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

Share with your friends

Rs Aggarwal 2015 Solutions for Class 10 Math Chapter 9 Mean, Median, Mode Of Grouped Data, Cumulative Frequency Graph And Ogive are provided here with simple step-by-step explanations. These solutions for Mean, Median, Mode Of Grouped Data, Cumulative Frequency Graph And Ogive are extremely popular among Class 10 students for Math Mean, Median, Mode Of Grouped Data, Cumulative Frequency Graph And Ogive Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2015 Book of Class 10 Math Chapter 9 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2015 Solutions. All Rs Aggarwal 2015 Solutions for class Class 10 Math are prepared by experts and are 100% accurate.

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