# RS Aggarwal Solutions Chapter 9 Mean, Mode and Median Exercise 9A Class 10 Maths

 Chapter Name RS Aggarwal Chapter 9 Mean, Mode and Median Book Name RS Aggarwal Mathematics for Class 10 Other Exercises Exercise 9BExercise 9CExercise 9DExercise 9EExercise 9F Related Study NCERT Solutions for Class 10 Maths

### Exercise 9A Solutions

1. If the mean of 5 observation x, x + 2, x + 4, x + 6 and x + 8, find the value of x.

Solution

Mean of given observations = (sum of given observations)/(total number of observation)

∴ 11 = [x + (x + 2) + (x + 4) + (x + 6) + (x + 8)]/5

⇒ 55 = 5x + 20

⇒ 5x = 55 – 20

⇒ 5x = 35

⇒ x = 35/5

⇒ x = 7

Hence, the value of x is 7.

2. If the mean of 25 observations is 27 and each observation is decreased by 7, what will be new mean?

Solution

Mean of given observations = (sum of given observations)/(total number of observations)

Mean of 25 observations = 27

∴ Sum of 25 observations = 27 × 25 = 675

If 7 is subtracted from every number, then the sum = 675 – (25 × 7)

= 675 – 175

= 500

Then, new mean = 500/25 = 20

Thus, the new mean will be 20.

3. Compute the mean for following data:
 Class 1 - 3 3 - 5 5 - 7 7 - 9 Frequency 12 22 27 19

Solution

The given data is shown as follows:

 Class Frequency (fi) Class mark (xi) 5 - 7 1 - 3 12 2 24 3 - 5 22 4 88 5 - 7 27 6 162 7 - 9 19 8 152 Total Æ©fi = 80 Æ© fi xi = 426

The mean of given data is given by

Thus, the mean of the following data is 5.325.

4. Find the mean using direct method:

 Class 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50 50 - 60 Frequency 7 5 6 12 8 2

Solution

5. Find the mean of the following data, using direct method:<

 Class 25 – 35 35 – 45 45 – 55 55 – 65 65 – 75 Frequency 6 10 8 12 4

Solution

 Class Frequency (fi) Mid values (xi) (fi × xi) 25 - 35 6 10 180 35 - 45 10 40 400 45 - 55 8 50 400 55 - 65 12 60 720 65 - 75 4 70 280 ∑ fi = 40 ∑(fi × xi)  = 1980

6. Find the mean of the following data, using direct method:

 Class 0 – 100 100-200 200-300 300-400 400-500 Frequency 6 9 15 12 8

Solution

 Class Frequency (fi) Mid values (xi) (fi × xi) 0 - 100 6 50 300 100 - 200 5 150 1350 200 - 300 6 250 3750 300 - 400 12 350 4200 400 - 500 8 450 3600 ∑ fi = 40 ∑(fi × xi) = 13200

7. Using an appropriate method, find the mean of the following frequency distribution:

 Class 84 - 90 90 - 96 96 - 102 102 - 108 108 - 114 114 - 120 Frequency 8 10 16 23 12 11

which method did you see, and why ?

Solution

 Class Frequency (fi) Mid values (xi) (fi × xi) 84 - 90 8 87 696 90 - 96 10 93 930 96 - 102 16 99 1584 102 - 108 23 105 2415 108 - 114 12 111 1332 114 – 120 11 117 1287 Total ∑ fi = 80 ∑(fi × xi) = 8244

The mean of the data is given by,

= 8244/80
= 103.05
Thus, the mean of the following data is 103.05.

8. If the mean of the following frequency distribution is 24, find the value of of p.
 Class 0 - 10 10 - 20 20 – 30 30 - 40 40 - 50 Frequency 3 4 P 3 2

Solution

The given data is shown as follows:

 Class Frequency (fi) Mid values (xi) (fi xi) 0 - 10 3 5 15 10 - 20 4 15 60 20 - 30 p 25 25p 30 - 40 3 35 105 40 - 50 2 45 90 Total ∑fi = 12 + p ∑fi xi = 270 + 25p

The mean of the given data is given by,

24 = (270 + 25p)/(12 + p)

24 (12 + p) = 270 + 25p

288 + 24p = 270 + 25p

25p – 24p = 288 – 270

p = 18

Hence, the value of p is 18.

