UPSC Insta–DART (Daily Aptitude and Reasoning Test) 18 Feb 2025
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 greatest number that will divide 64, 80 and 111 leaving remainders 4, 5 and 6 respectively is a) 12 b) 13 c) 14 d) 15 Correct Answer: D Explanation: Get all the common factors of 64 – 4, 80 – 5 and 111- 6, i.e., 60, 75 and 105and see that the common factors will divide them all. The greatest number is the H.C.F of 60, 75 and 105. 60=2 2 3 5, 75=3 5 5, 105=3 5 7 Hence, the HCF is 3 x 5=15, which is the greatest number that will divide 64, 80, 111 leaving remainders 4, 5 and 6 respectively. Incorrect Answer: D Explanation: Get all the common factors of 64 – 4, 80 – 5 and 111- 6, i.e., 60, 75 and 105and see that the common factors will divide them all. The greatest number is the H.C.F of 60, 75 and 105. 60=2 2 3 5, 75=3 5 5, 105=3 5 7 Hence, the HCF is 3 x 5=15, which is the greatest number that will divide 64, 80, 111 leaving remainders 4, 5 and 6 respectively.
#### 1. Question
The greatest number that will divide 64, 80 and 111 leaving remainders 4, 5 and 6 respectively is
Explanation:
Get all the common factors of 64 – 4, 80 – 5 and 111- 6, i.e., 60, 75 and 105and
see that the common factors will divide them all. The greatest number is the H.C.F of
60, 75 and 105.
60=2 2 3 5, 75=3 5 5, 105=3 5 * 7
Hence, the HCF is 3 x 5=15, which is the greatest number that will divide 64, 80,
111 leaving remainders 4, 5 and 6 respectively.
Explanation:
Get all the common factors of 64 – 4, 80 – 5 and 111- 6, i.e., 60, 75 and 105and
see that the common factors will divide them all. The greatest number is the H.C.F of
60, 75 and 105.
60=2 2 3 5, 75=3 5 5, 105=3 5 * 7
Hence, the HCF is 3 x 5=15, which is the greatest number that will divide 64, 80,
111 leaving remainders 4, 5 and 6 respectively.
• Question 2 of 5 2. Question What is the least number which when divided by 12, 18, 24 and 30 leaves the same remainder 5 in each case. a) 722 b) 730 c) 825 d) 725 Correct Answer: D Explanation: Solution: LCM of 12, 18,24 and 30 is 360. According to the 725 is the answer. Because 720 is multiple of 360 (3602) If we divide 725 by any given number remainder is ‘5’ Incorrect Answer: D Explanation: Solution: LCM of 12, 18,24 and 30 is 360. According to the 725 is the answer. Because 720 is multiple of 360 (3602) If we divide 725 by any given number remainder is ‘5’
#### 2. Question
What is the least number which when divided by 12, 18, 24 and 30 leaves the same remainder 5 in each case.
Explanation:
Solution: LCM of 12, 18,24 and 30 is 360.
According to the 725 is the answer. Because 720 is multiple of 360 (360*2)
If we divide 725 by any given number remainder is ‘5’
Explanation:
Solution: LCM of 12, 18,24 and 30 is 360.
