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

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

Solution 1

(i)

(ii)

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

Numbers between 3 to 12 are 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12.

Here n = 10

Solution 4

(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 5

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 6

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 7
 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 8

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 9
 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 10

 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 11

 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 12

Solution 13

Solution 14

Solution 15

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

Solution 1

 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 2

(i) Direct Method

 Weekly Wages (Rs) Mid-Value xi No. of Workers (fi) fixi 50-55 52.5 5 262.5 55-60 57.5 20 1150.0 60-65 62.5 10 625.0 65-70 67.5 10 675.0 70-75 72.5 9 652.5 75-80 77.5 6 465.0 80-85 82.5 12 990.0 85-90 87.5 8 700.0 Total 80 5520.00

(ii) Short - cut method

 Weekly wages (Rs) No. of workers (fi) Mid-value xi A = 72.5 di=x-A fidi 50-55 5 52.5 -20 -100 55-60 20 57.5 -15 -300 60-65 10 62.5 -10 -100 65-70 10 67.5 -5 -50 70-75 9 A=72.5 0 0 75-80 6 77.5 5 30 80-85 12 82.5 10 120 85-90 8 87.5 15 120 Total 80 -280

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

Let the assumed mean A= 72.5

 C.I fi Mid value (xi) di=xi -; A fidi 50-55 5 52.5 -20 -100 55-60 20 57.5 -15 -300 60-65 10 62.5 -10 -100 65-70 10 67.5 -5 -50 70-75 9 72.5 0 0 75-80 6 77.5 5 30 80-85 12 82.5 10 120 85-90 8 87.5 15 120 Total 80 -280

Solution 7

 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 8

 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 9

 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 10

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

Solution 1

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 2

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 3

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 4

Arrange in ascending order:

0, 7, 10, 18, 25, 36, 38, 40, 45, 56, 60, 65, 77, 83, 88, 95

(i) Median is the mean of 8th and 9th term

(ii) Upper quartile =

(iii) Interquartile range =

Solution 5

 Age (in years) Frequency Cumulative Frequency 11 2 2 12 4 6 13 6 12 14 10 22 15 8 30 16 7 37

Number of terms = 37

Median =

Median = 14

Solution 6

 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 7
 Class Frequency Cumulative Frequency 0-10 4 4 10-20 9 13 20-30 15 28 30-40 14 42 40-50 8 50

Number of terms = 50

Through mark of 25.5 on the y-axis, draw a line parallel to x-axis which meets the curve at A. From A, draw a perpendicular to x-axis, which meets x-axis at B.

The value of B is the median which is 28.

Solution 8

 Weight (kg) No. of boys Cumulative Frequency 10-15 11 11 15-20 25 36 20-25 12 48 25-30 5 53 30-35 2 55

Number of terms = 55

Through mark of 28 on the y-axis, draw a line parallel to x-axis which meets the curve at A. From A, draw a perpendicular to x-axis, which meets x-axis at B.

The value of B is the median which is 18.4 kg

Solution 9

 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 10

 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

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

Solution 1

(i) Mode = 7

Since 7 occurs 4 times

(ii) Mode = 11

Since it occurs 4 times

Solution 2

Mode is 122 cm because it occur maximum number of times. i.e. frequency is 18.

Solution 3

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

Solution 4

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

Solution 5

which is 5.

Mode = 5 because it occurs maximum number of times.

Solution 6

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 7

(i)

(ii) Median = mean of 8th and 9th term

(iii) Mode = 5 as it occurs maximum number of times.

Solution 8
 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)

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

Solution 1

Taking Height of student along x-axis and cumulative frequency along y-axis we will draw an ogive.

(i)

Through mark for 80, draw a parallel line to x-axis which meets the curve; then from the curve draw a vertical line which meets the x-axis at the mark of 157.5.

(ii)Since, number of terms = 160

(iii)Through mark for 172 on x-axis, draw a vertical line which meets the curve; then from the curve draw a horizontal line which meets the y-axis at the mark of 145.

