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Page No 23.12:

Question 1:

A die was thrown 20 times and the following scores were recorded:
5, 2, 1, 3, 4, 4, 5, 6, 2, 2, 4, 5, 5, 6, 2, 2, 4, 5, 5, 1
Prepare the frequency table of the scores on the upper face of the die and find the mean score.

Answer:

The frequency table for the given data is as follows:
x: 1           2           3          4              5                 6
f:  2           5           1          4              6                 2

In order to compute the arithmetic mean, we prepare the following table:

                                  Computation of Arithmetic Mean

Scores (xi) Frequency (fi) xifi
1 2 2
2 5 10
3 1 3
4 4 16
5 6 30
6 2 12
Total fi = 20 fixi =73

We have, fi = 20 and fixi =73
∴ The mean score = fixifi = 7320 = 3.65.

Page No 23.12:

Question 2:

The daily wages (in Rs) of 15 workers in a factory are given below:
200, 180, 150, 150, 130, 180, 180, 200, 150, 130, 180, 180, 200, 150, 180
Prepare the frequency table and find the mean wage.

Answer:

The frequency table for the given data is as follows:
          Wagesxi ) :                               130        150          180         200
Number of workersfi ):                  2           4              6           3 

In order to compute the mean wage, we prepare the following table:

Mean wages of the workers
xi  fi  fi xi
130 2  260
150 4  600
180 6 1080
200 3  600
Total fi =N = 15  fi xi =  2540

Mean  = fi xifi  = 254015 = Rs. 169.33 .

Page No 23.12:

Question 3:

The following table shows the weights (in kg) of 15 workers in a factory:

Weight (in kg): 60 63 66 72 75
Numbers of workers: 4 5 3 1 2
Calculate the mean weight.

Answer:

                                    Calculation of Mean

 xi fi fi xi
 60  4 240
63  5 315
66  3 198
72  1 72
75  2 150
Total fi = 15 fixi = 975

∴ Mean weight = ​ fixifi  = 97515 = 65 kg.

Page No 23.12:

Question 4:

The ages (in years) of 50 students of a class in a school are given below:

Age (in years): 14 15 16 17 18
Numbers of students: 15 14 10 8 3
Find the mean age.

Answer:

        Calculation of Mean
 xi fi fi xi
 14  15 210
15  14 210
16  10 160
17  8 136
18  3 54
Total fi = 50 fixi = 770

∴ Mean age = ​ fixifi  = 77050 = 15.4 yrs.

Page No 23.12:

Question 5:

Calculate the mean for the following distribution:

x : 5 6 7 8 9
f : 4 8 14 11 3

Answer:

                       Calculation of Mean
 xi fi fi xi
 5  4 20
6  8 48
7  14 98
8  11 88
9  3 27
Total fi = 40 fixi = 281

∴ Mean = ​ fixifi  = 28140 = 7.025 .



Page No 23.13:

Question 6:

Find the mean of the following data:

x: 19 21 23 25 27 29 31
f: 13 15 16 18 16 15 13

Answer:

                    Calculation of Mean

xi fi fixi
19 13  247
21 15 315
23 16 368
25 18 450
27 16 432
29 15 435
31 13 403
Total  fi= N = 106 fixi = 2650

 ∴ Mean  =fixifi = 2650106 = 25.

Page No 23.13:

Question 7:

The mean of the following data is 20.6. Find the value of p.

x: 10 15 p 25 35
f: 3 10 25 7 5

Answer:

            Calculation of Mean

xi fi fi xi
10 3 30
15 10 150
p 25 25p
25 7 175
35 5 175
Total fi = 50 fixi = 530 + 25p

We have: 

fi = 50, ​fixi = 530 +25p

∴ Mean =fixifi  

20.6 = 530 +25p5020.6×50 = 530 +25p 1030 = 530 +25p 1030 - 530 = 25p 500 = 25p p = 50025 p = 20 

Page No 23.13:

Question 8:

If the mean of the following data is 15, find p.

x: 5 10 15 20 25
f: 6 p 6 10 5

Answer:

                                 Calculation of Mean

xi fi fi xi
5 6 30
10 p 10p
15 6 90
20 10 200
25 5 125
Total fi = 27 + p fixi = 445 + 10p

We have:

fi = 27 + p,  ​fixi = 445 +10p

∴ Mean =fixifi  

15= 445 +10p27 +p15 (27 +p) = 445 +10p 405 + 15p =445 +10p 15p - 10p = 445 -405 5p = 40 p = 40÷5

Therefore, p =  8.

