UPSC Insta–DART (Daily Aptitude and Reasoning Test) 2 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 X and Y start a business with capitals in the ratio 6:5. After 4 months, X withdraws one-third of his capital for the remaining 8 months, while Y increases his capital by 50% for the remaining 8 months. If the total profit at year end is ₹34,000, what is Y’s share? (a) ₹18,000 (b) ₹19,000 (c) ₹20,000 (d) ₹22,000 Correct Answer: (c) Solution: Let initial capitals be X = 6k, Y = 5k. For first 4 months: • X = 6k × 4 • Y = 5k × 4 For next 8 months: • X = (6k − ⅓·6k) = 4k ⇒ 4k × 8 • Y = (5k + 50%) = 7.5k ⇒ 7.5k × 8 Total: • X = 24k + 32k = 56k • Y = 20k + 60k = 80k Ratio = 56 : 80 = 7 : 10 Total parts = 17 Y’s share = (10/17) × 34,000 = ₹20,000. Hence, option (c). Incorrect Answer: (c) Solution: Let initial capitals be X = 6k, Y = 5k. For first 4 months: • X = 6k × 4 • Y = 5k × 4 For next 8 months: • X = (6k − ⅓·6k) = 4k ⇒ 4k × 8 • Y = (5k + 50%) = 7.5k ⇒ 7.5k × 8 Total: • X = 24k + 32k = 56k • Y = 20k + 60k = 80k Ratio = 56 : 80 = 7 : 10 Total parts = 17 Y’s share = (10/17) × 34,000 = ₹20,000. Hence, option (c).

#### 1. Question

X and Y start a business with capitals in the ratio 6:5. After 4 months, X withdraws one-third of his capital for the remaining 8 months, while Y increases his capital by 50% for the remaining 8 months. If the total profit at year end is ₹34,000, what is Y’s share?

• (a) ₹18,000

• (b) ₹19,000

• (c) ₹20,000

• (d) ₹22,000

Answer: (c)

Solution: Let initial capitals be X = 6k, Y = 5k. For first 4 months: • X = 6k × 4 • Y = 5k × 4 For next 8 months: • X = (6k − ⅓·6k) = 4k ⇒ 4k × 8 • Y = (5k + 50%) = 7.5k ⇒ 7.5k × 8 Total: • X = 24k + 32k = 56k • Y = 20k + 60k = 80k Ratio = 56 : 80 = 7 : 10 Total parts = 17 Y’s share = (10/17) × 34,000 = ₹20,000. Hence, option (c).

Answer: (c)

• Question 2 of 5 2. Question In a class, 80% of students are boys and 35% of the total are scholarship holders. If ¼ of the boys are scholarship holders, what fraction of the girls do not have a scholarship? (a) 1/2 (b) 2/3 (c) 3/5 (d) 1/4 Correct Answer – D Solution: Total students = 100. Boys = 80, Girls = 20. Total scholarship holders = 35. Scholarship boys = ¼ × 80 = 20. Scholarship girls = 35 – 20 = 15. Non-scholarship girls = 20 – 15 = 5. Fraction = 5 / 20 = 1/4. Incorrect Answer – D Solution: Total students = 100. Boys = 80, Girls = 20. Total scholarship holders = 35. Scholarship boys = ¼ × 80 = 20. Scholarship girls = 35 – 20 = 15. Non-scholarship girls = 20 – 15 = 5. Fraction = 5 / 20 = 1/4.

#### 2. Question

In a class, 80% of students are boys and 35% of the total are scholarship holders. If ¼ of the boys are scholarship holders, what fraction of the girls do not have a scholarship?

Answer – D Solution: Total students = 100. Boys = 80, Girls = 20. Total scholarship holders = 35. Scholarship boys = ¼ × 80 = 20. Scholarship girls = 35 – 20 = 15. Non-scholarship girls = 20 – 15 = 5. Fraction = 5 / 20 = 1/4.

• Question 3 of 5 3. Question A water tank can supply a family for 20 days. If their daily water usage increases by 25%, how long will the water last? (a) 12 days (b) 14 days (c) 15 days (d) 16 days Correct Answer – D Solution: Tank lasts 20 days at normal use. Let daily use = 100 litres. Total water = 20 × 100 = 2000 litres. If usage rises by 25%, daily use = 125 litres. Number of days = 2000 ÷ 125 = 16. Hence, the water lasts for 16 days. Incorrect Answer – D Solution: Tank lasts 20 days at normal use. Let daily use = 100 litres. Total water = 20 × 100 = 2000 litres. If usage rises by 25%, daily use = 125 litres. Number of days = 2000 ÷ 125 = 16. Hence, the water lasts for 16 days.

