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
There are 14 points in a plane, out OF which 4 are collinear. Find the number of triangles made by these points.
Correct
Answer: b
The required number of triangles = nC3 – mC3
Here, n = 14 , m = 4
= 14C3 – 4C3
= (14* 13 *12* 11!)/(3! *11!) – (4!/(3!*1!)
=[(14 *13 *12)/6 ] – (4/1)
= 14 *26 – 4= 364 – 4=360
Incorrect
Answer: b
The required number of triangles = nC3 – mC3
Here, n = 14 , m = 4
= 14C3 – 4C3
= (14* 13 *12* 11!)/(3! *11!) – (4!/(3!*1!)
=[(14 *13 *12)/6 ] – (4/1)
= 14 *26 – 4= 364 – 4=360
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Question 2 of 5
2. Question
There are 10 points in a plane, out of which 5 are collinear. Find the number of straight lines formed by joining them.
Correct
Answer : a
= nC2 – Mc2 +1
Here, n = 10, m = 5
=10C2 – 5C2 + 1 = 45 -10 +1=36
Incorrect
Answer : a
= nC2 – Mc2 +1
Here, n = 10, m = 5
=10C2 – 5C2 + 1 = 45 -10 +1=36
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Question 3 of 5
3. Question
Find the number of diagonals formed hexagon.
Correct
Answer : d
Hexagon has 6 sides.
N = 6
∴Required number of diagonals =n C2 –n
=6 C2-6
(6!)/2!(6-2)! – 6
= [6!/(2!4!)] – 6
[(6* 5 *4!)/(2* 4!)] – 6 = 15 – 6 = 9
Incorrect
Answer : d
Hexagon has 6 sides.
N = 6
∴Required number of diagonals =n C2 –n
=6 C2-6
(6!)/2!(6-2)! – 6
= [6!/(2!4!)] – 6
[(6* 5 *4!)/(2* 4!)] – 6 = 15 – 6 = 9
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Question 4 of 5
4. Question
In how many different ways, 5 boys and 5 girls can sit on a circular table, so that the boys and girls are alternate?
Correct
Answer : a
After fixing up one boy on the table, the remaining can be arranged in ways, but boys and girls have to be alternate. There will be 5 places, one place each between two boys. These 5 places can be filled by
5 girls in 5! ways.
Hence, by the principle of multiplication, the required number of ways= 4 X5! = 2880
Incorrect
Answer : a
After fixing up one boy on the table, the remaining can be arranged in ways, but boys and girls have to be alternate. There will be 5 places, one place each between two boys. These 5 places can be filled by
5 girls in 5! ways.
Hence, by the principle of multiplication, the required number of ways= 4 X5! = 2880
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Question 5 of 5
5. Question
20 persons were invited to a party. In how many ways, they and the host can be seated at a circular table?
Correct
Answer : c
Total persons on the circular table = 20 guests + 1 host = 21
They can be seated in (21-1) ways=factorial 20 ways.
Incorrect
Answer : c
Total persons on the circular table = 20 guests + 1 host = 21
They can be seated in (21-1) ways=factorial 20 ways.
