# Frank Solutions for Chapter 24 Measure of Central Tendency Class 10 ICSE Mathematics

### Exercise 24.1

1. Find the mean of first 12 even numbers.

We know that, the first 12 even numbers are,

2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24

Then, x̅ = (x1 + x2 + x3 + … + xn)/n

where n is the total numbers,

n = 12

x̅ = (2 + 4 + 6 + 8 + 10 + 12 + 14 + 16 + 18 + 20 + 22 + 24)/12

⇒ x̅ = 156/12

⇒ x̅ = 13

Hence, mean of first 12 even numbers is 13.

2. Find the mean of first 10 prime numbers.

We know that, the first 10 prime numbers are,

2, 3, 5, 7, 11, 13, 17, 19, 23, 29

Then, x̅ = (x1 + x2 + x3 + … + xn)/n

Where n is the total numbers,

n = 10

x̅ = (2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 + 29)/10

⇒ x̅ = 129/10

⇒ x̅ = 12.9

Hence, mean of first 10 prime numbers is 12.9.

3. Find the mean of all numbers from 7 to 17.

All numbers from 7 to 17 are,

7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17

Then, x̅ = (x1 + x2 + x3 + … + xn)/n

Where n is the total numbers,

n = 11

x̅ = (7 + 8 + 9 + 10 + 11 + 12 + 13 + 14 + 15 +16 +17)/11

⇒ x̅ = 132/11

⇒ x̅ = 12

Hence, mean of all numbers from 7 to 17.

4. Find the mean of all odd numbers from 5 to 20. Find the new mean when each number is multiplied by 4.

All odd numbers from 5 to 20 are,

5, 7, 9, 11, 13, 15, 17, 19

Then, x̅ = (x1 + x2 + x3 + … + xn)/n

Where n is the total numbers,

n = 8

x̅ = (5 + 7 + 9 + 11 + 13 + 15 + 17 + 19)/11

⇒ x̅ = 96/8

⇒ x̅ = 12

Hence, mean of all odd numbers from 5 to 20 is 12.

Then, all odd numbers from 5 to 20 multiplied by 4 are,

20, 28, 36, 44, 52, 60, 68, 76

n = 8

x̅ = (20 + 28 + 36 + 44 + 52 + 60 + 68 + 76)/8

⇒ x̅ = 384/8

⇒ x̅ = 48

Hence, mean all odd numbers from 5 to 20 multiplied by 4 is 48.

5. Find the mean of all natural numbers from 32 to 46. Find the new mean when each number is diminished by 5.

All natural numbers from 32 to 46 are,

32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46.

Then, x̅ = (x1 + x2 + x3 + … + xn)/n

Where n is the total numbers,

n = 15

x̅ = (32 + 33 +34 + 35 + 36 + 37 + 38 + 39 + 40 + 41 + 42 + 43 + 44 + 45 + 46)/15

⇒ x̅ = 585/15

⇒ x̅ = 39

Hence, mean of all natural numbers from 32 to 46 is 39.

Then, all natural numbers from 32 to 46 diminished by 5 are,

27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41

n = 15

x̅ = (27 + 28 + 29 + 30 + 31 + 32 + 33 + 34 + 35 + 36 + 37 + 38 + 39 + 40 + 41)/15

⇒ x̅ = 510/15

⇒ x̅ = 34

Hence, mean all natural numbers from 32 to 46 diminished by 5 is 34.

6. If the mean of 8, 14, 20, x and 12 is 13, find x.

Form the question it is given that, 8, 14, 20, x, 12

Mean = 13

We have to find the value of x,

Then, x̅ = (x1 + x2 + x3 + … + xn)/n

Where n is the total numbers,

n = 5

13 = (8 + 14 + 20 + x + 12)/5

⇒ 13 × 5 = (54 + x)

⇒ 65 = 54 + x

⇒ x = 65 – 54

⇒ x = 11

Therefore, the value of x is 11.

7. If the mean of 11, 14, p, 26, 10, 12, 18 and 6 is 15, find p.

Form the question it is given that, 11, 14, p, 26, 10, 12, 18 and 6.

Mean = 15

We have to find the value of p,

Then, x̅ = (x1 + x2 + x3 + … + xn)/n

Where n is the total numbers,

n = 8

15 = (11 + 14 + p + 26 + 10 + 12 + 18 + 6)/8

⇒ 15 × 8 = (97 + p)

⇒ 120 = 97 + p

⇒ p = 120 – 97

⇒ p = 23

Therefore, the value of p is 23.

8. The mean monthly income of 10 persons is Rs 8,670. If a new member with a monthly income of Rs 9,000 joins the group, find the new monthly income.

From the question it is given that,

The mean monthly income of 10 persons is ₹8,670.