#### 3. Question

A water tank can supply a family for 20 days. If their daily water usage increases by 25%, how long will the water last?

• (a) 12 days

• (b) 14 days

• (c) 15 days

• (d) 16 days

Answer – D Solution: Tank lasts 20 days at normal use. Let daily use = 100 litres. Total water = 20 × 100 = 2000 litres. If usage rises by 25%, daily use = 125 litres. Number of days = 2000 ÷ 125 = 16. Hence, the water lasts for 16 days.

• Question 4 of 5 4. Question Two persons plan to hire a taxi. Person A has ₹9, which he finds is 60% of the taxi fare for both. His friend B gives him ₹7. Will the combined money be enough to pay the fare, and how much will be left or still needed? (a) Just sufficient, nothing left (b) ₹1 left after paying (c) ₹2 short of the fare (d) ₹3 left after paying Correct Answer – B Solution: Let taxi fare for both = F. 60% of F = 9 ⇒ F = (9 × 100) / 60 = ₹15. Total money = 9 + 7 = ₹16. Balance = 16 – 15 = ₹1 left. Hence, option (b) is correct. Incorrect Answer – B Solution: Let taxi fare for both = F. 60% of F = 9 ⇒ F = (9 × 100) / 60 = ₹15. Total money = 9 + 7 = ₹16. Balance = 16 – 15 = ₹1 left. Hence, option (b) is correct.

#### 4. Question

Two persons plan to hire a taxi. Person A has ₹9, which he finds is 60% of the taxi fare for both. His friend B gives him ₹7. Will the combined money be enough to pay the fare, and how much will be left or still needed?

• (a) Just sufficient, nothing left

• (b) ₹1 left after paying

• (c) ₹2 short of the fare

• (d) ₹3 left after paying

Answer – B Solution: Let taxi fare for both = F. 60% of F = 9 ⇒ F = (9 × 100) / 60 = ₹15. Total money = 9 + 7 = ₹16. Balance = 16 – 15 = ₹1 left. Hence, option (b) is correct.

• Question 5 of 5 5. Question A company has to deliver 15,000 bulbs. It produces 800 bulbs daily, but 12.5% are rejected. After 5 days, due to additional workers, daily production becomes 1,200 bulbs with only 10% rejected. Find the total days required to finish the order. (a) 14 (b) 15 (c) 16 (d) 17 Correct Correct option = (c). Solution: First 5 days: Daily good bulbs = 800 – 12.5% of 800 = 800 – 100 = 700. In 5 days = 5 × 700 = 3,500. Remaining = 15,000 – 3,500 = 11,500. After workers: Daily good bulbs = 1,200 – 10% of 1,200 = 1,080. Days required = 11,500 ÷ 1,080 = 10.65 ≈ 11 days. Total days = 5 + 11 = 16. Incorrect Correct option = (c). Solution: First 5 days: Daily good bulbs = 800 – 12.5% of 800 = 800 – 100 = 700. In 5 days = 5 × 700 = 3,500. Remaining = 15,000 – 3,500 = 11,500. After workers: Daily good bulbs = 1,200 – 10% of 1,200 = 1,080. Days required = 11,500 ÷ 1,080 = 10.65 ≈ 11 days. Total days = 5 + 11 = 16.

#### 5. Question

A company has to deliver 15,000 bulbs. It produces 800 bulbs daily, but 12.5% are rejected. After 5 days, due to additional workers, daily production becomes 1,200 bulbs with only 10% rejected. Find the total days required to finish the order.

Correct option = (c).

Solution: First 5 days: Daily good bulbs = 800 – 12.5% of 800 = 800 – 100 = 700. In 5 days = 5 × 700 = 3,500.

Remaining = 15,000 – 3,500 = 11,500.

After workers: Daily good bulbs = 1,200 – 10% of 1,200 = 1,080.

Days required = 11,500 ÷ 1,080 = 10.65 ≈ 11 days.

Total days = 5 + 11 = 16.