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Daily CSAT Practice Test
Everyday 5 Questions from Aptitude, Logical Reasoning, and Reading Comprehension will be covered from Monday to Saturday.
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Question 1 of 5
1. Question
If n2 = 12345654321, what is the value of n?
Correct
Solution (d)
Observe the pattern given below
112 = 121
1112 = 12321
11112 = 1234321 so on
Therefore, 1111112 = 12345654321
Incorrect
Solution (d)
Observe the pattern given below
112 = 121
1112 = 12321
11112 = 1234321 so on
Therefore, 1111112 = 12345654321

Question 2 of 5
2. Question
The MRP of a chocolate is Rs 65. A customer purchased the chocolate for Rs 56.16. He got two successive discounts of which one is 10%. Find the other discount of the discount scheme that was given by the seller.
Correct
Solution (b)
Here MP = Rs 65 and SP = Rs 56.16.
Hence, the flat discount = [(65 – 56.16)/65]*100 = 13.6%
Now, the first discount is 10%. Let us try the question using options.
Let the second discount is 4%.
So, the flat discount = 10 + 4 – 40/100 = 13.6%, which is same as the actual value.
Hence, the second discount is 4%
Incorrect
Solution (b)
Here MP = Rs 65 and SP = Rs 56.16.
Hence, the flat discount = [(65 – 56.16)/65]*100 = 13.6%
Now, the first discount is 10%. Let us try the question using options.
Let the second discount is 4%.
So, the flat discount = 10 + 4 – 40/100 = 13.6%, which is same as the actual value.
Hence, the second discount is 4%

Question 3 of 5
3. Question
The probability that a builder will get a construction contract is 3/4 and the probability that he will not get the electrical contract is 7/9. If the probability of getting at least one contract is 5/6, what is the probability that he will get both?
Correct
Solution (a)
Let P(A) = Probability of getting construction contract = 3/4
P(B) = Probability of getting electrical contract = 1 – 7/9 = 2/9
P(A U B) = Probability of getting at least one contract = 5/6
Now P (A U B) = P(A) + P(B) – P (A ∩ B)
5/6 = 3/4 + 2/9 – P (A ∩ B)
P (A ∩ B) = 5/36
Incorrect
Solution (a)
Let P(A) = Probability of getting construction contract = 3/4
P(B) = Probability of getting electrical contract = 1 – 7/9 = 2/9
P(A U B) = Probability of getting at least one contract = 5/6
Now P (A U B) = P(A) + P(B) – P (A ∩ B)
5/6 = 3/4 + 2/9 – P (A ∩ B)
P (A ∩ B) = 5/36

Question 4 of 5
4. Question
In how many ways can 21 identical books on Literature and 19 identical books on Nature be placed in a row on a shelf so that two books on Nature may not be together?
Correct
Solution (c)
When the 21 books of Literature are arranged in a row, there will be 22 places.
Now, we have to place 19 books of Nature at these 22 places which can be done in 22C19 ways
22C19 = 22!/19!*3! = 1540 ways. [nCr = n!/(nr)!*r!]
Incorrect
Solution (c)
When the 21 books of Literature are arranged in a row, there will be 22 places.
Now, we have to place 19 books of Nature at these 22 places which can be done in 22C19 ways
22C19 = 22!/19!*3! = 1540 ways. [nCr = n!/(nr)!*r!]

Question 5 of 5
5. Question
Q is twice efficient as P and P can do a piece of work in 15 days. P started the work and after a few days Q joined him. They completed the work in 11 days, from the starting. For how many days did they work together?
Correct
Solution (b)
P can do a piece of work in 15 days.
Q is twice as efficient as P.
So, Q can do the same piece of work in 7.5 days.
Assume the total work to be 15 units
P does 1 unit per day and Q does 2 units per day.
For the first x days, P worked alone and hence work done per day is
1 * x = x units.
For the next (11 – x) days, P and Q worked together. In 1 day, they will do (1 + 2) = 3 units.
So, in 11 – x days, they will do 33 – 3x units.
33 – 3x + x = 15
2x = 18 and x = 9.
Thus, P worked alone for 9 days whereas P and Q worked together for 2 days.
Incorrect
Solution (b)
P can do a piece of work in 15 days.
Q is twice as efficient as P.
So, Q can do the same piece of work in 7.5 days.
Assume the total work to be 15 units
P does 1 unit per day and Q does 2 units per day.
For the first x days, P worked alone and hence work done per day is
1 * x = x units.
For the next (11 – x) days, P and Q worked together. In 1 day, they will do (1 + 2) = 3 units.
So, in 11 – x days, they will do 33 – 3x units.
33 – 3x + x = 15
2x = 18 and x = 9.
Thus, P worked alone for 9 days whereas P and Q worked together for 2 days.
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