Page No 23.13:

Question 9:

Find the value of p for the following distribution whose mean is 16.6

x: 8 12 15 p 20 25 30
f: 12 16 20 24 16 8 4

Answer:

​                      Calculation of Mean

xi fi fixi
8 12 96
12 16 192
15 20 300
p 24 24p
20 16 320
25 8 200
30 4 120
Total  fi= N = 100 fixi = 1228 + 24p

We have:

fi = 100, fixi = 1228 +24p  

∴ Mean  =fixifi 

16.6 = 1228 +24p10016.6×100 = 1228 +24p1660 = 1228 +24p1660 - 1228 = 24p432 = 24pp = 43224p =18.

Page No 23.13:

Question 10:

Find the missing value of p for the following distribution whose mean is 12.58

x: 5 8 10 12 p 20 25
f: 2 5 8 22 7 4 2

Answer:

​                      Calculation of Mean

xi fi fixi
5 2 10
8 5 40
10 8 80
12 22 264
p 7 7p
20 4 80
25 2 50
Total  fi= N = 50 fixi = 524 + 7p

We have:

fi = 50, fixi = 524 +7p  

∴ Mean  =fixifi 

12.58 = 524 +7p5012.58×50 = 524 +7p629 = 524 +7p629 - 524 = 7p105 = 7pp = 1057p =15.

Page No 23.13:

Question 11:

Find the missing frequency (p) for the following distribution whose mean is 7.68

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

Answer:

                                 Calculation of Mean
xi fi fi xi
3 6 18
5 8 40
7 15 105
9 p 9p
11 8 88
13 4 52
Total fi = 41 + p fixi = 303 + 9p

We have:

fi = 41+ p, ​ fixi = 303 +9p

∴ Mean =fixifi  

7.68 = 303 +9p41+p7.68 × (41 +p) =303 +9p314.88 + 7.68p = 303 +9p 314.88 -303 = 9p -7.68p11.88 = 1.32pp = 11.881.32  p =9.

Page No 23.13:

Question 12:

Find the value of p, if the mean of the following distribution is 20

x: 15 17 19 20 + p 23
f: 2 3 4 5 p 6

Answer:

                                 Calculation of Mean
xi fi fi xi
15 2 30
17 3 51
19 4 76
20 + p 5p (20+p)5p
23 6 138
Total fi = 15 + 5p fixi = 295 + (20 + p) 5p

We have:

fi = 15 + 5p, ​fixi = 295 + (20 + p5p​.

∴ Mean =fixifi  

20= 295 + (20+p)5p15+5p20 × (15 +5p) =295 + (20+p)5p300+ 100p = 295 +100p + 5p

⇒ 300 295 100-1005p2
⇒ 5p2
⇒ p2 1
⇒ 1.



Page No 23.16:

Question 1:

Find the median of the following data (1-8)
83, 37, 70, 29, 45, 63, 41, 70, 34, 54

Answer:

 Arranging the data in ascending order, we have:
 

              29, 34, 37, 41, 45, 54, 63, 70, 70, 83

Here, the number of observations, n = 10 (Even).

  Median=n2th observation +n2 +1th observation2Median=Value of 5th observation +Value of 6th observation2Median=(45 + 54)2 = 49.5
Hence, the median of the given data is 49.5.

Page No 23.16:

Question 2:

Find the median of the following data (1-8)
133, 73, 89, 108, 94, 104, 94, 85, 100, 120

Answer:

 Arranging the data in ascending order, we have:

 73, 85, 89, 94, 94, 100, 104, 108, 120, 133.

Here, the number of observations n = 10 (Even).

  Median=n2th observation +n2 +1th observation2Median=Value of 5th observation +Value of 6th observation2Median=(94 + 100)2 = 1942Median=97
Hence, the median of the given data is 97.

