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# Selina Concise Mathematics - Part II Solution for Class 10 Mathematics Chapter 24 - Measures of Central Tendency (Mean, Median, Quartiles and Mode)

## Selina Textbook Solutions Chapter 24 - Measures of Central Tendency (Mean, Median, Quartiles and Mode)

Selina Textbook Solutions are a perfect way to ace your examination with high marks. These Textbook Solutions are extremely helpful for solving difficult questions in the ICSE Class 10 Mathematics exam. Our Selina Textbook Solutions are written by our subject experts. Find all the answers to the Selina textbook questions of Chapter 24 - Measures of Central Tendency (Mean, Median, Quartiles and Mode).

All Selina textbook questions of Chapter 24 - Measures of Central Tendency (Mean, Median, Quartiles and Mode) solutions are created in accordance with the latest ICSE syllabus. These free Textbook Solutions for ICSE Class 10 Selina Concise Mathematics will give you a deeper insight on the fundamentals in this chapter and will help you to score more marks in the final examination. ICSE Class 10 students can refer to these solutions while doing their homework and while studying and revising for the Mathematics exam.

Exercise/Page

## Selina Concise Mathematics - Part II Solution for Class 10 Mathematics Chapter 24 - Measures of Central Tendency (Mean, Median, Quartiles and Mode) Page/Excercise 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

Solution 11

## Selina Concise Mathematics - Part II Solution for Class 10 Mathematics Chapter 24 - Measures of Central Tendency (Mean, Median, Quartiles and Mode) Page/Excercise 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

## Selina Concise Mathematics - Part II Solution for Class 10 Mathematics Chapter 24 - Measures of Central Tendency (Mean, Median, Quartiles and Mode) Page/Excercise 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)

## Selina Concise Mathematics - Part II Solution for Class 10 Mathematics Chapter 24 - Measures of Central Tendency (Mean, Median, Quartiles and Mode) Page/Excercise 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

 Weekly wages (in Rs) No. of workers (f) Cumulative frequency Class Marks (x) fx 50-55 5 5 52.5 262.5 55-60 20 25 57.5 1150.0 60-65 10 35 62.5 625.0 65-70 10 45 67.5 675.0 70-75 9 54 72.5 652.5 75-80 6 60 77.5 465.0 80-85 12 72 82.5 990.0 85-90 8 80 87.5 700.0 Total 80 5520.0

(i)

(ii) Modal class = 55-60 as it has maximum frequencies.

(iii) Number of workers getting wages below Rs.80 = 60

(iv) Number of workers getting Rs.65 or more and less than Rs.85 = 72 - 35 = 37

Solution 3

 Marks No. of Students Cumulative frequency 0.5-9.5 5 5 9.5-19.5 9 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

Total number of terms = 120

(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 =

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

Solution 4

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 5

 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 6

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 7

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 8

 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 9

 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 10

 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 11

 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 12

 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 13

(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 14

 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 15

Solution 16

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 17

Solution 18

Solution 19

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 20

 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 21

Solution 22

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

 Weight in Kg Number of workers (f) Cumulative frequency 50-60 4 4 60-70 7 11 70-80 11 22 80-90 14 36 90-100 6 42 100-110 5 47 110-120 3 50

Plot the points (60, 4), (70, 11), (80, 22), (90, 36), (100, 42), (110, 47) and (120, 50) on a graph paper and join them to get an ogive.

Solution 23

Solution 24

 Scores f c.f. 0 - 10 9 9 10 - 20 13 22 20 - 30 20 42 30 - 40 26 68 40 - 50 30 98 50 - 60 22 120 60 - 70 15 135 70 - 80 10 145 80 - 90 8 153 90 - 100 7 160 n = 160

The ogive is shown below:

Solution 25

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.

## Selina Concise Mathematics - Part II Solution for Class 10 Mathematics Chapter 24 - Measures of Central Tendency (Mean, Median, Quartiles and Mode) Page/Excercise 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

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