Rs Aggarwal 2015 for Class 10 Math Chapter 9 - Mean, Median, Mode Of Grouped Data, Cumulative Frequency Graph And Ogive

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Page No 359:

Question 1:

Find the mean, using direct method:

Class	0-10	10-20	20-30	30-40	40-50
Frequency	3	5	9	5	3

Answer:

Class	Frequency $(f_{i})$	Mid Values $(x_{i})$	*$(f_{i} \times x_{i})$*
0-10	3	5	15
10-20	5	15	75
20-30	9	25	225
30-40	5	35	175
40-50	3	45	135
	$\sum_{f_{i}} = 25$		$\sum (f_{i} \times x_{i}) = 625$

$∴ Mean, \bar{x} = \frac{\sum (f_{i} \times x_{i})}{\sum f_{i}} = \frac{625}{25} =25 ∴ \bar{x} = 25$

Page No 359:

Question 2:

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

Answer:

Class	Frequency $(f_{i})$	Mid Values $(x_{i})$	*$(f_{i} \times x_{i})$*
0-10	7	5	35
10-20	5	15	75
20-30	6	25	150
30-40	12	35	420
40-50	8	45	360
50-60	2	55	110
	$\sum_{f_{i}} = 40$		$\sum (f_{i} \times x_{i}) = 1150$

$∴ Mean, \bar{x} = \frac{\sum (f_{i} \times x_{i})}{\sum f_{i}} = \frac{1150}{40} =28 .75 ∴ \bar{x} = 28 .75$

Page No 359:

Question 3:

Find the mean, using direct method:

Class	10-20	20-30	30-40	40-50	50-60	60-70
Frequency	11	15	20	30	14	10

Answer:

Class	Frequency $(f_{i})$	Mid Values $(x_{i})$	*$(f_{i} \times x_{i})$*
10-20	11	15	165
20-30	15	25	375
30-40	20	35	700
40-50	30	45	1350
50-60	14	55	770
60-70	10	65	650
	$\sum f_{i} = 100$		$\sum (f_{i} \times x_{i}) = 4010$

$∴ Mean, \bar{x} = \frac{\sum (f_{i} \times x_{i})}{\sum f_{i}} = \frac{4010}{100} =40 .1 ∴ \bar{x} = 40.1$

Page No 360:

Question 4:

Find the mean, using direct method:

Marks	10-20	20-30	30-40	40-50	50-60	60-70	70-80
Number of students	6	8	13	7	3	2	1

Answer:

Class	Frequency $(f_{i})$	Mid Values $(x_{i})$	*$(f_{i} \times x_{i})$*
10-20	6	15	90
20-30	8	25	200
30-40	13	35	455
40-50	7	45	315
50-60	3	55	165
60-70	2	65	130
70-80	1	75	75
	$\sum f_{i} = 40$		$\sum (f_{i} \times x_{i}) = 1430$

$∴ Mean, \bar{x} = \frac{\sum (f_{i} \times x_{i})}{\sum f_{i}} = \frac{1430}{40} =35 .75 ∴ \bar{x} = 35.75$

Page No 360:

Question 5:

Find the mean, using direct method:

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

Answer:

Class	Frequency $(f_{i})$	Mid values $(x_{i})$	*$(f_{i} \times x_{i})$*
25-35	6	30	180
35-45	10	40	400
45-55	8	50	400
55-65	12	60	720
65-75	4	70	280
	$\sum f_{i} = 40$		$\sum (f_{i} \times x_{i}) = 1980$

$∴ Mean, \bar{x} = \frac{\sum (f_{i} \times x_{i})}{\sum f_{i}} = \frac{1980}{40} =49 .5 ∴ \bar{x} = 49.5$

Page No 360:

Question 6:

Find the mean, using direct method:

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

Answer:

Class	Frequency $(f_{i})$	Mid values $(x_{i})$	*$(f_{i} \times x_{i})$*
0-100	6	50	300
100-200	9	150	1350
200-300	15	250	3750
300-400	12	350	4200
400-500	8	450	3600
	$\sum f_{i} = 50$		$\sum (f_{i} \times x_{i}) = 13200$

$∴ Mean, \bar{x} = \frac{\sum (f_{i} \times x_{i})}{\sum f_{i}} = \frac{13200}{50} =264 ∴ \bar{x} = 264$

Page No 360:

Question 7:

The mean of the following frequency distribution is 24. Find the value of p.

