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?
Correct
Answer: D
Explanation
LCM = 20 HCF
HCF＋20HCF = 525
21HCF = 525
HCF =25
LCM = 20*25 = 500
Second number =( LCM*HCF)/First number
= (500*25)/100 = 125
Incorrect
Answer: D
Explanation
LCM = 20 HCF
HCF＋20HCF = 525
21HCF = 525
HCF =25
LCM = 20*25 = 500
Second number =( LCM*HCF)/First number
= (500*25)/100 = 125
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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
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)
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Question 3 of 5
3. Question
The HCF and LCM of two 2-digit numbers are 5 and 200 respectively. The numbers are?
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)
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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?
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
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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
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
