UPSC Insta–DART (Daily Aptitude and Reasoning Test) 5 Feb 2025
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 1090! is exactly divisible by (990)^n. What is the value of n ? a) 101 b) 109 c) 100 d) 108 Correct Answer: D Explanation 990= 11x 3x3x2x5 The highest power of 990, which would divide 1090! would be the power of 11 available in 1090. This can be found by getting quotient for : (1090/11) + {1090/(11)^2} = 99 + 9 = 108 Hence (990)^108 will exactly divide 1090! Incorrect Answer: D Explanation 990= 11x 3x3x2x5 The highest power of 990, which would divide 1090! would be the power of 11 available in 1090. This can be found by getting quotient for : (1090/11) + {1090/(11)^2} = 99 + 9 = 108 Hence (990)^108 will exactly divide 1090!
#### 1. Question
1090! is exactly divisible by (990)^n. What is the value of n ?
Explanation
990= 11x 3x3x2x5
The highest power of 990, which would divide 1090! would be the power of 11 available in 1090.
This can be found by getting quotient for : (1090/11) + {1090/(11)^2} = 99 + 9 = 108
Hence (990)^108 will exactly divide 1090!
Explanation
990= 11x 3x3x2x5
The highest power of 990, which would divide 1090! would be the power of 11 available in 1090.
This can be found by getting quotient for : (1090/11) + {1090/(11)^2} = 99 + 9 = 108
Hence (990)^108 will exactly divide 1090!
• Question 2 of 5 2. Question In a set of alphabet first four letters move back three places, next four letters move ahead three places. Again, the next four letters move back three places and the next four letters move ahead three places and so on. And the last two letters move back six places. Which letter is at the middle of the series now? a) PQ b) ST c) KL d) OP Correct Answer: A Explanation The set of letters A B C D E F G H I J K L M N O P Q R S T U V W X Y Z We have to find the letters at the middle of the series. Implies we have to find the letter at the 13th and 14th position. We observe that M and N are the 13th and 14th letters. Given that first four moves back (ABCD) Next four moves ahead (EFGH) Next four moves back(IJKL) Next four moves ahead (MNOP) So these two letters will move ahead three places . M+3 = P N+3= Q So, the middle letter in the new series are PQ Incorrect Answer: A Explanation The set of letters A B C D E F G H I J K L M N O P Q R S T U V W X Y Z We have to find the letters at the middle of the series. Implies we have to find the letter at the 13th and 14th position. We observe that M and N are the 13th and 14th letters. Given that first four moves back (ABCD) Next four moves ahead (EFGH) Next four moves back(IJKL) Next four moves ahead (MNOP) So these two letters will move ahead three places . M+3 = P N+3= Q So, the middle letter in the new series are PQ
#### 2. Question
In a set of alphabet first four letters move back three places, next four letters move ahead three places. Again, the next four letters move back three places and the next four letters move ahead three places and so on. And the last two letters move back six places. Which letter is at the middle of the series now?
Explanation
The set of letters
A B C D E F G H I J K L M N O P Q R S T U V W X Y Z
We have to find the letters at the middle of the series. Implies we have to find the letter at the 13th and 14th position.
We observe that M and N are the 13th and 14th letters.
Given that first four moves back (ABCD)
Next four moves ahead (EFGH)
Next four moves back(IJKL)
Next four moves ahead (MNOP)
So these two letters will move ahead three places .
So, the middle letter in the new series are PQ
Explanation
The set of letters
A B C D E F G H I J K L M N O P Q R S T U V W X Y Z
We have to find the letters at the middle of the series. Implies we have to find the letter at the 13th and 14th position.
We observe that M and N are the 13th and 14th letters.
Given that first four moves back (ABCD)
Next four moves ahead (EFGH)
Next four moves back(IJKL)
Next four moves ahead (MNOP)
So these two letters will move ahead three places .
So, the middle letter in the new series are PQ
• Question 3 of 5 3. Question In a quiz competition the host shakes hand with each participant once. While he shakes hands with each qualifier amongst the participants twice more. Besides, the participants are required to shake hands once with each other, while the winner, and the host shakes hands with all the guests once. How many handshakes are there if there are 10 participants in all, three finalist, and 60 spectators? a) 181 b) 152 c) 167 d) 122 Correct Answer: A Explanation If there are n people and they shake hand with each other. Number of handshakes = n(n-1)/2 1) the host shakes hand with each participant once in 10 ways. 2) The host shakes hands with each qualifier (3 finalists) amongst the participants twice more in 3+3 ways = 6 ways 3) The participants are required to shake hands once with each other, this is done in 10×10/2= 45 ways 4) the winner, and the host shakes hands with all the guests (60 spectators) once in 60+60 ways =120 ways 5) Total number of handshakes = 10+6+45+120 =181 ways Incorrect Answer: A Explanation If there are n people and they shake hand with each other. Number of handshakes = n(n-1)/2 1) the host shakes hand with each participant once in 10 ways. 2) The host shakes hands with each qualifier (3 finalists) amongst the participants twice more in 3+3 ways = 6 ways 3) The participants are required to shake hands once with each other, this is done in 10×10/2= 45 ways 4) the winner, and the host shakes hands with all the guests (60 spectators) once in 60+60 ways =120 ways 5) Total number of handshakes = 10+6+45+120 =181 ways
#### 3. Question
In a quiz competition the host shakes hand with each participant once. While he shakes hands with each qualifier amongst the participants twice more. Besides, the participants are required to shake hands once with each other, while the winner, and the host shakes hands with all the guests once. How many handshakes are there if there are 10 participants in all, three finalist, and 60 spectators?