Page No 23.16:

Question 3:

Find the median of the following data (1-8)
31, 38, 27, 28, 36, 25, 35, 40

Answer:

 Arranging the data in ascending order, we have:

              25,27, 28, 31, 35, 36, 38, 40

Here, the number of observations n = 8 (Even).

  Median=n2th observation +n2 +1th observation2Median=Value of 4th observation +Value of 5th observation2Median=(31 + 35)2 = 662Median=33
Hence, the median of the given data is 33.

Page No 23.16:

Question 4:

Find the median of the following data (1-8)
15, 6, 16, 8, 22, 21, 9, 18, 25

Answer:

 Arranging the data in ascending order, we have:

             6, 8, 9, 15,16,18, 21, 22, 25

Here, the number of observations n = 9 (Odd).

  Median=Value of n+12th observation, i.e., value of the 5th observation =16. 
Hence, the median of the given data is 16.

Page No 23.16:

Question 5:

Find the median of the following data (1-8)
41, 43, 127, 99, 71, 92, 71, 58, 57

Answer:

 Arranging the given data in ascending order, we have:

  41, 43, 57, 58, 71,71, 92, 99, 127 
Here, n = 9, which is odd.

∴ Median = Value of 9+12th observation, i.e., the 5th observation = 71.

Page No 23.16:

Question 6:

Find the median of the following data (1-8)
25, 34, 31, 23, 22, 26, 35, 29, 20, 32

Answer:

 Arranging the given data in ascending order, we have:

  20, 22, 23, 25, 26, 29, 31, 32, 34, 35

Here, n = 10, which is even.

 Median = Value of n2th observation + Value of n2 +1th observation2Median =Value of 5th observation + Value of 6th observation2Median =26 +292 = 552Median =27.5
Hence, the median is 27.5 for the given data.

Page No 23.16:

Question 7:

Find the median of the following data (1-8)
12, 17, 3, 14, 5, 8, 7, 15

Answer:

 Arranging the given data in ascending order, we have:

 3,5,7,8,12,14,15,17

Here, n = 8, which is even.

 Median = Value of n2th observation + Value of n2 +1th observation2Median =Value of 4th observation + Value of 5th observation2Median =8 +122 Median =10.
Hence, the median of the given data is 10.

Page No 23.16:

Question 8:

Find the median of the following data (1-8)
92, 35, 67, 85, 72, 81, 56, 51, 42, 69

Answer:

 Arranging the given data in ascending order, we have:

 35, 42, 51, 56, 67, 69, 72, 81, 85, 92

Here, n = 10, which is even.

 Median = Value of n2th observation + Value of n2 +1th observation2Median =Value of 5th observation + Value of 6th observation2Median =67 +692  = 1362Median =68.
Hence, the median of the given data is 68.

Page No 23.16:

Question 9:

Numbers 50, 42, 35, 2x + 10, 2x − 8, 12, 11, 8, 6 are written in descending order and their median is 25, find x.

Answer:

Here, the number of observations n is 9. Since n is odd , the median is the n+12th observation, i.e. the 5th  observation.
As the numbers are arranged in the descending order, we therefore observe from the last.
Median =  ​5th  observation.
 25 = 2x -8
 2x = 25 +8
 2x = 33
x = 332
x = 16.5
Hence, ​x = 16.5.

Page No 23.16:

Question 10:

Find the median of the following observations : 46, 64, 87, 41, 58, 77, 35, 90, 55, 92, 33. If 92 is replaced by 99 and 41 by 43 in the above data, find the new median?

Answer:

 Arranging the given data in ascending order, we have:

      33, 35, 41, 46, 55, 58, 64, 77, 87, 90, 92

  Here, the number of observations n is 11 (odd).

Since the number of observations is odd, therefore, 
 
Median = Value of n+12th observation = Value of the 6th observation = 58.
Hence, median = 58.
           
If 92 is replaced by 99 and 41 by 43, then the new observations arranged in ascending order are:

 33, 35, 43, 46, 55, 58, 64, 77, 87, 90, 99.

∴ New median =  Value of the 6th observation = 58.

