Request a call back

# Class 10 SELINA Solutions Maths Chapter 24 - Measures of Central Tendency (Mean, Median, Quartiles and Mode)

## Measures of Central Tendency (Mean, Median, Quartiles and Mode) Exercise TEST YOURSELF

### Solution 1(a)

Correct option: (i)

Given A.P. is 2, 4, 6, 8, ….., 40

Here, first term, a = 2 and common difference, d = 2

Last term, 40 = 2 + (n – 1)(2)

38 = (n – 1)(2)

n = 20

### Solution 1(b)

Correct option: (i) 12

Data arranged in ascending order,

8, 8, 9, 10, 12, 12, 12, 13, 15

Number of observations, n = 9 (odd)

Then, median =

### Solution 1(c)

Correct option: (iv) 21

Data arranged in ascending order,

10, 12, 14, 16, 17, x

Number of observations, n = 6 (even)

Median =

Now, Mean = Median

### Solution 1(d)

Correct option: (iii) 6

First six prime numbers are as follows:

2, 3, 5, 7, 11, 13

Number of observations, n = 6 (even)

Median =

### Solution 1(e)

Correct option: (iii) 44

n = 80 (even)

From an ogive,

Inter-quartile range = Q3 – Q1 = 54 – 10 = 44

### Solution 2

Mean of 1, 7, 5, 3, 4 and 4 =

m=4

Now, mean of 3, 2, 4, 2, 3, 3 and p = m-1 = 4-1 = 3

Therefore, 17+p = 3 x n …. Where n = 7

17+p = 21

p = 4

Arranging in ascending order:

2, 2, 3, 3, 3, 3, 4, 4

Mean = 4th term = 3

Therefore, q = 3

### Solution 3

 Date Number C.f. 1 5 5 2 12 17 3 20 37 4 27 64 5 46 110 6 30 140 7 31 171 8 18 189 9 11 200 10 5 205 11 0 205 12 1 206

(i) Mode = 5th July as it has maximum frequencies.

(ii) Total number of terms = 206

Upper quartile =

Lower quartile =

### Solution 4

 Marks No. of students c.f. 0-10 5 5 10-20 9 14 20-30 16 30 30-40 22 52 40-50 26 78 50-60 18 96 60-70 11 107 70-80 6 113 80-90 4 117 90-100 3 120

(i)

Through mark 60.5, draw a parallel line to x-axis which meets the curve at A, From A draw a perpendicular to x-axis meeting it at B.

The value of point B is the median = 43

(ii) Number of students who obtained up to 75% marks in the test = 110

Number of students who obtained more than 75% marks in the test = 120 - 110 = 10

(iii) Number of students who obtained less than 40% marks in the test = 52 (from the graph; x=40, y=52)

(iv) Lower quartile = Q1 =

### Solution 5

 Weight Frequency C. f. 40-45 5 5 45-50 17 22 50-55 22 44 55-60 45 89 60-65 51 140 65-70 31 171 70-75 20 191 75-80 9 200

(i) Number of students weighing more than 55 kg = 200-44 = 156

Therefore, percentage of students weighing 55 kg or more

(ii) 30% of students =

Heaviest 60students in weight = 9 + 21 + 30 = 60

weight = 65 kg ( from table)

(iii) (a) underweight students when 55.70 kg is standard = 46 (approx) from graph

(b) overweight students when 55.70 kg is standard = 200- 55.70 = 154 (approx) from graph

### Solution 6

 Marks obtained(x) No. of students (f) c.f. fx 5 3 3 15 6 9 12 54 7 6 18 42 8 4 22 32 9 2 24 18 10 1 25 10 Total 25 171

Number of terms = 25

(i) Mean =

(ii)

(iii) Mode = 6 as it has maximum frequencies i.e. 6

### Solution 7

 Monthly Income (thousands) No. of employees (f) Cumulative frequency 6-7 20 20 7-8 45 65 8-9 65 130 9-10 95 225 10-11 60 285 11-12 30 315 12-13 5 320 Total 320

Number of employees = 320

(i)

Through mark 160, draw a parallel line to x-axis which meets the curve at A, From A draw a perpendicular to x-axis meeting it at B.