Explanation
If there are n people and they shake hand with each other.
Number of handshakes = n(n-1)/2
- 1.the host shakes hand with each participant once in 10 ways.
- 1.The host shakes hands with each qualifier (3 finalists) amongst the participants twice more in 3+3 ways = 6 ways
- 1.The participants are required to shake hands once with each other, this is done in 10×10/2= 45 ways
- 1.the winner, and the host shakes hands with all the guests (60 spectators) once in 60+60 ways =120 ways
- 1.Total number of handshakes = 10+6+45+120 =181 ways
Explanation
If there are n people and they shake hand with each other.
Number of handshakes = n(n-1)/2
- 1.the host shakes hand with each participant once in 10 ways.
- 1.The host shakes hands with each qualifier (3 finalists) amongst the participants twice more in 3+3 ways = 6 ways
- 1.The participants are required to shake hands once with each other, this is done in 10×10/2= 45 ways
- 1.the winner, and the host shakes hands with all the guests (60 spectators) once in 60+60 ways =120 ways
- 1.Total number of handshakes = 10+6+45+120 =181 ways
• Question 4 of 5 4. Question Ram is given three sticks measuring 51cm, 68cm, 85cm to measure laces at a fabric shop. What is the minimum length of the lace which Ram can measure exactly by each one of these sticks? a) 9.88m b) 10.20 m c) 8.4m d) 12.5m Correct Answer: B Explanation To find the minimum length of lace that can be measured exactly by using each one of these stick, we need to find the LCM of the given length of the sticks. Hence LCM of 51, 68, 85 = 1020 cm Hence, the minimum length of the less which RAM can measure exactly by each one of the sticks is 1020 cm that is 10.20 meter. Incorrect Answer: B Explanation To find the minimum length of lace that can be measured exactly by using each one of these stick, we need to find the LCM of the given length of the sticks. Hence LCM of 51, 68, 85 = 1020 cm Hence, the minimum length of the less which RAM can measure exactly by each one of the sticks is 1020 cm that is 10.20 meter.
#### 4. Question
Ram is given three sticks measuring 51cm, 68cm, 85cm to measure laces at a fabric shop. What is the minimum length of the lace which Ram can measure exactly by each one of these sticks?
• b) 10.20 m
Explanation
To find the minimum length of lace that can be measured exactly by using each one of these stick, we need to find the LCM of the given length of the sticks.
Hence LCM of 51, 68, 85 = 1020 cm
Hence, the minimum length of the less which RAM can measure exactly by each one of the sticks is 1020 cm that is 10.20 meter.
Explanation
To find the minimum length of lace that can be measured exactly by using each one of these stick, we need to find the LCM of the given length of the sticks.
Hence LCM of 51, 68, 85 = 1020 cm
Hence, the minimum length of the less which RAM can measure exactly by each one of the sticks is 1020 cm that is 10.20 meter.
• Question 5 of 5 5. Question A simple mathematical operation in each number of sequence 1, 19, 109, 271, 505, 929, __ results in a sequence with respect to prime numbers. Which one of the following is the next number in the sequence? a) 1424 b) 1534 c) 1482 d) 1639 Correct Answer: D Explanation The given expression has the following relation: n^2 – n – 1 Where n is the alternative prime number beginning from 2. List of prime numbers for our utility: 2,5,11,17,23,31,41 (2^2)-2-1=1 (5^2)-5-1=19 (11^2)-11-1=109 (17^2)-17-1=271 (23^2)-23-1=505 (31^2)-31-1=929 Next term is (41^2)-41-1= 1639. Incorrect Answer: D Explanation The given expression has the following relation: n^2 – n – 1 Where n is the alternative prime number beginning from 2. List of prime numbers for our utility: 2,5,11,17,23,31,41 (2^2)-2-1=1 (5^2)-5-1=19 (11^2)-11-1=109 (17^2)-17-1=271 (23^2)-23-1=505 (31^2)-31-1=929 Next term is (41^2)-41-1= 1639.
#### 5. Question
A simple mathematical operation in each number of sequence 1, 19, 109, 271, 505, 929, __ results in a sequence with respect to prime numbers. Which one of the following is the next number in the sequence?
Explanation
The given expression has the following relation: n^2 – n – 1
Where n is the alternative prime number beginning from 2.
List of prime numbers for our utility: 2,5,11,17,23,31,41
(2^2)-2-1=1
(5^2)-5-1=19
(11^2)-11-1=109
(17^2)-17-1=271
(23^2)-23-1=505
(31^2)-31-1=929
Next term is (41^2)-41-1= 1639.
Explanation
The given expression has the following relation: n^2 – n – 1
Where n is the alternative prime number beginning from 2.
List of prime numbers for our utility: 2,5,11,17,23,31,41
(2^2)-2-1=1
(5^2)-5-1=19
(11^2)-11-1=109
(17^2)-17-1=271
(23^2)-23-1=505
(31^2)-31-1=929
Next term is (41^2)-41-1= 1639.
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