KC Sinha Mathematics Solution Class 10 Chapter 6 Statistics Exercise 6.1

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11 Q12 Q13 Q14 Q15 Q16 Q17 Q18 Q19 Q20 Q21 Q22 Q23 Q24 Q25 Q26 Q27 Q28 Q29

Exercise 6.1 Exercise 6.2 Exercise 6.3 Exercise 6.4

Exercise 6.1

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Question 1 

The mean of 11 results is 30. If the mean of the first 6 results is 28 and that of last 6 results is 32, find the 6th result.

Sol :

Let the 6^th number be x
Given that mean of 11 results = 30
∴ sum of 11 numbers = 11 × 30 = 330
Mean of the first 6 results = 28
Sum of first 6 numbers = 6 × 28 = 168
Mean of the last 6 results = 32
Sum of the last 6 results = 6 × 32 = 192
Therefore,
Sum of first 6 numbers + sum of last 6 numbers – 6^th number = sum of 11 numbers
168 + 192 - x = 330
⇒ 360 – x = 330
⇒ x = 30

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Question 2 

The mean of 17 observations is 20. If the mean of the first 9 observation is 23 and that of last 9 observations is 18, find the 9th observation.

Sol :
Let the 9^th observation be x
Given that mean of 17 observations = 20
∴ sum of 17 observations = 17 × 20 = 340
Mean of the first 9 observations= 23
Sum of first 9 observations = 9 × 23 = 207
Mean of the last 9 observations = 18
Sum of the last 9 observations = 9 × 18 = 162
Therefore,
Sum of first 9 observations + sum of last 9 observations – 9^th observation = sum of 17 observations
207 + 162 - x = 340
⇒ 369 – x = 340
⇒ x = 29

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Question 3 

The mean weight of 21 students of a class is 52 kg. If the mean weight of the first 11 students of the class is 50 kg and that of the last 11 students is 54 kg, find the weight of the 11th student.

Sol :
Let the weight of 11^th student be x
Given that mean weight of 21 students = 52kg
∴ sum of 21 students weight = 21 × 52 = 1092kg
Mean weight of the first 11 students = 50kg
Sum of first 11 students weight = 11 × 50 = 550kg
Mean weight of the last 11 students = 54kg
Sum of the last 11 students weight = 11 × 54 = 594kg
Therefore,
Sum of first 11 students weight + sum of last 11 students weight – weight of the 11^th student = sum of 21 students weight
550+594 - x = 1092
⇒ 1144 – x = 1092
⇒ x = 52
Hence, weight of 11^th student is 52kg

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Question 4 

The mean weight of 25 students of a class is 60 kg. If the mean weight of the first 13 students of the class is 57 kg and that of the last 13 students is 63 kg, find the weight of the 13th student.

Sol :
Let the weight of 13^th student be x
Given that mean weight of 25 students = 60kg
∴ sum of 25 students weight = 25 × 60 = 1500kg
Mean weight of the first 13 students = 57kg
Sum of first 13 students weight = 13 × 57 = 741kg
Mean weight of the last 13 students = 63kg
Sum of the last 13 students weight = 13 × 63 = 819kg
Therefore,
Sum of first 13 students weight + sum of last 13 students weight – weight of the 13^th student = sum of 25 students weight
741 + 819 - x = 1500
⇒ 1560 – x = 1500
⇒ x = 60
Hence, weight of 13^th student is 60kg

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Question 5

The mean of 23 observations is 34. If the mean of the first 12 observations is 32 and that of the last 12 observations is 38, find the 12th observation.

Sol :
Let the 12^th observation be x
Given that mean of 23 observations = 34
∴ sum of 23 observations = 23 × 34 = 782
Mean of the first 12 observations = 32
Sum of first 12 observations = 12 × 32 = 384
Mean of the last 12 observations = 38
Sum of the last 12 observations = 12 × 38 = 456
Therefore,
Sum of first 12 observations + sum of last 12 observations – 12^th observation = sum of 23 observations
384 + 456 - x = 782
⇒ 840 – x = 782
⇒ x = 58

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Question 6 

The mean of 11 numbers is 35. If the mean of first 6 numbers is 32 and that of last 6 numbers is 37, find the 6th number.

