UPSC Insta–DART (Daily Aptitude and Reasoning Test) 27 Feb 2026

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).

#### Quiz-summary

0 of 5 questions completed

Questions:

#### Information

Best of Luck! 🙂

You have already completed the quiz before. Hence you can not start it again.

Quiz is loading...

You must sign in or sign up to start the quiz.

You have to finish following quiz, to start this quiz:

0 of 5 questions answered correctly

Your time:

Time has elapsed

You have reached 0 of 0 points, (0)

#### Categories

• Not categorized 0%

• Question 1 of 5 1. Question Pipe A can fill a tank in 8 hours, Pipe B can empty it in 12 hours, and Pipe C can fill it in 24 hours. All three are opened together at 10:00 a.m. After some time, Pipe B is closed, and the tank gets completely filled at 8:00 p.m. When was Pipe B closed? (a) 12:00 p.m. (b) 2:00 p.m. (c) 4:00 p.m. (d) 6:00 p.m. Correct Answer: (d) Explanation: A = 8 h (fill), B = 12 h (empty), C = 24 h (fill) LCM = 24 units Efficiency: A = 3 u/hr, B = –2 u/hr, C = 1 u/hr With all three open: net = 3 – 2 + 1 = 2 u/hr After closing B: net = 3 + 1 = 4 u/hr Total time from 10 a.m. to 8 p.m. = 10 hours Let B was open for x hours. Total work = 24 units = (2 × x) + [4 × (10 – x)] ⇒ 24 = 2x + 40 – 4x = 40 – 2x ⇒ 2x = 16 ⇒ x = 8 So B was closed 8 hours after 10 a.m. ⇒ 6:00 p.m. Hence, option (d) is correct. Incorrect Answer: (d) Explanation: A = 8 h (fill), B = 12 h (empty), C = 24 h (fill) LCM = 24 units Efficiency: A = 3 u/hr, B = –2 u/hr, C = 1 u/hr With all three open: net = 3 – 2 + 1 = 2 u/hr After closing B: net = 3 + 1 = 4 u/hr Total time from 10 a.m. to 8 p.m. = 10 hours Let B was open for x hours. Total work = 24 units = (2 × x) + [4 × (10 – x)] ⇒ 24 = 2x + 40 – 4x = 40 – 2x ⇒ 2x = 16 ⇒ x = 8 So B was closed 8 hours after 10 a.m. ⇒ 6:00 p.m. Hence, option (d) is correct.

#### 1. Question

Pipe A can fill a tank in 8 hours, Pipe B can empty it in 12 hours, and Pipe C can fill it in 24 hours. All three are opened together at 10:00 a.m. After some time, Pipe B is closed, and the tank gets completely filled at 8:00 p.m. When was Pipe B closed?

• (a) 12:00 p.m.

• (b) 2:00 p.m.

• (c) 4:00 p.m.

• (d) 6:00 p.m.

Answer: (d) Explanation: A = 8 h (fill), B = 12 h (empty), C = 24 h (fill) LCM = 24 units Efficiency: A = 3 u/hr, B = –2 u/hr, C = 1 u/hr With all three open: net = 3 – 2 + 1 = 2 u/hr After closing B: net = 3 + 1 = 4 u/hr Total time from 10 a.m. to 8 p.m. = 10 hours Let B was open for x hours. Total work = 24 units = (2 × x) + [4 × (10 – x)] ⇒ 24 = 2x + 40 – 4x = 40 – 2x ⇒ 2x = 16 ⇒ x = 8 So B was closed 8 hours after 10 a.m. ⇒ 6:00 p.m. Hence, option (d) is correct.

