Insta–DART (Daily Aptitude and Reasoning Test) 2020 - 21
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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).
Wish you all the best ! 🙂
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Question 1 of 5
1. Question
In how many different ways can the letters of the word ‘LEADING’ be arranged in such a way that the vowels always come together?
Correct
Answer: Option C
Explanation:
The word ‘LEADING’ has 7 different letters.
When the vowels EAI are always together, they can be supposed to form one letter.
Then, we have to arrange the letters LNDG (EAI).
Now, 5 (4 + 1 = 5) letters can be arranged in 5! = 120 ways.
The vowels (EAI) can be arranged among themselves in 3! = 6 ways.
Required number of ways = (120 x 6) = 720.
Incorrect
Answer: Option C
Explanation:
The word ‘LEADING’ has 7 different letters.
When the vowels EAI are always together, they can be supposed to form one letter.
Then, we have to arrange the letters LNDG (EAI).
Now, 5 (4 + 1 = 5) letters can be arranged in 5! = 120 ways.
The vowels (EAI) can be arranged among themselves in 3! = 6 ways.
Required number of ways = (120 x 6) = 720.
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Question 2 of 5
2. Question
In how many different ways can the letters of the word ‘CORPORATION’ be arranged so that the vowels always come together?
Correct
Answer: Option D
Explanation:
In the word ‘CORPORATION’, we treat the vowels OOAIO as one letter.
Thus, we have CRPRTN (OOAIO).
This has 7 (6 + 1) letters of which R occurs 2 times and the rest are different.
Number of ways arranging these letters =7!= 2520.2!
Now, 5 vowels in which O occurs 3 times and the rest are different, can be arranged
in5!= 20 ways.3!
Required number of ways = (2520 x 20) = 50400
Incorrect
Answer: Option D
Explanation:
In the word ‘CORPORATION’, we treat the vowels OOAIO as one letter.
Thus, we have CRPRTN (OOAIO).
This has 7 (6 + 1) letters of which R occurs 2 times and the rest are different.
Number of ways arranging these letters =7!= 2520.2!
Now, 5 vowels in which O occurs 3 times and the rest are different, can be arranged
in5!= 20 ways.3!
Required number of ways = (2520 x 20) = 50400
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Question 3 of 5
3. Question
There are 18 stations between Hyderabad and Bangalore. How many second class tickets have to be printed, so that a passenger can travel from any station to any other station?
Correct
Answer: C. 380
Explanation: The total number of stations = 20
From 20 stations we have to choose any two stations and the direction of travel (i.e., Hyderabad to Bangalore is different from Bangalore to Hyderabad) in 20P2 ways.
20P2 = 20 * 19 = 380
Incorrect
Answer: C. 380
Explanation: The total number of stations = 20
From 20 stations we have to choose any two stations and the direction of travel (i.e., Hyderabad to Bangalore is different from Bangalore to Hyderabad) in 20P2 ways.
20P2 = 20 * 19 = 380
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Question 4 of 5
4. Question
six points are marked on a straight line and five points are marked on another line which is parallel to the first line. How many straight lines, including the first two, can be formed with these points?
Correct
Answer: B. 32
Explanation: We know that, the number of straight lines that can be formed by the 11 points in which 6 points are collinear except those that can be selected out of these 6 points are collinear.
Hence, the required number of straight lines
= 11C2 – 6C2 – 5C2 + 1 + 1
= 55 – 15 – 10 + 2 = 32
Incorrect
Answer: B. 32
Explanation: We know that, the number of straight lines that can be formed by the 11 points in which 6 points are collinear except those that can be selected out of these 6 points are collinear.
Hence, the required number of straight lines
= 11C2 – 6C2 – 5C2 + 1 + 1
= 55 – 15 – 10 + 2 = 32
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Question 5 of 5
5. Question
How many 3-digit numbers can be formed from the digits 2, 3, 5, 6, 7 and 9, which are divisible by 5 and none of the digits is repeated?
Correct
Answer: D. 20
Explanation: Since each desired number is divisible by 5, so we must have 5 at the unit place. So, there is 1 way of doing it.
The tens place can now be filled by any of the remaining 5 digits (2, 3, 6, 7, 9). So, there are 5 ways of filling the tens place.
The hundreds place can now be filled by any of the remaining 4 digits. So, there are 4 ways of filling it.
Required number of numbers = (1 * 5 * 4) = 20
Incorrect
Answer: D. 20
Explanation: Since each desired number is divisible by 5, so we must have 5 at the unit place. So, there is 1 way of doing it.
The tens place can now be filled by any of the remaining 5 digits (2, 3, 6, 7, 9). So, there are 5 ways of filling the tens place.
The hundreds place can now be filled by any of the remaining 4 digits. So, there are 4 ways of filling it.
Required number of numbers = (1 * 5 * 4) = 20