Sol :
Let the 6^th number be x
Given that mean of 11 results = 35
∴ sum of 11 numbers = 11 × 35 = 385
Mean of the first 6 results = 32
Sum of first 6 numbers = 6 × 32 = 192
Mean of the last 6 results = 37
Sum of the last 6 results = 6 × 37 = 222
Therefore,
Sum of first 6 numbers + sum of last 6 numbers – 6^th number = sum of 11 numbers
192 + 222 - x = 385
⇒ 414 – x = 385
⇒ x = 29

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Question 7 

The mean of 25 observations is 36. If the mean of the first 13 observations is 32 and that of the last 13 observations is 39, find the 13th observation.

Sol :
Let the 13^th observation be x
Given that mean of 25 observations = 36
∴ sum of 25 observations = 25 × 36 = 900
Mean of the first 13 observations = 32
Sum of first 13 observations = 13 × 32 = 416
Mean of the last 13 observations = 39
Sum of the last 13 observations = 13 × 39 = 507
Therefore,
Sum of first 13 observations + sum of last 13 observations – 13^th observation = sum of 25 observations
416 + 507 - x = 900
⇒ 923 – x = 900
⇒ x = 23

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Question 8 

If the mean of the following data is 25, find the value of k.

x	5	15	25	35	45
f	3	k	3	6	2

Sol :

xi	fi	xifi
5	5	15
15	K	15k
25	3	75
35	6	210
45	2	90
Total	Σfi=14+k	Σfixi=390+15k

Now, $\overline{\mathrm{x}}=\frac{\sum \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}}{\sum \mathrm{f}_{\mathrm{i}}}$
$\Rightarrow 25=\frac{390+15 k}{14+k}$
⇒25(14+k) = 390+ 15k
⇒350 + 25k = 390 + 15k
⇒ 25k – 15k = 390 – 350
⇒ 10k = 40
⇒ k = 4

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Question 9 

Find the arithmetic mean of the following distribution:

Marks obtained	10	15	20	25	30
No. of students	2	4	6	8	10

Sol :

Marks obtained (xi)	No. of student (fi)	xifi
10	2	20
15	4	60
20	6	120
25	8	200
30	10	300
Total	Σfi=30	Σfixi=700

Now, $\overline{\mathrm{x}}=\frac{\sum \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}=\frac{700}{30}=23.33$

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Question 10 

The mean of the following frequency distribution is 62.8. Find the missing frequency x:

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

Sol :

Now, $\overline{\mathrm{x}}=\frac{\sum \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}}{\sum \mathrm{f}_{\mathrm{i}}}$
$\Rightarrow 62.8=\frac{2640+50 \mathrm{x}}{40+\mathrm{x}}$
⇒62.8 (40 + x) = 2640 + 50x
⇒2512 + 62.8x = 2640 + 50x
⇒ 62.8x – 50x = 2640 – 2512
⇒ 12.8x = 128
⇒ x = 10

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Question 11 

The arithmetic mean of the following data is 14. Find the value of p:

x	5	10	15	20	25
f	7	p	8	4	5

Sol :

(xi)	(fi)	xifi
5	7	35
10	p	10p
15	8	120
20	4	80
25	5	125
Total	Σfi=24+p	Σfixi=360+10p

Now, $\overline{\mathrm{x}}=\frac{\sum \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}}{\sum \mathrm{f}_{\mathrm{i}}}$
$\Rightarrow 14=\frac{360+10 p}{24+p}$
⇒14 (24+p) = 360 + 10p
⇒336 + 14p = 360 + 10p
⇒ 14p – 10p = 360 – 336
⇒ 4p = 24
⇒ p = 6

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Question 12 

If the mean of the following data is 18, find the missing frequency p:

x	10	15	20	25
f	5	10	p	8

Sol :

(xi)	(fi)	xifi
10	5	50
15	10	150
20	p	20p
25	8	200
Total	Σfi=23+p	Σfixi=400+20p

Now, $\overline{\mathrm{x}}=\frac{\sum \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}}{\sum \mathrm{f}_{\mathrm{i}}}$
$\Rightarrow 18=\frac{400+20 p}{23+p}$
⇒18 (23 + p) = 400 + 20p
⇒414 + 18p = 400 + 20p
⇒ 18p – 20p = 400 – 414
⇒ -2p = -14
⇒ p = 7