Marks	0-10	10-20	20-30	30-40	40-50
Number of students	15	20	35	p	10

Answer:

Class	Frequency $(f_{i})$	Mid Values $(x_{i})$	*$(f_{i} \times x_{i})$*
0-10	15	5	75
10-20	20	15	300
20-30	35	25	875
30-40	p	35	35 p
40-50	10	45	450
	$\sum f_{i} = 80 + p$		$\sum (f_{i} \times x_{i}) = 1700 + 35 p$

$∴ Mean, \bar{x} = \frac{\sum (f_{i} \times x_{i})}{\sum f_{i}} \Rightarrow 24 = \frac{1700 + 35 p}{80 + p} [∵ Mean=24] \Rightarrow 1920+24 p = 1700 + 35 p \Rightarrow 11 p = 220 \Rightarrow p = 20 ∴ p = 20$

Page No 360:

Question 8:

Find the missing frequencies f₁ and f₂ in the table in given below, it is being given that the mean of the given frequency distribution is 50.

Class	0-20	20-40	40-60	60-80	80-100	Total
Frequency	17	f₁	32	f₂	19	120

Answer:

Class	Frequency $(f_{i})$	Mid values $(x_{i})$	*$(f_{i} \times x_{i})$*
0-20	17	10	170
20-40	f₁	30	30 f₁
40-60	32	50	1600
60-80	52- f₁	70	3640-70 f₁
80-100	19	90	1710
	$\sum f_{i} = 120$		_{$\sum (f_{i} \times x_{i}) = 7120 - 40 f_{1}$}

$We have: 17+ f_{1} + 32 + f_{2} + 19 = 120 \Rightarrow f_{1} + f_{2} = 52 \Rightarrow f_{2} = 52 - f_{1} ∴ Mean, \bar{x} = \frac{\sum (f_{i} \times x_{i})}{\sum f_{i}} \Rightarrow 50= \frac{7120 - 40 f_{1}}{120} [∵ Mean=50] \Rightarrow 40 f_{1} = 1120 ∴ f_{1} = 28 And f_{2} = 52 - 28 \Rightarrow f_{2} =24 ∴ The missing frequencies are f_{1} = 28 and f_{2} =24.$

Page No 360:

Question 9:

The mean of the following frequency distribution is 57.6 and the sum of the observations is 50

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

Answer:

Class	Frequency $(f_{i})$	Mid values $(x_{i})$	*$(f_{i} \times x_{i})$*
0-20	7	10	70
20-40	f₁	30	30 f₁
40-60	12	50	600
60-80	18- f₁	70	1260-70 f₁
80-100	8	90	720
100-120	5	110	550
	$\sum f_{i} = 50$		$\sum (f_{i} \times x_{i}) = 3200 - 40 f_{1}$

$We have: 7+ f_{1} + 12 + f_{2} + 8 + 5 = 50 \Rightarrow f_{1} + f_{2} = 18 \Rightarrow f_{2} = 18 - f_{1} ∴ Mean, \bar{x} = \frac{\sum (f_{i} \times x_{i})}{\sum f_{i}} \Rightarrow 57 .6= \frac{3200 - 40 f_{1}}{50} [∵ Mean=57 .6] \Rightarrow 40 f_{1} = 320 ∴ f_{1} = 8 And f_{2} = 18 - 8 \Rightarrow f_{2} =10 ∴ The missing frequencies are f_{1} = 8 and f_{2} =10.$

Page No 360:

Question 10:

Find the mean, 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

Answer:

Class	Frequency $(f_{i})$	Mid values $(x_{i})$	Deviation $(d_{i})$ $d_{i} = (x_{i} - 25)$	$(f_{i} \times d_{i})$
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
	$\sum f_{i} = 100$			$\sum (f_{i} \times d_{i}) = 300$