9. The following distribution shows the daily pocket allowance of children of a locality. If the mean pocket allowance is ₹ 18, find the missing frequency f.

 Daily pocket allowance (in ₹) 11 - 13 13 - 15 15 - 17 17 - 19 19 - 21 21 - 23 23 - 25 Number of children 7 6 9 13 f 5 4

Solution

The given data is shown as follows:

 Daily pocket allowance (in ₹) Number of children (fi) Class mark (xi) fi xi 11 - 13 7 12 84 13 - 15 6 14 84 15 - 17 9 16 144 17 - 19 13 18 234 19 - 21 f 20 20f 21 - 23 5 22 110 23 - 25 4 24 96 Total ∑fi = 44 + f ∑fi xi = 752 + 20f

The mean of the given data is given by,

⇒ 18 = (750 + 20f)/(44 + f)

⇒ 18 (44 + f) = 752 + 20f

⇒ 792 + 18f = 752 + 20f

⇒ 20f – 18f = 792 – 752

⇒ 2f = 40

⇒ f = 20

Hence, the value of f is 20.

10. The mean of following frequency distribution is 54. Find the value of p.

 Class 0 - 20 20 - 40 40 - 60 60 - 80 80 - 100 Frequency 7 p 10 9 13

Solution

The given data is shown as follows:

 Class Frequency (fi) Class mark (xi) fi xi 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 ∑fi = 39 + p ∑fi xi = 2370 + 30p

The mean of the given data is given by,

⇒ 54 = (2370 + 30p)/(39 + p)

⇒ 54 (39 + p) = 2370 + 30p

⇒ 2106 + 54p = 2370 – 2106

⇒ 24p = 264

⇒ p = 11

Hence, the value of p is 11.

11. The mean of the following frequency data is 42, Find the missing frequencies x and y if the sum of frequencies is 100.

 Class interval 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 60 - 70 70 – 80 Frequency 7 10 x 13 y 10 14 9

Find x and y.

Solution

The given data is shown as follows:

 Class interval Frequency (fi) Class mark (xi) fi xi 0 – 10 7 5 35 10 – 20 10 15 150 20 – 30 x 25 25x 30 – 40 13 35 455 f40 – 50 y 45 45y 50 – 60 10 55 550 60 – 70 14 65 910 70 – 80 9 75 675 Total Æ© fi = 63 + x + y Æ© fi xi = 2775 + 25x + 45y

Sum of the frequencies = 100

⇒ ∑i  fi = 100

⇒ 63 + x + y = 100

⇒ x + y = 100 – 63

⇒ x + y = 37

⇒ y = 37 – x ...(1)

Now, the mean of the given data is given by,

⇒ 42 = (2775 + 25x + 45y)/100

⇒ 4200 = 2775 + 25x + 45y

⇒ 4200 – 2775 = 25x + 45y

⇒ 1425 = 25x + 45(37 – x) [from (1)]

⇒ 1425 = 25x + 1665 – 45x

⇒ 20x = 1665 – 1425

⇒ 20x = 240

⇒ x = 12

If x = 12, then y = 37 – 12 = 25

Thus, the value of x is 12 and y is 25.