According to the 725 is the answer. Because 720 is multiple of 360 (360*2)
If we divide 725 by any given number remainder is ‘5’
• Question 3 of 5 3. Question The sum of two numbers is 45. Their difference is 1/9 of their sum. Their L.C.M is a) 100 b) 150 c) 200 d) 250 Correct Answer: A Explanation: Solution: Let the number be x and y where x > y According to the question, x + y= 45——-(1) x-y= 1/9 (x+ y) = (1/9) 45= 5——-(2) From (1) and (2) x = 20 Hence, LCM of 20 and 25 LCM=100 Incorrect Answer: A Explanation: Solution: Let the number be x and y where x > y According to the question, x + y= 45——-(1) x-y= 1/9 (x+ y) = (1/9) 45= 5——-(2) From (1) and (2) x = 20 Hence, LCM of 20 and 25 LCM=100
#### 3. Question
The sum of two numbers is 45. Their difference is 1/9 of their sum. Their L.C.M is
Explanation:
Solution: Let the number be x and y where x > y
According to the question,
x + y= 45——-(1)
x-y= 1/9 (x+ y) = (1/9) *45= 5——-(2)
From (1) and (2)
Hence, LCM of 20 and 25
Explanation:
Solution: Let the number be x and y where x > y
According to the question,
x + y= 45——-(1)
x-y= 1/9 (x+ y) = (1/9) *45= 5——-(2)
From (1) and (2)
Hence, LCM of 20 and 25
• Question 4 of 5 4. Question A merchant has three different types of milk: 870 liters, 986 liters and 1102 liters. Find the least number of casks of equal size required to store all the milk without mixing. a) 51 b) 47 c) 45 d) 61 Correct Answer: A Explanation: Solution: Finding the HCF of three quantities – 870=3 x 5 x 29 x 2, 986= 17 x 29 x 2 ,1102= 19 x 29 x 2 therefore, the HCF will be 58 (292) as it is common in all the numbers. Now, Number of caskets = 870/58 + 986/58 + 1102/58 =15 + 17 + 19 =51 Incorrect Answer: A Explanation: Solution: Finding the HCF of three quantities – 870=3 x 5 x 29 x 2, 986= 17 x 29 x 2 ,1102= 19 x 29 x 2 therefore, the HCF will be 58 (292) as it is common in all the numbers. Now, Number of caskets = 870/58 + 986/58 + 1102/58 =15 + 17 + 19 =51
#### 4. Question
A merchant has three different types of milk: 870 liters, 986 liters and 1102 liters. Find the least number of casks of equal size required to store all the milk without mixing.
Explanation:
Solution: Finding the HCF of three quantities –
870=3 x 5 x 29 x 2, 986= 17 x 29 x 2 ,1102= 19 x 29 x 2
therefore, the HCF will be 58 (29*2) as it is common in all the numbers.
Now, Number of caskets = 870/58 + 986/58 + 1102/58
=15 + 17 + 19 =51
Explanation:
Solution: Finding the HCF of three quantities –
870=3 x 5 x 29 x 2, 986= 17 x 29 x 2 ,1102= 19 x 29 x 2
therefore, the HCF will be 58 (29*2) as it is common in all the numbers.
Now, Number of caskets = 870/58 + 986/58 + 1102/58
=15 + 17 + 19 =51
• Question 5 of 5 5. Question The maximum number of students among whom 3003 pens and 2730 pencils can be distributed in such a way that each student gets same number of pens and same number of pencils, is : a) 273 b) 910 c) 300 d) 293 Correct Answer: A Explanation: Solution: Total number of pens=3003, Total number of pencils= 2730 maximum number of students who get the same number of pens and same number of pencils HCF (3003 and 2730) using Prime factorization: 2730=2x5x7x13 x2, 3003 = 7 x 11 x 13 x2 Now HCF (3003 and 2730) = 273 Incorrect Answer: A Explanation: Solution: Total number of pens=3003, Total number of pencils= 2730 maximum number of students who get the same number of pens and same number of pencils HCF (3003 and 2730) using Prime factorization: 2730=2x5x7x13 x2, 3003 = 7 x 11 x 13 x2 Now HCF (3003 and 2730) = 273
#### 5. Question
The maximum number of students among whom 3003 pens and 2730 pencils can be distributed in such a way that each student gets same number of pens and same number of pencils, is :
Explanation:
Solution: Total number of pens=3003, Total number of pencils= 2730
maximum number of students who get the same number of pens and same number of pencils HCF (3003 and 2730)
using Prime factorization: 2730=2x5x7x13 x2, 3003 = 7 x 11 x 13 x2
Now HCF (3003 and 2730) = 273
Explanation:
Solution: Total number of pens=3003, Total number of pencils= 2730
maximum number of students who get the same number of pens and same number of pencils HCF (3003 and 2730)
using Prime factorization: 2730=2x5x7x13 x2, 3003 = 7 x 11 x 13 x2
Now HCF (3003 and 2730) = 273
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