These Class 7 Maths Extra Questions Chapter 3 Data Handling will help students prepare well for the exams.
Class 7 Maths Chapter 3 Extra Questions Data Handling
Class 7 Maths Data Handling Extra Questions – 2 Marks
Question 1.
Define mean. Find the mean of –1, 2, 0, 4.
Solution:
Mean = \(\frac{\text { Sum of all observations }}{\text { Number of observations }}\)
= \(\frac{-1+2+0+4}{4}\) = \(\frac{6-1}{4}\) = \(\frac{5}{4}\) = 1.25
Question 2.
Find the mean of \(\frac{1}{2}\), \(\frac{1}{3}\), \(\frac{1}{4}\), \(\frac{1}{24}\)
Solution:
Question 3.
Find the mean of –6\(\frac{1}{2}\), 0, 7\(\frac{1}{3}\), 8\(\frac{1}{2}\), 4.
Solution:
Question 4.
Find the mean of a – 2, a, a + 2.
Solution:
Mean = \(\frac{\text { Sum of all observations }}{\text { Number of observations }}\)
= \(\frac{a-2+a+a+2}{3}\) = \(\frac{3 a}{3}\) = a.
Question 5.
Find the average of 1.3, 3.1, 8.5, 6.2, 3.0, 9.1
Solution:
Average
= \(\frac{1.3+3.1+8.5+6.2+3.0+9.1}{6}\)
= \(\frac{4.4+14.7+12.1}{6}\) = \(\frac{31.2}{6}\) = 5.2
Question 6.
i)What does IMG represent?
ii) How will you represent 17 using Tally marks?
Solution:
Question 7.
The mean 11 observations is 11. If one observation 11 is deleted from it. Find the mean of the remaining observations.
Solution:
Mean = 11
\(\frac{\text { Sum of all observations }}{11}\) = 11
Sum of all observations = 11 × 11 = 121
One observation 11 is deleted.
Mean of remaining observation = \(\frac{121-11}{11-1}\) = \(\frac{110}{10}\) = 11
Question 8.
How do you find median of a data.
Solution:
The observations (or) items given a data are to be arranged either in Ascending order or Descending order, then the middle most value is taken as median of the data.
If the number of observations of a data is odd then median is \(\left(\frac{\mathrm{n}+1}{2}\right) \mathrm{th}\) observation.
If the number of observations of a data is even then median is average of \(\left(\frac{\mathrm{n}}{2}\right) \mathrm{th}\), \(\left(\frac{\mathrm{n}+1}{2}\right) \mathrm{th}\) observations.
Question 8.
Find the median of the scores given below 45, 32, 43, 37, 31, 46, 25
Solution:
The Ascending order of the data is 25, 31, 32, 37, 43, 45, 46
Number of observations = 7
Median = \(\left(\frac{7+1}{2}\right) \mathrm{th}\)th Observation.
Median = 4th observation = 37.
Question 10.
Find the median of the following data
15,6,16,8,22,21,25,9,18
Solution:
Ascending order of the given observation is 6,8,9,15,16,18,21,22,25
Number of observations = 9
Median = \(\left(\frac{9+1}{2}\right) \mathrm{th}\) observation.
Median = \(\left(\frac{10}{2}\right) \mathrm{th}\) = 5th observation
∴ Median = 16
Question 11.
Find the median of the following data 85,62,31,0,7,14,19,1
Solution:
Method 1 :
Ascending 0, 1, 7, 14, 19, 31, 62, 85
n = Number of observations = 8
\(\frac { n }{ 2 }\) = \(\frac { 8 }{ 2 }\) =4th observation = 14
\(\frac { n }{ 2 }\) + 1 = \(\frac { 8 }{ 2 }\) + 1 = (4+1)th = 5th observation = 19
Median = \(\frac{14+19}{2}\) = \(\frac { 33 }{ 2 }\) = 16.5
Method 2:
Ascending 0,1,7,14,19,31,62,85
Middle most values = 14,19
Median = \(\frac{14+19}{2}\) = \(\frac{33}{2}\) = 16.5
Question 12.
Defind mode of a data and give an example.
Solution:
In a data an observation which occurs most frequently is called as median of the data.
Ex: Find the mode of 9,8,9,6,7,0,9,9,9,1
In the above the number 9 occurs 5 times in a data.
∴ Mode = 9
Question 13.
‘A data may have two modes’. Do you agree justify your answer.
Solution:
Yes, a data may have two modes. If two number of a data are repeated same number of times, then it will have two modes.
Ex : Find the mode of the following data
16,12,9,7,0,12,12,16,16,1,9.
Here 16 is repeated 3 times.
12 is repeated 3 times
∴ Mode = 12 and 16.
Question 14.
Find the mode of first 100 natural number.
Solution:
The number from 1 to 100 are 1,2,3,4, …. 100.
There is no number repeated twice or twice in the data.
∴ The data has no mode.