The number of students whose height is above 172 cm

= 160 - 144 = 16

Solution 2

 Marks No. of students Cumulative frequency 9.5 - 19.5 14 14 19.5 - 29.5 16 30 29.5 - 39.5 22 52 39.5 - 49.5 26 78 49.5 - 59.5 18 96 59.5 - 69.5 11 107 69.5 - 79.5 6 113 79.5 - 89.5 4 117 89.5 - 99.5 3 120

Scale:

1cm = 10 marks on X axis

1cm = 20 students on Y axis

(i)

Through mark 60, 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) Total marks = 100

75% of total marks =  marks

The number of students getting more than 75% marks = 120 - 111 = 9 students.

Solution 3

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 4

 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 5

We plot the points (8, 8), (16, 43), (24, 78), (32, 92) and (40, 100) to get the curve as follows:

At y = 50, affix A.

Through A, draw a horizontal line meeting the curve at B.

Through B, a vertical line is drawn which meets OX at M.

OM = 17.6 units

Hence, median income = 17.6 thousands

Solution 6

Arranging the terms in ascending order:

2, 6, 8, 9, 10, 11, 11, 12, 13, 13, 14, 14, 15, 15, 15, 16, 16, 18, 19, 20

Number of terms = 20

(i)

(ii)

(iii) Mode = 15 as it has maximum frequencies i.e. 3

Solution 7

 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 8

 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 9

 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 10

 C.I. Frequency(f) Mid value (x) fx 10-20 5 15 75 20-30 3 25 75 30-40 f 35 35f 40-50 7 45 315 50-60 2 55 110 60-70 6 65 390 70-80 13 75 975 Total 36+f 1940+35f

Solution 11

 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 12

(i)Draw the histogram

(ii) In the highest rectangle which represents modal class draw two lines AC and BD intersecting at P.

(iii) From P, draw a perpendicular to x-axis meeting at Q.

(iv) Value of Q is the mode = 82 (approx)

Solution 13
 Marks No. of students Cumulative frequency 0-10 5 5 10-20 11 16 20-30 10 26 30-40 20 46 40-50 28 74 50-60 37 111 60-70 40 151 70-80 29 180 80-90 14 194 90-100 6 200

Number of students = 200

(i)

Through mark 100, 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 = 57 marks (approx)

(ii) The number of students who failed (if minimum marks required to pass is 40)= 46 (approx from the graph)

(iii) The number of students who secured grade one in the examination = 200 - 188 = 12 (approx from the graph)

Solution 14

Solution 15

Since the frequency for x = 14 is maximum.

So Mode = 14.

According to the table it can be observed that the value of x from the 13th term to the 17th term is 13.

So the median = 13.

Solution 16

Solution 17

Solution 18

Histogram is as follows:

In the highest rectangle which represents modal class draw two lines AC and BD intersecting at E.

From E, draw a perpendicular to x-axis meeting at L.

Value of L is the mode. Hence, mode = 21.5

Solution 19

 Marks Number of students (Frequency) Cumulative Frequency 0-10 3 3 10-20 7 10 20-30 12 22 30-40 17 39 40-50 23 62 50-60 14 76 60-70 9 85 70-80 6 91 80-90 5 96 90-100 4 100

The ogive is as follows:

Solution 20

Solution 21

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 22

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

 Wages (Rs.) Upper limit No. of workers C.f. 400-450 450 2 2 450-500 500 6 8 500-550 550 12 20 550-600 600 18 38 600-650 650 24 62 650-700 700 13 75 700-750 750 5 80

The ogive is as follows:

Number of workers = n = 80

1) Median = term = 40th term, draw a horizontal line which meets the curve at point A.

Draw vertical line parallel to y axis from A to meet x axis at B.

The value of point B is 605.

2) Lower quartile (Q1)= term=20th term = 550

3) Through mark of point 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, the number of workers that earn more than Rs. 625 daily = 80 - 50 = 30

Solution 23

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.

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