$Let A = 25 be the assumed mean . Then we have: Mean, \bar{x} = A + \frac{\sum (f_{i} \times d_{i})}{\sum f_{i}} = 25+ \frac{300}{100} =28 ∴ \bar{x} = 28$

Page No 360:

Question 11:

Find the mean, using assumed-mean method:

Class	0-40	40-80	80-120	120-160	160-200
Frequency	12	20	35	30	23

Answer:

Class	Frequency $(f_{i})$	Mid values $(x_{i})$	Deviation $(d_{i})$ $(d_{i}) = (x_{i} - 100)$	$(f_{i} \times d_{i})$
0-40	12	20	-80	-960
40-80	20	60	-40	-800
80-120	35	100=A	0	0
120-160	30	140	40	1200
160-200	23	180	80	1840
	$\sum f_{i} = 120$			$\sum (f_{i} \times d_{i}) = 1280$

$Let A = 100 be the assumed mean . Then we have: Mean, \bar{x} = A + \frac{\sum (f_{i} \times d_{i})}{\sum f_{i}} = 100+ \frac{1280}{120} =100+10 .67 ∴ x = 110.67$

Page No 360:

Question 12:

Find the mean, using assumed-mean method:

Class	100-120	120-140	140-160	160-180	180-200
Frequency	10	20	30	15	5

Answer:

Class	Frequency $(f_{i})$	Mid values $(x_{i})$	Deviation $(d_{i})$ $d_{i} = (x_{i} - 150)$	$(f_{i} \times d_{i})$
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
	$\sum f_{i} = 80$			$\sum (f_{i} \times d_{i}) = - 300$

$Let A = 150 be the assumed mean . Then we have: Mean, \bar{x} = A + \frac{\sum (f_{i} \times d_{i})}{\sum f_{i}} = 150 - \frac{300}{80} =150 - 3 .75 ∴ \bar{x} = 146.25$

Page No 360:

Question 13:

Find the mean, using assumed-mean method:

Class	0-20	20-40	40-60	60-80	80-100	100-120
Frequency	20	35	52	44	38	31

Answer:

Class	Frequency $(f_{i})$	Mid Values $(x_{i})$	Deviation $(d_{i})$ $d_{i} = (x_{i} - 50)$	$(f_{i} \times d_{i})$
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
	$\sum f_{i} = 220$			$\sum (f_{i} \times d_{i}) = 2760$

$Let A = 50 be the assumed mean . Then we have: Now, m ean, \bar{x} = A + \frac{\sum (f_{i} \times d_{i})}{\sum f_{i}} =50+ \frac{2760}{220} =50+12 .55 ∴ \bar{x} = 62.55$

Page No 361:

Question 14:

Find the arithmetic mean of each of the following frequency distributions using step-deviation method:

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

Answer:

Class	Frequency $(f_{i})$	Mid Values $(x_{i})$	$u_{i} = \frac{(x_{i} - A)}{h}$ $= \frac{(x_{i} - 25)}{10}$	$(f_{i} \times u_{i})$
0-10	12	5	−2	−24
10-20	18	15	−1	−18
20-30	27	25=A	0	0
30-40	20	35	1	20
40-50	17	45	2	34
50-60	6	55	3	18
	$\sum f_{i} = 100$			$\sum (f_{i} \times u_{i}) = 30$

$Now, A = 25, h = 10, \sum f_{i} = 100 and \sum (f_{_{i}} \times u_{i}) = 30 ∴ Mean, \bar{x} = A + (h \times \frac{\sum (f_{_{i}} \times u_{i})}{\sum f_{i}}) =25+ (10 \times \frac{30}{100}) =25+3 =28 ∴ \bar{x} = 28$

Page No 361:

Question 15:

Find the arithmetic mean of each of the following frequency distributions using step-deviation method:

Class	Number of students
4-8	2
8-12	12
12-16	15
16-20	25
20-24	18
24-28	12
28-32	13
32-36	3