Page No 23.16:

Question 11:

Find the median of the following data : 41, 43, 127, 99, 61, 92, 71, 58, 57, If 58 is replaced by 85, what will be the new median?

Answer:

 Arranging the given data in ascending order, we have:

  41, 43, 57, 58, 61, 71, 92, 99,127

  Here, the number of observations, n, is 9(odd).

∴ Median = Value of n+12th observation = Value of the 5th observation = 61.
Hence, the median = 61.
If 58 is replaced by 85 , then the new observations arranged in ascending order are:
 41, 43, 57,  61, 71, 85, 92, 99,12 .
∴ New median =  Value of the 5th observation = 71.

Page No 23.16:

Question 12:

The weights (in kg) of 15 students are : 31, 35, 27, 29, 32, 43, 37, 41, 34, 28, 36, 44, 45, 42, 30. Find the median. If the weight 44 kg is replaced by 46 kg and 27 kg by 25 kg, find the new median.

Answer:

Arranging the given data in ascending order, we have:

  27, 28, 29, 30, 31, 32, 34, 35, 36, 37, 41, 42, 43, 44, 45.

  Here, the number of observations n is 15(odd).

   Since the number of observations is odd, therefore, 
   Median = Value of n+12th observation = Value of the 8th observation = 35.
  Hence, median = 35 kg.

If 44 kg is replaced by 46 kg and 27 kg by 25 kg , then the new observations arranged in ascending order are:
25, 28, 29, 30, 31, 32, 34, 35, 36, 37, 41, 42, 43, 45, 46.
 ∴ New median =  Value of the 8th observation = 35 kg.

Page No 23.16:

Question 13:

The following observations have been arranged in ascending order. If the median of the data is 63, find the value of x :
29, 32, 48, 50, x, x + 2, 72, 78, 84, 95

Answer:

Here, the number of observations n is 10. Since n is even,

 Median = n2th observation+ n2+1thobservation2Median=Value of 5th observation+Value of 6th observation263 =x+ (x+2)263 =2x+22 = 2(x+1)2 63 = x+1x = 63-1 = 62.
Hence, ​x = 62.



Page No 23.20:

Question 1:

Find the mode and median of the data: 13, 16, 12, 14, 19, 12, 14, 13, 14
By using the empirical relation also find the mean.

Answer:

Arranging the data in ascending order such that same numbers are put together, we get:


12,12,13,13, 14,14,14, 16, 19
Here, n = 9.
∴ Median = Value of n+12th observation = Value of the 5th observation = 14.
Here, 14 occurs the maximum number of times, i.e., three times. Therefore, 14 is the mode of the data.

Now,
Mode = 3 Median - 2 Mean
 14 = 3 x 14 - 2 Mean
2 Mean  = 42 - 14 = 28
 Mean = 28 ÷ 2 = 14.
 

Page No 23.20:

Question 2:

Find the median and mode of the data: 35, 32, 35, 42, 38, 32, 34

Answer:

Arranging the data in ascending order such that same numbers are put together, we get:

32, 32, 34,35,35, 38,42.

Here, n = 7
∴ Median = Value of n+12th observation = Value of the 4th observation = 35.
Here, 32 and 35, both occur twice. Therefore, 32 and 35 are the two modes.

Page No 23.20:

Question 3:

Find the mode of the data: 2, 6, 5, 3, 0, 3, 4, 3, 2, 4, 5, 2, 4

Answer:

Arranging the data in ascending order such that same values are put together, we get:

0, 2, 2, 2, 3, 3,3,4,4,4,5,5,6.

Here, 2,3 and 4 occur three times each. Therefore, 2 ,3 and 4 are the three modes.

Alternate Solution
​Arranging the data in the form of a frequency table, we have:
Values Tally Bars Frequency
0 1
2 ∣∣∣ 3
3 ∣∣∣ 3
4 ∣∣∣ 3
5 ∣∣ 2
6 1
Total   13

Clearly, the values 2,3 and 4 occur the maximum number of times, i.e., three times.
Hence, the mode is 2,3 and 4.