12. The daily expenditure of 100 families are given below. Calculate fand fif the mean daily expenditure is ₹ 188.

 Expenditure (in ₹) 140 - 160 160 - 180 180 - 200 200 - 220 220 - 240 Number of families 5 25 f1 f2 5

Solution

The given data is shown as follows:

 Expenditure (in ₹) Number of families (fi) Class mark (xi) fi xi 140 - 160 5 150 750 160 - 180 25 170 4250 180 - 200 F1 190 190f1 200 - 220 F2 210 210f2 220 – 240 5 230 1150 Total ∑fi = 35 + f1 + f2. ∑fi  xi = 6150 + 190f1 + 210f2

Sum of the frequencies = 100

⇒ ∑i fi = 100

⇒ 35 + f1 + f2 = 100

⇒ f1 + f2 = 100 – 35

⇒ f1 + f2 = 65

⇒ f2 = 65 – f….(1)

Now, the mean of the given data is given by,

⇒ 188 = (6150 + 190f1 + 210f2)/100

⇒ 1880 = 6150 + 190f1 + 210f2

⇒ 18800 - 6150 = 190f1 + 210f2

⇒ 12650 = 190f1 + 210(65 – f1) [From (1)]

⇒ 12650 = 190f1 – 210f1 + 13650

⇒ 20f1 = 13650 – 12650

⇒ 20f1 = 1000

⇒ f1 = 50

If f1 = 50, then f2 = 65 – 50 = 15

Thus, the value of f1 is 50 and f2 is 15.

13. Find the mean of the following frequency distribution is 57.6 and the total number of observation is 50.

 Class 0 - 20 20 - 40 40 - 60 60 - 80 80 - 100 100 - 120 Frequency 7 f1 12 f2 8 5

Solution

 Class Frequency (fi) Mid values (xi) (fi × xi) 0 - 20 7 10 70 20 - 40 F2 30 30f1 40 – 60 12 50 600 60 – 80 18 - f1 70 1260 – 70f1 80 – 100 8 90 720 100 - 120 5 110 550 Total ∑fi = 50 ∑(fi × xi) = 3200 – 40f1

We have:

7 + f1 + 12 + f2 + 8 + 5 = 50

⇒ f1 + f2 = 18

⇒ f2 = 18 – f1

⇒ 57.6 = (3200 – 40f1)/50

⇒ 40f1 = 320

∴ f1 = 8

And f2 = 18 – 8

⇒ f2 = 10

∴ The missing frequencies are f1 = 8 and f2 = 10.

14. During a medical check-up, the number of heartbeats per minute of 30 patients were recorded and summarized as follows:

 Number of heartbeats 65 - 68 68 - 71 71 - 74 74 - 77 77 - 80 80 - 83 83 - 86 Number of patients 2 4 3 8 7 4 2

Find the mean heartbeats per minute for these patients, choosing a suitable method.

Solution

Using direct method, the given data is shown as follows:

 Number of heartbeats per minute Number of patients (fi) Class mark (xi) fi xi 65 - 68 2 66.5 133 68 – 71 4 69.5 278 71 - 74 3 72.5 217.5 74 - 77 8 75.5 604 77 - 80 7 78.5 549.5 80 - 83 4 81.5 326 83 – 86 2 84.5 169 Total ∑fi = 30 ∑fi xi = 2277

The mean of the data is given by,

= 2277/30

= 75.9

Thus, the mean heartbeats per minute for these patients is 75.9.

15. Find the mean marks per student, using assumed-mean method:

 Marks 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50 50 - 60 Number of students 12 18 27 20 17 6

Solution

 Class Frequency (fi) Mid values (xi) Deviation (di) di = (xi – 25) (fi × di) 0 - 10 12 5 -20 -240 10 - 20 18 15 -10 -180 20 - 30 27 25 = A 0 0 30 - 40 20 35 10 200 40 - 50 17 45 20 340 50 - 60 6 55 30 180 Total ∑fi = 100 ∑(fi × di) = 300

16. Find the mean of the following frequency distribution, using the assumed-mean method:
 Class 100 - 120 120 - 140 140 - 160 160 - 180 180 - 200 Frequency 10 20 30 15 5

Solution

 Class Frequency (fi) Mid values (xi) Deviation (di) di = (xi – 150) (fi × di) 100 - 120 10 110 -40 -400 120 - 140 20 130 -20 -400 140 - 160 30 150 = A 0 0 160 - 180 15 170 20 300 180 - 200 5 190 40 200 ∑fi = 80 ∑(fi × di) = -300

