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
Find the greatest 4 digit number which when divided by 12,15,20 and 35 leaves no remainder.
Correct
Solution:-
LCM of 12, 15, 20 and 35 is 420. Divide 9999 greatest 4 digit number by 420, and subtract the remainder. Answer would be 9660.
Incorrect
Solution:-
LCM of 12, 15, 20 and 35 is 420. Divide 9999 greatest 4 digit number by 420, and subtract the remainder. Answer would be 9660.
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Question 2 of 5
2. Question
A, B, and C start jogging around a circular field and complete a single round in 18 seconds, 22seconds and 30 seconds respectively. In how much time will they meet again at the starting point?
Correct
Solution: LCM of 18, 22 and 30 is 990. Hence they meet in 990 sec which is 16min 30sec.
Incorrect
Solution: LCM of 18, 22 and 30 is 990. Hence they meet in 990 sec which is 16min 30sec.
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Question 3 of 5
3. Question
Both the HCF and the difference of two numbers is 6. If the LCM of the two numbers is a 4 digit number, then what is the maximum possible value of the bigger number?
Correct
Answer:
Let the numbers be 6k and 6 k(k + 1).
For the HCF to be 6, k and k + 1 are always co-prime.
LCM of 6k and 6(k + 1) is 6 k (k + 1).
6 k (k + 1) is a 5-digit number for k ≥ 41.
Hence, the maximum possible value of k is 40 for which the LCM is a 4 digit number.
Hence, the required number = 6(k + 1) = 6 × 41 = 246.
Incorrect
Answer:
Let the numbers be 6k and 6 k(k + 1).
For the HCF to be 6, k and k + 1 are always co-prime.
LCM of 6k and 6(k + 1) is 6 k (k + 1).
6 k (k + 1) is a 5-digit number for k ≥ 41.
Hence, the maximum possible value of k is 40 for which the LCM is a 4 digit number.
Hence, the required number = 6(k + 1) = 6 × 41 = 246.
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Question 4 of 5
4. Question
Three planets revolve round the Sun once in 200, 250 and 300 days, respectively in their own orbits. When do they all come relatively to the same position as at a certain point of time in their orbits?
Correct
Answer: (A) After 3000 days.
Solution:
Given that, three planets revolves the Sun once in 200, 250 and 300 days, respectively in their own orbits.
Required time = LCM of (200, 250 and 300) = 3000 days
Hence, after 3000 days they all come relatively to the same position as at a certain point of time in their orbits.
Hence, option A is correct.
Incorrect
Answer: (A) After 3000 days.
Solution:
Given that, three planets revolves the Sun once in 200, 250 and 300 days, respectively in their own orbits.
Required time = LCM of (200, 250 and 300) = 3000 days
Hence, after 3000 days they all come relatively to the same position as at a certain point of time in their orbits.
Hence, option A is correct.
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Question 5 of 5
5. Question
The greatest five digit number which is exactly divisible by each one of the number 12, 18, 21 and 28 is
Correct
Answer: (D) 99792
Explain it
LCM of 12, 18, 21 and 28 = 252
When we divide greatest five digit no. 99,999 by 252,
we get 207 remainder
Now, subtracting 207 by 99,999,
we get 99999-207 = 99,792.
99,792 is largest five digit number divisible by 12, 18, 21 and 28.
Incorrect
Answer: (D) 99792
Explain it
LCM of 12, 18, 21 and 28 = 252
When we divide greatest five digit no. 99,999 by 252,
we get 207 remainder
Now, subtracting 207 by 99,999,
we get 99999-207 = 99,792.
99,792 is largest five digit number divisible by 12, 18, 21 and 28.
