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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.
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Question 1 of 5
1. Question
There are 6 tasks and 6 people. Task 1 cannot be assigned either to person 1 or to person 2; task 2 must be assigned to either person 3 or person 4. Every person is to be assigned one task. In how many ways can the assignment be done?
Correct
Correct Answer : A
Explanation:
Task 1 can not be assigned to either person 1 or 2 i.e. there are 4 options. Task 2 can be assigned to 3 or 4.
So, there are only 2 options for task 2.
So required no. of ways =
2 options for task 2
3 options for task 1
4 options for task 3
3 options for task 4
2 options for task 5
1 option for task 6.
Total ways = 2 × 3 × 4 × 3 × 2 × 1 = 144
Incorrect
Correct Answer : A
Explanation:
Task 1 can not be assigned to either person 1 or 2 i.e. there are 4 options. Task 2 can be assigned to 3 or 4.
So, there are only 2 options for task 2.
So required no. of ways =
2 options for task 2
3 options for task 1
4 options for task 3
3 options for task 4
2 options for task 5
1 option for task 6.
Total ways = 2 × 3 × 4 × 3 × 2 × 1 = 144
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Question 2 of 5
2. Question
In a chess competition involving some boys and girls of a school, every student had to play exactly one game with every other student. It was found that in 45 games both the players were girls and in 190 games both were boys. The number of games in which one player was a boy and the other was a girl is
Correct
Correct Answer : A
Explanation:
Let number of girls = x
The number of boys = y
45 games in which both the players were girls xC2 = 45
x! / 2!(x-2)! = x(x-1) = 90
Therefore, x = 10.
Incorrect
Correct Answer : A
Explanation:
Let number of girls = x
The number of boys = y
45 games in which both the players were girls xC2 = 45
x! / 2!(x-2)! = x(x-1) = 90
Therefore, x = 10.
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Question 3 of 5
3. Question
Each of two women and three men is to occupy one chair out of eight chairs, each of which numbered from 1 to 8. First, women are to occupy any two chairs from those numbered 1 to 4; and then the three men would occupy any, three chairs out of the remaining six chairs. What is the maximum number of different ways in which this can be done?
Correct
Correct Answer : C
Explanation:
2 Women can occupy 2 chairs out of the first four chairs in 4P2 ways. 3 men can be arranged in the remaining 6 chairs in 6P3 ways. Hence, total no. of ways = 4P2 × 6P3 = 1440
Incorrect
Correct Answer : C
Explanation:
2 Women can occupy 2 chairs out of the first four chairs in 4P2 ways. 3 men can be arranged in the remaining 6 chairs in 6P3 ways. Hence, total no. of ways = 4P2 × 6P3 = 1440
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Question 4 of 5
4. Question
-‘X’ is twice as old as ‘Y’ 3 years ago, when ‘X’ was as old as ‘Y’ today. If the difference between their ages as present is 3 years, how old is ‘X’ at present?
Correct
Correct Answer : C
Explanation:
Let present age of X = x years Present age of Y = (x – 3) years
3 years ago, age of X = (x – 3) years Age of Y = (x – 6) years
According to the question, x – 3 = 2(x – 6)
x – 3 = 2x – 12
12 – 3 = 2x – x
x = 9 years
Incorrect
Correct Answer : C
Explanation:
Let present age of X = x years Present age of Y = (x – 3) years
3 years ago, age of X = (x – 3) years Age of Y = (x – 6) years
According to the question, x – 3 = 2(x – 6)
x – 3 = 2x – 12
12 – 3 = 2x – x
x = 9 years
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Question 5 of 5
5. Question
A candidate attempted 12 questions and secured full marks in all of them. If he obtained 60% in the test and all questions carry equal marks, then what is the number of questions in the test?
Correct
Ans-D
Let total no. of questions be x.
Now, Right questions = 12 = 60% of x. 0.6x = 12
x = 20
Incorrect
Ans-D
Let total no. of questions be x.
Now, Right questions = 12 = 60% of x. 0.6x = 12
x = 20
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