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

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 9B Exercise 9C Exercise 9D Exercise 9E Exercise 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	Ʃfi = 80		Ʃ 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	Ʃ fi = 63 + x + y	Ʃ fi xi = 2775 + 25x + 45y

Sum of the frequencies = 100 

⇒ ∑_i  f_i = 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 f₁ and f₂ if 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 f_i = 100 

⇒ 35 + f₁ + f₂ = 100 

⇒ f₁ + f₂ = 100 – 35 

⇒ f₁ + f₂ = 65

⇒ f₂ = 65 – f₁ ….(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 d_i = x_i – 25 and u_i = (x_i – 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 d_i = x_i – 40 and u_i = (x_i – 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

21. The 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_i = x_i – 202.5 and u_i = (x_i – 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, d_i = x_i – 45 and u_i = (x_i – 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 d_i = x_i – 52.5 and u_i = (x_i – 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, ∑f_i = 40 and ∑(f_i × u_i) = 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_i × u_i) = - 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, ∑ f_i = 70 and ∑ (f_i × u_i) = - 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 ∑ (f_i × u_i) = 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_i = x_i – 92 and u_i = (x_i – 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 d_i = x_i – 17.5 and u_i = (x_i – 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.