UPSC Insta–DART (Daily Aptitude and Reasoning Test) 17 Nov 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 A shopkeeper prints pages numbered from 1 to 18,000. He uses a special ink that costs ₹0.40 per digit only when digit 5 appears in the units or tens place. What is the total extra cost incurred? (a) ₹1,200 (b) ₹1,360 (c) ₹1,440 (d) ₹1,500 Correct Answer: (c) Solution: We only count 5s in units and tens place. Units place: Every 10 numbers → units digit = 5 once → 18,000 ÷ 10 = 1,800 occurrences Tens place: In every 100 numbers → tens digit = 5 in 10 numbers → 18,000 → 180 blocks → 180 × 10 = 1,800 occurrences Total 5s in units or tens = 1,800 + 1,800 = 3,600 Cost = 3,600 × ₹0.40 = ₹1,440 Digit Place | Occurrences of Digit 5 | Cost per Digit (₹) | Total Cost (₹) Units | 18,000 / 10 = 1,800 | 0.40 | 1,800 × 0.40 = 720 Tens | 180 × 10 = 1,800 | 0.40 | 1,800 × 0.40 = 720 Total | 3,600 | | 720 + 720 = 1,440 Hence, option (c) is correct. Incorrect Answer: (c) Solution: We only count 5s in units and tens place. Units place: Every 10 numbers → units digit = 5 once → 18,000 ÷ 10 = 1,800 occurrences Tens place: In every 100 numbers → tens digit = 5 in 10 numbers → 18,000 → 180 blocks → 180 × 10 = 1,800 occurrences Total 5s in units or tens = 1,800 + 1,800 = 3,600 Cost = 3,600 × ₹0.40 = ₹1,440 Digit Place | Occurrences of Digit 5 | Cost per Digit (₹) | Total Cost (₹) Units | 18,000 / 10 = 1,800 | 0.40 | 1,800 × 0.40 = 720 Tens | 180 × 10 = 1,800 | 0.40 | 1,800 × 0.40 = 720 Total | 3,600 | | 720 + 720 = 1,440 Hence, option (c) is correct.

#### 1. Question

A shopkeeper prints pages numbered from 1 to 18,000. He uses a special ink that costs ₹0.40 per digit only when digit 5 appears in the units or tens place. What is the total extra cost incurred?

• (a) ₹1,200

• (b) ₹1,360

• (c) ₹1,440

• (d) ₹1,500

Answer: (c) Solution: We only count 5s in units and tens place. Units place: Every 10 numbers → units digit = 5 once → 18,000 ÷ 10 = 1,800 occurrences Tens place: In every 100 numbers → tens digit = 5 in 10 numbers → 18,000 → 180 blocks → 180 × 10 = 1,800 occurrences Total 5s in units or tens = 1,800 + 1,800 = 3,600 Cost = 3,600 × ₹0.40 = ₹1,440

Digit Place | Occurrences of Digit 5 | Cost per Digit (₹) | Total Cost (₹) Units | 18,000 / 10 = 1,800 | 0.40 | 1,800 × 0.40 = 720 Tens | 180 × 10 = 1,800 | 0.40 | 1,800 × 0.40 = 720 Total | 3,600 | | 720 + 720 = 1,440

Hence, option (c) is correct.

• Question 2 of 5 2. Question Consider the following series and identify how many statements are correct: In the series 1, 4, 9, 16, 27, 36, 49, the wrong term is 27. In the series 5, 11, 23, 47, 95, 191, the wrong term is 191. In the series 3, 5, 9, 17, 33, 65, the wrong term is 33. How many of the above statements is/are correct? (a) Only one (b) Only two (c) Only three (d) None Correct Answer: (a) Solution • (1) Squares: 1², 2², 3², 4², 5²(25), 6²(36), 7²(49). Here 27 should be 25 → wrong term correctly identified ⇒ correct. • (2) Pattern ×2 + 1: 5→11→23→47→95→191 (all correct). Saying 191 is wrong ⇒ incorrect statement. • (3) Differences double each step: +2, +4, +8, +16, +32 → 3,5,9,17,33,65 is consistent. Saying 33 is wrong ⇒ incorrect. Hence only statement (1) is correct. Incorrect Answer: (a) Solution • (1) Squares: 1², 2², 3², 4², 5²(25), 6²(36), 7²(49). Here 27 should be 25 → wrong term correctly identified ⇒ correct. • (2) Pattern ×2 + 1: 5→11→23→47→95→191 (all correct). Saying 191 is wrong ⇒ incorrect statement. • (3) Differences double each step: +2, +4, +8, +16, +32 → 3,5,9,17,33,65 is consistent. Saying 33 is wrong ⇒ incorrect. Hence only statement (1) is correct.

