[Mission 2022] Insta–DART (Daily Aptitude and Reasoning Test) 24 September 2021
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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
Q1-Let x and y be positive integers such that x˃y. The expression 3x + 2y and 2x +3y, when divided by 5 leave remainders 2 and 3, respectively. What is the remainder when (x-y), is divided by 5?
Correct
Ans: a
We have, 3x+2y is divided by 5 remainder is 2.
∴3x+2y=5q+2 ……………………..(1)
And 2x +3y is divided by 5 remainder is 3.
∴2x + 3y = 5m+3 ……………………..(2)
Subtract Eq. (2 ) from Eq.(1) , we get
x-y = 5(q-m)-1
x-y=5(q-m)-5+4
x-y=5(q-m-1)+4
∴x-y is divided by 5 remainder is 4.
Incorrect
Ans: a
We have, 3x+2y is divided by 5 remainder is 2.
∴3x+2y=5q+2 ……………………..(1)
And 2x +3y is divided by 5 remainder is 3.
∴2x + 3y = 5m+3 ……………………..(2)
Subtract Eq. (2 ) from Eq.(1) , we get
x-y = 5(q-m)-1
x-y=5(q-m)-5+4
x-y=5(q-m-1)+4
∴x-y is divided by 5 remainder is 4.
Question 2 of 5
2. Question
If (x + 1) is the HCF of Ax2 + Bx +C and Bx2+Ax + C, where A # B, then the value of C is
Correct
Ans: C
Since, (x + 1) is the HCF of Ax2 + Bx + C and Bx2+ Ax +C
∴A(-1)++ B(-1)+C = 0
⇒ A-B+C = 0 ⇒ C = B- A
or B(-1)2+ A(-1) + C = 0
⇒ B – A+C=0 ⇒C = A-B
Because A# B
∴C = A-B
Incorrect
Ans: C
Since, (x + 1) is the HCF of Ax2 + Bx + C and Bx2+ Ax +C
∴A(-1)++ B(-1)+C = 0
⇒ A-B+C = 0 ⇒ C = B- A
or B(-1)2+ A(-1) + C = 0
⇒ B – A+C=0 ⇒C = A-B
Because A# B
∴C = A-B
Question 3 of 5
3. Question
In a fire range, 4 shooters are firing at the respective targets. The first, the second, the third and fourth shooters hit the target once in every 5 s, 6 s, 7s and 8 s, respectively. If all of them hit their target at 9 : 00 am, when will they hit their target together again?
Correct
Ans-c
Time after which they will hit the target again together
= LCM (5, 6, 7 and 8)
= 5 × 3 × 7 × 2 × 4
= 840 s
They will hit target together = 840/60 = 14 min.
So, they will hit together again at 9:14 am.
Incorrect
Ans-c
Time after which they will hit the target again together
= LCM (5, 6, 7 and 8)
= 5 × 3 × 7 × 2 × 4
= 840 s
They will hit target together = 840/60 = 14 min.
So, they will hit together again at 9:14 am.
Question 4 of 5
4. Question
The number 58129745812974 is divisible by
Correct
Ans: a
Number 58129745812974 is divisible by 11 as the difference of sum of all digits at even and odd places is 0.
Incorrect
Ans: a
Number 58129745812974 is divisible by 11 as the difference of sum of all digits at even and odd places is 0.
Question 5 of 5
5. Question
When a ball bounces, it rise to 2/3 of the height from which it fell. If the ball is dropped from a height of 36 m, then how high will it rise at the third bounce?
Correct
Ans: a
After first bounce, height of ball = (2/3)1 x36
And after third bounce, height of ball = (2/3)3 x36
= (8/27)x36 =[ (8/4)/3] = 32/3m = =10(2/3)m
Hence, the required height at third bounce is 10(2/3)m.
Incorrect
Ans: a
After first bounce, height of ball = (2/3)1 x36
And after third bounce, height of ball = (2/3)3 x36
= (8/27)x36 =[ (8/4)/3] = 32/3m = =10(2/3)m
Hence, the required height at third bounce is 10(2/3)m.
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