Answer:

Class	Frequency $(f_{i})$	Mid values $(x_{i})$	$u_{i} = \frac{(x_{i} - A)}{h}$ $= \frac{(x_{i} - 18)}{4}$	$(f_{i} \times u_{i})$
4-8	2	6	-3	-6
8-12	12	10	-2	-24
12-16	15	14	-1	-15
16-20	25	18=A	0	0
20-24	18	22	1	18
24-28	12	26	2	24
28-32	13	30	3	39
32-36	3	34	4	12
	$\sum f_{i} = 100$			$\sum (f_{i} \times u_{i}) = 48$

$Now, A = 18, h = 4, \sum f_{i} = 100 and \sum (f_{_{i}} \times u_{i}) = 48 ∴ Mean, \bar{x} = A + {h \times \frac{\sum (f_{_{i}} \times u_{i})}{\sum f_{i}}} =18+ {4 \times \frac{48}{100}} =18+1 .92 =19 .92 ∴ \bar{x} = 19.92$

Page No 361:

Question 16:

Find the arithmetic mean of each of the following frequency distributions using step-deviation method:

Class	0-30	30-60	60-90	90-120	120-150	150-180
Frequency	12	21	34	52	20	11

Answer:

Class	Frequency $(f_{i})$	Mid values $(x_{i})$	$u_{i} = \frac{(x_{i} - A)}{h}$ $= \frac{(x_{i} - 75)}{30}$	$(f_{i} \times u_{i})$
0-30	12	15	−2	−24
30-60	21	45	−1	−21
60-90	34	75 = A	0	0
90-120	52	105	1	52
120-150	20	135	2	40
150-180	11	165	3	33
	_{$\sum f_{i} = 150$}			$\sum (f_{i} \times u_{i}) = 80$

$Now, A = 75, h = 30, \sum f_{i} = 150 and \sum (f_{_{i}} \times u_{i}) = 80 ∴ Mean, \bar{x} = A + {h \times \frac{\sum (f_{_{i}} \times u_{i})}{\sum f_{i}}} =75+ {30 \times \frac{80}{150}} =75+16 =91 ∴ \bar{x} = 91$

Page No 361:

Question 17:

Find the arithmetic mean of each of the following frequency distributions using step-deviation method:

Class	0-20	20-40	40-60	60-80	80-100	100-120	120-140
Frequency	12	18	15	25	26	15	9

Answer:

Class	Frequency $(f_{i})$	Mid values $(x_{i})$	$u_{i} = \frac{(x_{i} - A)}{h} = \frac{(x_{i} - 70)}{20}$	$(f_{i} \times u_{i})$
0-20	12	10	−3	−36
20-40	18	30	−2	−36
40-60	15	50	−1	−15
60-80	25	70 = A	0	0
80-100	26	90	1	26
100-120	15	110	2	30
120-140	9	130	3	27
	$\sum f_{i} = 120$			$\sum (f_{i} \times u_{i}) = - 4$

$Now, A = 70, h = 20, \sum f_{i} = 120 and \sum (f_{_{i}} \times u_{i}) = - 4 ∴ Mean, \bar{x} = A + {h \times \frac{\sum (f_{_{i}} \times u_{i})}{\sum f_{i}}} =70+ {20 \times \frac{(- 4)}{120}} =70-0 .67 =69 .33 ∴ \bar{x} = 69.33$

Page No 361:

Question 18:

Find the arithmetic mean of each of the following frequency distributions using step-deviation method:

Marks	0-14	14-28	28-42	42-56	56-70
Number of students	7	21	35	11	16

Answer:

Class	Frequency $(f_{i})$	Mid values $(x_{i})$	$u_{i} = \frac{(x_{i} - A)}{h} = \frac{(x_{i} - 35)}{14}$	$(f_{i} \times u_{i})$
0-14	7	7	−2	−14
14-28	21	21	−1	−21
28-42	35	35 = A	0	0
42-56	11	49	1	11
56-70	16	63	2	32
	$\sum f_{i} = 90$			$\sum (f_{i} \times u_{i}) = 8$