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Question 13  

Find the value of p if the mean of the following distribution is 7.5:

x	3	5	7	9	11	13
f	6	8	15	p	8	4

Sol :

Now, $\overline{\mathrm{x}}=\frac{\sum \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}}{\sum \mathrm{f}_{\mathrm{i}}}$
$\Rightarrow 7.5=\frac{303+9 p}{41+p}$
⇒7.5(41+p) = 303 + 9p
⇒307.5 + 7.5p = 303 + 9p
⇒ 7.5p – 9p = 303 – 307.5
⇒ -1.5p = -4.5
⇒ p = 3

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Question 14 

Find the mean of the following data:

Class interval	0-10	10-20	20-30	30-40	40-50
Frequency	12	11	8	10	9

Sol :

Now, $\overline{\mathrm{x}}=\frac{\sum \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}}{\sum \mathrm{f}_{\mathrm{i}}}=\frac{1180}{50}=23.6$

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Question 15 

Find the mean of the following distribution:

Sol :

Now, $\overline{\mathrm{x}}=\frac{\sum \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}}{\sum \mathrm{f}_{\mathrm{i}}}=\frac{1992}{100}=19.92$

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Question 16 

The arithmetic mean of the following frequency distribution is 53. Find the value of p:

Class	0-20	20-40	40-60	60-80	80-100
Frequency	12	15	32	p	13

Sol :

Now, $\overline{\mathrm{x}}=\frac{\sum \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}}{\sum \mathrm{f}_{\mathrm{i}}}$
$\Rightarrow 53=\frac{1900+70 p}{72+p}$
⇒53 (72+p) = 3340 + 70p
⇒3816 + 53p = 3340 + 70p
⇒ 53p – 70p = 3340 – 3816
⇒ -17p = -476
⇒ p = 28

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Question 17 

If the mean of the following distribution is 5. Find the value of f₁:

Class	0-20	20-40	40-60	60-80	80-100
Frequency	17	28	32	f1	19

Sol :

Now, $\overline{\mathrm{x}}=\frac{\sum \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}}{\sum \mathrm{f}_{\mathrm{i}}}$
$\Rightarrow 5=\frac{4320+70 f_{1}}{96+f_{1}}$
⇒ 5(96+ f₁) = 4320 + 70f₁
⇒480 + 5f₁ = 4320 + 70f₁
⇒ 5f₁ – 70f₁ = 4320 - 480
⇒ -65f₁ = +3840
⇒ f₁ = - 59.07
This is not possible as frequency can not be negative.

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Question 18 

Find the mean of the following frequency distribution:

Class	0-20	20-40	40-60	60-80	80-100
Frequency	15	18	21	29	17

Sol :

Now, $\overline{\mathrm{x}}=\frac{\sum \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}=\frac{5300}{100}=53$

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Question 19  

The mean of the following frequency distribution is 62.8 and the sum of all frequency is 50. Compute the missing frequency f₁ and f₂ :

Class	0-20	2040	40-60	60-80	80-100	100-120	Total
Frequency	5	f1	10	f2	7	8	50

Sol :

Now, $\overline{\mathrm{x}}=\frac{\sum \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}}{\sum \mathrm{f}_{\mathrm{i}}}$
$\Rightarrow 62.8=\frac{2060+30 f_{1}+70 f_{2}}{50}$ [given: ∑f_i = 50]
⇒ 62.8(50) = 2060 + 30f₁ +70f₂
⇒ 3140 = 2060 + 30f₁ +70f₂
⇒ 3140 - 2060 = 30f₁ +70f₂
⇒ 1080 = 30f₁ +70f₂
⇒ 108 = 3f₁ +7f₂ …(i)
and 30 + f₁ +f₂ = 50
⇒ f₁ +f₂ = 50 – 30
⇒ f₁ +f₂ = 20
⇒ f₁ = 20 – f₂ …(ii)
Now, putting the value of f₁ in eq. (i), we get
3(20 –f₂) + 7f₂ = 108
⇒ 60 – 3f₂ + 7f₂ = 108
⇒ 4f₂ = 108 – 60
⇒ 4f₂ = 48
⇒ f₂ = 12
Now, substitute the value of f₂ in eq. (ii), we get
f₁ = 20 – 12
⇒ f₁ = 8

