UPSC Insta–DART (Daily Aptitude and Reasoning Test) 29 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 There are some hens and some men in a farm. If total heads are 50 and total legs are 140, then how many hens are there? (a) 20 (b) 25 (c) 30 (d) 35 Correct Answer: C Solution: Let hens = H, men = M. Equation 1: H + M = 50. Equation 2: 2H + 4M = 140. Simplify: divide Eqn 2 by 2 → H + 2M = 70. Subtract Eqn 1: (H + 2M) – (H + M) = 70 – 50 → M = 20. Then H = 30. Hence, option (c). Incorrect Answer: C Solution: Let hens = H, men = M. Equation 1: H + M = 50. Equation 2: 2H + 4M = 140. Simplify: divide Eqn 2 by 2 → H + 2M = 70. Subtract Eqn 1: (H + 2M) – (H + M) = 70 – 50 → M = 20. Then H = 30. Hence, option (c).

#### 1. Question

There are some hens and some men in a farm. If total heads are 50 and total legs are 140, then how many hens are there?

Answer: C

Solution: Let hens = H, men = M.

Equation 1: H + M = 50. Equation 2: 2H + 4M = 140.

Simplify: divide Eqn 2 by 2 → H + 2M = 70. Subtract Eqn 1: (H + 2M) – (H + M) = 70 – 50 → M = 20. Then H = 30.

Hence, option (c).

Answer: C

Solution: Let hens = H, men = M.

Equation 1: H + M = 50. Equation 2: 2H + 4M = 140.

Simplify: divide Eqn 2 by 2 → H + 2M = 70. Subtract Eqn 1: (H + 2M) – (H + M) = 70 – 50 → M = 20. Then H = 30.

Hence, option (c).

• Question 2 of 5 2. Question How many integers are there between 201 and 300 which have 7 as a digit but only at unit place, and are also divisible by 9? (a) There are less than 20 but more than 8 such integers. (b) There are less than 5 such integers. (c) There are only 7 such integers. (d) There are only 22 such integers Correct Answer: B Explanation Integers having 7 as a digit placed only at the unit place are: 207, 217, 227, 237, 247, 257, 267, 277, 287, 297. So, there are a total of 10 such integers. Out of these, the integers which are divisible by 9 are 207 and 297. Hence, option (b) is the correct answer. Incorrect Answer: B Explanation Integers having 7 as a digit placed only at the unit place are: 207, 217, 227, 237, 247, 257, 267, 277, 287, 297. So, there are a total of 10 such integers. Out of these, the integers which are divisible by 9 are 207 and 297. Hence, option (b) is the correct answer.

#### 2. Question

How many integers are there between 201 and 300 which have 7 as a digit but only at unit place, and are also divisible by 9?

• (a) There are less than 20 but more than 8 such integers.

• (b) There are less than 5 such integers.

• (c) There are only 7 such integers.

• (d) There are only 22 such integers

Answer: B Explanation Integers having 7 as a digit placed only at the unit place are: 207, 217, 227, 237, 247, 257, 267, 277, 287, 297. So, there are a total of 10 such integers. Out of these, the integers which are divisible by 9 are 207 and 297. Hence, option (b) is the correct answer.

• Question 3 of 5 3. Question What is the minimum length that can be measured using two scales of length 6 cm and 10 cm respectively? a) 1 b) 2 c) 4 d) 0.5 Correct Answer: B Exp) Option b is the correct answer. We need the highest common factor (HCF) of 6 and 10. Factors of 6: 1, 2, 3, 6. Factors of 10: 1, 2, 5, 10. HCF = 2. Therefore, the minimum measurable length is 2 cm. Incorrect Answer: B Exp) Option b is the correct answer. We need the highest common factor (HCF) of 6 and 10. Factors of 6: 1, 2, 3, 6. Factors of 10: 1, 2, 5, 10. HCF = 2. Therefore, the minimum measurable length is 2 cm.

#### 3. Question

What is the minimum length that can be measured using two scales of length 6 cm and 10 cm respectively?

Answer: B Exp) Option b is the correct answer. We need the highest common factor (HCF) of 6 and 10. Factors of 6: 1, 2, 3, 6. Factors of 10: 1, 2, 5, 10. HCF = 2. Therefore, the minimum measurable length is 2 cm.

• Question 4 of 5 4. Question Each digit of a five-digit number is either 1 or 0 in an alternating manner starting from 1. It is multiplied by itself. What is the sum of the digits of the resulting number? a) 7 b) 8 c) 9 d) 10 Correct Answer: C Explanation Number = 10101. Using the same pattern: 101 × 101 = 10201 (10101) × (10101) = 102030201 Sum of digits = 1 + 2 + 3 + 2 + 1 = 9 (equivalently, 2 × (1 + 2) + 3 = 9). Incorrect Answer: C Explanation Number = 10101. Using the same pattern: 101 × 101 = 10201 (10101) × (10101) = 102030201 Sum of digits = 1 + 2 + 3 + 2 + 1 = 9 (equivalently, 2 × (1 + 2) + 3 = 9).

#### 4. Question

Each digit of a five-digit number is either 1 or 0 in an alternating manner starting from 1. It is multiplied by itself. What is the sum of the digits of the resulting number?

Answer: C Explanation Number = 10101. Using the same pattern: 101 × 101 = 10201 (10101) × (10101) = 102030201 Sum of digits = 1 + 2 + 3 + 2 + 1 = 9 (equivalently, 2 × (1 + 2) + 3 = 9).

• Question 5 of 5 5. Question Let a be a positive integer such that (a + 11)(a + 13) is divisible by a. How many values of a are possible? a) 3 b) 4 c) 5 d) 6 Correct Answer: B Explanation (a + 11)(a + 13) = a² + 24a + 143. Dividing by a gives a + 24 + 143/a. So a | 143. 143 = 11 × 13 ⇒ divisors are {1, 11, 13, 143} → 4 values. Hence, option (b) is correct. Incorrect Answer: B Explanation (a + 11)(a + 13) = a² + 24a + 143. Dividing by a gives a + 24 + 143/a. So a | 143. 143 = 11 × 13 ⇒ divisors are {1, 11, 13, 143} → 4 values. Hence, option (b) is correct.

#### 5. Question

Let a be a positive integer such that (a + 11)(a + 13) is divisible by a. How many values of a are possible?

Answer: B Explanation (a + 11)(a + 13) = a² + 24a + 143. Dividing by a gives a + 24 + 143/a. So a | 143. 143 = 11 × 13 ⇒ divisors are {1, 11, 13, 143} → 4 values. Hence, option (b) is correct.

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