Grade 7 : Social Science (Exploring Society) - Geographical Diversity (Set 2) | Concept of Maths

Question 1

1. If the local time at the Prime Meridian ($0^{\circ}$) is $12:00$ Noon, what will be the local time at a city located at $45^{\circ} E$ longitude?

A) 9:00 AM

B) 2:00 PM

C) 3:00 PM

D) 4:00 PM

Answer

Correct Answer: C

Solution: The Earth rotates $360^{\circ}$ in 24 hours. Therefore, $1^{\circ}$ of longitude equals $\frac{24 \times 60}{360} = 4$ minutes. Difference in longitude = $45^{\circ} - 0^{\circ} = 45^{\circ}$. Time difference = $45 \times 4 = 180$ minutes = 3 hours. Since the city is East (E) of the Prime Meridian, we add the time: $12:00 + 3$ hours = $3:00$ PM.

Question 2

2. On a map with a scale of $1 : 2,500,000$, two mountain peaks are $4\text{ cm}$ apart. What is the actual ground distance between these peaks?

A) $10\text{ km}$

B) $100\text{ km}$

C) $250\text{ km}$

D) $1,000\text{ km}$

Answer

Correct Answer: B

Solution: Scale $1 : 2,500,000$ means $1\text{ cm} = 2,500,000\text{ cm}$ in reality. Convert cm to km: $2,500,000 \div 100,000 = 25\text{ km}$. Actual distance = $4\text{ cm} \times 25\text{ km/cm} = 100\text{ km}$.

Question 3

3. The normal lapse rate states that temperature decreases by $6.5^{\circ}\text{C}$ for every $1,000\text{ m}$ of ascent. If the temperature at sea level is $30^{\circ}\text{C}$, what is the temperature at a hill station at $2,000\text{ m}$?

A) $17^{\circ}\text{C}$

B) $23.5^{\circ}\text{C}$

C) $13^{\circ}\text{C}$

Question 4

4. A region has a population of $4.5$ million people living in an area of $15,000\text{ km}^2$. Calculate the population density.

A) $30\text{ persons/km}^2$

B) $450\text{ persons/km}^2$

C) $150\text{ persons/km}^2$

D) $300\text{ persons/km}^2$

Answer

Correct Answer: D

Solution: Population Density = $$\frac{\text{Total Population}}{\text{Total Area}}$$ Density = $$\frac{4,500,000}{15,000} = \frac{4500}{15} = 300\text{ persons/km}^2$$.

Question 5

5. If the Indian Standard Time (IST) is based on $82^{\circ}30' E$, what is the time difference between IST and Greenwich Mean Time (GMT)?

A) $4$ hours $30$ mins

B) $5$ hours $30$ mins

C) $6$ hours

D) $5$ hours $15$ mins

Answer

Correct Answer: B

Solution: Longitudinal difference = $82.5^{\circ}$. Time difference = $82.5 \times 4\text{ minutes} = 330\text{ minutes}$. $330 \div 60 = 5.5$ hours, which is $5$ hours and $30$ minutes ahead of GMT.

Question 6

6. Which of the following best describes the 'Rain Shadow' effect mathematically?

A) Precipitation on windward side $<$ Leeward side

B) Precipitation on windward side $=$ Leeward side

C) Precipitation on windward side $>$ Leeward side

D) Humidity increases as air descends the leeward side

Answer

Correct Answer: C

Solution: As moist air rises up the windward side of a mountain, it cools and precipitates. By the time it crosses to the leeward side, it has lost most of its moisture. Hence, Windward Precipitation $>$ Leeward Precipitation.

Question 7

7. If a point A is at $20^{\circ}N$ and point B is at $30^{\circ}N$ on the same longitude, approximately how many kilometers apart are they? (Assume $1^{\circ} \approx 111\text{ km}$)

A) $1,110\text{ km}$

B) $111\text{ km}$

C) $2,220\text{ km}$

D) $555\text{ km}$

Answer

Correct Answer: A

Solution: Difference in latitude = $30^{\circ} - 20^{\circ} = 10^{\circ}$. Distance = $10 \times 111\text{ km} = 1,110\text{ km}$.

Question 8

8. In a certain region, the land drops $200\text{ meters}$ in height over a horizontal distance of $5\text{ km}$. What is the gradient (slope) of this land?

A) $10\text{ m/km}$

B) $20\text{ m/km}$

C) $1,000\text{ m/km}$

D) $40\text{ m/km}$

Answer

Correct Answer: D

Solution: Gradient = $$\frac{\text{Vertical Change}}{\text{Horizontal Distance}}$$ Gradient = $$\frac{200\text{ m}}{5\text{ km}} = 40\text{ m/km}$$.

Question 9

9. If the Earth's axis were not tilted ($0^{\circ}$ tilt), which of the following would be true regarding seasonal diversity?

A) Seasons would be more extreme

B) The poles would be the hottest

C) There would be no change in seasons

D) Days would always be longer than nights

Answer

Correct Answer: C

Solution: The tilt of $23.5^{\circ}$ is the primary reason for the variation in the angle of sun rays throughout the year, causing seasons. Without a tilt, the sun would always be directly over the equator, leading to no seasonal variation.

Question 10

10. Fresh water accounts for only $2.5\%$ of total water on Earth. If the total water is $1.4$ billion cubic kilometers, how much is fresh water?

A) $3.5$ million $\text{km}^3$

B) $35$ million $\text{km}^3$

C) $14$ million $\text{km}^3$

D) $0.35$ million $\text{km}^3$

Answer

Correct Answer: B

Solution: Fresh water = $2.5\% \text{ of } 1,400,000,000\text{ km}^3$ Fresh water = $0.025 \times 1,400,000,000 = 35,000,000\text{ km}^3 = 35$ million $\text{km}^3$.

Accepted Answer

Correct Answer: A

Solution: Total ascent = $2,000\text{ m}$. Temperature drop = $2 \times 6.5^{\circ}\text{C} = 13^{\circ}\text{C}$. Final temperature = $30^{\circ}\text{C} - 13^{\circ}\text{C} = 17^{\circ}\text{C}$.

Grade 7 : Social Science (Exploring Society) - Geographical Diversity (Set 2) | Concept of Maths