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Question 20 

The mean of the following frequency distribution is 57.6 and the sum of the frequencies is 50. Find the missing frequencies f₁ and f₂ :

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

Sol :

Now, $\overline{\mathrm{x}}=\frac{\sum \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{j}}}{\sum \mathrm{f}_{\mathrm{i}}}$
$\Rightarrow 57.6=\frac{2060+30 f_{1}+70 f_{2}}{50}$ [given: ∑f_i = 50]
⇒ 57.6(50) = 1940 + 30f₁ +70f₂
⇒ 2880 = 1940 + 30f₁ +70f₂
⇒ 2880 – 1940 = 30f₁ +70f₂
⇒ 940 = 30f₁ +70f₂
⇒ 94 = 3f₁ +7f₂ …(i)
and 32 + f₁ +f₂ = 50
⇒ f₁ +f₂ = 50 – 32
⇒ f₁ +f₂ = 18
⇒ f₁ = 18 – f₂ …(ii)
Now, putting the value of f₁ in eq. (i), we get
3(18 –f₂) + 7f₂ = 94
⇒ 54 – 3f₂ + 7f₂ = 94
⇒ 4f₂ = 94 – 54
⇒ 4f₂ = 40
⇒ f₂ = 10
Now, substitute the value of f₂ in eq. (ii), we get
f₁ = 18 – 10
⇒ f₁ = 8

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Question 21 

Find the mean of the following data:

Class Interval	50-60	60-70	70-80	80-90	90-100
Frequency	8	6	12	11	13

Sol :

Now, $\overline{\mathrm{x}}=\frac{\sum \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}}{\sum \mathrm{f}_{\mathrm{i}}}=\frac{3900}{50}=78$

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Question 22 

Find the mean of the following frequency distribution:

Class Interval	15-25	25-35	35-45	45-55	55-56
Frequency	60	35	22	18	15

Sol :

Now, $\overline{\mathrm{x}}=\frac{\sum \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}}{\sum \mathrm{f}_{\mathrm{i}}}=\frac{4930}{150}=32.87$

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Question 23 

Find the mean of the following frequency distribution:

Class	50-60	70-90	90-110	110-130	130-150	150-170
Frequency	18	12	13	27	8	22

Sol :
Here, the class size varies, and x_i’s are large. Now, we apply the step deviation method with a = 120 and h = 20

Now, $\overline{\mathrm{x}}=\mathrm{a}+\mathrm{h}\left(\frac{\sum \mathrm{f}_{\mathrm{i}} \mathrm{u}_{\mathrm{i}}}{\sum \mathrm{f}_{\mathrm{i}}}\right)$
$\Rightarrow \overline{\mathrm{x}}=120+20\left(\frac{-39}{100}\right)$
$\Rightarrow \overline{\mathrm{x}}=120+\left(\frac{-78}{10}\right)$
$\Rightarrow \overline{\mathrm{x}}=\frac{1200-78}{10}$
⇒ $\overline{\mathrm{x}}$= 112.2

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Question 24 

Find the mean of the following frequency distribution:

Class	25-29	30-34	35-39	40-44	45-49	50-54	55-59
Frequency	14	22	16	6	5	3	4

Sol :
Here, we can see that the class interval is not continuous. So, we make it continuous.

Now, $\overline{\mathrm{x}}=\mathrm{a}+\mathrm{h}\left(\frac{\sum \mathrm{f}_{\mathrm{i}} \mathrm{u}_{\mathrm{i}}}{\sum \mathrm{f}_{\mathrm{i}}}\right)$
$\Rightarrow \overline{\mathrm{x}}=42+5\left(\frac{-79}{70}\right)$
$\Rightarrow \bar{x}=\frac{2940-395}{70}$
⇒ x̄ = 36.36

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Question 25 

The following table gives the marks scored by 50 students in a class-test:

Marks	0-100	100-200	200-300	300-400	400-500	500-600
No. of students	2	8	12	20	5	3

Find the mean marks scored by a student in the class-test.
Sol :
Here, the x_i’s are large. Now, we apply the step deviation method with a = 350 and h = 100