Number of persons, n = 10

We know that,

x̅ = (x1 + x2 + x3 + … + xn)/n

₹8,670 = ∑xn/10

⇒ ∑xn = 8,670 × 10

⇒ ∑xn = ₹86,700

Also it is given that, a new member with a monthly income of ₹ 9,000.

So, ∑xn = ₹ (86,700 + 9,000)

⇒ ∑xn = ₹ 95,700

Then, n = 11

x̅ = (x1 + x2 + x3 + … + xn)/n

⇒ x̅ = ₹ 95,700/11

⇒ x̅ = ₹ 8,700

Therefore, the new mean monthly income is ₹ 8,700.

9. The height of 9 persons are 142 cm, 158 cm, 152 cm, 143 cm, 139 cm, 144 cm, 148 cm and 151 cm. Find the mean height.

From the question it is given that,

The height of 9 persons are, 142 cm, 158 cm, 152 cm, 143 cm, 139 cm, 144 cm, 148 cm and 151 cm.

We know that,

x̅ = (x1 + x2 + x3 + … + xn)/n

Where n is the total numbers,

n = 9

x̅ = (142 + 158 + 152 + 143 + 139 + 144 + 146 + 148 + 151)/9

⇒ x̅ = 1323/9

⇒ x̅ = 147 cm

Therefore, the mean height is 147 cm.

10. Find the mean of the following frequency distribution:

(i)

 Class 0 - 10 10 – 20 20 - 30 30 – 40 40 - 50 Frequency 4 7 6 3 5

(ii)

 Class 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50 50 - 60 60 - 70 Frequency 4 4 7 10 12 8 5
(iii)
 Class 0 - 6 6 - 12 12 – 18 18 - 24 24 - 30 Frequency 7 5 10 12 6

(iv)
 Class 25 - 35 35 - 45 45 - 55 55 - 65 65 – 75 Frequency 6 10 8 12 4

(v)
 Class 50 – 60 60 - 70 70 - 80 80 – 90 90 - 100 Frequency 8 6 12 11 13
(vi)
 Class 1 - 10 11 - 20 21 - 30 31 - 40 41 - 50 Frequency 9 12 15 10 14

(vii)
 Class 101 - 110 111 - 120 121 – 130 131 – 140 141 - 150 151 - 160 Frequency 9 12 15 10 14
(i)  So, now we have to prepare the frequency distribution table

we know that,

x̅ = ∑fixi/∑fi

⇒ x̅ = 605/25

⇒ x̅ = 24.2

Therefore, the mean is 24.2.

(ii) So, now we have to prepare the frequency distribution table,

x̅ = ∑fixi/∑fi

we know that,

x̅ = 1910/50

⇒ x̅ = 38.2

Therefore, the mean is 38.2.

(iii) So, now we have to prepare the frequency distribution table,

x̅ = ∑fixi/∑fi

we know that,

x̅ = 630/40

⇒ x̅ = 15.75

Therefore, the mean is 15.75.

(iv) So, now we have to prepare the frequency distribution table,

So, now we have to prepare the frequency distribution table,

we know that,

x̅ = ∑fixi/∑fi

⇒ x̅ = 1980/40

⇒ x̅ = 49.5

Therefore, the mean is 49.5.

(v) So, now we have to prepare the frequency distribution table,

we know that,

x̅ = ∑fixi/∑fi

⇒ x̅ = 3900/50

⇒ x̅ = 78

Therefore, the mean is 78.

(vi) So, now we have to prepare the frequency distribution table,

we know that,

x̅ = ∑fixi/∑fi

⇒ x̅ = 1610/60

⇒ x̅ = 26.83

Therefore, the mean is 26.83.

(vii) So, now we have to prepare the frequency distribution table,

we know that,

x̅ = ∑fixi/∑fi

⇒ x̅ = 13020/100

⇒ x̅ = 130.2

Therefore, the mean is 130.2.

11. The mean of the following frequency distribution is 25.8 and the sum of all the frequencies is 50. Find x and y.

 Class 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50 Frequency 7 x 15 y 10
So, now we have to prepare the frequency distribution table

We know that,

∑fi = x1 + x2 + … + xn

⇒ 50 = 7 + x + 15 + y + 10

⇒ x + y + 32 = 50

⇒ x + y = 18 …(i)

Also we know that, x̅ = ∑fixi/∑fi

So,

25.8 = (860 + 15x + 35y)/50

By cross multiplication we get,

15x + 35y + 860 = 1290

⇒ 15x + 35y = 1290 – 860

⇒ 15x + 35y = 430  [divide both side by 5]

⇒ 3x + 7y = 86 …(ii)

Now multiplying equation (i) by 3 we get,

3x + 3y = 54 …(iii)

Subtract equation (ii) from equation (iii) we get,

4y = 32

⇒ y = 32/4

⇒ y = 8

Substitute value of y in equation (i) to get the value of x,

x + y = 18

⇒ x + 8 = 18

⇒ x = 18 – 8

⇒ x = 10

Hence, the value of x = 10 and y = 8.

