Data Interpretation Questions and Answers

Answer: B

First, find the urban population of State D.

Rural population % for State D = 55%. Therefore, Urban population % = 100% - 55% = 45%.

Total population of State D = 25 million.

Urban Population of D = 45% of 25 million = 0.45 * 25 = 11.25 million.

Total population of State B = 25 million.

Required Percentage = \(\frac{\text{Urban Pop. D}}{\text{Total Pop. B}} \times 100 = \frac{11.25}{25} \times 100 = 0.45 \times 100 = 45\)%.

Answer: A

First, calculate the required populations.

Urban Population of State E:
Rural % = 40%, so Urban % = 60%.
Total Population = 20 million.
Urban Population = 60% of 20 million = 12 million.

Rural Population of State A:
Rural % = 60%.
Total Population = 20 million.
Rural Population = 60% of 20 million = 12 million.

The ratio is 12 million : 12 million, which simplifies to 1:1.

Answer: C

The caselet states the total directly: the college has 2000 students, made up of 900 in Science, 350 in Commerce and the remaining 750 in Arts.

\[ 900 + 350 + 750 = 2000 \]

Answer: B

Correct answer: 27:23

Take the streams one at a time. Arts has \(2000 - 900 - 350 = 750\) students.

Science (900). 30% are girls, so girls \(= 270\) and boys \(= 630\).

Commerce (350). The ratio 3:4 makes 7 parts of 50, so boys \(= 150\) and girls \(= 200\).

Arts (750). 40% are boys, so boys \(= 300\) and girls \(= 450\).

\[\text{Boys} = 630 + 150 + 300 = 1080, \qquad \text{Girls} = 270 + 200 + 450 = 920\]

\[1080 : 920 = 27 : 23\]

The two totals add to 2000, which is the check that no stream has been missed.

Answer: A

First, we need to find the specific numbers from the caselet data.

Number of girls in Science = 30% of 900 = 0.30 * 900 = 270.

Number of boys in Arts = 60% of 750 = 0.60 * 750 = 450.

Required Percentage = \(\frac{\text{Girls in Science}}{\text{Boys in Arts}} \times 100\)

= \(\frac{270}{450} \times 100 = \frac{3}{5} \times 100 = 60\)%.

Answer: B

Add the three segments of the 2018 bar.

\[ \text{wheat } 60 + \text{rice } 50 + \text{maize } 50 = 160 \text{ lakh tonnes} \]

Answer: C

Production of Wheat in 2016 = 40 lakh tonnes.

Production of Rice in 2020 = 50 lakh tonnes.

Required Percentage = \(\frac{\text{Wheat 2016}}{\text{Rice 2020}} \times 100\)

= \(\frac{40}{50} \times 100 = 80\)%.

Answer: A

Correct answer: Wheat in 2018

Each option needs its increase measured against the previous year, then expressed as a percentage of that previous year.

Wheat in 2018: from 40 to 60, so \(\tfrac{20}{40}\times 100 = 50\%\).
Rice in 2017: from 30 to 40, so \(\tfrac{10}{30}\times 100 = 33.33\%\).
Maize in 2019: from 50 to 55, so \(\tfrac{5}{50}\times 100 = 10\%\).
Wheat in 2020: from 65 to 70, so \(\tfrac{5}{65}\times 100 = 7.69\%\).

Wheat in 2018 is the largest at 50%. Note that it also has the largest absolute rise, 20 lakh tonnes, but that is coincidence - percentage growth depends on the base, which is why the smaller 2017 Rice rise of 10 still outranks the equal-sized 2019 Maize and 2020 Wheat rises of 5.

Answer: B

Correct answer: 45.33 lakh tonnes

Read the Maize series across all six years and add.

\[35 + 40 + 50 + 50 + 55 + 42 = 272\]

\[\text{Average} = \frac{272}{6} = 45.33 \text{ lakh tonnes}\]

The 2020 figure of 42 is the only fall in the series, and it is what pulls the average below 46.

Answer: B

Read the two bars: 125 million tonnes in 2018 and 140 in 2019.

\[ \frac{140 - 125}{125} \times 100 = \frac{15}{125} \times 100 = 12\% \]

The increase is measured against the earlier year, which is the base.