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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).
Wish you all the best ! 🙂
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Question 1 of 5
1. Question
In an examination, it is required to get 296 marks out of aggregate marks to pass. A student gets 222 marks and is declared failed by 10% marks. What are aggregate marks the student can get?
Correct
Ans. (d)
Let the maximum aggregate marks be x.
According to the question,
10% of x = 296 – 222
x/10 = 74
x=7410= 740
Incorrect
Ans. (d)
Let the maximum aggregate marks be x.
According to the question,
10% of x = 296 – 222
x/10 = 74
x=7410= 740
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Question 2 of 5
2. Question
The ratio of the number of boys and girls in a school is 3: 2. If 20% of the boys and 25% of the girls are scholarship holders, then the percentage of the students who do not get the scholarship, is
Correct
Ans. (a)
Let number of boys be 300.
Number of girls 200
Boys holding scholarship = (20/100)300 = 60
Girls holding scholarship = (25/100)200 = 50
Total students holding scholarship
= 60 + 50 = 110
Percentage of students not holding
Scholarship = [(500-110)/500]100%
=(390/500)100% = 78%
Incorrect
Ans. (a)
Let number of boys be 300.
Number of girls 200
Boys holding scholarship = (20/100)300 = 60
Girls holding scholarship = (25/100)200 = 50
Total students holding scholarship
= 60 + 50 = 110
Percentage of students not holding
Scholarship = [(500-110)/500]100%
=(390/500)100% = 78%
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Question 3 of 5
3. Question
A child has four pockets and three marbles. In how many ways, the child can put the marbles in the pockets?
Correct
Ans. (b)
The first marble can be put into pockets in 4 ways, so can the second and third. Thus, the number of ways in which the child can put the marbles
= (4 4 4 )= 64 ways
Incorrect
Ans. (b)
The first marble can be put into pockets in 4 ways, so can the second and third. Thus, the number of ways in which the child can put the marbles
= (4 4 4 )= 64 ways
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Question 4 of 5
4. Question
A and B can complete a work in 8 days, working together. B alone can do it in 12 days. After working for 4 days, B left the work. How many days will A take to complete the remaining work?
Correct
Ans. (a)
(A+B)’s 1 day’s work = (1/8)
B’s 1 day’s work = (1/12)
A’s 1 day’s work = (1/8)-(1/12)=[(3-2)/24]=(1/24)
A can complete the work in 24 days.
Now, B’s 4 day’s work = (4/12) = 1/3)
Remaining work =1 – (1/3) = (2/3)
As, time taken by A to complete the work is 24 days
Time taken by A to do (2/3) of the work= (2/3) 24 = 16 days
Incorrect
Ans. (a)
(A+B)’s 1 day’s work = (1/8)
B’s 1 day’s work = (1/12)
A’s 1 day’s work = (1/8)-(1/12)=[(3-2)/24]=(1/24)
A can complete the work in 24 days.
Now, B’s 4 day’s work = (4/12) = 1/3)
Remaining work =1 – (1/3) = (2/3)
As, time taken by A to complete the work is 24 days
Time taken by A to do (2/3) of the work= (2/3) 24 = 16 days
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Question 5 of 5
5. Question
4 goats or sheeps can graze a field in 50 days. 2 goats and 9 sheeps can graze the field in
Correct
Ans. (d)
Part of field grazed by 4 goats in 1 day = (1/50)
Part of field grazed by 1 goat in 1 day = (1/(50 4)=(1/200)
4g=6s [here, g=goats, and s= sheeps ]
1s = (4/6)g=(2/3)g
Now, 2g + 6g=8g
8 goats can graze the field in
[(1/8) /200]= 25 days
Incorrect
Ans. (d)
Part of field grazed by 4 goats in 1 day = (1/50)
Part of field grazed by 1 goat in 1 day = (1/(50 4)=(1/200)
4g=6s [here, g=goats, and s= sheeps ]
1s = (4/6)g=(2/3)g
Now, 2g + 6g=8g
8 goats can graze the field in
[(1/8) /200]= 25 days
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