UPSC Insta–DART (Daily Aptitude and Reasoning Test) 3 Jan 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).

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• Question 1 of 5 1. Question A shopkeeper sells 1 kg of tea to two customers, Ritu and Neetu. For Ritu, he charges 10% above the cost price but under-weighs the quantity by 5%. For Neetu, he sells at 15% less than the cost price but gives 5% extra quantity. If each is billed for 1 kg, what is his overall profit/loss percentage? (a) 2.5% loss (b) 2.5% profit (c) 5% loss (d) 5% profit Correct Answer: (a) Explanation: Assume cost price = Rs.100 per kg. For Ritu: Selling price for 1 kg = 110. Quantity actually given = 0.95 kg. Cost to him = 0.95 × 100 = 95. Profit = 110 – 95 = 15. For Neetu: Selling price for 1 kg = 85 (15% less than 100). Quantity given = 1.05 kg. Cost to him = 1.05 × 100 = 105. Profit = 85 – 105 = –20 (loss). Total cost = 95 + 105 = 200. Total profit = 15 – 20 = –5. Overall profit % = (–5 ÷ 200) × 100 = –2.5% → 2.5% loss. Incorrect Answer: (a) Explanation: Assume cost price = Rs.100 per kg. For Ritu: Selling price for 1 kg = 110. Quantity actually given = 0.95 kg. Cost to him = 0.95 × 100 = 95. Profit = 110 – 95 = 15. For Neetu: Selling price for 1 kg = 85 (15% less than 100). Quantity given = 1.05 kg. Cost to him = 1.05 × 100 = 105. Profit = 85 – 105 = –20 (loss). Total cost = 95 + 105 = 200. Total profit = 15 – 20 = –5. Overall profit % = (–5 ÷ 200) × 100 = –2.5% → 2.5% loss.

#### 1. Question

A shopkeeper sells 1 kg of tea to two customers, Ritu and Neetu. For Ritu, he charges 10% above the cost price but under-weighs the quantity by 5%. For Neetu, he sells at 15% less than the cost price but gives 5% extra quantity. If each is billed for 1 kg, what is his overall profit/loss percentage?

• (a) 2.5% loss

• (b) 2.5% profit

• (c) 5% loss

• (d) 5% profit

Answer: (a)

Explanation: Assume cost price = Rs.100 per kg. For Ritu: Selling price for 1 kg = 110. Quantity actually given = 0.95 kg. Cost to him = 0.95 × 100 = 95. Profit = 110 – 95 = 15. For Neetu: Selling price for 1 kg = 85 (15% less than 100). Quantity given = 1.05 kg. Cost to him = 1.05 × 100 = 105. Profit = 85 – 105 = –20 (loss). Total cost = 95 + 105 = 200. Total profit = 15 – 20 = –5. Overall profit % = (–5 ÷ 200) × 100 = –2.5% → 2.5% loss.

Answer: (a)

• Question 2 of 5 2. Question A cube is painted on five faces, leaving only one face unpainted. The cube is then cut into 64 small cubes of equal size. How many small cubes will have at least one face painted? (a) 52 (b) 48 (c) 62 (d) 36 Correct Answer: (b) Explanation: 64 cubes → original cube was 4 × 4 × 4. An unpainted face contains: 4 × 4 = 16 cubes with zero painted faces. All other cubes have at least one painted face. So cubes with at least one painted face: 64 − 16 = 48 Incorrect Answer: (b) Explanation: 64 cubes → original cube was 4 × 4 × 4. An unpainted face contains: 4 × 4 = 16 cubes with zero painted faces. All other cubes have at least one painted face. So cubes with at least one painted face: 64 − 16 = 48

#### 2. Question

A cube is painted on five faces, leaving only one face unpainted. The cube is then cut into 64 small cubes of equal size. How many small cubes will have at least one face painted?

Answer: (b)

Explanation:

64 cubes → original cube was 4 × 4 × 4.

An unpainted face contains: 4 × 4 = 16 cubes with zero painted faces.

All other cubes have at least one painted face.

So cubes with at least one painted face: 64 − 16 = 48

Answer: (b)

Explanation:

64 cubes → original cube was 4 × 4 × 4.

An unpainted face contains: 4 × 4 = 16 cubes with zero painted faces.

All other cubes have at least one painted face.

So cubes with at least one painted face: 64 − 16 = 48

• Question 3 of 5 3. Question All the six faces of a cube are painted blue. The cube is then cut into 216 identical small cubes. If N is the number of small cubes that do not have any paint on their faces, then which of the following is true? (a) 40 < N ≤ 50 (b) 50 < N ≤ 60 (c) 60 < N ≤ 70 (d) 70 < N ≤ 80 Correct Answer: C Explanation: 216 = 6³, so the original cube is 6 × 6 × 6. Cubes with no painted face are the inner cubes. Their count = (6 − 2)³ = 4³ = 64. Now check the range: 60 < 64 ≤ 70 → option (c). So, the correct option is (c). Incorrect Answer: C Explanation: 216 = 6³, so the original cube is 6 × 6 × 6. Cubes with no painted face are the inner cubes. Their count = (6 − 2)³ = 4³ = 64. Now check the range: 60 < 64 ≤ 70 → option (c). So, the correct option is (c).

#### 3. Question

All the six faces of a cube are painted blue. The cube is then cut into 216 identical small cubes. If N is the number of small cubes that do not have any paint on their faces, then which of the following is true?

