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 boat goes 30 km upstream and 44 km downstream in 10 hours. In 13 hours, it can go 40 km upstream and 55 km down-stream. The speed of the boat in still water is
Correct
Solution (c)
Let us assume the upstream speed of the boat = u km/hr and the downstream speed = v km/hr
30/u+44/v=10
40/u+55/v=13
Consider 1/u=a, 1/v=b, these equations become
30a+44b=10à (1)
40a+55b=13à (2)
On solving equations (1) and (2)
We get a=1/5 and b=1/11 Hence, 1/u=1/5 and 1/v =1/11 => u=5 and v=11
Assuming speed of boat in still water is x and the speed of stream=y
U = x-y = 5
v = x+y = 11
From these equations, we get x = 8 km/hr
Incorrect
Solution (c)
Let us assume the upstream speed of the boat = u km/hr and the downstream speed = v km/hr
30/u+44/v=10
40/u+55/v=13
Consider 1/u=a, 1/v=b, these equations become
30a+44b=10à (1)
40a+55b=13à (2)
On solving equations (1) and (2)
We get a=1/5 and b=1/11 Hence, 1/u=1/5 and 1/v =1/11 => u=5 and v=11
Assuming speed of boat in still water is x and the speed of stream=y
U = x-y = 5
v = x+y = 11
From these equations, we get x = 8 km/hr
-
Question 2 of 5
2. Question
A cylindrical overhead tank is filled by two pumps P1 and P2. P1 can fill the tank in 8 hours while P2 can fill the tank in 12 hours. There is a pipe P3 which can empty the tank in 8 hours. Both the pumps are opened simultaneously. The supervisor of the tank, before going out on a work, sets a timer to open P3 when the tank is half filled so that tank is exactly filled up by the time he is back. Due to technical fault P3 opens when tank is one third filled. If the supervisor comes back as per the plan what percentage of the tank is still empty?
Correct
Solution (a)
P1 can fill the tank in 8 hours while P2 can fill the tank in 12 hours. Therefore when both the pumps are opened simultaneously they can fill 1/8 + 1/12 = 5/24 of tank in an hour.
Therefore time required by P1 and P2 to fill full tank is 24/5 hr
Hence they can fill half the tank in (24/5)*1/2 = 2.4 hours.
After all the 3 pipes are opened, the tank will be filled at the rate of 1/8 + 1/12 – 1/8 = 1/12. Therefore for filling the remaining tank we will need 6 more hours.
If supervisor was planning to come exactly when the tank is fully filled then he must come after 8.4 hours
Time required by P1 and P2 to fill 1/3rd of the tank is (24/5)*1/3 = 1.6 hours.
Therefore supervisor will come after = 8.4 – 1.6 = 6.8 hours
In 6.8 hours all the three pipe will fill 6.8/12 = 0.56 of the capacity of tank
Therefore remaining tank to be filled when the supervisor comes = 1- 0.56 – 0.33 0.1 i.e. 10%
Incorrect
Solution (a)
P1 can fill the tank in 8 hours while P2 can fill the tank in 12 hours. Therefore when both the pumps are opened simultaneously they can fill 1/8 + 1/12 = 5/24 of tank in an hour.
Therefore time required by P1 and P2 to fill full tank is 24/5 hr
Hence they can fill half the tank in (24/5)*1/2 = 2.4 hours.
After all the 3 pipes are opened, the tank will be filled at the rate of 1/8 + 1/12 – 1/8 = 1/12. Therefore for filling the remaining tank we will need 6 more hours.
If supervisor was planning to come exactly when the tank is fully filled then he must come after 8.4 hours
Time required by P1 and P2 to fill 1/3rd of the tank is (24/5)*1/3 = 1.6 hours.
Therefore supervisor will come after = 8.4 – 1.6 = 6.8 hours
In 6.8 hours all the three pipe will fill 6.8/12 = 0.56 of the capacity of tank
Therefore remaining tank to be filled when the supervisor comes = 1- 0.56 – 0.33 0.1 i.e. 10%
-
Question 3 of 5
3. Question
Pointing to a photograph Jai says, “He is the only brother of the only daughter of my sister’s maternal grandmother.” Pointing to another photograph Surya says, “he is the only brother of the only daughter of my sister’s maternal grandmother.” If among the two photographs, one was either of Surya or Jai, and the photograph, towards which Jai was pointing, was not of Surya, then how is Jai related to Surya?
Correct
Solution (b)
“only brother of the only daughter of my sister’s maternal grandmother”
The daughter of my sister’s maternal grandmother is sister’s mother. Mother’s brother is maternal uncle of the person.
If Jai is not pointing towards Surya then Surya is pointing towards Jai.
Thus, Jai is the maternal uncle of Surya.
Hence, option b is the correct answer.
Incorrect
Solution (b)
“only brother of the only daughter of my sister’s maternal grandmother”
The daughter of my sister’s maternal grandmother is sister’s mother. Mother’s brother is maternal uncle of the person.
If Jai is not pointing towards Surya then Surya is pointing towards Jai.
Thus, Jai is the maternal uncle of Surya.
Hence, option b is the correct answer.
-
Question 4 of 5
4. Question
In a certain language ‘to be polite’ is coded as ‘fa so la’, ‘she is polite’ is coded as ‘so me pa’ and ‘to have manners’ is coded as ‘na la ma’. Which of the following is the code for ‘be’ in that language?
Correct
Solution (b)
From the given information, we can write ‘to be polite’ is coded as ‘fa so la’
‘she is polite’ is coded as ‘so me pa’ and ‘to have manners’ is coded as ‘na la ma’.
From 1 st and 2 nd statements common word is polite which codes for so
And from 1 st and third statements common word is to which codes for la.
Hence be is coded as fa.
Incorrect
Solution (b)
From the given information, we can write ‘to be polite’ is coded as ‘fa so la’
‘she is polite’ is coded as ‘so me pa’ and ‘to have manners’ is coded as ‘na la ma’.
From 1 st and 2 nd statements common word is polite which codes for so
And from 1 st and third statements common word is to which codes for la.
Hence be is coded as fa.
-
Question 5 of 5
5. Question
Statement 1: All guilty politicians were arrested
Statement 2: Kishan and Chandru were among those arrested
Which of the following conclusions can be drawn from the above statements?
Correct
Solution (d)
We can see that Kishan and Chandru being guilty is a possibility as we can’t place them inside the subset of those who are guilty for sure.
Incorrect
Solution (d)
We can see that Kishan and Chandru being guilty is a possibility as we can’t place them inside the subset of those who are guilty for sure.
