## Introduction to Data Interpretation

#### Data Interpretation

Direction: The table shows the percentage of total population of a city in different age groups. Study the table and answer the questions.

1. If the difference between the number of people in the age groups (46 – 55) and (16 – 25) is 0.975 million, then the total population (in millions) of the city is

1. On the basis of given table in question ,
∵ Difference between the number of people in the age groups (46 – 55) and (16 – 25) = 0.975 million
Percent of people in the age groups (46 – 55) = 18.25%
And Percent of people in the age groups (16 – 25) = 15%
⇒ (18.25 – 15)% ≡ 0.975 million

##### Correct Option: B

On the basis of given table in question ,
∵ Difference between the number of people in the age groups (46 – 55) and (16 – 25) = 0.975 million
Percent of people in the age groups (46 – 55) = 18.25%
And Percent of people in the age groups (16 – 25) = 15%
⇒ (18.25 – 15)% ≡ 0.975 million

 ∴ 100 % ≡ 0.975 × 100 ≡ 30 millions 3.25

1. If there are 22 million people below 36 years, then the number of people (in millions) in the age group (56 – 65) is

1. As per the given above table , we see
Percentage of people below 36 years = (20 + 18.25 + 16.75)% = 55%
Percent of people in the age group (56 – 65) = 12.50%
∵ 55% ≡ 22 million

##### Correct Option: A

As per the given above table , we see
Percentage of people below 36 years = (20 + 18.25 + 16.75)% = 55%
Percent of people in the age group (56 – 65) = 12.50%
∵ 55% ≡ 22 million

 ∴ 12.50 % ≡ 22 × 12.50 = 5 millions 55

Direction: Study the table and answer the questions

1. Average height (in cm) of the girls whose heights are 155cm and above is about

1. On the basis of given table in question ,
Class interval (155 – 160) = 6 girls

 Mid-value = 155 + 160 = 157.5 2

Class interval (160 – 165) = 5 girls
 Mid-value = 160 + 165 = 162.5 2

 ∴ Required average = 6 × 157.5 + 5 × 162.5 11

##### Correct Option: B

On the basis of given table in question ,
Class interval (155 – 160) = 6 girls

 Mid-value = 155 + 160 = 157.5 2

Class interval (160 – 165) = 5 girls
 Mid-value = 160 + 165 = 162.5 2

 ∴ Required average = 6 × 157.5 + 5 × 162.5 11

 Required average = 945 + 812.5 = 1757.5 = 159.8 cm 3 3

1. The number of girls whose height is above 150 cm is

1. According to given table, we have

 Height No. of girls 135-140 04 140-145 07 145-150 18 150-155 11 155-160 06 160-165 05

##### Correct Option: A

According to given table, we have

 Height No. of girls 135-140 04 140-145 07 145-150 18 150-155 11 155-160 06 160-165 05

Required answer = 11 + 6 + 5 = 22

Direction: Study the following table and answer the questions.

1. The difference in the average number of candidates qualified in Science discipline per year from 2006 to 2008 and the average number of candidates qualified in the same discipline from 2009 to 2011 was

1. As per the given above table , we see

 Candidates qualified under science discipline in year 2006 ⇒ 780 × 40 = 312 100

 Candidates qualified under science discipline in year 2007 ⇒ 650 × 42 = 273 100

 Candidates qualified under science discipline in year 2008 ⇒ 500 × 45 = 225 100

##### Correct Option: A

As per the given above table , we see

 Candidates qualified under science discipline in year 2006 ⇒ 780 × 40 = 312 100

 Candidates qualified under science discipline in year 2007 ⇒ 650 × 42 = 273 100

 Candidates qualified under science discipline in year 2008 ⇒ 500 × 45 = 225 100

 Candidates qualified under science discipline in year 2009 ⇒ 45 × 620 = 279 100

 Candidates qualified under science discipline in year 2010 ⇒ 35 × 900 = 315 100

 Candidates qualified under science discipline in year 2011 ⇒ 42 × 850 = 357 100

Required average difference = Average number of candidates qualified in Science year from 2009 to 20011 - Average number of candidates qualified in Science year from 2006 to 2008
 Required average difference = 1 [(279 + 315 + 357) – (312 + 273 + 225)] = 1 (951 – 810) 3 3

 = 1 × 141 = 47 3