Introduction to Data Interpretation
Direction: Study the bar diagram and answer the given questions.
 Ratio of highest and lowest marks obtained in first term among all the subjects is

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According to given bar graph , we have
Highest marks obtained in first term subject = 100 ( Math )
Lowest marks obtained in first term subject = 70 ( English )
Required ratio = Highest marks obtained in first term subject : Lowest marks obtained in first term subjectCorrect Option: C
According to given bar graph , we have
Highest marks obtained in first term subject = 100 ( Math )
Lowest marks obtained in first term subject = 70 ( English )
Required ratio = Highest marks obtained in first term subject : Lowest marks obtained in first term subject
Required ratio = 100 : 70 = 10 : 7
Direction: In the following bar diagram sales of books (in thousand numbers) from six branches – B1, B2, B3, B4, B5 and B6 of a publishing company in 2009 and 2010 have been shown. Study the graph and answer the questions.
 Total sales of branches B1, B3 and B5 together for both the years is

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From the given bar diagram , we see
Sales of branches B1 in both years = 80 + 105 = 185 thousand
Sales of branches B3 in both years = 95 + 110 = 205 thousand
Sales of branches B5 in both years = 75 + 95 = 170 thousandCorrect Option: D
From the given bar diagram , we see
Sales of branches B1 in both years = 80 + 105 = 185 thousand
Sales of branches B3 in both years = 95 + 110 = 205 thousand
Sales of branches B5 in both years = 75 + 95 = 170 thousand
Total sales of branches B1, B3 and B5 for both the years = ( 185 + 205 + 170 ) thousand = 560 thousand
 The average sales of all the branches for the year 2009 is

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According to given bar graph , we have
Total sales of all the branches in 2009 = 80 + 75 + 95 + 85 + 75 + 70 = 480
Number of given branches = 6Required Average = Total sales of all the branches in 2009 Number of given branches
Correct Option: C
According to given bar graph , we have
Total sales of all the branches in 2009 = 80 + 75 + 95 + 85 + 75 + 70 = 480
Number of given branches = 6Required Average = Total sales of all the branches in 2009 Number of given branches Average sales of all the branches in 2009 = 480 = 80 thousand 6
 x% of the average sales of branches B1, B2 and B3 in 2010 is the average sales of branches B1, B3 and B6 in 2009. The value of x is

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As per the given bar diagram , we see
Total sales of branches B1, B2 and B3 in 2010 = 105 + 65 + 110 = 280 thousand
Total sales of branches B1, B3 and B6 in 2009 = 80 + 95 + 70 = 245 thousand
Number of given branches = 3
From question ,∴ 280 × x = 245 3 100 3
Correct Option: B
As per the given bar diagram , we see
Total sales of branches B1, B2 and B3 in 2010 = 105 + 65 + 110 = 280 thousand
Total sales of branches B1, B3 and B6 in 2009 = 80 + 95 + 70 = 245 thousand
Number of given branches = 3
From question ,∴ 280 × x = 245 3 100 3 ⇒ x = 245 × 100 = 87.5% 280
 Total sales of branch B6 for both the years is x percent of the total sales of branch B3 for both the years. The value of x is

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From the given bar diagram , we see
Total sales of branch B6 for both the years = 70 + 80 = 150 thousand
Total sales of branch B3 for both the years = 95 + 110 = 205 thousand
According to question ,
x% of Total sales of branch B3 for both the years = Total sales of branch B6 for both the years∴ 205 × x = 150 100
Correct Option: B
From the given bar diagram , we see
Total sales of branch B6 for both the years = 70 + 80 = 150 thousand
Total sales of branch B3 for both the years = 95 + 110 = 205 thousand
According to question ,
x% of Total sales of branch B3 for both the years = Total sales of branch B6 for both the years∴ 205 × x = 150 100 ⇒ x = 150 × 100 = 73.17% 205