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
Out of the two boxes, if one can contain maximum 35 books and the other cannot contain less than 20 books and more than 50 books, then which of the following can be concluded:
I. The total number of books in both the boxes can never be less than 20.
II. Both of the boxes together can contain more than 90 books.Which one of the following is correct with respect to above?
Correct
Answer: (a)
Explanation:
The first box can contain at most 35 books.
The second box cannot contain less than 20 books and more than 50 books, so it can contain from 20 to 50 books.For Statement I:
Even if the first box contains 0 books, the second box must contain at least 20 books.
So the total can never be less than 20.
Hence, Statement I is correct.For Statement II:
Maximum total = 35 + 50 = 85 books.
So together, the two boxes cannot contain more than 90 books.
Hence, Statement II is incorrect.So the correct answer is (a).
Incorrect
Answer: (a)
Explanation:
The first box can contain at most 35 books.
The second box cannot contain less than 20 books and more than 50 books, so it can contain from 20 to 50 books.For Statement I:
Even if the first box contains 0 books, the second box must contain at least 20 books.
So the total can never be less than 20.
Hence, Statement I is correct.For Statement II:
Maximum total = 35 + 50 = 85 books.
So together, the two boxes cannot contain more than 90 books.
Hence, Statement II is incorrect.So the correct answer is (a).
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Question 2 of 5
2. Question
A, B, C, D, and E are standing in a straight line facing the north. E is on the immediate left of D. B is between E and A. If D and C are at the ends, which of the following is second to the right of C?
Correct
Answer: (b)
Explanation:
Since D and C are at the ends, and E is on the immediate left of D, D cannot be at the left end. So D must be at the right end and C at the left end.Arrangement starts as:
C _ _ E DNow, B is between E and A. So B must be immediately to the left of E, and A must be to the left of B.
Thus, the arrangement becomes:
C A B E DC is at the left end. The second to the right of C is B.
Hence, the correct answer is (b).Incorrect
Answer: (b)
Explanation:
Since D and C are at the ends, and E is on the immediate left of D, D cannot be at the left end. So D must be at the right end and C at the left end.Arrangement starts as:
C _ _ E DNow, B is between E and A. So B must be immediately to the left of E, and A must be to the left of B.
Thus, the arrangement becomes:
C A B E DC is at the left end. The second to the right of C is B.
Hence, the correct answer is (b). -
Question 3 of 5
3. Question
Let p, q, r and s be natural numbers such that
p + 2 = q – 1 = r + 4 = s – 3Which one of the following is the smallest number?
Correct
Answer: (c)
Explanation:
Let
p + 2 = q – 1 = r + 4 = s – 3 = kThen,
p = k – 2
q = k + 1
r = k – 4
s = k + 3Comparing all four numbers:
r = k – 4 is the smallest.Hence, the correct answer is (c).
Incorrect
Answer: (c)
Explanation:
Let
p + 2 = q – 1 = r + 4 = s – 3 = kThen,
p = k – 2
q = k + 1
r = k – 4
s = k + 3Comparing all four numbers:
r = k – 4 is the smallest.Hence, the correct answer is (c).
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Question 4 of 5
4. Question
Five students – A, B, C, D and E – scored different marks in a class. B scored more than D but less than C. C did not score the highest marks. A scored more than B but neither scored the highest or the second highest marks. The one who scored the second least marks got 52 and the one who scored the second highest marks got 94. What is the possible score of A?
Correct
Answer: (b)
Explanation:
Order from lowest to highest: _ _ _ _ _Given:
- B > D and B < C ⇒ D < B < C
- A > B, but A is neither highest nor second highest ⇒ A must be in middle position (3rd)
- C is not highest
So arrangement becomes:
D < B < A < C < ENow:
- 2nd least = 52 ⇒ B = 52
- 2nd highest = 94 ⇒ C = 94
So A lies between 52 and 94
Hence, A = 60 is possible
Incorrect
Answer: (b)
Explanation:
Order from lowest to highest: _ _ _ _ _Given:
- B > D and B < C ⇒ D < B < C
- A > B, but A is neither highest nor second highest ⇒ A must be in middle position (3rd)
- C is not highest
So arrangement becomes:
D < B < A < C < ENow:
- 2nd least = 52 ⇒ B = 52
- 2nd highest = 94 ⇒ C = 94
So A lies between 52 and 94
Hence, A = 60 is possible
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Question 5 of 5
5. Question
Megha is decorating a rod. She painted 40% of the rod in green and 50% of the remaining part in white. Thereafter, 50% of the remaining part is painted in blue. The uncoloured part left after this is 30 cm long and is cut off. What is the length of the coloured part of the rod?
Correct
Answer: (b)
Explanation:
Start with 100%.
First, 40% is painted green.
Remaining = 60%.Now 50% of 60% is painted white.
That is 30%.So total painted = 70%.
Remaining = 30%.
Now 50% of this 30% is painted blue.
That is 15%.So total painted = 85%.
Uncoloured part = 15%.
This 15% is 30 cm.
So total length = 30 × 100 / 15 = 200 cm
Coloured part = 85% of 200 = 170 cm
Incorrect
Answer: (b)
Explanation:
Start with 100%.
First, 40% is painted green.
Remaining = 60%.Now 50% of 60% is painted white.
That is 30%.So total painted = 70%.
Remaining = 30%.
Now 50% of this 30% is painted blue.
That is 15%.So total painted = 85%.
Uncoloured part = 15%.
This 15% is 30 cm.
So total length = 30 × 100 / 15 = 200 cm
Coloured part = 85% of 200 = 170 cm