12. Find the mean of the following frequency distribution by the short cut method.

 Class 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50 Frequency 9 12 15 10 14
So, now we have to prepare the frequency distribution table,

we know that,

x̅ = A + ∑fid/∑fi

⇒ x̅ = 25 + 80/60

⇒ x̅ = 25 + 1.33

⇒ x̅ = 26.33

Therefore, the value of mean is 26.33.

13. Find the mean of the following frequency distribution by the short cut method:

 Class 1 - 10 11 - 20 21 – 30 31 - 40 41 - 50 51 – 60 61 – 70 Frequency 7 10 14 17 15 11 6

So, now we have to prepare the frequency distribution table,

we know that,

x̅ = A + ∑fid/∑f

⇒ x̅ = 35.5 + 0/80

⇒ x̅ = 35.5 + 0

⇒ x̅ = 35.5

Therefore, the value of mean is 35.5.

14. Find the mean of the following frequency distribution by the step deviation method:

 Class 1 - 10 11 - 20 21 - 30 31 – 40 41 – 50 51 – 60 61 – 70 Frequency 7 10 14 17 15 11 6

So, from the table A = 125 and hi = 10

We know that,

x̅ = A + h × ∑fiu/∑fi

⇒ x̅ = 125 + 10 × (-3/100)

⇒ x̅ = 125 – 0.3

⇒ x̅ = 124.70

Therefore, the mean is 124.70.

15. Find the mean of the following frequency distribution by the step deviation method:

 Class 0 - 20 20 – 40 40 – 60 60 – 80 80 - 100 100 - 120 120 – 140 Frequency 12 24 52 88 66 42 16

So, from the table A = 70 and hi = 20

we know that,

x̅ = A + h × ∑fiu/∑fi

⇒ x̅ = 70 + 20 × (62/300)

⇒ x̅ = 70 + 4.13

⇒ x̅ = 74.13

Therefore, the mean is 74.13.

### Exercise 24.2

1. The weights of 11 students in a class are 36 kg, 45 kg, 44 kg, 37 kg, 36 kg, 41 kg, 45 kg, 43 kg, 39 kg, 42 kg and 40 kg. Find the median of their weights.

Arranging the given data in descending order:

45 kg, 45 kg, 44 kg, 43 kg, 42 kg, 41 kg, 40 kg, 39 kg, 37 kg, 36 kg, 36 kg

The middle term is 41 kg which is 6th term

Therefore, Median of weights = 41 kg.

2. The percentage marks obtained in 10 subjects by a student are 84, 88, 72, 91, 68, 75, 98, 96, 79 and 86. Find the median of the marks obtained.

Arranging the given data in descending order:

98, 96, 91, 88, 86, 84, 79, 75, 72, 68

The middle term are 86 and 84 which are 5th and 6th terms

Median = (86 + 84)/2 = 170/2 = 85

Therefore, Median of marks = 85

3. Find the median of the first 15 whole numbers.

The first 15 whole numbers are :

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14

Arranging the given data in descending order:

14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0

The middle term is 7 which is 8th term.

Therefore, Median of numbers = 7

4. Find the median of all prime numbers between 20 and 50.

The prime number between 20 and 50 are:

23, 29, 31, 37, 41, 43, 47

Arranging the given data in descending order:

47, 43, 41, 37, 31, 29, 23

The middle term is 37 which is 4th term

Therefore, Median of numbers = 37.

5. Find the median of the following:

(i) 11, 8, 15, 5, 9, 4, 19, 6, 18

(ii) 25, 34, 31, 23, 22, 26, 35, 29, 20, 32

(iii) 3x, x + 5, x + 7, x + 9, x + 11, x + 13

(i) 11, 8, 15, 5, 9, 4, 19, 6, 18

Arranging the given data in descending order:

19, 18, 15, 11, 9, 8, 6, 5, 4

The middle term is 9 which is 5th term

Therefore, Median of numbers = 9.

(ii) 25, 34, 31, 23, 22, 26, 35, 29, 20, 32

Arranging the given data in descending order:

35, 34, 32, 31, 29, 26, 25, 23, 22, 20

The middle terms are 29 and 26 which are 5th and 6th terms

Median = (29 + 26)/2 = 55/2 = 27.5

Therefore, Median of numbers = 27.5

(iii) 3x, x + 5, x + 7, x + 9, x + 11, x + 13

Arranging the given data in descending order:

x + 13, x + 11, x + 9, x + 7, x + 5, 3x

The middle terms are x + 9 and x + 7 which are 3rd and 4th terms

median = (x + 9 + x + 7)/2

= (2x + 16)/2

= x + 8

Therefore, Median = x + 8.

