UPSC Insta–DART (Daily Aptitude and Reasoning Test) 19 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 LCM of two numbers is 20 times of their HCF. The sum of LCM and HCF is 525.If one of the number is 100 then the other number is? a) 355 b) 150 c) 225 d) 125 Correct Answer: D Explanation LCM = 20 HCF HCF＋20HCF = 525 21HCF = 525 HCF =25 LCM = 2025 = 500 Second number =( LCMHCF)/First number = (50025)/100 = 125 Incorrect Answer: D Explanation LCM = 20 HCF HCF＋20HCF = 525 21HCF = 525 HCF =25 LCM = 2025 = 500 Second number =( LCMHCF)/First number = (50025)/100 = 125
#### 1. Question
The LCM of two numbers is 20 times of their HCF. The sum of LCM and HCF is 525.If one of the number is 100 then the other number is?
Explanation
LCM = 20 HCF
HCF＋20HCF = 525
21HCF = 525
LCM = 20*25 = 500
Second number =( LCM*HCF)/First number
= (500*25)/100 = 125
Explanation
LCM = 20 HCF
HCF＋20HCF = 525
21HCF = 525
LCM = 20*25 = 500
Second number =( LCM*HCF)/First number
= (500*25)/100 = 125
• Question 2 of 5 2. Question The HCF and product of two numbers are 24 and 13824 respectively. The number of possible pairs of the numbers are/is a) 0 b) 1 c) 2 d) 3 Correct Answer: C Explanation HCF = 24 Numbers will be 24x and 24y Where x and y are prime to each other 24x 24y = 13824 xy = 24 Possible pairs = (1, 24) and (3, 8) Incorrect Answer: C Explanation HCF = 24 Numbers will be 24x and 24y Where x and y are prime to each other 24x 24y = 13824 xy = 24 Possible pairs = (1, 24) and (3, 8)
#### 2. Question
The HCF and product of two numbers are 24 and 13824 respectively. The number of possible pairs of the numbers are/is
Explanation
Numbers will be 24x and 24y
Where x and y are prime to each other
24x * 24y = 13824
Possible pairs = (1, 24) and (3, 8)
Explanation
Numbers will be 24x and 24y
Where x and y are prime to each other
24x * 24y = 13824
Possible pairs = (1, 24) and (3, 8)
• Question 3 of 5 3. Question The HCF and LCM of two 2-digit numbers are 5 and 200 respectively. The numbers are? a) (25, 40) b) (40, 96) c) (16, 40) d) None of the above Correct Answer: A Explanation Let numbers are 5x and 5y. 5xy = 200 xy = 40 Possible pairs = (1, 40), (5, 8) Possible numbers = (5, 200), (25, 40) Answer = (25, 40) Incorrect Answer: A Explanation Let numbers are 5x and 5y. 5xy = 200 xy = 40 Possible pairs = (1, 40), (5, 8) Possible numbers = (5, 200), (25, 40) Answer = (25, 40)
#### 3. Question
The HCF and LCM of two 2-digit numbers are 5 and 200 respectively. The numbers are?
• a) (25, 40)
• b) (40, 96)
• c) (16, 40)
• d) None of the above
Explanation
Let numbers are 5x and 5y.
Possible pairs = (1, 40), (5, 8)
Possible numbers = (5, 200), (25, 40)
Answer = (25, 40)
Explanation
Let numbers are 5x and 5y.
Possible pairs = (1, 40), (5, 8)
Possible numbers = (5, 200), (25, 40)
Answer = (25, 40)
• Question 4 of 5 4. Question What is the smallest number which leaves remainder 3 in each case when the number is divided by 10, 12, and 16 but leaves no remainder when it is divided by 9? a) 243 b) 343 c) 293 d) 178 Correct Answer: A Explanation Required number gives remainder 3 when divided by (10, 12, 16) and zero remainder when divided by 9. LCM of (10, 12, 16) = 240 (240k+3)/9 at k= 1 Remainder = 0 Hence answer = 243 Incorrect Answer: A Explanation Required number gives remainder 3 when divided by (10, 12, 16) and zero remainder when divided by 9. LCM of (10, 12, 16) = 240 (240k+3)/9 at k= 1 Remainder = 0 Hence answer = 243
#### 4. Question
What is the smallest number which leaves remainder 3 in each case when the number is divided by 10, 12, and 16 but leaves no remainder when it is divided by 9?
Explanation
Required number gives remainder 3 when divided by (10, 12, 16) and zero remainder when divided by 9.
LCM of (10, 12, 16) = 240
(240k+3)/9 at k= 1
Remainder = 0
Hence answer = 243
Explanation
Required number gives remainder 3 when divided by (10, 12, 16) and zero remainder when divided by 9.
LCM of (10, 12, 16) = 240
(240k+3)/9 at k= 1
Remainder = 0
Hence answer = 243
• Question 5 of 5 5. Question Let N be the greatest number that will divide 66, 88, 110 leaving the same remainder in each case. then sum of the digits in N is a) 5 b) 4 c) 6 d) 3 Correct Answer: B Explanation 66, 88, 110 are three numbers. To find the greatest number which leaves common remainder in each case. The HCF of (88-66), (110-88), (110-66) = 22 In this case the number will be 22 And sum of digits will be = 4 Incorrect Answer: B Explanation 66, 88, 110 are three numbers. To find the greatest number which leaves common remainder in each case. The HCF of (88-66), (110-88), (110-66) = 22 In this case the number will be 22 And sum of digits will be = 4
#### 5. Question
Let N be the greatest number that will divide 66, 88, 110 leaving the same remainder in each case. then sum of the digits in N is
Explanation
66, 88, 110 are three numbers.
To find the greatest number which leaves common remainder in each case.
The HCF of (88-66), (110-88), (110-66) = 22
In this case the number will be 22
And sum of digits will be = 4
Explanation
66, 88, 110 are three numbers.
To find the greatest number which leaves common remainder in each case.
The HCF of (88-66), (110-88), (110-66) = 22
In this case the number will be 22
And sum of digits will be = 4
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