∴ There is no mode for the number from 1 to 100.
Question 15.
Find the mean of the first 5 prime number.
Solution:
The first 5 prime number are 2,3,5,7, 11
Mean = \(\frac{\text { Sum of observation }}{\text { No.of observation }}\)
= \(\frac{2+3+5+7+11}{5}\) = \(\frac{28}{5}\) = 5.6
Data Handling Extra Questions Class 7 – 3 Marks
Question 1.
If the mean of 13, 17, 19, 16, x , 0 is 12 find x.
Solution:
Mean = \(\frac{\text { Sum of observation }}{\text { No.of observation }}\)
12 = \(\frac{13+17+19+16+x+0}{6}\)
72 = 65 + x
x = 72 – 65
x = 7
Question 2.
Find the mean and median of the first 5 multiples of 6? What do you notice.
Solution:
The first 5 multiples of 6 are 6, 12, 18, 24, 30
Mean = \(\frac{6+12+18+24+30}{5}\)
= \(\frac{90}{5}\) = 18
Middle most value = 18
∴ Median = 18
Here, Mean = Median.
Question 3.
Write the relation among mean, median and mode. Also find mode if median of a data is 30, and mean is 20 .
Solution:
The relation between mean, median and mode is = 3 median – 2 mean
Mode = 3 × 30 – 2 × 20
Mode = 90 – 40
Mode = 50
Question 4.
If the mean 6,7, x, 10,15 and 7 is 9 find x.
Solution:
Mean = \(\frac{\text { Sum of all observations }}{\text { No.of observations }}\)
9 = \(\frac{6+7+x+10+15+7}{6}\)
54 = 45 + x
x = 54 – 45
x = 9
Question 5.
The mean of 10 observation is 19. If one observation 8 is added to it, then find the new mean.
Solution:
Mean = \(\frac{\text { Sum of all observations }}{\text { No.of observations }}\)
19 = \(\frac{\text { Sum of all observations }}{10}\)
Sum of observations = 19 × 10 = 190
One observation 8 is added
Sum of observation = 190 + 8 = 198
New mean = \(\frac{198}{10+1}\) = \(\frac{198}{11}\) = 18.
Question 6.
There are 6 observations in the data and their mean is 10. If each observation is multiplied by 2, then find the mean of new observation.
Solution:
Let a, b, c, d, e, f are the observation
Mean = 10
\(\frac{a+b+c+d+e+f}{6}\) = 10
a+b+c+d+e+f = 60
Each observation is multiplied by 2 then new observations are
2a, 2b, 2c, 2c, 2d, 2e, 2f
New mean = \(\frac{2 a+2 b+2 c+2 d+2 e+2 f}{6}\)
= \(\frac{2(a+b+c+d+e+f)}{6}\)
= \(\frac{2 \times 60}{6}\) = 2 × 10 = 20
Question 7.
The mean weight of 20 students is 20 kg. If one student Rathode weighing 20 kg is removed from the group than find the mean of remaining students.
Solution:
Mean = 20
\(\frac{\text { Sum of all observations }}{20}\) = 20
Sum of all obseryations = 20 × 20 = 400 kgs
One student, weight 20 kg is removed
New sum of observation = (400 – 20)kg = 380 kg
New mean = \(\frac{380}{20-1}\) = \(\frac{380}{19}\) = 20 kg.
Question 8.
Using Tally mark find the mode of the data.
1 , 2 , 3 , 3 , 2 , 7 , 9 , 8 , 4 , 4 , 3 , 2 , 3 , 3 , 5 , 5 , 3
Solution:
Mode = 3
Question 9.
Find the median, mean and mode of the following data
–1,2,3,7,–4,6,0
Solution:
Mean = \(\frac{-1+2+3+7-4+6+0}{7}\)
= \(\frac{18-5}{7}\) = \(\frac{13}{7}\)
Ascending order = –4,–1,0,2,3,6,7
Median = 2
There is no value repeated
∴ there is no mode.
Question 10.
What is a bi-modal data. Illustrate with an example.
Solution:
A data having 2 modes is called Bi-modal data.
Example: Find the mode of 7, 9, 16, 9, 16,10,7,7,9,13,-1,8,9
7 is repeated 3 times
9 is repeated 3 times
∴ Mode = 7 and 9
∴ This is a Bi-modal data.
Question 11.
If the median of the observations 12,12,14,18, a+2, a+4,30,32,35,41 which are arranged in ascending order is 24. Find the value of a.
Solution:
Question 12.
Find
i) Range of first 10 even natural numbers.
ii) AM of 0,1,2,–1,–2
iii) Mode of –30, –29, –28, …… 0.
Solution:
i) The first 10 even natural number are 2, 4,6, ……. 20
Range = Max. value – Mini. value = 20 – 2 = 18
ii) AM = \(\frac{0+1+2-1-2}{5}\) = \(\frac{0}{5}\) = 0
∴ AM = 0
iii) There is no number repeated
∴ There is no mode
Question 13.