Page No 23.20:

Question 4:

The runs scored in a cricket match by 11 players are as follows:
6, 15, 120, 50, 100, 80, 10, 15, 8, 10, 10
Find the mean, mode and median of this data.

Answer:

Arranging the data in ascending order such that same values are put together, we get:

6,8,10,10,10,15,15,50,80,100,120.

Here, n = 11
∴ Median = Value of n+12th observation = Value of the 6th observation = 15.
Here, 10 occurs three times. Therefore, 10 is the mode of the given data.
Now, 
Mode = 3 Median - 2 Mean
⇒ 10 = 3 x 15 - 2 Mean
⇒2 Mean  = 45 - 10 = 35
⇒ Mean = 35 ÷ 2 = 17.5.

Page No 23.20:

Question 5:

Find the mode of the following data:
12, 14, 16, 12, 14, 14, 16, 14, 10, 14, 18, 14

Answer:

Arranging the data in ascending order such that same values are put together, we get:

10,12,12,14,14,14,14,14,14, 16, 16, 18.

Here, clearly, 14 occurs the most number of times.
Therefore, 14 is the mode of the given data.

Alternate solution:
Arranging the data in the form of a frequency table, we get:
  Values Tally Bars Frequency
10 1
12 ∣∣ 2
14 IIIII 6
16 ∣∣ 2
18 1
Total   12

Clearly, 14 has maximum frequency. So, the mode of the given data is 14. 

Page No 23.20:

Question 6:

Heights of 25 children (in cm) in a school are as given below:
168, 165, 163, 160, 163, 161, 162, 164, 163, 162, 164, 163, 160, 163, 163, 165, 163, 162, 163, 164, 163, 160, 165, 163, 162
What is the mode of heights?
Also, find the mean and median.

Answer:

Arranging the data in tabular form, we get:

Height of Children (cm)  Tally Bars Frequency
160 ∣∣ 3
161 1
162 ∣∣∣∣ 4
163 10
164 ∣∣ 3
165 ∣∣∣ 3
168 1
Total   25

Here, n = 25
∴ Median = Value of n+12th observation = Value of the 13th observation = 163 cm.
Here, clearly, 163 cm occurs the most number of times. Therefore, the mode of the given data is 163 cm.
Now, 
Mode = 3 Median - 2 Mean
 163 = 3 x 163 - 2 Mean
2 Mean  =  326
 Mean = 326 ÷ 2 = 163 cm.

Page No 23.20:

Question 7:

The scores in mathematics test (out of 25) of 15 students are as follows:
19, 25, 23, 20, 9, 20, 15, 10, 5, 16, 25, 20, 24, 12, 20
Find the mode and median of this data. Are they same?

Answer:

Arranging the data in ascending order such that same values are put together, we get:

5,9,10,12,15,16, 19, 20, 20, 20, 20,  23, 24, 25, 25.
Here, n = 15
∴ Median = Value of n+12th observation = Value of the 8th observation = 20.
Here, clearly, 20 occurs the most number of times, i.e., 4 times. Therefore, the mode of the given data is 20.
Yes, the median and mode of the given data are the same.

Page No 23.20:

Question 8:

Calculate the mean and median for the folllowing data:

Marks : 10 11 12 13 14 16 19 20
Number of students : 3 5 4 5 2 3 2 1
Using empirical formula, find its mode.

Answer:

                               Calculation of Mean
Marks (xi) 10 11 12 13 14 16 19 20 Total
Number of Students (fi) 3 5 4 5 2 3 2 1 fi =  25
fixi 30 55 48 65 28 48 38 20  fixi = 332

Mean  = fixifi = 33225 = 13.28
Here, n = 25, which is an odd number. Therefore, 
Median = Value of n+12th observation = the 13th observation = 13.
Now,
Mode = 3 Median - 2 Mean
Mode = 3 x 13 - 2 x (13.28) 
Mode = 39 - 26.56
Mode = 12.44.

Page No 23.20:

Question 9:

The following table shows the weights of 12 persons.

Weight (in kg): 48 50 52 54 58
Number of persons: 4 3 2 2 1
Find the median and mean weights. Using empirical relation, calculate its mode.

