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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
A number of students are standing in a row facing north is such a way that Deepak is nineteenth from both the ends. So find the number of students in the class.
Correct
Solution: B
By the above given information it is clear that, Deepak is 19th from both the ends which means there are 18 students standing left to Deepak and another 18 students standing right to Deepak.
Therefore, total number of students = 18 + 18+ 1(Deepak) = 37.
Hence, option (b) is correct.
Incorrect
Solution: B
By the above given information it is clear that, Deepak is 19th from both the ends which means there are 18 students standing left to Deepak and another 18 students standing right to Deepak.
Therefore, total number of students = 18 + 18+ 1(Deepak) = 37.
Hence, option (b) is correct.
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Question 2 of 5
2. Question
Consider the sequence given below:
126, 64, 34, 20, 14, ‘?’
What is the next term of the series?
Correct
Solution: A
First number in the series is 126.
Second number = 126/2 + 1 = 64.
Third number = 64/2 + 2 = 34.
Fourth number = 34/2 + 3 = 20
Fifth number = 20/2 + 4 = 14.
Therefore, sixth number = 14/2 + 5 = 7+5= 12
Hence, option (a) is correct.
Incorrect
Solution: A
First number in the series is 126.
Second number = 126/2 + 1 = 64.
Third number = 64/2 + 2 = 34.
Fourth number = 34/2 + 3 = 20
Fifth number = 20/2 + 4 = 14.
Therefore, sixth number = 14/2 + 5 = 7+5= 12
Hence, option (a) is correct.
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Question 3 of 5
3. Question
If ‘Q’ denotes ‘×’; ‘R’ denotes ‘-’, ‘T’ denotes ‘÷’ and ‘W’ denotes ‘+’, then find the value of the expression 20R12T4Q6W5?
Correct
Solution: C
On substituting the letters as per the given rules, the expression becomes:
20 – 12 ÷ 4 × 6 + 5
Solving the above expression using BODMAS rule:
è 20 – [(12 ÷ 4) × 6] + 5
è 20 – [3 × 6] + 5
è 20 – [18] + 5
è 25 – 18 = 7
Hence, option (c) is correct.
Incorrect
Solution: C
On substituting the letters as per the given rules, the expression becomes:
20 – 12 ÷ 4 × 6 + 5
Solving the above expression using BODMAS rule:
è 20 – [(12 ÷ 4) × 6] + 5
è 20 – [3 × 6] + 5
è 20 – [18] + 5
è 25 – 18 = 7
Hence, option (c) is correct.
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Question 4 of 5
4. Question
If the code for ‘HEN’ is calculated as 27, what will be the code for ‘ASK’?
Correct
Solution: D
As per English Alphabet, H is 8th letter, E is 5th letter and N is 14th letter.
By adding all the positions i.e 8 + 5 + 14 , we get sum of 27.
Therefore, code for ASK can be calculated as A= 1 , S = 19 and K = 11 th letter.
Therefore, sum = 1+19+11 = 31
Hence, option (d) is correct.
Incorrect
Solution: D
As per English Alphabet, H is 8th letter, E is 5th letter and N is 14th letter.
By adding all the positions i.e 8 + 5 + 14 , we get sum of 27.
Therefore, code for ASK can be calculated as A= 1 , S = 19 and K = 11 th letter.
Therefore, sum = 1+19+11 = 31
Hence, option (d) is correct.
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Question 5 of 5
5. Question
Consider the sequence given below:
JAK, KBL, LCM, MDN, ‘?’ :
What is the next term of the series?
Correct
Solution: C
Every position in the word is increasing by the next letter in English alphabet.
Thus, M becomes N, D becomes E and N becomes O
Hence, option (c) is correct.
Incorrect
Solution: C
Every position in the word is increasing by the next letter in English alphabet.
Thus, M becomes N, D becomes E and N becomes O
Hence, option (c) is correct.