Let A = 150 be the assumed mean, Then we have:

17. Find the mean of the following data, using assumed-mean method:
 Class 0 - 20 20 - 40 40 - 60 60 - 80 80 - 100 Frequency 20 35 52 44 38

Solution

 Class Frequency (fi) Mid values (xi) Deviation (di) di = (xi – 50) (fi × di) 0 - 20 20 10 -40 -800 20 - 40 35 30 -20 -700 40 - 60 52 50 = A 0 0 60 - 80 44 70 20 880 80 - 100 38 90 40 1520 100 - 120 31 110 60 1860 ∑fi = 220 ∑(fi × di) = 2760

Let A = 50 be the assumed mean. Then we have:

18. The following table gives the literacy rate (in percentage) in 40 cities. Find the mean literacy rate, choosing a suitable method.

 Literacy rate (%) 45 - 55 55 - 65 65 - 75 75 - 85 85 - 95 Number of cities 4 11 12 9 4

Solution

Using Direct method, the given data is shown as follows:

 Literacy rate (%) Number of cities (fi) Class mark (xi) (fi xi) 45 - 65 4 50 200 55 - 65 11 60 660 65 - 75 12 70 840 75 – 85 9 80 720 85 - 95 4 90 360 Total ∑fi = 40 ∑fi xi = 2780

The mean of the data is given by,

= 2780/40

= 69.5

Thus, the mean literacy is 69.5 %.

19. Find the mean of the following frequency distribution using step-deviation method.

 Class 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50 Frequency 7 10 15 8 10

Solution

Let us choose a = 25, h = 10, then di = x– 25 and u= (xi – 25)/10

Using step-deviation method, the given data is shown as follows:

 Class Frequency (fi) Class mark (xi) di = xi – 25 ui = (xi – 25)/10 (fi ui) 0 - 10 7 5 -20 -2 -14 10 - 20 10 15 -10 -1 -10 20 - 30 15 25 0 0 0 30 - 40 8 35 10 1 8 40 - 50 10 45 20 2 20 Total ∑fi = 50 ∑fi ui = 4

The mean of the data is given by,

= 25 + 4/50 × 10

= 25 + 4/5

= (125 + 4)/5

= 129/5

= 25.8

Thus, the mean is 25.8.

20. Find the mean of the following data, using step-deviation method:

 Class 5 - 15 15 - 25 25 - 35 35 - 45 45 - 55 55 - 65 65 - 75 Frequency 6 10 16 15 24 8 7

Solution

Let us choose a = 40, h = 10, then di = xi – 40 and u= (x– 40)/10

Using step-deviation method, the given data is shown as follows:

 Class Frequency (fi) Class mark (xi) di = xi – 40 ui = (xi – 40)/10 (fi ui) 5 - 15 6 10 -30 -3 -18 15 - 25 10 20 -20 -2 -20 25 - 35 16 30 -10 -1 -16 35 – 45 15 40 0 0 0 45 – 55 24 50 10 1 24 55 – 65 8 60 20 2 16 65 – 75 7 70 30 3 21 Total ∑fi = 86 ∑fi ui = 7

The mean of the data is given by,

= 40 + 7/86 × 10

= 40 + 70/86

= 40 + 0.81

= 40.81

21The weights of tea in 70 packets are shown in the following table:

 Weight 200 - 201 201 - 202 202 - 203 203 - 204 204-205 205-206 Number of packets 13 27 18 10 1 1

Find the mean weight of packets using step deviation method.