Now, $\overline{\mathrm{x}}=\mathrm{a}+\mathrm{h}\left(\frac{\sum \mathrm{f}_{\mathrm{i}} \mathrm{u}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}\right)$
$\Rightarrow \bar{x}=350+100\left(\frac{-23}{50}\right)$
⇒x̄ = 350 - 46
⇒x̄ = 304
Hence, the mean marks scored by a student in the class-test is 304

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Question 26 

Find the mean of the following data:

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

Sol :

Now, $\overline{\mathrm{x}}=\frac{\sum \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}}{\sum \mathrm{f}_{\mathrm{i}}}=\frac{625}{25}=25$

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Question 27 

Find the mean of the following data:

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

Sol :
Here, the x_i’s are large. Now, we apply the step deviation method with a = 250 and h = 100

Now, $\overline{\mathrm{x}}=\mathrm{a}+\mathrm{h}\left(\frac{\sum \mathrm{f}_{\mathrm{i}} \mathrm{u}_{\mathrm{i}}}{\sum \mathrm{f}_{\mathrm{i}}}\right)$
$\Rightarrow \overline{\mathrm{x}}=250+100\left(\frac{7}{50}\right)$
⇒x̄ = 250 + 14
⇒x̄ = 264

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Question 28 

The following table gives the marks scored by 80 students in a class-test:

Marks	0-50	50-100	100-150	150-200	200-250	250-300
No. of students	8	12	20	25	10	5

Find the mean marks scored by a student in the class-test.
Sol :
Here, the x_i’s are large. Now, we apply the step deviation method with a = 175 and h = 50

Now, $\overline{\mathrm{x}}=\mathrm{a}+\mathrm{h}\left(\frac{\sum \mathrm{f}_{\mathrm{i}} \mathrm{u}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}\right)$
$\Rightarrow \bar{x}=175+50\left(\frac{-48}{80}\right)$
⇒x̄ = 175 - 30
⇒x̄ = 145
Hence, the mean marks scored by a student in the class-test is 145

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Question 29 

The following table gives the distribution of expenditure of different families on education. Find the mean expenditure on education of a family:

Expenditure (in Rs)	Number of Families
1000-1500	24
1500-2000	40
2000-2500	33
2500-3000	28
3000-3500	30
3500-4000	22
4000-4500	16
4500-5000	7

Sol :
Here, the x_i’s are large. Now, we apply the step deviation method with a = 3250 and h = 500

Now, $\overline{\mathrm{x}}=\mathrm{a}+\mathrm{h}\left(\frac{\sum \mathrm{f}_{\mathrm{i}} \mathrm{u}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}\right)$
$\Rightarrow \bar{x}=3250+500\left(\frac{-235}{200}\right)$
⇒x̄ = 3250 – 587.5
⇒x̄ = 2662.5
Hence, the mean expenditure on education of a family is Rs 2662.5

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S.no	Chapters	Links
1	Real numbers	Exercise 1.1 Exercise 1.2 Exercise 1.3 Exercise 1.4
2	Polynomials	Exercise 2.1 Exercise 2.2 Exercise 2.3
3	Pairs of Linear Equations in Two Variables	Exercise 3.1 Exercise 3.2 Exercise 3.3 Exercise 3.4 Exercise 3.5
4	Trigonometric Ratios and Identities	Exercise 4.1 Exercise 4.2 Exercise 4.3 Exercise 4.4
5	Triangles	Exercise 5.1 Exercise 5.2 Exercise 5.3 Exercise 5.4 Exercise 5.5
6	Statistics	Exercise 6.1 Exercise 6.2 Exercise 6.3 Exercise 6.4
7	Quadratic Equations	Exercise 7.1 Exercise 7.2 Exercise 7.3 Exercise 7.4 Exercise 7.5
8	Arithmetic Progressions (AP)	Exercise 8.1 Exercise 8.2 Exercise 8.3 Exercise 8.4
9	Some Applications of Trigonometry: Height and Distances	Exercise 9.1
10	Coordinates Geometry	Exercise 10.1 Exercise 10.2 Exercise 10.3 Exercise 10.4
11	Circles	Exercise 11.1 Exercise 11.2
12	Constructions	Exercise 12.1
13	Area related to Circles	Exercise 13.1
14	Surface Area and Volumes	Exercise 14.1 Exercise 14.2 Exercise 14.3 Exercise 14.4
15	Probability	Exercise 15.1