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
The number of times the digit 5 will appear while writing the integers from 1 to 1000 is?
Correct
Answer: c
solution-
Number of times 5 will come at the unitplace = (5, 15, …………25, 95)*10
= 100
Number of times 5 will come to the tens place = ( 50, 51, ………. 58, 59 ) *10
= 100
Number of times 5 will come at the hundred place = ( 500, 501, 502………..599)
= 100
Hence, Required Number = 100 * 3
= 300
Incorrect
Answer: c
solution-
Number of times 5 will come at the unitplace = (5, 15, …………25, 95)*10
= 100
Number of times 5 will come to the tens place = ( 50, 51, ………. 58, 59 ) *10
= 100
Number of times 5 will come at the hundred place = ( 500, 501, 502………..599)
= 100
Hence, Required Number = 100 * 3
= 300
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Question 2 of 5
2. Question
In a school every student is assigned a unique identification number. A student is a football player if and only if the identification number is divisible by 4, whereas a student is a cricketer if and only if the identification number is divisible by 6. If every number from 1 to 100 is assigned to a student, then how many of them play cricket as well as football?
Correct
Answer: b
Solution-
The Number assigned to the students who play both cricket as well as football should be multiplies of both 4 and 6 which is Nothing but multiplies of 12
Hence, the required answer will be 8 as there are 8 multiplies of 12 in first 100 Natural Numbers.
Incorrect
Answer: b
Solution-
The Number assigned to the students who play both cricket as well as football should be multiplies of both 4 and 6 which is Nothing but multiplies of 12
Hence, the required answer will be 8 as there are 8 multiplies of 12 in first 100 Natural Numbers.
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Question 3 of 5
3. Question
While writing all the numbers from 700 to 1000, how many numbers occur in which the digit at hundred’s place is greater than the digit at ten’s place, and the digit at ten’s place is greater than the digit at unit’s place?
Correct
Answer- C
Solution-
Hundreds digit > Tens digit > unit digit
From 700 to 800, total 21 Numbers are there in which the digit at Hundred’s place is greater than the digit at ten’s place, and the digit at ten’s place is greater than the digit at unit’s place.(ex-721,731,743,754,765,……..)
From 801 to 900, total such Numbers are = 28(821,831,842……)
From 901 to 1000, 36 Numbers are possible
∴Total 21 + 28 + 36 = 85 Numbers are possible.
Incorrect
Answer- C
Solution-
Hundreds digit > Tens digit > unit digit
From 700 to 800, total 21 Numbers are there in which the digit at Hundred’s place is greater than the digit at ten’s place, and the digit at ten’s place is greater than the digit at unit’s place.(ex-721,731,743,754,765,……..)
From 801 to 900, total such Numbers are = 28(821,831,842……)
From 901 to 1000, 36 Numbers are possible
∴Total 21 + 28 + 36 = 85 Numbers are possible.
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Question 4 of 5
4. Question
Find the Number of divisiors of 1420
Correct
answer –d
solution-Number of divisiors of 1420-
1420 = 142 * 10 = 71 * 2 * 2 * 5 = 71*2*2*5
71 comes here only one time (1)
2 comes here two times (2)
And 5 comes here only once(1)
So factors(divisors) of 1420 will be-
Number of divisor = ( 1 + 1 )(2+1) (1 + 1 )(just add one to all factors and then multiply)
=2*3*2
=12
Incorrect
answer –d
solution-Number of divisiors of 1420-
1420 = 142 * 10 = 71 * 2 * 2 * 5 = 71*2*2*5
71 comes here only one time (1)
2 comes here two times (2)
And 5 comes here only once(1)
So factors(divisors) of 1420 will be-
Number of divisor = ( 1 + 1 )(2+1) (1 + 1 )(just add one to all factors and then multiply)
=2*3*2
=12
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Question 5 of 5
5. Question
Find the least no which being divided by 9, 12, 16, and 30 leaves in each case a remainder 3
Correct
Answer-b
Solution-
For least number we will take lcm so now-
LCM of 9, 12, 16, 30
LCM = 720
So required number is LCM + 3 = 723
(in each case we are getting same reainder so wewill add 3 in LCM)
Incorrect
Answer-b
Solution-
For least number we will take lcm so now-
LCM of 9, 12, 16, 30
LCM = 720
So required number is LCM + 3 = 723
(in each case we are getting same reainder so wewill add 3 in LCM)