Solution

Let us choose a = 202.5, h = 1, then d= x– 202.5 and ui = (x– 202.5)/1

Using step-deviation method, the given data is shown as follows:

 Weight Number of packets (fi) Class mark (xi) di = xi – 202.5 ui = (xi – 202.5)/1 (fi ui) 200 - 201 13 200.5 -2 -2 -26 201 – 202 27 201.5 -1 -1 -27 202 – 203 18 202.5 0 0 0 203 - 204 10 203.5 1 2 10 204 – 205 1 204.5 2 3 2 205 - 206 1 205.5 3 3 Total ∑fi = 70 ∑fi ui = - 38

The mean of the given data is given by,

= 202.5 + (-38/70) × 1

= 202.5 – 0.542

= 201.96

Hence, the mean is 201.96 g.

22. Find the mean of the following frequency distribution table using a suitable method:

 Class 20 - 30 30 - 40 40 - 50 50 - 60 60 - 70 Frequency 25 40 42 33 10

Solution

Let us choose a = 45, h = 10, di = x– 45 and u= (xi – 45)/10

Using step-deviation method, the given data is shown as follows:

 Weight Number of packets (fi) Class mark (xi) di = xi - 45 ui = (xi – 45)/10 (fi ui) 20 - 30 25 35 -20 -2 -50 30 - 40 40 35 -10 -1 -40 40 - 50 42 45 0 0 0 50 - 60 33 55 10 1 33 60 – 70 10 65 20 2 20 Total ∑fi = 150 ∑fi ui = - 37

The mean of the given data is given by,

= 45 – (37/150) × 10

= 45 – 37/15

= 45 – 2.466

= 42.534

Hence, the mean is 42.534.

23. In an annual examination, marks (out of 90) obtained by students of Class X in mathematics are given below:

 Marks Obtained 0 - 15 15 - 30 30 - 45 45 - 60 60 - 75 75 - 90 Number of students 2 4 5 20 9 10

Find the mean marks.

Solution

Let us choose a = 52.5, h = 15, then di = xi – 52.5 and ui = (xi – 52.5)/15

Using step-deviation method, the given data is shown as follows:

 Weight Number of students (fi) Class mark (xi) di = xi – 37.5 ui = (xi – 52.5)/15 (fi ui) 0 - 15 2 7.5 -45 -3 -6 15 - 30 4 22.5 -30 -2 -8 30 - 45 5 37.5 -15 -1 -5 45 - 60 20 52.5 0 0 0 60 – 75 9 67.5 15 1 9 75 – 90 10 82.5 30 2 20 Total ∑fi = 50 ∑fi ui = 10

The mean of the given data is given by,

= 52.5 + (10/50) × 15

= 52.5 + 3

= 55.5

Thus, the mean is 55.5.

24. Find the arithmetic mean of the following frequency distribution using step-deviation method:

 Age (in years) 18 - 24 24 - 30 30 - 36 36 - 42 42 – 48 48 - 54 Number of workers 6 8 12 8 4 2

Solution

 Class Frequency (fi) Mid values (xi) ui = (xi – A)/h = (xi – 33)/6 (fi × ui) 18 - 24 6 21 -2 -12 24 - 30 8 27 -1 -8 30 - 36 12 33 = A 0 0 36 - 42 8 39 1 8 42 – 48 4 45 2 8 48 – 54 2 51 3 6 Total ∑fi = 40 ∑(fi × ui) = 2

Now, A = 33, h = 6, ∑fi = 40 and ∑(fi × ui) = 2

25. Find the mean of the following data using step-deviation method:

 Class 500 - 520 520 - 540 540 - 560 560 - 580 580 - 600 600 – 620 Frequency 14 9 5 4 3 5

Solution

 Class Frequency (fi) Mid values (xi) ui = (xi - A)/h = (xi – 550)/20 (fi × ui) 500 - 520 14 510 -2 -28 520 - 540 9 530 -1 -9 540 - 560 5 550 = A 0 0 560 - 580 4 570 1 4 580 - 600 3 590 2 6 600 - 620 5 610 3 15 ∑fi = 40 ∑(fi × ui) = - 12

Now, a = 550, h = 20∑fi = 40 and ∑(f× ui) = - 12

26. Find the mean age from the following frequency distribution:

 Age (in years) 25 - 29 30 - 34 35 - 39 40 - 44 45 - 49 50 - 54 55 - 59 Number of persons 4 14 22 16 6 5 3