Answer:

Calculation of Mean
Weight (xi) 48 50 52 54 58 Total
Number of Persons (fi) 4 3 2 2 1     fi=12
fixi 192 150 104 108 58 fixi = 612

Mean  = fixifi = 61212 = 51 kg.
Here, n = 12
 Median = value ofn2th observation  + value ofn2 +1th observation2 Median= value of6th observation  + value of7th observation2  Median=50 +502 Median= 50 kg. 

Now, 
     Mode = 3 Median - 2 Mean
⇒ Mode = 3 x 50 - 2 x 51
⇒Mode  = 150 - 102 
⇒ Mode = 48 kg.
Thus, Mean = 51 kg, Median = 50 kg and Mode = 48 kg.



Page No 23.6:

Question 1:

Ashish studies for 4 hours, 5 hours and 3 hours on three consecutive days. How many hours does he study daily on an average?

Answer:

Average number of  study hours =  ( 4  + 5 + 3) ÷ 3 
                                                 =  12 ÷ 3

                                                 = 4 hours           
           Thus, Ashish studies for 4 hours on an average.

Page No 23.6:

Question 2:

A cricketer scores the following runs in 8 innings: 58, 76, 40, 35, 48, 45, 0, 100. Find the mean score.

Answer:

We have:
The mean score = (58 + 76  +40 +35 +48 +45 + 0+ 100)8 =4028   =   50.25 runs.

Thus, the mean score of the cricketer is 50.25 runs.

Page No 23.6:

Question 3:

The marks (out of 100) obtained by a group of students in science test are 85, 76, 90, 84, 39, 48, 56, 95, 81 and 75. Find the
(i) highest and the lowest marks obtained by the students.
(ii) range of marks obtained.
(iii) mean marks obtained by the group.

Answer:

In order to find the highest and lowest marks, let us arrange the marks in ascending order as follows:
39, 48, 56, 75, 76, 81, 84, 85, 90, 95
(i) Clearly, the highest mark is 95 and the lowest is 39.
(ii) The range of the marks obtained is: ( 95 - 39) = 56.
(iii) We have:
      Mean marks = Sum of the marks ​÷ Total number of students

⇒ Mean marks = (39 +48 + 56 +75 +76 +81 +84 +85 +90 +95)10 = 72910 = 72.9.

Hence, the mean marks of the students is 72.9.

Page No 23.6:

Question 4:

The enrolment of a school during six consecutive years was as follows:
1555, 1670, 1750, 2019, 2540, 2820
Find the mean enrollment of the school for this period.

Answer:

The mean enrolment = Sum of the enrolments in each year ÷ Total number of years

The mean enrolment = (1555 +1670 +1750 +2019 +2540 +2820) 6 = 123546 = 2059.

Thus, the mean enrolment of the school for the given period is 2059.

Page No 23.6:

Question 5:

The rainfall (in mm) in a city on 7 days of a certain week was recorded as follows:

Day Mon Tue Wed Thu Fri Sat Sun
Rainfall (in mm) 0.0 12.2 2.1 0.0 20.5 5.3 1.0
(i) Find the range of the rainfall from the above data.
(ii) Find the mean rainfall for the week.
(iii) On how many days was the rainfall less than the mean rainfall.

Answer:

(i) The range of the rainfall = Maximum rainfall - Minimum rainfall

                                    =    20.5  - 0.0
                                    =    20.5 mm .            

(ii) The mean rainfall = (0.0+12.2 +2.1 +0.0+20.5 + 5.3 +1.0)7 = 41.17 = 5.87 mm.

(iii) Clearly, there are 5 days (Mon, Wed, Thu, Sat, and Sun), when the rainfall was less than the mean, i.e., 5.87 mm.

Page No 23.6:

Question 6:

If the heights of 5 persons are 140 cm, 150 cm, 152 cm, 158 cm and 161 cm respectively, find the mean height.

Answer:

The mean height = Sum of the heights ÷ Total number of persons

                        = (140 + 150 +152 +158 +161)5 = 7615= 152.2 cm

Thus, the mean height of 5 persons is 152.2 cm.



Page No 23.7:

Question 7:

Find the mean of 994, 996, 998, 1002 and 1000.