#### 2. Question

Consider the following series and identify how many statements are correct:

• In the series 1, 4, 9, 16, 27, 36, 49, the wrong term is 27.

• In the series 5, 11, 23, 47, 95, 191, the wrong term is 191.

• In the series 3, 5, 9, 17, 33, 65, the wrong term is 33.

How many of the above statements is/are correct?

• (a) Only one

• (b) Only two

• (c) Only three

Answer: (a) Solution • (1) Squares: 1², 2², 3², 4², 5²(25), 6²(36), 7²(49). Here 27 should be 25 → wrong term correctly identified ⇒ correct. • (2) Pattern ×2 + 1: 5→11→23→47→95→191 (all correct). Saying 191 is wrong ⇒ incorrect statement. • (3) Differences double each step: +2, +4, +8, +16, +32 → 3,5,9,17,33,65 is consistent. Saying 33 is wrong ⇒ incorrect. Hence only statement (1) is correct.

• Question 3 of 5 3. Question Let a, b, c, d be distinct positive integers such that a, b, c are odd and d is even. Consider the following statements (a + b + c + d) is even abc + d is odd a⋅b⋅c⋅d is even Which of the statements given above are correct? (a) 1 and 2 only (b) 2 and 3 only (c) 1 and 3 only (d) 1, 2 and 3 Correct Answer: (b) Solution: a, b, c odd; d even a + b + c + d = odd + odd + odd + even = odd → False abc = odd × odd × odd = odd → odd + d(even) = odd → True Product includes even (d) → overall even → True Hence, 2 and 3 only → option (b). Incorrect Answer: (b) Solution: a, b, c odd; d even a + b + c + d = odd + odd + odd + even = odd → False abc = odd × odd × odd = odd → odd + d(even) = odd → True Product includes even (d) → overall even → True Hence, 2 and 3 only → option (b).

#### 3. Question

Let a, b, c, d be distinct positive integers such that a, b, c are odd and d is even. Consider the following statements

• (a + b + c + d) is even

• abc + d is odd

• a⋅b⋅c⋅d is even

Which of the statements given above are correct?

• (a) 1 and 2 only

• (b) 2 and 3 only

• (c) 1 and 3 only

• (d) 1, 2 and 3

Answer: (b) Solution:

a, b, c odd; d even

• a + b + c + d = odd + odd + odd + even = odd → False

• abc = odd × odd × odd = odd → odd + d(even) = odd → True

• Product includes even (d) → overall even → True Hence, 2 and 3 only → option (b).

Answer: (b) Solution:

a, b, c odd; d even

• a + b + c + d = odd + odd + odd + even = odd → False

• abc = odd × odd × odd = odd → odd + d(even) = odd → True

• Product includes even (d) → overall even → True Hence, 2 and 3 only → option (b).

