UPSC Insta–DART (Daily Aptitude and Reasoning Test) 12 Mar 2026
Kartavya Desk Staff
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 test has 7 questions, each to be answered as True or False. No two students gave the same set of answers, and no student gave exactly 4 True answers. What is the maximum number of students who could have appeared? (a) 88 (b) 93 (c) 96 (d) 128 Correct Incorrect
#### 1. Question
A test has 7 questions, each to be answered as True or False. No two students gave the same set of answers, and no student gave exactly 4 True answers. What is the maximum number of students who could have appeared?
• Question 2 of 5 2. Question A bag contains 7 marbles: 2 black and 5 white. If 4 marbles are drawn with replacement, what is the probability of getting exactly 3 white and 1 black? (a) 500/2401 (b) 1000/2401 (c) 750/2401 (d) 1250/2401 Correct Incorrect
#### 2. Question
A bag contains 7 marbles: 2 black and 5 white. If 4 marbles are drawn with replacement, what is the probability of getting exactly 3 white and 1 black?
• (a) 500/2401
• (b) 1000/2401
• (c) 750/2401
• (d) 1250/2401
• Question 3 of 5 3. Question A person has 5 distinct keys and 5 distinct locks. If 2 keys are already matched correctly to their locks, in how many ways can the remaining 3 keys be put into the remaining 3 locks such that none of them open their respective locks? (a) 2 (b) 6 (c) 9 (d) 44 Correct Answer: (a) Solution: 2 keys are already correctly matched, so we are left with 3 distinct keys and 3 distinct locks. We need the number of ways to assign these 3 keys to the 3 locks such that no key goes to its own correct lock. This is exactly the number of derangements of 3 items, denoted . Total permutations of 3 items: 3!=6 Count the invalid ones (where at least one key is in its correct lock): If key 1 is correct, remaining 2 can be arranged in ways = 2 Similarly for key 2 correct: 2 ways Similarly for key 3 correct: 2 ways But this overcounts the case where two keys are correct (and then the third is also forced correct), so we use derangement formula directly: Derangements of 3: !3=2 So, the remaining 3 keys can be put into the remaining 3 locks in 2 ways such that none matches correctly. Incorrect Answer: (a) Solution: 2 keys are already correctly matched, so we are left with 3 distinct keys and 3 distinct locks. We need the number of ways to assign these 3 keys to the 3 locks such that no key goes to its own correct lock. This is exactly the number of derangements of 3 items, denoted . Total permutations of 3 items: 3!=6 Count the invalid ones (where at least one key is in its correct lock): If key 1 is correct, remaining 2 can be arranged in ways = 2 Similarly for key 2 correct: 2 ways Similarly for key 3 correct: 2 ways But this overcounts the case where two keys are correct (and then the third is also forced correct), so we use derangement formula directly: Derangements of 3: !3=2 So, the remaining 3 keys can be put into the remaining 3 locks in 2 ways such that none matches correctly.
#### 3. Question
A person has 5 distinct keys and 5 distinct locks. If 2 keys are already matched correctly to their locks, in how many ways can the remaining 3 keys be put into the remaining 3 locks such that none of them open their respective locks?
Answer: (a)
Solution:
2 keys are already correctly matched, so we are left with 3 distinct keys and 3 distinct locks.
We need the number of ways to assign these 3 keys to the 3 locks such that no key goes to its own correct lock. This is exactly the number of derangements of 3 items, denoted .
Total permutations of 3 items:
Count the invalid ones (where at least one key is in its correct lock):
• If key 1 is correct, remaining 2 can be arranged in ways = 2 Similarly for key 2 correct: 2 ways Similarly for key 3 correct: 2 ways
But this overcounts the case where two keys are correct (and then the third is also forced correct), so we use derangement formula directly:
Derangements of 3:
So, the remaining 3 keys can be put into the remaining 3 locks in 2 ways such that none matches correctly.
Answer: (a)
Solution:
2 keys are already correctly matched, so we are left with 3 distinct keys and 3 distinct locks.
We need the number of ways to assign these 3 keys to the 3 locks such that no key goes to its own correct lock. This is exactly the number of derangements of 3 items, denoted .
Total permutations of 3 items:
Count the invalid ones (where at least one key is in its correct lock):
• If key 1 is correct, remaining 2 can be arranged in ways = 2 Similarly for key 2 correct: 2 ways Similarly for key 3 correct: 2 ways
But this overcounts the case where two keys are correct (and then the third is also forced correct), so we use derangement formula directly:
Derangements of 3:
So, the remaining 3 keys can be put into the remaining 3 locks in 2 ways such that none matches correctly.
• Question 4 of 5 4. Question (a) 1/120 (b) 1/30 (c) 1/15 (d) 1/24 Correct Incorrect
#### 4. Question
• Question 5 of 5 5. Question The table below shows the age-wise distribution of employees in Company X and Company Y. Total employees in Company X = 500, Company Y = 750. Age Group (years) Company X Company Y Below 30 30% 20% 30–49 50% 40% 50 and above 20% 40% Find the ratio of employees aged 30 years and above but less than 50 years in Company X to those in Company Y. (a) 5 : 6 (b) 2 : 3 (c) 3 : 4 (d) 4 : 5 Correct Incorrect
#### 5. Question
The table below shows the age-wise distribution of employees in Company X and Company Y. Total employees in Company X = 500, Company Y = 750.
Age Group (years) | Company X | Company Y
Below 30 | 30% | 20%
30–49 | 50% | 40%
50 and above | 20% | 40%
Find the ratio of employees aged 30 years and above but less than 50 years in Company X to those in Company Y.
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