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Considering the alarming importance of CSAT in UPSC CSE Prelims exam and with enormous requests we received recently, InsightsIAS has started Daily CSAT Test to ensure students practice CSAT Questions on a daily basis. Regular Practice would help one overcome the fear of CSAT too.
We are naming this initiative as Insta– DART – Daily Aptitude and Reasoning Test. We hope you will be able to use DART to hit bull’s eye in CSAT paper and comfortably score 100+ even in the most difficult question paper that UPSC can give you in CSP-2021. Your peace of mind after every step of this exam is very important for us.
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Question 1 of 5
1. Question
Direction for the following 5 (five) items: Consider the given information and answer the five items that follow:
The following graph shows the percentage growth in population of six
cities from 1990 to 2000 and 2000 to 2010.
If the population of City F in year 1990 was 12 lakh, what will be its population in year 2010?
Correct
Solution: B
It is given that population of City F in year 1990 was 12 lakh.
It has increased by 75 % from 1990 to 2000 for city F. Thus population of city F in 2000 = 1.75 * 12 = 21 lakh.
Population of city F has further increased by 55 % from 2000 to 2010.
Therefore, population of City F in year 2010 was 1.55 * 21 = 32.55 lakh.
Hence, option (b) is correct.
Incorrect
Solution: B
It is given that population of City F in year 1990 was 12 lakh.
It has increased by 75 % from 1990 to 2000 for city F. Thus population of city F in 2000 = 1.75 * 12 = 21 lakh.
Population of city F has further increased by 55 % from 2000 to 2010.
Therefore, population of City F in year 2010 was 1.55 * 21 = 32.55 lakh.
Hence, option (b) is correct.
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Question 2 of 5
2. Question
Direction for the following 5 (five) items: Consider the given information and answer the five items that follow:
The following graph shows the percentage growth in population of six
cities from 1990 to 2000 and 2000 to 2010.The population of City D in year 2000 was what per cent of its population in year 2010?
Correct
Solution: C
Let the population in year 2000 be ‘x’.
Therefore, its population in year 2010 = (160/100) * x = 8x/5.
Thus, the population of city D in 2000 is ((x)/(8x/5)) * 100 % of its population in year 2010.
= 62.5 %
Hence, option (c) is correct.
Incorrect
Solution: C
Let the population in year 2000 be ‘x’.
Therefore, its population in year 2010 = (160/100) * x = 8x/5.
Thus, the population of city D in 2000 is ((x)/(8x/5)) * 100 % of its population in year 2010.
= 62.5 %
Hence, option (c) is correct.
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Question 3 of 5
3. Question
Direction for the following 5 (five) items: Consider the given information and answer the five items that follow:
The following graph shows the percentage growth in population of six
cities from 1990 to 2000 and 2000 to 2010.In year 1990 the population of City A and City B are equal and the population of City A in year 2010 is 37.7 lakh. What is the population of City B in year 2010?
Correct
Solution: B
It is given that, the population of City A in year 2010 is 37.7 lakh.
Therefore, the population of City A in year in 1990
è 37.7 *(100/145)*(100/130) = 20 lakh
Therefore, the population of City B in year 2010 is 20 lakh.
The population of city B in 2010 = 20 * (140/100)*(150/100) = 42 lakh.
Hence, option (b) is correct.
Incorrect
Solution: B
It is given that, the population of City A in year 2010 is 37.7 lakh.
Therefore, the population of City A in year in 1990
è 37.7 *(100/145)*(100/130) = 20 lakh
Therefore, the population of City B in year 2010 is 20 lakh.
The population of city B in 2010 = 20 * (140/100)*(150/100) = 42 lakh.
Hence, option (b) is correct.
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Question 4 of 5
4. Question
Direction for the following 5 (five) items: Consider the given information and answer the five items that follow:
The following graph shows the percentage growth in population of six
cities from 1990 to 2000 and 2000 to 2010.If the population of City C in year 2010 and that of City D in year 2000 are equal and they are 27.2 lakh each. The population of City C in year 1990 is what percentage of population of City D in the same year?
Correct
Solution: A
It is given that, population of City C in year 2010 = population of City D in year 2000 = 27.2 lakh.
Therefore, population of City C in year 1990 = (100/170) * (100/160) * 27.2
= 10 lakh
The population of City D in year 1990 = (100/136) * 27.2 lakh = 20 lakh.
The population of City C in year 1990 is 50% of population of City D in the same year.
Hence, option (a) is correct.
Incorrect
Solution: A
It is given that, population of City C in year 2010 = population of City D in year 2000 = 27.2 lakh.
Therefore, population of City C in year 1990 = (100/170) * (100/160) * 27.2
= 10 lakh
The population of City D in year 1990 = (100/136) * 27.2 lakh = 20 lakh.
The population of City C in year 1990 is 50% of population of City D in the same year.
Hence, option (a) is correct.
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Question 5 of 5
5. Question
Direction for the following 5 (five) items: Consider the given information and answer the five items that follow:
The following graph shows the percentage growth in population of six
cities from 1990 to 2000 and 2000 to 2010.The population of City E in year 1990 was what fraction of its population in 2010?
Correct
Solution: D
Let the population of City E in year 1990 be ‘x’.
Therefore, the population of City E in year 2010 = x * (150/100) * (140/100)
= 2.1 x
Thus, the population of City E in year 1990 was 10/21 of its population in 2010.
Hence, option (d) is correct.
Incorrect
Solution: D
Let the population of City E in year 1990 be ‘x’.
Therefore, the population of City E in year 2010 = x * (150/100) * (140/100)
= 2.1 x
Thus, the population of City E in year 1990 was 10/21 of its population in 2010.
Hence, option (d) is correct.