• Question 4 of 5 4. Question A Question is given followed by two Statements I and II. Consider the Question and the Statements. Question: What are the values of two natural numbers, a and b? Statement I: The sum of a and b is 12. Statement II: The product of a and b is 32. (a) Either statement alone is sufficient (b) Only one of the statements is sufficient (c) Both statements together are necessary (d) Even both statements together are not sufficient Correct Answer: (c) Explanation: Statement I: a + b = 12 → Possible pairs: (1,11), (2,10), (3,9), (4,8), (5,7), (6,6) → Not unique Statement II: ab = 32 → Possible pairs: (1,32), (2,16), (4,8) → Not unique Combining both: Only (4,8) satisfies both ⇒ Unique. Hence, option (c). Incorrect Answer: (c) Explanation: Statement I: a + b = 12 → Possible pairs: (1,11), (2,10), (3,9), (4,8), (5,7), (6,6) → Not unique Statement II: ab = 32 → Possible pairs: (1,32), (2,16), (4,8) → Not unique Combining both: Only (4,8) satisfies both ⇒ Unique. Hence, option (c).

#### 4. Question

A Question is given followed by two Statements I and II. Consider the Question and the Statements. Question: What are the values of two natural numbers, a and b? Statement I: The sum of a and b is 12. Statement II: The product of a and b is 32.

• (a) Either statement alone is sufficient

• (b) Only one of the statements is sufficient

• (c) Both statements together are necessary

• (d) Even both statements together are not sufficient

Answer: (c) Explanation: Statement I: a + b = 12 → Possible pairs: (1,11), (2,10), (3,9), (4,8), (5,7), (6,6) → Not unique Statement II: ab = 32 → Possible pairs: (1,32), (2,16), (4,8) → Not unique Combining both: Only (4,8) satisfies both ⇒ Unique.

Hence, option (c).

• Question 5 of 5 5. Question X can complete one-half of a certain work in 15 days, Y can complete one-fourth of the same work in 5 days and Z can complete two-thirds of the same work in 16 days. All of them work together for n days and then X and Z quit and Y alone finishes the remaining work in 7½ days. What is n equal to? (a) 3 (b) 4 (c) 5 (d) 6 Correct Answer: (c) X can complete total work in 15 × 2 = 30 days Y can complete total work in 5 × 4 = 20 days Z can complete total work in 16 ÷ (2/3) = 24 days Let total work = LCM(30, 20, 24) = 120 units Efficiency of X = 120/30 = 4 units/day Efficiency of Y = 120/20 = 6 units/day Efficiency of Z = 120/24 = 5 units/day Together = (4 + 6 + 5) = 15 units/day Work done by Y alone in 7½ days = 6 × 7.5 = 45 units Remaining work before Y’s solo = 120 − 45 = 75 units Time for all three to do 75 units = 75/15 = 5 days ⇒ n = 5 Incorrect Answer: (c) X can complete total work in 15 × 2 = 30 days Y can complete total work in 5 × 4 = 20 days Z can complete total work in 16 ÷ (2/3) = 24 days Let total work = LCM(30, 20, 24) = 120 units Efficiency of X = 120/30 = 4 units/day Efficiency of Y = 120/20 = 6 units/day Efficiency of Z = 120/24 = 5 units/day Together = (4 + 6 + 5) = 15 units/day Work done by Y alone in 7½ days = 6 × 7.5 = 45 units Remaining work before Y’s solo = 120 − 45 = 75 units Time for all three to do 75 units = 75/15 = 5 days ⇒ n = 5

#### 5. Question

X can complete one-half of a certain work in 15 days, Y can complete one-fourth of the same work in 5 days and Z can complete two-thirds of the same work in 16 days. All of them work together for n days and then X and Z quit and Y alone finishes the remaining work in 7½ days. What is n equal to?

Answer: (c) X can complete total work in 15 × 2 = 30 days Y can complete total work in 5 × 4 = 20 days Z can complete total work in 16 ÷ (2/3) = 24 days Let total work = LCM(30, 20, 24) = 120 units Efficiency of X = 120/30 = 4 units/day Efficiency of Y = 120/20 = 6 units/day Efficiency of Z = 120/24 = 5 units/day Together = (4 + 6 + 5) = 15 units/day Work done by Y alone in 7½ days = 6 × 7.5 = 45 units Remaining work before Y’s solo = 120 − 45 = 75 units Time for all three to do 75 units = 75/15 = 5 days ⇒ n = 5

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