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.
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
P, Q, and R can complete a work in 42 days, 35 days, and 28 days respectively. P started the work and was assisted by Q and R on alternate days starting with Q. After 8 days P took a break, Q and R worked together during that period. Neither Q nor R worked for the last 7 days. Find the total time taken to complete the work if P resumed his work for the last 7 days
Correct
Solution (a)
Let the total work be 42 units (LCM of 42, 35, and 28)
The efficiency of P = 10 units/day
The efficiency of Q = 12 units/day
The efficiency of R = 15 units/day
In 8 days Q and R worked for 4 days each starting with Q
The total work done in 8 days = (10*8) + (12*4) + (15*4) = 188 units
The amount of work done by P in the last 7 days = 10*7 = 70 units
Therefore, work done by Q and R together = 420 – (188+70) = 162 units
Time taken by Q and R to complete the work = 162/(12+15) = 6 days
Therefore, the total time taken to complete the work = 8+6+7 = 21 days
Hence, option a is correct.
Incorrect
Solution (a)
Let the total work be 42 units (LCM of 42, 35, and 28)
The efficiency of P = 10 units/day
The efficiency of Q = 12 units/day
The efficiency of R = 15 units/day
In 8 days Q and R worked for 4 days each starting with Q
The total work done in 8 days = (10*8) + (12*4) + (15*4) = 188 units
The amount of work done by P in the last 7 days = 10*7 = 70 units
Therefore, work done by Q and R together = 420 – (188+70) = 162 units
Time taken by Q and R to complete the work = 162/(12+15) = 6 days
Therefore, the total time taken to complete the work = 8+6+7 = 21 days
Hence, option a is correct.

Question 2 of 5
2. Question
Sinchana brought 30 books on History, Polity, and Geography, priced at Rs 17, Rs 19, and Rs 23 per book respectively, for distribution among poor students preparing for UPSC. The polity books were more in number than the History books but less than the Geography books, the difference being more than one. The total cost amounted to Rs 620. How many books on History, Polity, and Geography could have been bought respectively?
Correct
Solution (c)
As it is mentioned that difference in the number of books is more than one
Hence, option b and d can be eliminated.
Now, total amount is = 620 = 17H + 19P + 23G
We can check by putting values from option a and c.
Option c satisfies the above equation. Hence option c is correct
Incorrect
Solution (c)
As it is mentioned that difference in the number of books is more than one
Hence, option b and d can be eliminated.
Now, total amount is = 620 = 17H + 19P + 23G
We can check by putting values from option a and c.
Option c satisfies the above equation. Hence option c is correct

Question 3 of 5
3. Question
The last time Rahul bought invitation cards for New Year, he found that the four types of cards that he liked were priced Rs 2, Rs 3.50, Rs 4.50 and Rs 5.00 each. As Rahul wanted 30 cards, he took five each of two kinds and ten each of the other two, putting down the exact number of 10 rupees notes on the counter payment. How many notes did Rahul give?
Correct
Solution (d)
According to question Rahul put an exact number of 10 Rs notes, hence the total price will be a multiple of 10. And Rahul wants 30 cards, where he took 5 each of two kinds and 10 each of the other two kinds.
So, the summation of (the price of one type card multiplied by a number of that type of card) should be a multiple of 10.
By looking at the prices of cards and to make the sum a multiple of 10, we can say that two 5 cards were of Rs 3.5 and 4.5 and two 10 cards were of prices 2 Rs and 5 Rs each respectively.
Hence, total sum will be 5 (3.5+4.5) + 10 (2+5) = 110
So, Rahul gave 11 notes of 10 Rs.
Incorrect
Solution (d)
According to question Rahul put an exact number of 10 Rs notes, hence the total price will be a multiple of 10. And Rahul wants 30 cards, where he took 5 each of two kinds and 10 each of the other two kinds.
So, the summation of (the price of one type card multiplied by a number of that type of card) should be a multiple of 10.
By looking at the prices of cards and to make the sum a multiple of 10, we can say that two 5 cards were of Rs 3.5 and 4.5 and two 10 cards were of prices 2 Rs and 5 Rs each respectively.
Hence, total sum will be 5 (3.5+4.5) + 10 (2+5) = 110
So, Rahul gave 11 notes of 10 Rs.

Question 4 of 5
4. Question
The price of a product is P. A shopkeeper raises its price by X% and then offers a discount of Y% on the raised price. The discounted price again becomes P. If Y is the difference between X and Y, then find X.
Correct
Solution (b)
Let the cost price of the article be Rs 100.
The shopkeeper raises the price by X% and then decreases it by Y%.
As a result, he reaches the cost price of the article.
Also, it has been given that Y is the difference between Y% and X%.
Y = X – Y
2Y = X
We know that, (1+2Y) (1Y)*100 = 100
Or, (1+2Y) (1Y) = 1
1Y+2Y2Y2 = 1
2Y2Y = 0
2Y = 1
Y = 1/2 or 0.5
X = 2Y
=> X = 1 or 100 %
Therefore, option c is the correct.
Incorrect
Solution (b)
Let the cost price of the article be Rs 100.
The shopkeeper raises the price by X% and then decreases it by Y%.
As a result, he reaches the cost price of the article.
Also, it has been given that Y is the difference between Y% and X%.
Y = X – Y
2Y = X
We know that, (1+2Y) (1Y)*100 = 100
Or, (1+2Y) (1Y) = 1
1Y+2Y2Y2 = 1
2Y2Y = 0
2Y = 1
Y = 1/2 or 0.5
X = 2Y
=> X = 1 or 100 %
Therefore, option c is the correct.

Question 5 of 5
5. Question
X dealt some cards to Y and himself from a full pack of playing cards and laid the rest aside. X then said to Y. “If you give me a certain number of your cards, I will have four times as many cards as you will have. If I give you the same number of cards, I will have thrice as many cards as you will have”. Of the given choices, which could represent the number of cards with X?
Correct
Solution (b)
Let’s say X had ‘a’ cards initially and Y had ‘b’ cards initially.
So, in the first instance, Y gave t cards to X, hence (a+t) = 4(bt)
Now in the second instance, X gave t cards to Y, hence at = 3(b+t)
Solving the above two equations, we will get a = 31t and b = 9t
And we know that a+b < 52 [because a full pack of cards contain a maximum of 52 cards and they have kept some cards aside]
Therefore, 40t<52
Since ‘t’ should be a whole number, it will be 1 here and a=31 and b=9
Incorrect
Solution (b)
Let’s say X had ‘a’ cards initially and Y had ‘b’ cards initially.
So, in the first instance, Y gave t cards to X, hence (a+t) = 4(bt)
Now in the second instance, X gave t cards to Y, hence at = 3(b+t)
Solving the above two equations, we will get a = 31t and b = 9t
And we know that a+b < 52 [because a full pack of cards contain a maximum of 52 cards and they have kept some cards aside]
Therefore, 40t<52
Since ‘t’ should be a whole number, it will be 1 here and a=31 and b=9