Solution

 Class Frequency (fi) Mid values (xi) ui = (xi – A)/h = (xi – 42)/5 (fi × ui) 24.5 – 29.5 4 27 -3 -12 29.5 – 34.5 14 32 -2 -28 34.5 – 39.5 22 37 -1 -22 39.5 – 44.5 16 42 = A 0 0 44.5 – 49.5 6 47 1 6 49.5 – 54.5 5 52 2 10 54.5 – 59.5 3 57 3 9 ∑fi = 70 ∑ (fi × ui) = - 37

Now, A = 42, h = 5, ∑ fi = 70 and ∑ (fi × ui) = - 37

27. The following table shows the age distribution of patients of malaria in a village during a particular month:

 Age in (years) 5 - 14 15 - 24 25 - 34 35 - 44 45 - 54 55 - 64 No. of cases 6 11 21 23 14 5

Find the average age of the patients.

Solution

 Class Frequency (fi) Mid values (xi) ui = (xi – A)/h = (xi – 29.5)/10 (fi × ui) 4.5 – 14.5 6 9.5 -2 -12 14.5 – 24.5 11 19.5 -1 -11 24.5 – 34.5 21 29.5 = A 0 0 34.5 – 44.5 23 39.5 1 23 44.5 – 54.5 14 49.5 2 28 54.5 – 64.5 5 59.5 3 15 ∑fi = 80 ∑(fi × ui) = 43

Now, A = 29.5, h = 10, ∑ fi = 80 and ∑ (fi × ui) = 43

28. Weight of 60 eggs were recorded as given below:
 Weight (in grams) 75 - 79 80 - 84 85 - 89 90 - 94 95 - 99 100 - 104 105 - 109 No. of eggs 4 9 13 17 12 3 2

Calculate their mean weight to the nearest gram.

Solution

Let us choose a = 92, h = 5, then d= xi – 92 and ui = (xi – 92)/5

Using step-deviation method, the given data is shown as follows

 Weight (in grams) Number of eggs (fi) Class mark (xi) di = xi - 92 ui = (xi – 92)/5 (fi ui) 74.5 – 79.5 4 77 -15 -3 -12 79.5 – 84.5 9 82 -10 -2 -18 84.5 – 89.5 13 87 -5 -1 -13 89.5 – 94.5 17 92 0 0 0 94.5 – 99.5 12 97 5 1 12 99.5 – 104.5 3 102 10 2 6 104.5 – 109.5 2 107 15 3 6 Total ∑fi = 60 ∑fi ui = - 19

The mean of the given data is given by,

Thus, the mean weight to the nearest gram is 90 g.

29. The following table shows the marks scored by 80 students in an examination.
 Marks 0 - 5 5 - 10 10 - 15 15 - 20 20 - 25 25 - 30 30 - 35 35 - 40 No. of students 3 10 25 49 65 73 78 80

Solution

Let us choose a = 17.5, h = 5, then di = xi – 17.5 and u= (xi – 17.5)/5

Using step-deviation method, the given data is shown as follows:

 Marks Number of students (cf) Frequency (fi) Class mark (xi) di = xi – 17.5 ui = (xi – 17.5)/3 (fi ui) 0 - 5 3 3 2.5 -15 -3 -9 5 - 10 10 7 7.5 -10 -2 -14 10 - 15 25 15 12.5 -5 -1 -15 15 - 20 49 24 17.5 0 0 0 20 - 25 65 16 22.5 5 1 16 25 - 30 73 8 27.5 10 2 16 30 – 35 78 5 32.5 15 3 15 35 – 40 80 2 37.5 20 4 8 Total ∑ fi = 80 ∑ fi ui = 17
The mean of the given data is given by,

= 17.5 + (17/80) × 5

= 17.5 + 1.06

= 18.56

Thus, the mean marks correct to 2 decimal places is 18.56.