QPR = 30°

Question 35.
From the top of a building 20 m high, the angle of elevation of the top of a monumenti is 45° and the angle of depression of its foot is 15°. Find the height of the monument.
Solution:
Let AB be the building and AB = 20 m and
let CD be the monument and let CD = x
The distance between the building and the monument be y,

Question 36.
The angle of elevation of the top of an unfinished tower at a point distant 120 m from its base is 45°. How much higher must the tower be raised so that its angle of elevation at the same point may be 60°?
Solution:
Let AB be the unfinished tower and AB = 120 m
and angle of elevation = 45°
Let x be higher raised so that
the angle of elevation becomes 60°

Question 37.
In the adjoining figure, the shadow of a vertical tower on the level ground increases by 10 m, when the altitude of the sun changes from 45° to 30°. Find the height of the tower and give your answer, correct to $$\\ \frac { 1 }{ 10 }$$ of a metre.

[Remark. Altitude of the sun means angle of elevation of the sun.]
Solution:
Let TR be the tower and TR = h ;
Let BR = x,
AB = 10 m
Angles of elevation from the top of the tower
at A and B are 30° and 45° respectively.

Question 38.
An aircraft is flying at a constant height with a speed of 360 km/h. From a point on the ground, the angle of elevation of the aircraft at an instant was observed to be 45°. After 20 seconds, the angle of elevation was observed to be 30°. Determine the height at which the aircraft is flying (use √3 = 1.732)
Solution:
Speed of aircraft = 360 km/h
Distance covered in 20 seconds = $$\\ \frac { 360X20 }{ 60X60 }$$ = 2 km
E is the fixed point on the ground
and CD is the position of AB in height of aircraft

We hope the ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 20 Heights and Distances Ex 20 help you. If you have any query regarding ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 20 Heights and Distances Ex 20, drop a comment below and we will get back to you at the earliest.