Introduction to Data Interpretation

Introduction to Data Interpretation

Easy Questions
Moderate Questions
Difficult Questions

Data Interpretation

Introduction to Data Interpretation
Table chart
Line chart
Pie chart
Bar graph
Mixed Graph
Data Sufficiency

Direction: Sutdy the following bar graph and answer the questions.

  1. The number of years, the production of fertilizers was more than average production of the given years is :








  1. View Hint View Answer Discuss in Forum

    On the basis of given graph in question ,
    Sum of the production of fertilizers = 25 + 40 + 60 + 45 + 65 + 50 + 75 + 80 = 440
    Number of given years = 8

    Average production =
    Sum of the production of fertilizers
    Number of given years

    Average production = 25 + 40 + 60 + 45 + 65 + 50 + 75 + 80 ten thousand tonnes
    8

    Correct Option: A

    On the basis of given graph in question ,
    Sum of the production of fertilizers = 25 + 40 + 60 + 45 + 65 + 50 + 75 + 80 = 440
    Number of given years = 8

    Average production =
    Sum of the production of fertilizers
    Number of given years

    Average production = 25 + 40 + 60 + 45 + 65 + 50 + 75 + 80 ten thousand tonnes
    8

    Average production =
    440
    		= 55 ten thousand tonnes
    8

    Required years ⇒ 1997, 1999, 2001 and 2002

Direction: The following table shows the worldwide production of steel in 1920–1927. Study the table and answer the questions.

  1. The average production of steel is (in million tonnes)








  1. View Hint View Answer Discuss in Forum

    According to given bar graph , we have
    Average production of steel = 77.1 million tonnes

    Correct Option: C

    According to given bar graph , we have
    Average production of steel = 77.1 million tonnes
    Hence , required answer is 77.1 million tonnes .

  1. The number of years during which the company has its production less than the average production during 1920–1927 is approximately








  1. View Hint View Answer Discuss in Forum

    On the basis of given graph in question ,
    Sum of production during 1920–1927 = 71.30 + 43.51 + 67.66 + 76.23 + 77.23 + 88.93 + 91.75 + 100.17
    Number of given years = 8

    Average production during 1920 – 1927 =
    Sum of production during 1920–1927
    Number of given years

    Average production during 1920 – 1927 = 71.30 + 43.51 + 67.66 + 76.23 + 77.23 + 88.93 + 91.75 + 100.17 million tonnes
    8

    Correct Option: B

    On the basis of given graph in question ,
    Sum of production during 1920–1927 = 71.30 + 43.51 + 67.66 + 76.23 + 77.23 + 88.93 + 91.75 + 100.17
    Number of given years = 8

    Average production during 1920 – 1927 =
    Sum of production during 1920–1927
    Number of given years

    Average production during 1920 – 1927 = 71.30 + 43.51 + 67.66 + 76.23 + 77.23 + 88.93 + 91.75 + 100.17 million tonnes
    8

    Average production during 1920 – 1927 =
    616.78
    		 = 77.1 million tonnes
    8

    ∴ Required years = 1920, 1921, 1922 and 1923

  1. The ratio of production of steel in the year 1924 and 1925 to that of 1923 and 1927 is








  1. View Hint View Answer Discuss in Forum

    According to given bar graph , we have
    Total production of steel in the year 1924 and 1925 = 77.23 + 88.93 = 166.16
    Total production of steel in the year 1923 and 1927 = 76.23 + 100.17 = 176.40
    Required ratio = Total production of steel in the year 1924 and 1925 : Total production of steel in the year 1923 and 1927

    Correct Option: B

    According to given bar graph , we have
    Total production of steel in the year 1924 and 1925 = 77.23 + 88.93 = 166.16
    Total production of steel in the year 1923 and 1927 = 76.23 + 100.17 = 176.40
    Required ratio = Total production of steel in the year 1924 and 1925 : Total production of steel in the year 1923 and 1927
    Required ratio = 166.16 : 176.40 = 16616 : 17640
    ∴ Required ratio = 2077 : 2205

  1. The difference of the production of steel in the year 1923 and 1924 is X% of 1927. Then the value of X is approximately








  1. View Hint View Answer Discuss in Forum

    As per the given bar graph ,
    The production of steel in the year 1924 = 77.23
    and The production of steel in the year 1923 = 76.23
    The production of steel in the year 1927 = 100.17
    Difference of the production of steel in the year 1923 and 1924 = 77.23 - 76.23 = 1
    According to the question,
    X% of the production of steel in the year 1927 = Difference of the production of steel in the year 1923 and 1924

    X × 100.17
    		 = 1
    100

    ⇒
    X × 100.17
    		 = 1
    100

    Correct Option: D

    As per the given bar graph ,
    The production of steel in the year 1924 = 77.23
    and The production of steel in the year 1923 = 76.23
    The production of steel in the year 1927 = 100.17
    Difference of the production of steel in the year 1923 and 1924 = 77.23 - 76.23 = 1
    According to the question,
    X% of the production of steel in the year 1927 = Difference of the production of steel in the year 1923 and 1924

    X × 100.17
    		 = 1
    100

    ⇒
    X × 100.17
    		= 1
    100

    ⇒ X =
    100
    		≈ 1
    100.7