IASbaba’s Daily CSAT Practice Test
ARCHIVES
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.
Test-summary
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
A child while playing tore a page from a novel written by his father. It was found that the sum of the digits of remaining pages of a novel was 312. What was the page number of page torn from this novel?
Correct
Solution (b)
Let the total pages be ‘n’. We have
1 + 2 + 3 + ………+n = 312
⇒ n(n+1)/2 = 312
n (n+1) = 624
The square nearest to 624 is 625.This is the square of 25. Hence, we will take n= 25.
Therefore, sum of the pages = n(n+1)/2 = 25*26/2=25*13=325
Missing number = 325-312=13.
Now, we have to think of 2 consecutive numbers whose sum is 13.
x+x+1=13 ⇒2×=12
x=6.
Therefore, this was the page torn from the novel.
Incorrect
Solution (b)
Let the total pages be ‘n’. We have
1 + 2 + 3 + ………+n = 312
⇒ n(n+1)/2 = 312
n (n+1) = 624
The square nearest to 624 is 625.This is the square of 25. Hence, we will take n= 25.
Therefore, sum of the pages = n(n+1)/2 = 25*26/2=25*13=325
Missing number = 325-312=13.
Now, we have to think of 2 consecutive numbers whose sum is 13.
x+x+1=13 ⇒2×=12
x=6.
Therefore, this was the page torn from the novel.
-
Question 2 of 5
2. Question
3 identical cones with base radius ‘r’ are placed on their bases so that each is touching the other two. The radius of the circle drawn through their vertices is
Correct
Solution (c)
This is one of such questions where a student can get the answer based on his/her analysing power and symmetry depicted in the question.
Just try to visualize the three cones from their heads. The central point between the three is definitely at a distance, which is more than the radius; hence the third option is the answer. Otherwise; just check if you have taken three equal circles which are touching the other two, the circle which will cover those three circles will be the radius of the smaller circle plus something. Now it should be sufficient to conclude that it is more than r.
Incorrect
Solution (c)
This is one of such questions where a student can get the answer based on his/her analysing power and symmetry depicted in the question.
Just try to visualize the three cones from their heads. The central point between the three is definitely at a distance, which is more than the radius; hence the third option is the answer. Otherwise; just check if you have taken three equal circles which are touching the other two, the circle which will cover those three circles will be the radius of the smaller circle plus something. Now it should be sufficient to conclude that it is more than r.
-
Question 3 of 5
3. Question
A student is required to answer 6 out of 10 questions in an examination. The questions are divided into two groups, each containing 5 questions. He is not allowed to attempt more than 4 questions from each group. The number of different ways in which the student can choose the 6 questions is
Correct
Solution (c)
There are 3 ways in which he can answer the questions
- 3 from group 1 and 3 from group 2
She can select 3 questions from each group in 5C3 ways and thus the total number of ways 5C3 * 5C3 = 100 [nCr = n!/(n-r)! r!]
- 4 questions from group 1 and 2 questions from group 2.
She can select 4 questions from group 1 in 5C4 ways. i.e. 5 ways and 2 questions from group 2 in 5C2 ways. i.e. 10 ways and thus, total number of ways = 5*10 = 50
- Same will be the case for 4 questions from group 2 and 2 questions from group 1
Thus, the total number of ways = 100+50+50 = 200
Hence, option c is correct.
Incorrect
Solution (c)
There are 3 ways in which he can answer the questions
- 3 from group 1 and 3 from group 2
She can select 3 questions from each group in 5C3 ways and thus the total number of ways 5C3 * 5C3 = 100 [nCr = n!/(n-r)! r!]
- 4 questions from group 1 and 2 questions from group 2.
She can select 4 questions from group 1 in 5C4 ways. i.e. 5 ways and 2 questions from group 2 in 5C2 ways. i.e. 10 ways and thus, total number of ways = 5*10 = 50
- Same will be the case for 4 questions from group 2 and 2 questions from group 1
Thus, the total number of ways = 100+50+50 = 200
Hence, option c is correct.
-
Question 4 of 5
4. Question
A government school is to be constructed in 56 days with 104 men, each working 8 hours a day. If after 30 days, 2/5th of the work is finished, then how many additional men should be employed so that work may be completed on time, each man now working 9 hours per day?
Correct
Solution (a)
Let ‘W’ be the amount of work. It is given that,
2/5*W = 105*30*8 –> (1)
Let ‘X’ be the number of additional men required to finish the work on time
W – (2/5) W = (105+X)*9*(26-30) –> (2)
By dividing equation 1 by equation 2
2/3 = (105*30*8)/(105+X)*9*(26-30)
⇒ (104 + x) ∗ 9 ∗ 26 = 12 ∗ 104 ∗ 30
⇒ (104 + x) = 160
⇒ x = 56
Hence, option a is correct
Incorrect
Solution (a)
Let ‘W’ be the amount of work. It is given that,
2/5*W = 105*30*8 –> (1)
Let ‘X’ be the number of additional men required to finish the work on time
W – (2/5) W = (105+X)*9*(26-30) –> (2)
By dividing equation 1 by equation 2
2/3 = (105*30*8)/(105+X)*9*(26-30)
⇒ (104 + x) ∗ 9 ∗ 26 = 12 ∗ 104 ∗ 30
⇒ (104 + x) = 160
⇒ x = 56
Hence, option a is correct
-
Question 5 of 5
5. Question
In a certain Engineering college total expenses of a semester are partly fixed and partly varying linearly with the number of students. The average expense per student is Rs 400 when there are 20 students and Rs 300 when there are 40 students. When there are 80 students, what is the average expense per student?
Correct
Solution (a)
Let the total fees = T, the fixed fees = F and the number of students = n and let x be the varying fees.
Thus, T = F + nx
Given, when n = 20, T = 8000
Thus, 8000 = F + 20x à (1)
Also, when n = 40, T = 12000
Thus, 12000 = F + 40x à (2)
Solving (1) and (2) we get F = 4000 and x = 200
Thus, the average rent when 80 students are staying = 4000/80 + 200 = 250 Rs
Hence, option a is correct.
Incorrect
Solution (a)
Let the total fees = T, the fixed fees = F and the number of students = n and let x be the varying fees.
Thus, T = F + nx
Given, when n = 20, T = 8000
Thus, 8000 = F + 20x à (1)
Also, when n = 40, T = 12000
Thus, 12000 = F + 40x à (2)
Solving (1) and (2) we get F = 4000 and x = 200
Thus, the average rent when 80 students are staying = 4000/80 + 200 = 250 Rs
Hence, option a is correct.