• (a) 40 < N ≤ 50

• (b) 50 < N ≤ 60

• (c) 60 < N ≤ 70

• (d) 70 < N ≤ 80

Explanation: 216 = 6³, so the original cube is 6 × 6 × 6. Cubes with no painted face are the inner cubes. Their count = (6 − 2)³ = 4³ = 64. Now check the range: 60 < 64 ≤ 70 → option (c).

So, the correct option is (c).

Explanation: 216 = 6³, so the original cube is 6 × 6 × 6. Cubes with no painted face are the inner cubes. Their count = (6 − 2)³ = 4³ = 64. Now check the range: 60 < 64 ≤ 70 → option (c).

So, the correct option is (c).

• Question 4 of 5 4. Question Pinky was born on 29th February 2016 which happened to be a Monday. On what day would she celebrate her birthday next? (a) Monday (b) Tuesday (c) Friday (d) Saturday Correct Answer: (d). Explanation: 29 Feb 2016 = Monday. So 28 Feb 2016 = Sunday. 28 Feb 2017 = Tuesday (2016 is a leap year → 2 odd days). 28 Feb 2018 = Wednesday 28 Feb 2019 = Thursday 28 Feb 2020 = Friday 29 Feb 2020 = Saturday Therefore, she celebrates her next birthday on Saturday. Incorrect Answer: (d). Explanation: 29 Feb 2016 = Monday. So 28 Feb 2016 = Sunday. 28 Feb 2017 = Tuesday (2016 is a leap year → 2 odd days). 28 Feb 2018 = Wednesday 28 Feb 2019 = Thursday 28 Feb 2020 = Friday 29 Feb 2020 = Saturday Therefore, she celebrates her next birthday on Saturday.

#### 4. Question

Pinky was born on 29th February 2016 which happened to be a Monday. On what day would she celebrate her birthday next?

• (a) Monday

• (b) Tuesday

• (c) Friday

• (d) Saturday

Answer: (d).

Explanation: 29 Feb 2016 = Monday. So 28 Feb 2016 = Sunday. 28 Feb 2017 = Tuesday (2016 is a leap year → 2 odd days). 28 Feb 2018 = Wednesday 28 Feb 2019 = Thursday 28 Feb 2020 = Friday 29 Feb 2020 = Saturday

Therefore, she celebrates her next birthday on Saturday.

Answer: (d).

Therefore, she celebrates her next birthday on Saturday.

• Question 5 of 5 5. Question Three siblings—Karan, Ruchi and Manav—were born in different months: March, July, and November. Their ages are 5, 8, and 10 years. The clues are: (i) Manav is older than the child born in July. (ii) The child born in March is 5 years old. (iii) Karan is not the youngest. (iv) Ruchi was not born in November. Which of the following is true? (a) Manav is 10 years old, born in November. (b) Karan is 8 years old, born in July. (c) Ruchi is 5 years old, born in March. (d) Manav is 8 years old, born in March. Correct Answer: (a) Explanation: From (ii), March → 5 years old. From (iv), Ruchi ≠ November → So Ruchi is either March or July. But March child = 5 years → Ruchi can be 5. Check (iii), Karan is not the youngest → so he ≠ 5. Thus March (5 years) → Ruchi. Remaining ages: 8 and 10 for Karan & Manav. From (i), Manav older than the July child → So July child must be younger → must be 8. Thus July → Karan → 8 years. Remaining month November → Manav → 10 years. Hence option (a) Manav is 10 years old, born in November is correct. Incorrect Answer: (a) Explanation: From (ii), March → 5 years old. From (iv), Ruchi ≠ November → So Ruchi is either March or July. But March child = 5 years → Ruchi can be 5. Check (iii), Karan is not the youngest → so he ≠ 5. Thus March (5 years) → Ruchi. Remaining ages: 8 and 10 for Karan & Manav. From (i), Manav older than the July child → So July child must be younger → must be 8. Thus July → Karan → 8 years. Remaining month November → Manav → 10 years. Hence option (a) Manav is 10 years old, born in November is correct.

#### 5. Question

Three siblings—Karan, Ruchi and Manav—were born in different months: March, July, and November. Their ages are 5, 8, and 10 years. The clues are:

(i) Manav is older than the child born in July. (ii) The child born in March is 5 years old. (iii) Karan is not the youngest. (iv) Ruchi was not born in November.

Which of the following is true?

• (a) Manav is 10 years old, born in November.

• (b) Karan is 8 years old, born in July.

• (c) Ruchi is 5 years old, born in March.

• (d) Manav is 8 years old, born in March.

Answer: (a)

Explanation:

From (ii), March → 5 years old. From (iv), Ruchi ≠ November → So Ruchi is either March or July.

But March child = 5 years → Ruchi can be 5.

Check (iii), Karan is not the youngest → so he ≠ 5. Thus March (5 years) → Ruchi.

Remaining ages: 8 and 10 for Karan & Manav.

From (i), Manav older than the July child → So July child must be younger → must be 8.

Thus July → Karan → 8 years. Remaining month November → Manav → 10 years.

Hence option (a) Manav is 10 years old, born in November is correct.

Answer: (a)

Explanation:

From (ii), March → 5 years old. From (iv), Ruchi ≠ November → So Ruchi is either March or July.

But March child = 5 years → Ruchi can be 5.

Check (iii), Karan is not the youngest → so he ≠ 5. Thus March (5 years) → Ruchi.

Remaining ages: 8 and 10 for Karan & Manav.

From (i), Manav older than the July child → So July child must be younger → must be 8.

Thus July → Karan → 8 years. Remaining month November → Manav → 10 years.

Hence option (a) Manav is 10 years old, born in November is correct.

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