Answer:

Mean = Sum of the observations ÷ Total number of observations

 Mean  = (994 + 996 + 998 + 1002 +1000)5 = 49905 = 998.

Page No 23.7:

Question 8:

Find the mean of first five natural numbers.

Answer:

The first five natural numbers are 1, 2, 3, 4 and 5. Let X¯ denote their arithmetic mean. Then,

X¯ = 1+2+3+4+55 = 155 = 3.

Page No 23.7:

Question 9:

Find the mean of all factors of 10.

Answer:

The factors of 10 are 1, 2, 5 and 10 itself. Let ​X¯ denote their arithmetic mean. Then,

X¯ = 1+2+5 +104 = 184 = 4.5.

Page No 23.7:

Question 10:

Find the mean of first 10 even natural numbers.

Answer:

The first 10 even natural numbers are 2,4, 6, 8,10,12,14,16,18 and 20. Let ​X¯ denote their arithmetic mean. Then,
 
X¯ = 2+4+6+8+10+12+14+16+ 18+2010 = 11010 = 11.

Page No 23.7:

Question 11:

Find the mean of x, x + 2, x + 4, x + 6, x + 8.

Answer:

Mean =  Sum of observationsNumber of observations 

Mean = x +x+2 +x+4 +x+6 +x+85    Mean= 5x + 205 = 5(x+4)5 Mean= x+4.

Page No 23.7:

Question 12:

Find the mean of first five multiples of 3.

Answer:

The first five multiples of 3 are 3,6,9,12 and 15. Let X¯ denote their arithmetic mean. Then,

X¯ = 3+6+9+12+155 = 455 = 9.

Page No 23.7:

Question 13:

Following are the weights (in kg) of 10 new born babies in a hospital on a particular day:
3.4, 3.6, 4.2, 4.5, 3.9, 4.1, 3.8, 4.5, 4.4, 3.6. Find the mean X.

Answer:

We have:
        X¯ = Sum of the observationsNumber of observations   X¯ = 3.4 +3.6 +4.2 +4.5 + 3.9 +4.1 +3.8 +4.5 +4.4 +3.610X¯  = 4010 = 4 kg.

Page No 23.7:

Question 14:

The percentage of marks obtained by students of a class in mathematics are:
64, 36, 47, 23, 0, 19, 81, 93, 72, 35, 3, 1. Find their mean.

Answer:

We have :
 Mean = Sum of the marks obtained by studentsTotal number of students.Mean =64 +36 +47 +23+ 19+81+93 +72 +35 +3+112Mean =47412  = 39.5 %.

Page No 23.7:

Question 15:

The numbers of children in 10 families of a locality are:
2, 4, 3, 4, 2, 3, 5, 1, 1, 5. Find the mean number of children per family.

Answer:

The mean number of children per family = Sum of the total number of childrenTotal number of families
 

Mean = 2+4+3+4+2+3+5+1+1+510 = 3010 = 3.

Thus, on an average there are 3 children per family in the locality.

Page No 23.7:

Question 16:

The mean of marks scored by 100 students was found to be 40. Later on it was discovered that a score of 53 was misread as 83. Find the correct mean.

Answer:

We have:
 

n = The number of observations = 100,  Mean = 40

 Mean = Sum of the observationsTotal number of observations40 =  Sum of the observations100 40 ×100 = Sum of the observations

Thus, the incorrect sum of the observations = 40 x 100 = 4000.

Now,

The correct sum of the observations = Incorrect sum of the observations - Incorrect observation + Correct observation

⇒ The correct sum of the observations = 4000 - 83 + 53 
⇒ The correct sum of the observations = 4000 - 30 =  3970

∴ Correct mean = Correct sum of observationsNumber of observations = 3970100 = 39.7

Page No 23.7:

Question 17:

The mean of five numbers is 27. If one number is excluded, their mean is 25. Find the excluded number.

Answer:

We have:

 Mean = Sum of the five numbers5 = 27So, sum of the five numbers = 5 ×27 = 135.Now, The mean of four numbers = Sum of the four numbers4 = 25So, sum of the four numbers = 4 ×25  = 100.

