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: The following Pie Chart shows the export of different food grains from India in 2010. Study the chart and answer the questions :

  1. Of the total export of food grains, the percentage of crop B exported is








  1. View Hint View Answer Discuss in Forum

    On the basis of given pie chart in question ,
    The corresponding angle of crop B = 72°
    Total corresponding angle of food-grains = 360°

    Percentage of crop B exported =
    The corresponding angle of crop B
    		 × 100
    Total corresponding angle of food-grains

    Correct Option: B

    On the basis of given pie chart in question ,
    The corresponding angle of crop B = 72°
    Total corresponding angle of food-grains = 360°

    Percentage of crop B exported =
    The corresponding angle of crop B
    		 × 100
    Total corresponding angle of food-grains

    Percentage of crop B exported =
    72
    		× 100 = 20%
    360

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

  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

Direction: The bar graph shows the percentage distribution of the total expenditures of a company under various expense heads during 2005. Study the bar graph and answer the following four questions.

  1. The expenditure on the interest on loans is by what percent more than the expenditure on transport?








  1. View Hint View Answer Discuss in Forum

    On the basis of given graph in question ,
    The expenditure on the interest on loans = 17.5%
    The expenditure on the interest on transport = 12.5%
    Change in expenditure = 17.5 - 12.5 = 5

    Required percentage
    =
    Change in expenditure
    The expenditure on the interest on transport

    Required percentage = 5 × 100
    12.5

    Correct Option: B

    On the basis of given graph in question ,
    The expenditure on the interest on loans = 17.5%
    The expenditure on the interest on transport = 12.5%
    Change in expenditure = 17.5 - 12.5 = 5

    Required percentage
    =
    Change in expenditure
    The expenditure on the interest on transport

    Required percentage = 5 × 100
    12.5

    Required percentage =
    500
    		 = 40
    12.5