• Question 2 of 5 2. Question Three painters P, Q and R can finish a mural in 20, 30 and 60 days respectively. P alone works on Wednesday, Q alone on Thursday, R alone on Friday, then again P on Saturday, and so on (repeating P–Q–R). Consider the statements: (I) The work will be finished on Thursday. (II) The work will be finished in 29 days. Which of the above statement(s) is/are false? (a) Only (I) (b) Only (II) (c) Both (I) and (II) (d) Neither (I) nor (II) Correct Ans: (c) Explanation: Total work = LCM(20, 30, 60) = 60 units. Efficiencies: P = 3 u/d, Q = 2 u/d, R = 1 u/d. Each 3‑day cycle (P–Q–R) = 3 + 2 + 1 = 6 units. Cycles needed = 60/6 = 10 cycles = 30 days. Start on Wednesday (P). Every 7 days the weekday repeats; after 28 days it’s Wednesday again; add 2 more days → Friday. So it finishes on Friday and in 30 days. Therefore (I) (“Thursday”) is false and (II) (“29 days”) is false → both false. Incorrect Ans: (c) Explanation: Total work = LCM(20, 30, 60) = 60 units. Efficiencies: P = 3 u/d, Q = 2 u/d, R = 1 u/d. Each 3‑day cycle (P–Q–R) = 3 + 2 + 1 = 6 units. Cycles needed = 60/6 = 10 cycles = 30 days. Start on Wednesday (P). Every 7 days the weekday repeats; after 28 days it’s Wednesday again; add 2 more days → Friday. So it finishes on Friday and in 30 days. Therefore (I) (“Thursday”) is false and (II) (“29 days”) is false → both false.

#### 2. Question

Three painters P, Q and R can finish a mural in 20, 30 and 60 days respectively. P alone works on Wednesday, Q alone on Thursday, R alone on Friday, then again P on Saturday, and so on (repeating P–Q–R). Consider the statements:

(I) The work will be finished on Thursday. (II) The work will be finished in 29 days.

Which of the above statement(s) is/are false?

• (a) Only (I)

• (b) Only (II)

• (c) Both (I) and (II)

• (d) Neither (I) nor (II)

Ans: (c)

Explanation: Total work = LCM(20, 30, 60) = 60 units. Efficiencies: P = 3 u/d, Q = 2 u/d, R = 1 u/d. Each 3‑day cycle (P–Q–R) = 3 + 2 + 1 = 6 units. Cycles needed = 60/6 = 10 cycles = 30 days. Start on Wednesday (P). Every 7 days the weekday repeats; after 28 days it’s Wednesday again; add 2 more days → Friday. So it finishes on Friday and in 30 days. Therefore (I) (“Thursday”) is false and (II) (“29 days”) is false → both false.

Ans: (c)

• Question 3 of 5 3. Question P and Q can complete a handmade carpet in 8 days and 12 days respectively. The contract amount is ₹5,700. They work together for 2 days and then R joins, who alone can finish the carpet in 18 days. What amount is paid to R? (a) ₹630 (b) ₹700 (c) ₹760 (d) ₹840 Correct Answer: (b) Explanation: Work in 1 day: P = 1/8, Q = 1/12, R = 1/18. P + Q in 1 day = 1/8 + 1/12 = 5/24. In 2 days, P + Q complete 2 × 5/24 = 10/24 = 5/12. Remaining work = 1 − 5/12 = 7/12. Now P + Q + R in 1 day = 1/8 + 1/12 + 1/18 = (9 + 6 + 4)/72 = 19/72. Time to finish remaining 7/12 = (7/12) ÷ (19/72) = (7/12) × (72/19) = 42/19 days. R’s contribution = (1/18) × (42/19) = 7/57 of total work. R’s share = (7/57) × ₹5,700 = ₹700. Hence, option (b) is correct. Incorrect Answer: (b) Explanation: Work in 1 day: P = 1/8, Q = 1/12, R = 1/18. P + Q in 1 day = 1/8 + 1/12 = 5/24. In 2 days, P + Q complete 2 × 5/24 = 10/24 = 5/12. Remaining work = 1 − 5/12 = 7/12. Now P + Q + R in 1 day = 1/8 + 1/12 + 1/18 = (9 + 6 + 4)/72 = 19/72. Time to finish remaining 7/12 = (7/12) ÷ (19/72) = (7/12) × (72/19) = 42/19 days. R’s contribution = (1/18) × (42/19) = 7/57 of total work. R’s share = (7/57) × ₹5,700 = ₹700. Hence, option (b) is correct.