6. Find the median of the following frequency distribution:

(i)

 Weight (kg) 36 38 40 42 44 No. of Students 11 26 29 24 10

(ii)

 Weight (kg) 3500 4000 4500 5000 5500 6000 No. of people 9 17 23 15 6 5

(i)

(ii)

7. The frequency distribution table below shows the height of 50 students of grade 10.

 Heights (in cm) 138 139 140 141 142 Frequency 6 11 16 10 7

Find the median, the upper quartile and the lower quartile of the heights.

8. Find the lower quartile, the upper quartile, the interquartile range and the semi-interquartile range for the following frequency distributions:

(i)

 Shoe size 5 6 7 8 9 10 11 Frequency 8 1 7 14 11 5 4

(ii)

 Marks 25 30 35 40 45 50 No. of students 6 15 12 10 18 9

(iii)

 Variate 10 11 12 13 14 15 16 17 18 19 20 Frequency 1 2 3 1 2 4 2 1 1 2 1

(i)

(ii)

(iii)

9. Estimate the median, the lower quartile and the upper quartile of the following frequency distributions by drawing an ogive:

(i)

 Class Interval 0 - 10 10 – 20 20 - 30 30 - 40 40 - 50 50 - 60 60 - 70 Frequency 4 12 21 18 15 7 3

(ii)

 Marks 30 - 40 40 - 50 50 - 60 60 - 70 70 - 80 80 - 90 90 - 100 No. of boys 10 12 14 12 9 7 6

(iii)

 Marks (less than) 10 20 30 40 50 60 70 80 No. of Students 5 15 30 54 72 86 94 100

(iv)

 Age (in yrs) Under 10 Under 20 Under 30 Under 40 Under 50 Under 60 No. of males 6 10 25 32 43 50

(v)

 Marks (more than) 90 80 70 60 50 40 30 20 10 0 No. of students 6 13 22 34 48 60 70 78 80 80

(i)

(ii)

(iii)

(iv)

(v)

10. The marks obtained by 200 students in an examination are given below:

 Marks 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50 50 – 60 60 – 70 70 – 80 80 - 90 90 - 100 No. of students 5 10 11 20 27 38 40 29 14 6

Using a graph paper, draw an ogive for the above distribution and estimate

(i) the median

(ii) the lower quartile

(iii) the number of students who obtained more than 80% marks in the examination.

(i)

(ii)

(iii)

11. The marks of 200 students in a test is given below:

 Marks % 10 - 19 20 - 29 30 - 39 40 - 49 50 - 59 60 - 69 70 - 79 80 - 89 No. of Students 7 11 20 46 57 37 15 7

Draw an ogive and find the

(i) the median

(ii) the number of students who scored more than 35% marks

(i)

(ii)

### Exercise- 24.3

1. Find the mode of the following:

(i) 6, 7, 1, 8, 6, 5, 9, 4, 6, 7, 1, 3, 2, 6, 7, 8

(ii) 21, 22, 28, 23, 24, 21, 26, 22, 29, 27, 21, 21, 26, 24, 23

(iii) 3, 4, 5, 7, 6, 3, 5, 4, 3, 5, 6, 4, 7, 5, 4, 5, 4, 3, 4, 5, 7, 6, 5, 6, 6, 7

(iv) 15, 17, 16, 17, 10, 12, 14, 16, 19, 12, 16, 15, 16

(iv) 20, 20, 30, 30, 30, 30, 35, 40, 40, 45, 45, 45, 50, 55, 60, 60, 60, 65, 70, 70, 70

2. Find the mode of the following frequency distribution:

(i)

 Variate 20 21 22 23 24 25 26 Frequency 21 20 26 35 22 13 10

(ii)

 Pocket money per week in Rs 25 50 75 100 125 150 No. of students 4 7 13 18 6 2

(iii)

 Hrs. Spent daily in studies 3 3.5 4 4.5 5 5.5 6 6.5 No. of students 8 7 3 5 10 6 3 4

(i)

(ii)

(iii)

3. Draw a histogram for the following distribution and estimate the mode:

(i)

 Marks 0 - 10 10 - 20 20 - 30 30 – 40 40 - 50 50 - 60 60 - 70 70 - 80 No. of students 3 7 15 24 16 8 5 2

(ii)

 I. Q. Score 80 - 100 100 - 120 120 - 140 140 - 160 160 – 180 180 - 200 No. of Students 6 9 16 13 4 2

(iii)

 Mangoes per tree 0 - 9 10 - 19 20 - 29 30 - 39 40 - 49 50 - 59 No. of trees 10 16 20 14 6 4

(iv)

 Marks % 30 - 39 40 - 49 50 - 59 60 - 69 70 - 79 80 - 89 90 - 99 No. of students 14 26 40 92 114 78 36