IASbaba’s Daily CSAT Practice Test
Daily CSAT Practice Test
Everyday 5 Questions from Aptitude, Logical Reasoning, and Reading Comprehension will be covered from Monday to Saturday.
Make the best use of the initiative. All the best!
To Know More about Ace the Prelims (ATP) 2021 – CLICK HERE
Important Note:
 Don’t forget to post your marks in the comment section. Also, let us know if you enjoyed today’s test 🙂
 After completing the 5 questions, click on ‘View Questions’ to check your score, time taken and solutions.
Testsummary
0 of 5 questions completed
Questions:
 1
 2
 3
 4
 5
Information
To view Solutions, follow these instructions:
 Click on – ‘Start Test’ button
 Solve Questions
 Click on ‘Test Summary’ button
 Click on ‘Finish Test’ button
 Now click on ‘View Questions’ button – here you will see solutions and links.
You have already completed the test before. Hence you can not start it again.
Test is loading...
You must sign in or sign up to start the test.
You have to finish following test, to start this test:
Results
0 of 5 questions answered correctly
Your time:
Time has elapsed
You have scored 0 points out of 0 points, (0)
Average score 

Your score 

Categories
 Not categorized 0%
Pos.  Name  Entered on  Points  Result 

Table is loading  
No data available  
 1
 2
 3
 4
 5
 Answered
 Review

Question 1 of 5
1. Question
MNT has three sports in extracurricular activities: Cricket, Badminton and Football. Students can participate in one, two or even in all three sports. There are 180 students in the school. There are 90 students in Cricket, 100 in the Badminton, and 70 in the Football. Furthermore, 45 students are in both the Cricket and Badminton, 20 are in both the Cricket and the Football, and 10 students are in all three groups. 30 students are just in the Badminton, not in anything else.
How many students participate in only the Football?
Correct
Solution (d)
No. of students playing only badminton= No. of students playing badminton – No. of students playing both
Badminton & cricket –– No. of students playing both Badminton & Football+ No. of students playing all three games
n(only B) = n(B) – n(B∩C) – n(B∩F) + n(B∩C∩F)
30 = 100 – 45 – n(B∩F) + 10
n(B∩F) = 35
So, number of students playing Badminton and Football is 35.
Using the same result, the number of students playing only Football will
n(Only F) = n(F) – n(F∩C) – n(F∩B) + n(B∩C∩F)
n(Only F) = 70 – 20 – 35 + 10 = 25
Incorrect
Solution (d)
No. of students playing only badminton= No. of students playing badminton – No. of students playing both
Badminton & cricket –– No. of students playing both Badminton & Football+ No. of students playing all three games
n(only B) = n(B) – n(B∩C) – n(B∩F) + n(B∩C∩F)
30 = 100 – 45 – n(B∩F) + 10
n(B∩F) = 35
So, number of students playing Badminton and Football is 35.
Using the same result, the number of students playing only Football will
n(Only F) = n(F) – n(F∩C) – n(F∩B) + n(B∩C∩F)
n(Only F) = 70 – 20 – 35 + 10 = 25

Question 2 of 5
2. Question
Prakash travelled 1500km by air which formed 2/5 of his trip.Onethird of the whole trip, he travelled by train and the rest of the journey he performed by car. The distance travelled by car was?
Correct
Solution (a)
Let the total trip be x km.
Then, 2x/5=1500 which gives x=1500*5/2=3750km
Distance travelled by train =1/3*3750=1250km
Therefore, journey by car =[3750(1500+1250)]=1000km
Incorrect
Solution (a)
Let the total trip be x km.
Then, 2x/5=1500 which gives x=1500*5/2=3750km
Distance travelled by train =1/3*3750=1250km
Therefore, journey by car =[3750(1500+1250)]=1000km

Question 3 of 5
3. Question
15 men can complete a piece of work in 9 days. In how many days will 27 men complete the same work?
Correct
Solution (b)
Alternative1:
15 men can do the work in 9 days
1 man will do the same work in 15*9days = 135days
So, 27 men will do the same work in 135days/27 = 5days.
Alternative2:
We should understand the concept of mandays.
It takes 15 men to complete the work in 9days. It means the work needs 15men*9days i.e., 135mandays to get completed.
So, now whether number of days is given or number of people working is given or the total work to be done is mentioned, we can easily find out the other variable.
In this case, the number of men given is 27 and work to be done is 1 complete unit.
1 unit of work needs 135mandays.
We have 27men available so we will need 135mandays/27men = 5days.
Incorrect
Solution (b)
Alternative1:
15 men can do the work in 9 days
1 man will do the same work in 15*9days = 135days
So, 27 men will do the same work in 135days/27 = 5days.
Alternative2:
We should understand the concept of mandays.
It takes 15 men to complete the work in 9days. It means the work needs 15men*9days i.e., 135mandays to get completed.
So, now whether number of days is given or number of people working is given or the total work to be done is mentioned, we can easily find out the other variable.
In this case, the number of men given is 27 and work to be done is 1 complete unit.
1 unit of work needs 135mandays.
We have 27men available so we will need 135mandays/27men = 5days.

Question 4 of 5
4. Question
Anand gives 35% of the money he had to his wife and gave 50% of the money he had to his sons. Remaining amount of Rs.12,000 he kept for himself. What was the total amount of money Anand had?
Correct
Solution (c)
Anand gave away 35%+50% = 85% of his money to his wife and sons. So, he is left with 15% of the money.
Using unitary method,
15% of the money = Rs.12,000
1% of the money = Rs.12,000/15 = Rs.800
So, 100% of the money would had been = Rs.800*100 = Rs.80,000
Incorrect
Solution (c)
Anand gave away 35%+50% = 85% of his money to his wife and sons. So, he is left with 15% of the money.
Using unitary method,
15% of the money = Rs.12,000
1% of the money = Rs.12,000/15 = Rs.800
So, 100% of the money would had been = Rs.800*100 = Rs.80,000

Question 5 of 5
5. Question
Three pipes A, B and C can fill a cistern in 8 hours. After working at it together for 2 hours, C is closed and A and B fill it in 9 hours more. The time taken by C alone to fill the cistern is
Correct
Solution (a)
As it takes 8hrs to completely fill the cistern, in two hours they will together be filling 1/4th of the cistern.
Then, 3/4th of the cistern is filled in 9hrs by A+B. It means they can fill whole cistern in 12hrs.
In terms of percentage,
A+B does 100% work in 12hrs and A+B+C does 100% work in 8hrs.
So, amount of work done by A+B in 8hrs = (100/12)*8 = 200/3 %
So, it means remaining part is done by C which is (100200/3)% = 100/3%
100/6% work is done by C in 8hrs
1% work will be done by C in 8/(100/3)hrs
100% work will be done by C in 8/(100/3)*100 = 24hrs.
Incorrect
Solution (a)
As it takes 8hrs to completely fill the cistern, in two hours they will together be filling 1/4th of the cistern.
Then, 3/4th of the cistern is filled in 9hrs by A+B. It means they can fill whole cistern in 12hrs.
In terms of percentage,
A+B does 100% work in 12hrs and A+B+C does 100% work in 8hrs.
So, amount of work done by A+B in 8hrs = (100/12)*8 = 200/3 %
So, it means remaining part is done by C which is (100200/3)% = 100/3%
100/6% work is done by C in 8hrs
1% work will be done by C in 8/(100/3)hrs
100% work will be done by C in 8/(100/3)*100 = 24hrs.