UPSC Insta–DART (Daily Aptitude and Reasoning Test) 4 Oct 2025

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.

Looking forward to your enthusiastic participation (both in sending us questions and solving them on daily basis on this portal).

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• Question 1 of 5 1. Question Based on the passage, the following assumptions have been made: Monitoring frog calls can serve as an early warning system for environmental degradation. Human-induced changes in land use directly affect frog behaviour and breeding cycles. Select the correct answer using the code given below: (a) 1 only (b) 2 only (c) Both 1 and 2 (d) Neither 1 nor 2 Correct Solution: (c) Explanation: Statement 1 is valid — the passage suggests frog calls are ecological indicators, useful in tracking ecosystem health. Statement 2 is valid — habitat loss and urbanization are cited as factors altering frog breeding behaviour. Thus, both assumptions logically follow, making (c) correct. Incorrect Solution: (c) Explanation: Statement 1 is valid — the passage suggests frog calls are ecological indicators, useful in tracking ecosystem health. Statement 2 is valid — habitat loss and urbanization are cited as factors altering frog breeding behaviour. Thus, both assumptions logically follow, making (c) correct.

#### 1. Question

Based on the passage, the following assumptions have been made:

• Monitoring frog calls can serve as an early warning system for environmental degradation.

• Human-induced changes in land use directly affect frog behaviour and breeding cycles.

Select the correct answer using the code given below:

• (a) 1 only

• (b) 2 only

• (c) Both 1 and 2

• (d) Neither 1 nor 2

Solution: (c)

Explanation: Statement 1 is valid — the passage suggests frog calls are ecological indicators, useful in tracking ecosystem health. Statement 2 is valid — habitat loss and urbanization are cited as factors altering frog breeding behaviour. Thus, both assumptions logically follow, making (c) correct.

Solution: (c)

• Question 2 of 5 2. Question In 2018, number of males in a town are 20 percent less than the number of females. The male population decreases by 10 percent every year while the female population increases by 10 percent every year. Find how much percent the population in 2020 is more or less than the population in 2018. (a) 3 percent more (b) 7 percent more (c) 3 percent less (d) No change Correct Answer – A Solution: Given that, Males in 2018 are 20 percent less than females. Male change per year = minus 10 percent. Female change per year = plus 10 percent. Now, Let females in 2018 be x. Then males in 2018 = 0.8x. Total 2018 = x + 0.8x = 1.8x. In 2020 (two years later): Males = 0.8x × 0.9 × 0.9 = 0.8x × 0.81 = 0.648x. Females = x × 1.1 × 1.1 = 1.21x. Total 2020 = 0.648x + 1.21x = 1.858x. Percentage change = {(1.858x − 1.8x) ÷ 1.8x} × 100 = {0.058x ÷ 1.8x} × 100 ≈ 3.22 percent increase. Hence option (a) is correct. Incorrect Answer – A Solution: Given that, Males in 2018 are 20 percent less than females. Male change per year = minus 10 percent. Female change per year = plus 10 percent. Now, Let females in 2018 be x. Then males in 2018 = 0.8x. Total 2018 = x + 0.8x = 1.8x. In 2020 (two years later): Males = 0.8x × 0.9 × 0.9 = 0.8x × 0.81 = 0.648x. Females = x × 1.1 × 1.1 = 1.21x. Total 2020 = 0.648x + 1.21x = 1.858x. Percentage change = {(1.858x − 1.8x) ÷ 1.8x} × 100 = {0.058x ÷ 1.8x} × 100 ≈ 3.22 percent increase. Hence option (a) is correct.

#### 2. Question

In 2018, number of males in a town are 20 percent less than the number of females. The male population decreases by 10 percent every year while the female population increases by 10 percent every year. Find how much percent the population in 2020 is more or less than the population in 2018.

• (a) 3 percent more

• (b) 7 percent more

• (c) 3 percent less

• (d) No change

Answer – A Solution:

Given that,

Males in 2018 are 20 percent less than females. Male change per year = minus 10 percent. Female change per year = plus 10 percent.

Let females in 2018 be x. Then males in 2018 = 0.8x. Total 2018 = x + 0.8x = 1.8x.

In 2020 (two years later): Males = 0.8x × 0.9 × 0.9 = 0.8x × 0.81 = 0.648x. Females = x × 1.1 × 1.1 = 1.21x. Total 2020 = 0.648x + 1.21x = 1.858x.

Percentage change = {(1.858x − 1.8x) ÷ 1.8x} × 100 = {0.058x ÷ 1.8x} × 100 ≈ 3.22 percent increase.

Hence option (a) is correct.

Answer – A Solution:

Given that,

Males in 2018 are 20 percent less than females. Male change per year = minus 10 percent. Female change per year = plus 10 percent.

Let females in 2018 be x. Then males in 2018 = 0.8x. Total 2018 = x + 0.8x = 1.8x.

In 2020 (two years later): Males = 0.8x × 0.9 × 0.9 = 0.8x × 0.81 = 0.648x. Females = x × 1.1 × 1.1 = 1.21x. Total 2020 = 0.648x + 1.21x = 1.858x.

Percentage change = {(1.858x − 1.8x) ÷ 1.8x} × 100 = {0.058x ÷ 1.8x} × 100 ≈ 3.22 percent increase.

Hence option (a) is correct.