The value of point B is the median = Rs 9.3 thousands

(ii) The number of employees with income below Rs 8500 = 95 (approx from the graph)

(iii) Number of employees with income below Rs 11500 = 305 (approx from the graph)

Therefore number of employees (senior employees) = 320-305 =15

(iv) Upper quartile =

### Solution 11

Here the number of observations i. e is 10, which is even.'

So, the given data is 13, 35, 43, 46, 46, 50, 55, 61, 71, 80.

In the given data, 46 occurs most frequently.

Mode = 46

### Solution 12

i. The frequency distribution table is as follows:

 Class interval Frequency 0-10 2 10- 20 5 20-30 8 30-40 4 40-50 6

ii.

 Class interval Frequency (f) Mean value (x) fx 0-10 2 5 10 10- 20 5 15 75 20-30 8 25 200 30-40 4 35 140 40-50 6 45 270 Sf = 25 Sf = 695

iii. Here the maximum frequency is 8 which is corresponding to class 20 - 30.

Hence, the modal class is 20 - 30.

## Measures of Central Tendency (Mean, Median, Quartiles and Mode) Exercise Ex. 24(A)

### Solution 2

(a) Here n = 9

(b)

If marks of each student be increased by 4 then new arithmetic mean will be = 59 + 4 = 63

### Solution 3

(a) The mean of 7, 11, 6, 5 and 6

(b)

If we subtract 2 from each number, then the mean will be 7-2 = 5

### Solution 4

No. of terms = 5

Mean = 8

Sum of numbers = 8 x 5 = 40 .(i)

But, sum of numbers = 6+4+7+a+10 = 27+a ..(ii)

From (i) and (ii)

27+a = 40

a = 13

### Solution 5

No. of terms = 5 and mean = 8

Sum of numbers = 5 x 8 = 40 ..(i)

but sum of numbers = 6+y+7+x+14 = 27+y+x .(ii)

from (i) and (ii)

27 + y + x = 40

x + y = 13

y = 13 - x

### Solution 6

 Age in yrs xi Frequency (fi) fixi 12 2 24 13 4 52 14 6 84 15 9 135 16 8 128 17 7 119 18 4 72 Total 40 614

### Solution 7

No. of terms = 10

Mean = 69.5

Sum of the numbers = 69.5 x 10 = 695 ..........(i)

But sum of numbers = 72+70+x+62+ 50+71+90+64+58+82

= 619 + x ......(ii)

from (i) and (ii)

619 + x = 695

x = 76

### Solution 8

 Height (cm) xi No. of Plants fi fixi 50 2 100 55 4 220 58 10 580 60 f 60f 65 5 325 70 4 280 71 3 213 Total 28+f 1718 + 60f

Mean = 60.95

### Solution 9

 Wages (Rs/day) (x) No. of Workers (f) fx 50 2 100 60 4 240 70 8 560 80 12 960 90 10 900 100 6 600 Total 42 3360

(i) Mean remains the same if the number of workers in each category is doubled.

Mean = 80

(ii) Mean will be increased by 60% if the wages per day per worker is increased by 60%

New mean =

(iii) No change in the mean if the number of workers is doubled but if wages per worker is reduced by 40%, then

New mean =

### Solution 10

 No. of matches (x) No. of boxes (f) fx 35 6 210 36 10 360 37 18 666 38 25 950 39 21 819 40 12 480 41 8 328 Total 100 3813

(i)

(ii) In the second case,

New mean = 39 matches

Total contents = 39 x 100 = 3900

But total number of matches already given = 3813

Number of new matches to be added = 3900 - 3813 = 87

### Solution 1(a)

Correct option: (i) 38

Let the fifth variable be x.

