KC Sinha: Exercise 6.1 – Mathematics Solution Class 10 Chapter 6 Statistics

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

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

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

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

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

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

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

Question 8 

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

Sol :

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

Question 9 

Find the arithmetic mean of the following distribution:

Sol :

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

Question 10 

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

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

Question 11 

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

Sol :

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

Question 12 

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

Sol :

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

Question 13  

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

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

Question 14 

Find the mean of the following data:

Sol :

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

Question 15 

Find the mean of the following distribution:

Sol :

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

Question 16 

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

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

Question 17 

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

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.

Question 18 

Find the mean of the following frequency distribution:

Sol :

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

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 :

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

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 :

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

Question 21 

Find the mean of the following data:

Sol :

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

Question 22 

Find the mean of the following frequency distribution:

Sol :

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

Question 23 

Find the mean of the following frequency distribution:

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

Question 24 

Find the mean of the following frequency distribution:

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

Question 25 

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

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

Question 26 

Find the mean of the following data:

Sol :

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

Question 27 

Find the mean of the following data:

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

Question 28 

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

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

Question 29 

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

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