• Question 3 of 5 3. Question In an 800 metre race, A beats B by 80 metres. In a 1000 metre race, B can give a start of 200 metres to C. By how much distance should A give a start to C so that A beats C by 160 metres in an 800 metre race? (a) 56 m (b) 60 m (c) 64 m (d) 72 m Correct Answer: (c) Solution: From 800 metres: A:B = 800:720 = 10:9. From 1000 metres: B:C = 1000:800 = 5:4. So A:B:C = 50:45:36. When A runs 800, C runs (36/50) × 800 = 576. To beat by 160, C should be at 640. Required start = 640 − 576 = 64 metres. Incorrect Answer: (c) Solution: From 800 metres: A:B = 800:720 = 10:9. From 1000 metres: B:C = 1000:800 = 5:4. So A:B:C = 50:45:36. When A runs 800, C runs (36/50) × 800 = 576. To beat by 160, C should be at 640. Required start = 640 − 576 = 64 metres.

#### 3. Question

In an 800 metre race, A beats B by 80 metres. In a 1000 metre race, B can give a start of 200 metres to C. By how much distance should A give a start to C so that A beats C by 160 metres in an 800 metre race?

Answer: (c)

Solution: From 800 metres: A:B = 800:720 = 10:9. From 1000 metres: B:C = 1000:800 = 5:4. So A:B:C = 50:45:36. When A runs 800, C runs (36/50) × 800 = 576. To beat by 160, C should be at 640. Required start = 640 − 576 = 64 metres.

Answer: (c)

• Question 4 of 5 4. Question A box has x red balls, 20 blue balls, and y green balls. The probability of drawing a red ball is 3/10, and the probability of drawing a green ball is 2/5. Find the probability of getting 2 green balls when two balls are picked. (a) 29/190 (b) 31/190 (c) 51/190 (d) 39/190 Correct Answer: (c) Solution: Probability of red = x / (x + 20 + y) = 3/10 So, 10x = 3x + 60 + 3y → 7x − 3y = 60 …(1) Probability of green = y / (x + 20 + y) = 2/5 So, 5y = 2x + 40 + 2y → 3y − 2x = 40 …(2) Solving (1) and (2): x = 20, y = 40 Total = 20 + 20 + 40 = 80 Required probability = (40C2) / (80C2) = 780/3160 = 195/790 = 51/190. Incorrect Answer: (c) Solution: Probability of red = x / (x + 20 + y) = 3/10 So, 10x = 3x + 60 + 3y → 7x − 3y = 60 …(1) Probability of green = y / (x + 20 + y) = 2/5 So, 5y = 2x + 40 + 2y → 3y − 2x = 40 …(2) Solving (1) and (2): x = 20, y = 40 Total = 20 + 20 + 40 = 80 Required probability = (40C2) / (80C2) = 780/3160 = 195/790 = 51/190.

#### 4. Question

A box has x red balls, 20 blue balls, and y green balls. The probability of drawing a red ball is 3/10, and the probability of drawing a green ball is 2/5. Find the probability of getting 2 green balls when two balls are picked.

• (a) 29/190

• (b) 31/190

• (c) 51/190

• (d) 39/190

Answer: (c)

Solution: Probability of red = x / (x + 20 + y) = 3/10 So, 10x = 3x + 60 + 3y → 7x − 3y = 60 …(1)

Probability of green = y / (x + 20 + y) = 2/5 So, 5y = 2x + 40 + 2y → 3y − 2x = 40 …(2)

Solving (1) and (2): x = 20, y = 40 Total = 20 + 20 + 40 = 80

Required probability = (40C2) / (80C2) = 780/3160 = 195/790 = 51/190.

Answer: (c)

Solution: Probability of red = x / (x + 20 + y) = 3/10 So, 10x = 3x + 60 + 3y → 7x − 3y = 60 …(1)

Probability of green = y / (x + 20 + y) = 2/5 So, 5y = 2x + 40 + 2y → 3y − 2x = 40 …(2)

Solving (1) and (2): x = 20, y = 40 Total = 20 + 20 + 40 = 80

Required probability = (40C2) / (80C2) = 780/3160 = 195/790 = 51/190.

• Question 5 of 5 5. Question The probability that A, B, and C hit a target is 70%, 50%, and 30% respectively. Find the probability that A and B hit, but C misses. (a) 10.5% (b) 21% (c) 24.5% (d) 35% Correct Answer: (c) Solution: Probability (A hits) = 70/100 = 0.7 Probability (B hits) = 50/100 = 0.5 Probability (C misses) = 1 – 0.3 = 0.7 Required probability = 0.7 × 0.5 × 0.7 = 0.245 = 24.5% Correct answer = 24.5% Incorrect Answer: (c) Solution: Probability (A hits) = 70/100 = 0.7 Probability (B hits) = 50/100 = 0.5 Probability (C misses) = 1 – 0.3 = 0.7 Required probability = 0.7 × 0.5 × 0.7 = 0.245 = 24.5% Correct answer = 24.5%

#### 5. Question

The probability that A, B, and C hit a target is 70%, 50%, and 30% respectively. Find the probability that A and B hit, but C misses.

Answer: (c)

Solution: Probability (A hits) = 70/100 = 0.7 Probability (B hits) = 50/100 = 0.5 Probability (C misses) = 1 – 0.3 = 0.7

Required probability = 0.7 × 0.5 × 0.7 = 0.245 = 24.5%

Correct answer = 24.5%

Answer: (c)

Solution: Probability (A hits) = 70/100 = 0.7 Probability (B hits) = 50/100 = 0.5 Probability (C misses) = 1 – 0.3 = 0.7

Required probability = 0.7 × 0.5 × 0.7 = 0.245 = 24.5%

Correct answer = 24.5%

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