$Now, A = 35, h = 14, \sum f_{i} = 90 and \sum (f_{_{i}} \times u_{i}) = 8 ∴ Mean, \bar{x} = A + {h \times \frac{\sum (f_{_{i}} \times u_{i})}{\sum f_{i}}} =35+ {14 \times \frac{8}{90}} =35+1 .24 =36 .24 ∴ \bar{x} = 36.24 cm$

Page No 361:

Question 19:

Find the arithmetic mean of each of the following frequency distributions using step-deviation method:

Class	10-15	15-20	20-25	25-30	30-35	35-40
Frequency	5	6	8	12	6	3

Answer:

Class	Frequency $(f_{i})$	Mid values $(x_{i})$	$u_{i} = \frac{(x_{i} - A)}{h} = \frac{(x_{i} - 22.5)}{5}$	$(f_{i} \times u_{i})$
10-15	5	12.5	−2	−10
15-20	6	17.5	−1	−6
20-25	8	22.5 = A	0	0
25-30	12	27.5	1	12
30-35	6	32.5	2	12
35-40	3	37.5	3	9
	$\sum f_{i} = 40$			$\sum (f_{i} \times u_{i}) = 17$

$Now, A = 22.5, h = 5, \sum f_{i} = 40 and \sum (f_{_{i}} \times u_{i}) = 17 ∴ Mean, \bar{x} = A + {h \times \frac{\sum (f_{_{i}} \times u_{i})}{\sum f_{i}}} =22 .5+ {5 \times \frac{17}{40}} =22 .5+2 .125 =24 .625 ∴ x = 24.625$

Page No 361:

Question 20:

Find the arithmetic mean of each of the following frequency distributions 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

Answer:

Class	Frequency $(f_{i})$	Mid values $(x_{i})$	$u_{i} = \frac{(x_{i} - A)}{h} = \frac{(x_{i} - 33)}{6}$	$(f_{i} \times u_{i})$
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
	$\sum f_{i} = 40$			$\sum (f_{i} \times u_{i}) = 2$

$Now, A = 33, h = 6, \sum f_{i} = 40 and \sum (f_{_{i}} \times u_{i}) = 2 ∴ Mean, \bar{x} = A + {h \times \frac{\sum (f_{_{i}} \times u_{i})}{\sum f_{i}}} =33+ {6 \times \frac{2}{40}} =33+0 .3 =33 .3 ∴ x = 33.3 years$

Page No 361:

Question 21:

Find the arithmetic mean of each of the following frequency distributions using step-deviation method:

Class	84-90	90-96	96-102	102-108	108-114	114-120
Frequency	15	22	20	18	20	25

Answer:

Class	Frequency $(f_{i})$	Mid values $(x_{i})$	$u_{i} = \frac{(x_{i} - A)}{h} = \frac{(x_{i} - 99)}{6}$	$(f_{i} \times u_{i})$
84-90	15	87	−2	−30
90-96	22	93	−1	−22
96-102	20	99 = A	0	0
102-108	18	105	1	18
108-114	20	111	2	40
114-120	25	117	3	75
	$\sum f_{i} = 120$			$\sum (f_{i} \times u_{i}) = 81$

$\begin{array}{l} Now, A = 99, h = 6, \sum f_{i} = 120 and \sum (f_{_{i}} \times u_{i}) = 81 \\ ∴ Mean, \bar{x} = A + {h \times \frac{\sum (f_{_{i}} \times u_{i})}{\sum f_{i}}} \\ =99+ {6 \times \frac{81}{120}} \\ =99+4 .05 \\ =103 .05 \\ ∴ \bar{x} = 103.05 \end{array}$

Page No 362:

Question 22:

Find the arithmetic mean of each of the following frequency distributions using step-deviation method:

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

Answer:

Class	Frequency $(f_{i})$	Mid values $(x_{i})$	$u_{i} = \frac{(x_{i} - A)}{h} = \frac{(x_{i} - 550)}{20}$	$(f_{i} \times u_{i})$
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