## Introduction to Data Interpretation

#### Data Interpretation

Direction: Study the bar diagram and answer the given questions. 1. Ratio of highest and lowest marks obtained in first term among all the subjects is

1. 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

##### Correct 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. 1. Total sales of branches B1, B3 and B5 together for both the years is

1. 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

##### Correct 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

1. The average sales of all the branches for the year 2009 is

1. 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 = 6

 Required 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 = 6

 Required 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

1. 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

1. 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

1. 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

1. 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