IASbaba’s Daily CSAT Practice Test
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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
In how many ways can 15 indistinguishable fish be placed into 5 different ponds, so that each pond contains at least one fish?
Correct
Solution (b)
According to given question we have 15(n) fish and 5(r) pond:
We have to place fish in each pond so that there will be at least one fish in each pond.
i.e. x1 + x2 + x3 + x4 + x5 = 15
(n – 1)Cr – 1
= (15 – 1)C5 – 1
= 14C4
= 14! / (10!* 4!)
= 14 * 13 * 12 * 11 * 10! / (10! * 4!)
= 14 * 13 * 12 * 11 / 4!
= 1001
Incorrect
Solution (b)
According to given question we have 15(n) fish and 5(r) pond:
We have to place fish in each pond so that there will be at least one fish in each pond.
i.e. x1 + x2 + x3 + x4 + x5 = 15
(n – 1)Cr – 1
= (15 – 1)C5 – 1
= 14C4
= 14! / (10!* 4!)
= 14 * 13 * 12 * 11 * 10! / (10! * 4!)
= 14 * 13 * 12 * 11 / 4!
= 1001
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Question 2 of 5
2. Question
5 skilled workers can build a wall in 20 days: 8 semi-skilled workers can build a wall in 25 days; 10 unskilled workers can build a wall in 30 days. If a team has 2 skilled, 6 semi-skilled and 5 unskilled workers, how long will it take to build the wall?
Correct
Solution (b)
5 skilled workers can build a wall in 20 days
1 skilled worker can build the same wall in 100 days
Capacity of each skilled worker is 1/100
8 semi-skilled workers can build a wall in 25 days
1 semi-skilled worker can build the same wall in 200 days
Capacity of each semi-skilled worker is 1/200
Capacity of 1 unskilled worker is 1/300
Capacity of 2 skilled + 6 semi-skilled + 5 unskilled workers is 2(1/100) + 6(1/200) + 5/300 = 1/15
Incorrect
Solution (b)
5 skilled workers can build a wall in 20 days
1 skilled worker can build the same wall in 100 days
Capacity of each skilled worker is 1/100
8 semi-skilled workers can build a wall in 25 days
1 semi-skilled worker can build the same wall in 200 days
Capacity of each semi-skilled worker is 1/200
Capacity of 1 unskilled worker is 1/300
Capacity of 2 skilled + 6 semi-skilled + 5 unskilled workers is 2(1/100) + 6(1/200) + 5/300 = 1/15
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Question 3 of 5
3. Question
Consider a set A = {1, 2, 3……..1000}. How many numbers of A shall be divisible by 3 or by 5 or by both 3 and 5?
Correct
Solution (b)
From set A numbers {3,6,9,……..999} which are divisible by 3 are 999 / 3 (A)= 333
From set A numbers {5,10,……995,1000} which are divisible by 5 are 1000 / 5 (B)= 200
From set A numbers {15, 30..…990} which are divisible by 3 and 5 are 990 / 3 * 5 (A ∧ B)= 990 / 15 = 66
So, numbers divisible by 3 or by 5 or by both 3 and 5:
(A ∨ B) = A + B – (A ∧ B)
(A ∨ B) = 333 + 200 – 67
(A ∨ B) = 467
Incorrect
Solution (b)
From set A numbers {3,6,9,……..999} which are divisible by 3 are 999 / 3 (A)= 333
From set A numbers {5,10,……995,1000} which are divisible by 5 are 1000 / 5 (B)= 200
From set A numbers {15, 30..…990} which are divisible by 3 and 5 are 990 / 3 * 5 (A ∧ B)= 990 / 15 = 66
So, numbers divisible by 3 or by 5 or by both 3 and 5:
(A ∨ B) = A + B – (A ∧ B)
(A ∨ B) = 333 + 200 – 67
(A ∨ B) = 467
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Question 4 of 5
4. Question
Aishwarya studies either computer science or mathematics every day. If she studies computer science on a day, then the probability that she studies mathematics the next day is 0.6. If she studies mathematics on a day, then the probability that she studies computer science the next day is 0.4. Given that Aishwarya studies computer science on Monday, what is the probability that she studies computer science on Wednesday?
Correct
Solution (c)
Aishwarya studies computer science on Monday –> The probability that she studies mathematics on Tuesday is 0.6 and the probability that she studies computer science on Tuesday is 0.4
She studies mathematics on Tuesday and computer science on Wednesday = 0.6 x 0.4 = 0.24è(1)
She studies computer science on Tuesday and computer science on Wednesday = 0.4 x 0.4 = 0.16 è(2)
Adding (1) and (2), the required probability that she studies computer science on Wednesday = 0.24 + 0.16 = 0.40
Incorrect
Solution (c)
Aishwarya studies computer science on Monday –> The probability that she studies mathematics on Tuesday is 0.6 and the probability that she studies computer science on Tuesday is 0.4
She studies mathematics on Tuesday and computer science on Wednesday = 0.6 x 0.4 = 0.24è(1)
She studies computer science on Tuesday and computer science on Wednesday = 0.4 x 0.4 = 0.16 è(2)
Adding (1) and (2), the required probability that she studies computer science on Wednesday = 0.24 + 0.16 = 0.40
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Question 5 of 5
5. Question
Three people A, B and C working individually can finish a job in 10, 12 and 20 days respectively. They decided to work together but after 2 days, A left the work and after another one day, B also left work. If they got twenty thousand rupees collectively for the entire work, find the difference of the highest and lowest share.
Correct
Solution (c)
Let the total work be LCM (10, 12, 20) = 60 units
=> Efficiency of A = 60/10 = 6 units / day
=> Efficiency of B = 60/12 = 5 units / day
=> Efficiency of C = 60/20 = 3 units / day
Since the number of working days is different for each person, the share of each will be calculated in the ratio of the units of work done.
Now, A works for 2 days and B works for 3 days.
=> Work done by A = 2 x 6 = 12 units
=> Work done by B = 3 x 5 = 15 units
=> Work done by C = 60 – 12 – 15 = 33 units
Therefore, ratio of work done = 12:15:33 = 4:5:11
So, A’s share = (4/20) x 20,000 = Rs 4,000
B’s share = (5/20) x 20,000 = Rs 5,000
C’s share = (11/20) x 20,000 = Rs 11,000
Therefore, difference of the highest and lowest share = Rs 11,000 – 4,000 = Rs 7,000
Incorrect
Solution (c)
Let the total work be LCM (10, 12, 20) = 60 units
=> Efficiency of A = 60/10 = 6 units / day
=> Efficiency of B = 60/12 = 5 units / day
=> Efficiency of C = 60/20 = 3 units / day
Since the number of working days is different for each person, the share of each will be calculated in the ratio of the units of work done.
Now, A works for 2 days and B works for 3 days.
=> Work done by A = 2 x 6 = 12 units
=> Work done by B = 3 x 5 = 15 units
=> Work done by C = 60 – 12 – 15 = 33 units
Therefore, ratio of work done = 12:15:33 = 4:5:11
So, A’s share = (4/20) x 20,000 = Rs 4,000
B’s share = (5/20) x 20,000 = Rs 5,000
C’s share = (11/20) x 20,000 = Rs 11,000
Therefore, difference of the highest and lowest share = Rs 11,000 – 4,000 = Rs 7,000