Then,

62 + x = 100

x = 38

### Solution 1(b)

Correct option: (ii) 136 cm

Mean height of 5 students = 140 cm

Then, the total height of 5 students = 140 × 5 = 700 cm

Height of one student = 156 cm

Then, the total height of the remaining 4 students = 700 – 156 = 544

Hence, mean height of the remaining 4 students

### Solution 1(c)

Correct option: (iii) 16 years

Mean age of 8 boys = 16 years

Then, sum of the ages of 8 boys = 16 × 8 = 128 years

Sum of the ages of two more boys = 18 + 14 = 32 years

Now, total age of 10 boys = 128 + 32 = 160 years

Hence, the resulting mean age

### Solution 1(d)

Correct option: (iv) 18

Mean value of 15 numbers = 20

Then, the (incorrect) sum of values of 15 numbers = 20 × 15 = 300

Correct sum = 300 – 45 + 15 = 270

Therefore, correct mean

### Solution 1(e)

Correct option: (iii)

## Measures of Central Tendency (Mean, Median, Quartiles and Mode) Exercise Ex. 24(B)

### Solution 2

 Age in years C.I. xi Number of students (fi) xifi 16 - 18 17 2 34 18 - 20 19 7 133 20 - 22 21 21 441 22 - 24 23 17 391 24 - 26 25 3 75 Total 50 1074

### Solution 3

(i) Short - cut method

 Marks No. of boys (fi) Mid-value xi A = 65 di=x-A fidi 30 - 40 10 35 -30 -300 40 - 50 12 45 -20 -240 50 - 60 14 55 -10 -140 60 - 70 12 A = 65 0 0 70 - 80 9 75 10 90 80 - 90 7 85 20 140 90 - 100 6 95 30 180 Total 70 -270

(ii) Step - deviation method

 Marks No. of boys (fi) Mid-value xi A = 65 fiui 30 - 40 10 35 -3 -30 40 - 50 12 45 -2 -24 50 - 60 14 55 -1 -14 60 - 70 12 A = 65 0 0 70 - 80 9 75 1 9 80 - 90 7 85 2 14 90 - 100 6 95 3 18 Total 70 -27

Here A = 65 and h = 10

### Solution 4

 C. I. Frequency (fi) Mid-value xi A = 87.50 fiui 63 - 70 9 66.50 -3 -27 70 - 77 13 73.50 -2 -26 77 - 84 27 80.50 -1 -27 84 - 91 38 A = 87.50 0 0 91 - 98 32 94.50 1 32 98 - 105 16 101.50 2 32 105 - 112 15 108.50 3 45 Total 150 29

Here A = 87.50 and h = 7

### Solution 5

 C. I. frequency Mid-value (xi) fixi 0-10 8 5 40 10-20 22 15 330 20-30 31 25 775 30-40 f 35 35f 40-50 2 45 90 Total 63+f 1235+35f

### Solution 6

 C.I. Frequency Mid value x fx 15-25 10 20 200 25-35 20 30 600 35-45 25 40 1000 45-55 15 50 750 55-65 5 60 300 Total 75 2850

### Solution 7

 Class Frequency (f) Mid Value (x) fx 0 - 20 7 10 70 20 - 40 p 30 30p 40 - 60 10 50 500 60 - 80 9 70 630 80 - 100 13 90 1170 Total 39 + p 2370 + 30p

Here mean = 54 ..(ii)

from (i) and (ii)

### Solution 8

 Class Freq (f) Mid value fx 0-20 5 10 50 20-40 f1 30 30f1 40-60 10 50 500 60-80 f2 70 70f2 80-100 7 90 630 100-120 8 110 880 Total 30+f1+f2 2060+30f1+70f2

Now, and

from (i)

using (i) and (ii)

### Solution 1(a)

Correct option: (i) 41

 C.I. Class Mark (x) f fx 30-40 35 2 70 40-50 45 3 135 Σf= 5 Σfx = 205

Alternate Method:

The mean of the set of observations lies between the range of the given data.

Since, the range of observations is from 30 to 50, only option (i), that is, 41 lies between 30 and 50.

Hence, 41 is the mean.

### Solution 1(b)

Correct option: (i) 17

Let the assumed mean, A = 15

Class width, I = 10

 C.I. f x u = (x – A)/i f × u 0-10 5 5 –1 –5 10-20 10 15 0 0 20-30 10 25 1 10 Σf = 25 Σfu = 5

### Solution 1(c)

Correct option: () Back answer is 30

 x f fx 10 10 100 20 a 20a 30 10 300

Note: All four values of ‘a’ given in the options satisfy the equation.

In fact, a can take any value apart from this.

