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?
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 = ₹1,440Digit 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 = 1,440Hence, 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 = ₹1,440Digit 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 = 1,440Hence, option (c) is correct.
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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?
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. -
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?
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).
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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.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).
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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?
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 = 5Incorrect
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
