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 CSP2021. 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
Number 136 is added to 5B7 and the sum obtained is 7A3, where A are integers. It is given that 7A3 is exactly divisible by 3. The only possible value of B is
Correct
Answer : D
It is given 136 + 5B7 = 7A3.
Add the unit’s numbers to get 6 + 7 = 13. So, carry over 1.
=> 1 + 3 + B = 1A => 1 + 3 + B = 10 + A => B – A = 6.
Which means if A = 0, B = 6; if A = 1, B = 7, if A = 2, B = 8 and if A = 3, B = 9.
But given 7A3 is completely divisible by 3. So, as per rules of divisibility, 7 + A + 3 = 10 + A should also be completely divisible by 3.
So the possible values of A are 2, 5 and 8. (12 / 15 / 18 divisible by 3) Out of these, only 2 satisfies both the conditions so A = 2, so B = 8. Ans.(d)
Incorrect
Answer : D
It is given 136 + 5B7 = 7A3.
Add the unit’s numbers to get 6 + 7 = 13. So, carry over 1.
=> 1 + 3 + B = 1A => 1 + 3 + B = 10 + A => B – A = 6.
Which means if A = 0, B = 6; if A = 1, B = 7, if A = 2, B = 8 and if A = 3, B = 9.
But given 7A3 is completely divisible by 3. So, as per rules of divisibility, 7 + A + 3 = 10 + A should also be completely divisible by 3.
So the possible values of A are 2, 5 and 8. (12 / 15 / 18 divisible by 3) Out of these, only 2 satisfies both the conditions so A = 2, so B = 8. Ans.(d)

Question 2 of 5
2. Question
If X is between 3 and 1, and Y is between 1 and 1, then X2 – Y2 is in between which of the following?
Correct
Answer : d
X is between 3 and 1 so let x = – 2.9
And y is between 1 and 1 so let y = 0
∴x^{2}y^{2} = ( x+ y ) (x – y )
= ( 2.9 + 0 ) ( – 2.9 – 0 )
= 8.41
Which is between 0 and 9
Incorrect
Answer : d
X is between 3 and 1 so let x = – 2.9
And y is between 1 and 1 so let y = 0
∴x^{2}y^{2} = ( x+ y ) (x – y )
= ( 2.9 + 0 ) ( – 2.9 – 0 )
= 8.41
Which is between 0 and 9

Question 3 of 5
3. Question
X and Y are natural numbers other than 1, and Y is greater than X . Which of the following represents the largest number ?
Correct
Answer : A
Incorrect
Answer : A

Question 4 of 5
4. Question
Find the greatest 6digit number, which is a multiple of 12.
Correct
Answer: D
Greatest six – digit number is 999999. Divide this number by 12 and get remainder as 3. Since the remainder is 3, if you subtract 3 from the number, the remaining number will be a multiple of 12. So the greatest such number will be 999999 – 3 = 999996.
Incorrect
Answer: D
Greatest six – digit number is 999999. Divide this number by 12 and get remainder as 3. Since the remainder is 3, if you subtract 3 from the number, the remaining number will be a multiple of 12. So the greatest such number will be 999999 – 3 = 999996.

Question 5 of 5
5. Question
Find a positive number which when increased by 17 is equal to 60 times the reciprocal of the number.
Correct
Answer : A
Incorrect
Answer : A