### Solution 1(d)

Correct option: (iii) x1 – x2 = 0

 x f fx 10 x1 10x1 20 20 400 30 x2 30x2

### Solution 1(e)

Correct option: (ii) 47

 C.I. Class Mark (x) f fx 40-50 45 40 1800 50-60 55 10 550 Σf= 50 Σfx = 2350

## Measures of Central Tendency (Mean, Median, Quartiles and Mode) Exercise Ex. 24(C)

### Solution 2

Arranging the given data in descending order:

8, 7, 6, 5, 4, 3, 3, 1, 0

The middle term is 4 which is the 5th term.

Median = 4

### Solution 3

Arranging the given data in descending order:

28.5, 28, 27.5, 25.5, 24, 24, 22, 21, 21, 20.5

The middle terms are 24 and 24, 5th and 6th terms

### Solution 4

Arranging in ascending order:

22, 24, 25, 26, 26, 27, 27, 28, 28, 29, 21, 32, 32, 33, 35, 35, 36, 36, 37

(i) Middle term is 10th term i.e. 29

Median = 29

(ii) Lower quartile =

(iii) Upper quartile =

(iv) Interquartile range = q3 - q1 =35 - 26 = 9

### Solution 5

 Weight (kg) x no. of boys f cumulative frequency 37 10 10 38 14 24 39 18 42 40 12 54 41 6 60

Number of terms = 60

(i) median = the mean of the 30th and the 31st terms

(ii) lower quartile (Q1) =

(iii) upper quartile (Q3) =

(iv) Interquartile range = Q3 - Q1 = 40 - 38 = 2

### Solution 6

 Marks (less than) Cumulative frequency 10 5 20 24 30 37 40 40 50 42 60 48 70 70 80 77 90 79 100 80

Number of terms = 80

\Median = 40th term.

(i) Median = Through 40th term mark draw a line parallel to the x-axis which meets the curve at A. From A, draw a perpendicular to x-axis which meets it at B.

Value of B is the median = 40

(ii) Lower quartile (Q1) = 20th term = 18

(iii) Upper Quartile (Q3) = 60th term = 66

### Solution 7

 Height (in cm) No. of pupils Cumulative Frequency 121 - 130 12 12 131 - 140 16 28 141 - 150 30 58 151 - 160 20 78 161 - 170 14 92 171 - 180 8 100

Number of terms = 100

Through 50th term mark draw a line parallel to the x-axis which meets the curve at A. From A, draw a perpendicular to x-axis which meets it at B.

Value of B is the median = 148

Median height = 148cm

### Solution 8

Mode is in 20-30, because in this class there are 20 frequencies.

### Solution 9

Mode is in 30-35 because it has the maximum frequency.

### Solution 10

Arranging the given data in ascending order:

7, 10, 12, 12, 14, 15, 16, 16, 16, 17, 19

(i) Mode = 16 as it occurs maximum number of times.

(ii)

(iii)Total marks = 7+10+12+12+14+15+16+16+16+17+19 =

154

(iv)

### Solution 11

 Score x No. of shots f fx 0 0 0 1 3 3 2 6 12 3 4 12 4 7 28 5 5 25 Total 25 80

(i) Modal score = 4 as it has maximum frequency 7.

(ii)

(iii) Total score = 80

(iv)

### Solution 1(a)

Correct option: (ii) 18

Arranging data in ascending order,

14, 15, 18, 21, 29

Number of observations, n = 5 (odd)

Then, median =

### Solution 1(b)

Correct option: (iii) 5

Arranging data in ascending order,

0, 2, 3, 4, 6, 8, 11, 16

Number of observations, n = 8 (even)

Then, median =

### Solution 1(c)

Correct option: (iv) 23

Data arranged in ascending order,

5, 15, 20, x, 28, 30, 35

Number of observations, n = 7 (odd)

Then, median =

Therefore, x = 23

### Solution 1(d)

Correct option: (iii) 60 – 20

Data: 10, 20, 30, 40, 50, 60, 70 and 80

Number of observations, n = 8

Inter-quartile range = Q3 – Q1 = 60 – 20

### Solution 1(e)

Correct option: (iii) 50-60

Modal class is the class with the highest frequency.

Therefore, modal class is 50-60.