UPSC Insta–DART (Daily Aptitude and Reasoning Test) 8 Aug 2024
Kartavya Desk Staff
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 The least number of five digits which is exactly divisible by 12, 15 and 18 is (a) 10010 (b) 10015 (c) 10020 (d) 10080 Correct Solution: Least number of 5 digits is 10000. L.C.M. of 12, 15 and 18 is 180. On dividing 10000 by 180, the remainder is 100. Required number = 10000 + (180 – 100) = 10080. Incorrect Solution: Least number of 5 digits is 10000. L.C.M. of 12, 15 and 18 is 180. On dividing 10000 by 180, the remainder is 100. Required number = 10000 + (180 – 100) = 10080.
#### 1. Question
The least number of five digits which is exactly divisible by 12, 15 and 18 is
Solution:
Least number of 5 digits is 10000.
L.C.M. of 12, 15 and 18 is 180.
On dividing 10000 by 180, the remainder is 100.
Required number = 10000 + (180 – 100) = 10080.
Solution:
Least number of 5 digits is 10000.
L.C.M. of 12, 15 and 18 is 180.
On dividing 10000 by 180, the remainder is 100.
Required number = 10000 + (180 – 100) = 10080.
• Question 2 of 5 2. Question The number between 4000 and 5000 which is divisible by 12, 18, 21 and 32 is (a) 4023 (b) 4032 (c) 4203 D) 4302 Correct Solution: L.C.M. of 12, 18, 21 and 32 = 2 × 2 x 3 x 3 x 7 x 8= 2016.So, the required number is a multiple of 2016 and lies between 4000 and 5000. Hence, required number = 4032. Incorrect Solution: L.C.M. of 12, 18, 21 and 32 = 2 × 2 x 3 x 3 x 7 x 8= 2016.So, the required number is a multiple of 2016 and lies between 4000 and 5000. Hence, required number = 4032.
#### 2. Question
The number between 4000 and 5000 which is divisible by 12, 18, 21 and 32 is
Solution:
L.C.M. of 12, 18, 21 and 32
= 2 × 2 x 3 x 3 x 7 x 8= 2016.So, the required number is a multiple of 2016 and lies between 4000 and 5000. Hence, required number = 4032.
Solution:
L.C.M. of 12, 18, 21 and 32
= 2 × 2 x 3 x 3 x 7 x 8= 2016.So, the required number is a multiple of 2016 and lies between 4000 and 5000. Hence, required number = 4032.
• Question 3 of 5 3. Question A milkman has 3 jars containing 57 litres, 129 litres and 177 litres of pure milk respectively. A measuring can, after a different number of exact measurements of milk in each jar, leaves the same amount of milk unmeasured in each jar. What is the volume of the largest such can? (a) 12 litres (b) 16 litres (c) 24 litres (d) None of these Correct Solution: Required volume = [H.C.F. of (129 – 57), (177 – 129) and (177 – 57)] litres = (H.C.F. of 72, 48 and 120) litres = 24 litres. Incorrect Solution: Required volume = [H.C.F. of (129 – 57), (177 – 129) and (177 – 57)] litres = (H.C.F. of 72, 48 and 120) litres = 24 litres.
#### 3. Question
A milkman has 3 jars containing 57 litres, 129 litres and 177 litres of pure milk respectively. A measuring can, after a different number of exact measurements of milk in each jar, leaves the same amount of milk unmeasured in each jar. What is the volume of the largest such can?
• (a) 12 litres
• (b) 16 litres
• (c) 24 litres
• (d) None of these
Solution:
Required volume = [H.C.F. of (129 – 57), (177 – 129) and (177 – 57)] litres
= (H.C.F. of 72, 48 and 120) litres = 24 litres.
Solution:
Required volume = [H.C.F. of (129 – 57), (177 – 129) and (177 – 57)] litres
= (H.C.F. of 72, 48 and 120) litres = 24 litres.
• Question 4 of 5 4. Question Three electronic devices make a beep sound after every 30, 60, and 105 minutes respectively. It is given that all the devices beeped together at 3:00PM. When will they beep together again? (a) 7:00 PM (b) 8:00 PM (c) 9:00 PM (d) 10:00 PM Correct Solution: LCM of 30, 60, 105 = 420 minutes 420/60 =7 hours Therefore They will beep again at 3 + 7 = 10 PM Incorrect Solution: LCM of 30, 60, 105 = 420 minutes 420/60 =7 hours Therefore They will beep again at 3 + 7 = 10 PM
#### 4. Question
Three electronic devices make a beep sound after every 30, 60, and 105 minutes respectively. It is given that all the devices beeped together at 3:00PM. When will they beep together again?
• (a) 7:00 PM
• (b) 8:00 PM
• (c) 9:00 PM
• (d) 10:00 PM
Solution:
LCM of 30, 60, 105 = 420 minutes
420/60 =7 hours
Therefore They will beep again at 3 + 7 = 10 PM
Solution:
LCM of 30, 60, 105 = 420 minutes
420/60 =7 hours
Therefore They will beep again at 3 + 7 = 10 PM
• Question 5 of 5 5. Question Three girls start jogging from the same point around a circular track and each one completes one round in 24 seconds, 36 seconds and 48 seconds respectively. After how much time will they meet at one point a) 2 minutes 20 seconds b) 2 minutes 24 seconds c) 3 minutes 36 seconds d) 4 minutes 12 seconds Correct Solution: L.C.M. of 24, 36, 48 = 144. So, the three girls will meet at one point in 144 seconds i.e., 2 min 24 sec. Incorrect Solution: L.C.M. of 24, 36, 48 = 144. So, the three girls will meet at one point in 144 seconds i.e., 2 min 24 sec.
#### 5. Question
Three girls start jogging from the same point around a circular track and each one completes one round in 24 seconds, 36 seconds and 48 seconds respectively. After how much time will they meet at one point
• a) 2 minutes 20 seconds
• b) 2 minutes 24 seconds
• c) 3 minutes 36 seconds
• d) 4 minutes 12 seconds
Solution:
L.C.M. of 24, 36, 48 = 144.
So, the three girls will meet at one point in 144 seconds i.e., 2 min 24 sec.
Solution:
L.C.M. of 24, 36, 48 = 144.
So, the three girls will meet at one point in 144 seconds i.e., 2 min 24 sec.
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