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
Three persons A, B and C are standing in a queue not necessarily in the same order. There are 3 persons between A and B, and 6 persons between B and C. If there are 8 persons ahead of C and 10 behind A, what could be the minimum number of persons in the queue?
Correct
Answer: A
Solution:
Given that,
There are 3 persons between A and B
There are 6 persons between B and C
There are 8 persons ahead of C
There are 10 persons behind ANow,
Arrange as: (+1) B (+3) A (+2) C (+7)
Check:
Between A and B = 3 (ok)
Between B and C = 3 + 1 + 2 = 6 (ok)
Ahead of C = 1 + 1 + 3 + 1 = 6, add the 2 between A and C gives 8 total ahead (ok)
Behind A = 2 + 1 + 7 = 10 (ok)Total persons = 1 + 1 + 3 + 1 + 2 + 1 + 7 = 16
Hence option (a) is correct
Incorrect
Answer: A
Solution:
Given that,
There are 3 persons between A and B
There are 6 persons between B and C
There are 8 persons ahead of C
There are 10 persons behind ANow,
Arrange as: (+1) B (+3) A (+2) C (+7)
Check:
Between A and B = 3 (ok)
Between B and C = 3 + 1 + 2 = 6 (ok)
Ahead of C = 1 + 1 + 3 + 1 = 6, add the 2 between A and C gives 8 total ahead (ok)
Behind A = 2 + 1 + 7 = 10 (ok)Total persons = 1 + 1 + 3 + 1 + 2 + 1 + 7 = 16
Hence option (a) is correct
-
Question 2 of 5
2. Question
Two statements S1 and S2 are given below followed by a question.
S1: Ramesh and Suresh can complete a task together in 12 days.
S2: Ramesh alone can complete the same task in 20 days.Question: How many days will Suresh alone take to complete the task?
Which of the following is correct with respect to above?
Correct
Answer: (c)
Explanation:
From S1: Combined rate = 1/12
From S2: Ramesh’s rate = 1/20
⇒ Suresh’s rate = 1/12 – 1/20 = (5 – 3)/60 = 1/30
So, Suresh alone takes 30 days
Hence, both statements are needed.Incorrect
Answer: (c)
Explanation:
From S1: Combined rate = 1/12
From S2: Ramesh’s rate = 1/20
⇒ Suresh’s rate = 1/12 – 1/20 = (5 – 3)/60 = 1/30
So, Suresh alone takes 30 days
Hence, both statements are needed. -
Question 3 of 5
3. Question
A two-digit number is such that when its digits are reversed, the new number is 27 less than the original. How many such numbers are possible?
Correct
Answer: D
Solution:
Let the two-digit number be 10a + b, where a is tens digit and b is units digit.
Reversed number = 10b + a.
Difference = (10a + b) – (10b + a) = 9(a – b).
Given 9(a – b) = 27 ⇒ a – b = 3.Now check possible pairs:
(3,0)=30, (4,1)=41, (5,2)=52, (6,3)=63, (7,4)=74, (8,5)=85, (9,6)=96.
Total = 7 numbers.Hence, option (d) is correct.
Incorrect
Answer: D
Solution:
Let the two-digit number be 10a + b, where a is tens digit and b is units digit.
Reversed number = 10b + a.
Difference = (10a + b) – (10b + a) = 9(a – b).
Given 9(a – b) = 27 ⇒ a – b = 3.Now check possible pairs:
(3,0)=30, (4,1)=41, (5,2)=52, (6,3)=63, (7,4)=74, (8,5)=85, (9,6)=96.
Total = 7 numbers.Hence, option (d) is correct.
-
Question 4 of 5
4. Question
What is the rightmost digit preceding the zeros in the value of 1220?
Correct
Answer: B
Solution:
The value = 1220
12^20 = (3 × 4)20 = 320 × 240
The factor 240 will create zeros, so the deciding digit is from 320
Cyclicity of 3 is 4. 20 divided by 4 leaves remainder 0
So, 320 ends with 1
Therefore, the rightmost digit before zeros = 6
Hence, option B is correctIncorrect
Answer: B
Solution:
The value = 1220
12^20 = (3 × 4)20 = 320 × 240
The factor 240 will create zeros, so the deciding digit is from 320
Cyclicity of 3 is 4. 20 divided by 4 leaves remainder 0
So, 320 ends with 1
Therefore, the rightmost digit before zeros = 6
Hence, option B is correct -
Question 5 of 5
5. Question
Let p, q and r be digits (1 to 9) with p < q < r, and pp, qq and rr be the corresponding two-digit repeated-digit numbers. Suppose
pp + qq + rr = tt0,
where tt0 is a three-digit number ending with zero.- The number of possible ordered triples (p, q, r) is 8.
- The number of possible values of p is 4.
Which of the above statements is/are correct?
Correct
Answer: A
Solution:
Given that,
pp + qq + rr = tt0 and pp = 11p etc.Now,
11(p + q + r) = tt0, so p + q + r is 10 or 20.List the strictly increasing digit triples (p, q, r):
Sum 10 gives 4 ordered triples.
Sum 20 gives 4 ordered triples.
Total ordered triples = 8. So Statement 1 is correct.Distinct p values across these valid triples are {1, 2, 3, 4, 5} which is 5 values, not 4.
So Statement 2 is incorrect.Hence, only Statement 1 is correct. Therefore, option (a) is correct.
Incorrect
Answer: A
Solution:
Given that,
pp + qq + rr = tt0 and pp = 11p etc.Now,
11(p + q + r) = tt0, so p + q + r is 10 or 20.List the strictly increasing digit triples (p, q, r):
Sum 10 gives 4 ordered triples.
Sum 20 gives 4 ordered triples.
Total ordered triples = 8. So Statement 1 is correct.Distinct p values across these valid triples are {1, 2, 3, 4, 5} which is 5 values, not 4.
So Statement 2 is incorrect.Hence, only Statement 1 is correct. Therefore, option (a) is correct.