Therefore, the excluded number  = Sum of the five numbers - sum of the four numbers

⇒ The excluded number = 135 - 100 = 35.

Page No 23.7:

Question 18:

The mean weight per student in a group of 7 students is 55 kg. The individual weights of 6 of them (in kg) are 52, 54, 55, 53, 56 and 54. Find the weight of the seventh student.

Answer:

We have:

Mean = Sum of the weights of the studentsNumber of students 

Let the weight of the seventh student be x kg.

 Mean = 52+54 + 55 +53 +56 +54 +x7 = 55324 +x7 = 55324 +x = 385 x = 385 - 324 x = 61 kg.
Thus, the weight of the seventh student is 61 kg.

Page No 23.7:

Question 19:

The mean weight of 8 numbers is 15 kg. If each number is multiplied by 2, what will be the new mean?

Answer:

Let x1, x2, x3...x8 be the eight numbers whose mean is 15 kg. Then,

 15= x1+ x2+ x3 + ...+ x88

    x1 x2 x3 +...x15×8 
x1 x2 x3 +...x8 120.

Let the new numbers be 2x1 , 2x2, 2x3, ...2x8. Let M be the arithmetic mean of the new numbers.
Then,

M= 2x1+ 2x2+ 2x3 + ....+ 2x88=>M = 2(x1+ x2+ x3 + ....x8)8=>M=2×1208=>M=30  

Page No 23.7:

Question 20:

The mean of 5 numbers is 18. If one number is excluded, their mean is 16. Find the excluded number.

Answer:

Let x1,x2,x3,x4 & x5 be five numbers whose mean is 18. Then,
18 = Sum of five numbers ÷ 5
∴ Sum of five numbers = 18 × 5 = 90.

Now, if one number is excluded, then their mean is 16.
So,
​16= Sum of four numbers ​÷ 4
∴ Sum of four numbers = 16 × 4 = 64.

 
 The excluded number =  Sum of the five observations - Sum of the four observations
∴ The excluded number = 90 - 64
 ∴ The excluded number = 26.

Page No 23.7:

Question 21:

The mean of 200 items was 50. Later on, it was discovered that the two items were misread as 92 and 8 instead of 192 and 88. Find the correct mean.

Answer:

n = Number of observations = 200

Mean = Sum of the observationsNumber of observations50 = Sum of the observations200Sum of the observations = 50 × 200 = 10,000.

Thus, the incorrect sum of the observations = 50 x 200
Now, 
The correct sum of the observations = Incorrect sum of the observations - Incorrect observations + Correct observations
⇒​Correct sum of the observations =  10,000 - (92+ 8) + (192 + 88)

⇒ Correct sum of the observations = 10,000 - 100 + 280
⇒ Correct sum of the observations = 9900 +280
⇒ Correct sum of the observations = 10180.
 
∴ Correct Mean = Correct sum of the observationsNumber of observations = 10180200 = 50.9

Page No 23.7:

Question 22:

The mean of 5 numbers is 27. If one more number is included, then the mean is 25. Find the included number.

Answer:

We have:
Mean =  Sum of  five numbers ÷ 5
⇒ Sum of the five numbers = 27 x 5 = 135.

Now, New mean = 25
25 = Sum of six numbers ​÷ 6
⇒ Sum of the six numbers = 25 x 6 = 150.

The included number = Sum of the six numbers - Sum of the five numbers
⇒​The included number = 150 - 135
⇒The included number =  15.

Page No 23.7:

Question 23:

The mean of 75 numbers is 35. If each number is multiplied by 4, find the new mean.

Answer:

Let x1, x2, x3, ... x75 be 75 numbers with their mean equal to 35. Then,

35 = x1 +x2 +x3+... + x7575

x1x2 x3 +... + x75 = 35 ×75 

 x1 x2 x3 +... + x75  2625 

The new numbers are 4x1, 4x2, 4x3, ...4x75. Let M be the arithmetic mean of the new numbers. Then,

M = 4x1 +4x2 +4x3+... +4 x7575M = 4 (x1 +x2 +x3+...+ x75)75  M =4×262575M =35×4  M =140.



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