#### 3. Question

P and Q can complete a handmade carpet in 8 days and 12 days respectively. The contract amount is ₹5,700. They work together for 2 days and then R joins, who alone can finish the carpet in 18 days. What amount is paid to R?

Answer: (b) Explanation: Work in 1 day: P = 1/8, Q = 1/12, R = 1/18. P + Q in 1 day = 1/8 + 1/12 = 5/24. In 2 days, P + Q complete 2 × 5/24 = 10/24 = 5/12. Remaining work = 1 − 5/12 = 7/12. Now P + Q + R in 1 day = 1/8 + 1/12 + 1/18 = (9 + 6 + 4)/72 = 19/72. Time to finish remaining 7/12 = (7/12) ÷ (19/72) = (7/12) × (72/19) = 42/19 days. R’s contribution = (1/18) × (42/19) = 7/57 of total work. R’s share = (7/57) × ₹5,700 = ₹700. Hence, option (b) is correct.

• Question 4 of 5 4. Question A compiled note has 15 sheets, each with 120 lines, and each line has 72 characters. It is reformatted into sheets of 90 lines with 24 characters per line. The approximate percentage change in number of sheets is (a) 200 (b) 250 (c) 300 (d) 350 Correct A compiled note has 15 sheets, each with 120 lines, and each line has 72 characters. It is reformatted into sheets of 90 lines with 24 characters per line. The approximate percentage change in number of sheets is Incorrect A compiled note has 15 sheets, each with 120 lines, and each line has 72 characters. It is reformatted into sheets of 90 lines with 24 characters per line. The approximate percentage change in number of sheets is

#### 4. Question

A compiled note has 15 sheets, each with 120 lines, and each line has 72 characters. It is reformatted into sheets of 90 lines with 24 characters per line. The approximate percentage change in number of sheets is

• Question 5 of 5 5. Question In a firm, the number of employees reduces in the ratio of 4:3 and the wages increase in the ratio of 10:13. What is the percentage savings on total wages over the previous wages? (a) 1.50% (b) 2.50% (c) 3.33% (d) 5.00% Correct Answer: (b) Explanation: Let initial employees = 4a, initial wage per employee = 10b. Initial total = 4a × 10b = 40ab. After change: employees = (3/4) × 4a = 3a; wage per employee = (13/10) × 10b = 13b. New total = 3a × 13b = 39ab. Savings % = . Hence, option (b) is correct. Incorrect Answer: (b) Explanation: Let initial employees = 4a, initial wage per employee = 10b. Initial total = 4a × 10b = 40ab. After change: employees = (3/4) × 4a = 3a; wage per employee = (13/10) × 10b = 13b. New total = 3a × 13b = 39ab. Savings % = . Hence, option (b) is correct.

#### 5. Question

In a firm, the number of employees reduces in the ratio of 4:3 and the wages increase in the ratio of 10:13. What is the percentage savings on total wages over the previous wages?

Answer: (b) Explanation: Let initial employees = 4a, initial wage per employee = 10b. Initial total = 4a × 10b = 40ab.

After change: employees = (3/4) × 4a = 3a; wage per employee = (13/10) × 10b = 13b. New total = 3a × 13b = 39ab.

Savings % = . Hence, option (b) is correct.

Answer: (b) Explanation: Let initial employees = 4a, initial wage per employee = 10b. Initial total = 4a × 10b = 40ab.

After change: employees = (3/4) × 4a = 3a; wage per employee = (13/10) × 10b = 13b. New total = 3a × 13b = 39ab.

Savings % = . Hence, option (b) is correct.

• Official Facebook Page HERE

• Follow our Twitter Account HERE