Using the relation find the mean of the data of median is 40 and mode is 90.
Solution:
Mode = 3 median – 2 mean
90 = 3 × 40 – 2 mean
90 = 120 – 2 mean
90 – 120 = –2 mean
–30 = –2 mean
Mean = \(\frac{30}{2}\)
Mean = 15
Extra Questions of Data Handling Class 7 – 5 Marks
Question 1.
The data given below shows the production of cars in a factory for few months of two consecutive years.
| Month | Year 2009 | Year 2010 |
| June | 2000 | 2200 |
| July | 3100 | 4000 |
| August | 4200 | 4500 |
| December | 5000 | 5800 |
Study the table given above and answer the following questions.
a) Draw a double bar graph using appropriate scale to depict the above information and compare them.
b) In which year was the total output maximum.
c) Find mean production in the year 2009
Solution:
b) Production of no.of cars in year 2009
= 2000 + 3100 + 4200 + 5000 = 14,300
Production of no.of cars in year 2010
=2200 + 4000 + 4500 + 5800 = 16,500
∴ In the year 2010 production of cars is more than year 2009.
c) Mean product of number cars in year
2009 = \(\frac{2000+3100+4200+5000}{4}\)
= \(\frac{14300}{4}\) = 3575 cars.
Question 2.
Given below are the heights of 11 boys of a class measured in cm
132, 144, 110, 128, 116, 151, 142, 140, 130, 136, 134.
Find,
a) the height of a short boy.
b) the height of a tall boy.
c) Range of the data.
d) Median height of the boys.
Solution:
The Ascending order of heights of boys is 110,116,128,130,132,134,136,140,142,144,151.
a) Height of short boy = 110 cm
b) Height of tall boy = 151 cm
c) Range = max. value – min. value = 151 – 110 = 41 .
d) Median height : n = 11
\(\left(\frac{\mathrm{n}+1}{2}\right) \text { th }\) observation is the median
\(\left(\frac{11+1}{2}\right) \text { th }\) = \(\left(\frac{12}{2}\right) \mathrm{th}\) = 6th observation
∴ Median = 134.
Question 3.
Observe the following data and answer the following questions.
15,16,16,8,7,16,7,7,1
a) Which data value can be put in so that there is no change in the mode of the data?
b) How many times the number to be written to change the mode as 15 ?
c) Rahul wants to make the mode of the data as 7. Give your suggestion.
Solution:
a) Mode of the data is 16.
If we add 16 to the data
Mode remains 16.
b) 15 appears only one time
To convert it as mode, 15 should be written three more times in the data. So, that 15 repeats 4 times.
c) 7 appears to be 3 times in the data To make it mode, Rahul should add other 7 in the data the mode becomes 7.
Question 4.
Age (in years) of 5 children of two groups are recorded as below.
| Age | (in year) |
| Group A | Group B |
| 17 | 7 |
| 17 | 8 |
| 19 | 9 |
| 18 | 12 |
| 17 | 12 |
a) Find the Range and mode of each group.
b) Find the range and mode of two groups when combined together.
Solution:
a) Group A : 17,17,19,18,17.
Range = max. value – min , value = 19 – 17 = 2
Mode = 17
Group B : 7, 8, 9, 12, 12
Range = 12 – 7 = 5
Mode – 12
b) When two groups are combined the observations are as follows
17,17,19,18,17,7,8,9,12,12
Range – max. value – min. value = 19 – 7 = 12
Mode = 17.
Question 5.
Observe the given graph carefully and answer the following questions given.
a) In which month highest production of bikes happend by the company? and how many?
b) How many bikes produced in July? Is it is least (or) highest ?
c) In which two months the production of bikes is same?
d) Find the difference between the production of bikes in the months of April and June.
e) What is the median of the data ?
Solution:
a) February, 450 bikes
b) 100 , yes it is least.
c) January and June ( 300 bikes)
d) April – 350, June – 300
Difference = 350 – 300 = 50
e) Ascending order of the given number in the data is 100,150,250,300,300,350,450 Median = 300.
Question 6.
Number of sports persons in different categories of four major cities in India. in a particular year is given below.
| City | No.of sports persons |
| Indoor | 120 |
| Bhubaneswar | 93 |
| Gorakhpur | 80 |
| Delhi | 110 |
| Hyderabad | 75 |
Draw a bar graph for the above data.
Solution:
Question 7.
The results of pass percentage of class X and IX in an examination (like periodic test) for 5 years are given in the following table.
| Year | Class x | Class IX |
| 2004-2005 | 89 | 95 |
| 2005-2006 | 92 | 90 |
| 2006-2007 | 90 | 85 |
| 2007-2008 | 93 | 95 |
| 2008-2009 | 95 | 90 |
Draw a double bar graph for the above data.
Solution:
