UPSC Insta–DART (Daily Aptitude and Reasoning Test) 15 Oct 2024
Kartavya Desk Staff
Considering the alarming importance of CSAT in UPSC CSE Prelims exam and with enormous requests we received recently, InsightsIAS has started Daily CSAT Test to ensure students practice CSAT Questions on a daily basis. Regular Practice would help one overcome the fear of CSAT too.We are naming this initiative as Insta– DART – Daily Aptitude and Reasoning Test. We hope you will be able to use DART to hit bull’s eye in CSAT paper and comfortably score 100+ even in the most difficult question paper that UPSC can give you in CSP-2021. Your peace of mind after every step of this exam is very important for us.
Looking forward to your enthusiastic participation (both in sending us questions and solving them on daily basis on this portal).
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• Question 1 of 5 1. Question Two numbers u and v are chosen at random from the set of first 30 natural numbers. The probability that u2 -v2 is divisible by 3 is (a) 43/89 (b) 57/87 (c) 47/87 (d) 41/83 Correct Solution: Option (c) Explanation: Out of 30 numbers 2 numbers can be chosen in 30C2 ways. So, exhaustive number of cases = 30C2 = 435 Since u2 – v2 is divisible by 3 if either u and v are divisible by 3 or non of u and v is divisible by 3. Thus, the favourable numbers, of cases 10C2 + 20C2 = 235 ∴ required probability = 235/435 = 47/87 Hence option (c) is correct answer. Incorrect Solution: Option (c) Explanation: Out of 30 numbers 2 numbers can be chosen in 30C2 ways. So, exhaustive number of cases = 30C2 = 435 Since u2 – v2 is divisible by 3 if either u and v are divisible by 3 or non of u and v is divisible by 3. Thus, the favourable numbers, of cases 10C2 + 20C2 = 235 ∴ required probability = 235/435 = 47/87 Hence option (c) is correct answer.
#### 1. Question
Two numbers u and v are chosen at random from the set of first 30 natural numbers. The probability that u2 -v2 is divisible by 3 is
Solution: Option (c)
Explanation:
Out of 30 numbers 2 numbers can be chosen in 30C2 ways.
So, exhaustive number of cases = 30C2 = 435
Since u2 – v2 is divisible by 3 if either u and v are divisible by 3 or non of u and v is divisible by 3.
Thus, the favourable numbers, of cases 10C2 + 20C2 = 235
∴ required probability = 235/435 = 47/87 Hence option (c) is correct answer.
Solution: Option (c)
Explanation:
Out of 30 numbers 2 numbers can be chosen in 30C2 ways.
So, exhaustive number of cases = 30C2 = 435
Since u2 – v2 is divisible by 3 if either u and v are divisible by 3 or non of u and v is divisible by 3.
Thus, the favourable numbers, of cases 10C2 + 20C2 = 235
∴ required probability = 235/435 = 47/87 Hence option (c) is correct answer.
• Question 2 of 5 2. Question A committee of two members is to be chosen from a group of 5 men and 4 women. Out of the 9 people, there is one couple. What is the probability that the committee has one male and one female member if the couple cannot be in the committee simultaneously? (a) 5/18 (b) 19/36 (c) 21/36 (d) 1/20 Correct Solution: Option (b) Explanation: Number of ways of choosing a committee of two members = 9C2 = (9 × 8)/2 = 36 Number of ways of choosing a committee of one male and one female members = 5 × 4 = 20 Number of ways of choosing a committee such that the members are the couple = 1 Number of ways of choosing the committee of one male and one female such that the couple is not in the committee simultaneously = 20 – 1 = 19 Required probability = 19/36 Hence, option (b) is the correct answer. Incorrect Solution: Option (b) Explanation: Number of ways of choosing a committee of two members = 9C2 = (9 × 8)/2 = 36 Number of ways of choosing a committee of one male and one female members = 5 × 4 = 20 Number of ways of choosing a committee such that the members are the couple = 1 Number of ways of choosing the committee of one male and one female such that the couple is not in the committee simultaneously = 20 – 1 = 19 Required probability = 19/36 Hence, option (b) is the correct answer.
#### 2. Question
A committee of two members is to be chosen from a group of 5 men and 4 women. Out of the 9 people, there is one couple. What is the probability that the committee has one male and one female member if the couple cannot be in the committee simultaneously?
Solution: Option (b)
Explanation:
Number of ways of choosing a committee of two members = 9C2 = (9 × 8)/2 = 36
Number of ways of choosing a committee of one male and one female members = 5 × 4 = 20
Number of ways of choosing a committee such that the members are the couple = 1
Number of ways of choosing the committee of one male and one female such that the couple is not in the committee simultaneously = 20 – 1 = 19
Required probability = 19/36 Hence, option (b) is the correct answer.
Solution: Option (b)
Explanation:
Number of ways of choosing a committee of two members = 9C2 = (9 × 8)/2 = 36
Number of ways of choosing a committee of one male and one female members = 5 × 4 = 20
Number of ways of choosing a committee such that the members are the couple = 1
Number of ways of choosing the committee of one male and one female such that the couple is not in the committee simultaneously = 20 – 1 = 19
Required probability = 19/36 Hence, option (b) is the correct answer.
• Question 3 of 5 3. Question There are 5 Pink bowls, 4 Purple bowls and 3 Brown bowls in a basket. Four bowls are chosen at random. Quantity 1: The probability of being 2 Pink and 2 Purple bowl. Quantity 2: The probability of being 2 Pink, 1 Purple & 1 Brown. (a) Quantity 1 > Quantity 2 (b) Quantity 1 < Quantity 2 (c) Quantity 1 ≥ Quantity 2 (d) Quantity 1 ≤ Quantity 2 Correct Solution: Option (b) Explanation: Quantity 1 = (5C2 × 4C2)/12C4 = 60/495 = 4/33 Quantity 2 = (5C2 × 4C1 × 3C1)/12C4 =120/95 = 8/33 Hence Quantity 1 < Quantity 2 Hence, option (b) is the correct answer. Incorrect Solution: Option (b) Explanation: Quantity 1 = (5C2 × 4C2)/12C4 = 60/495 = 4/33 Quantity 2 = (5C2 × 4C1 × 3C1)/12C4 =120/95 = 8/33 Hence Quantity 1 < Quantity 2 Hence, option (b) is the correct answer.
#### 3. Question
There are 5 Pink bowls, 4 Purple bowls and 3 Brown bowls in a basket. Four bowls are chosen at random.
Quantity 1: The probability of being 2 Pink and 2 Purple bowl.
Quantity 2: The probability of being 2 Pink, 1 Purple & 1 Brown.
• (a) Quantity 1 > Quantity 2
• (b) Quantity 1 < Quantity 2
• (c) Quantity 1 ≥ Quantity 2
• (d) Quantity 1 ≤ Quantity 2
Solution: Option (b)
Explanation:
Quantity 1 = (5C2 × 4C2)/12C4 = 60/495 = 4/33
Quantity 2 = (5C2 × 4C1 × 3C1)/12C4 =120/95 = 8/33
Hence Quantity 1 < Quantity 2 Hence, option (b) is the correct answer.
Solution: Option (b)
Explanation:
Quantity 1 = (5C2 × 4C2)/12C4 = 60/495 = 4/33
Quantity 2 = (5C2 × 4C1 × 3C1)/12C4 =120/95 = 8/33
Hence Quantity 1 < Quantity 2 Hence, option (b) is the correct answer.
• Question 4 of 5 4. Question Three-unit squares are chosen at random on a chessboard. What is the probability that two of them are of same color while remaining one is of another color? (a) 16/21 (b) 8/21 (c) 5/21 (d) 4/7 Correct Solution: Option (a) Explanation: Total number of ways = ⁶⁴C₃ = 41664 Favorable ways = 2 × ³²C₂ × ³²C₁ = 2 × 496 × 32 = 31744 Required probability = 31744/41664 = 16/21 (³²C₂ × ³²C₁ is taken because if two are black, then one will be of white or if two are white, then one will be of black) Hence, option (a) is the correct answer. Incorrect Solution: Option (a) Explanation: Total number of ways = ⁶⁴C₃ = 41664 Favorable ways = 2 × ³²C₂ × ³²C₁ = 2 × 496 × 32 = 31744 Required probability = 31744/41664 = 16/21 (³²C₂ × ³²C₁ is taken because if two are black, then one will be of white or if two are white, then one will be of black) Hence, option (a) is the correct answer.
#### 4. Question
Three-unit squares are chosen at random on a chessboard. What is the probability that two of them are of same color while remaining one is of another color?
Solution: Option (a)
Explanation:
Total number of ways = ⁶⁴C₃ = 41664
Favorable ways = 2 × ³²C₂ × ³²C₁ = 2 × 496 × 32 = 31744
Required probability = 31744/41664 = 16/21
(³²C₂ × ³²C₁ is taken because if two are black, then one will be of white or if two are white, then one will be of black) Hence, option (a) is the correct answer.
Solution: Option (a)
Explanation:
Total number of ways = ⁶⁴C₃ = 41664
Favorable ways = 2 × ³²C₂ × ³²C₁ = 2 × 496 × 32 = 31744
Required probability = 31744/41664 = 16/21
(³²C₂ × ³²C₁ is taken because if two are black, then one will be of white or if two are white, then one will be of black) Hence, option (a) is the correct answer.
• Question 5 of 5 5. Question Tickets numbered 1 to 20 are mixed up and then a ticket is drawn at random. What is the probability that the ticket drawn has a number which is a multiple of 4 or 6? (a) 7/20 (b) 9/20 (c) 11/20 (d) 13/20 Correct Solution: Option (a) Explanation: Here, S = {1, 2, 3, 4, …. , 19, 20}. Let E = event of getting a multiple of 4 or 6 = {4, 6, 8, 12, 16, 18, 20}. P(E) = n(E)/n(S) = 7/20. Hence Option (a) is correct answer Incorrect Solution: Option (a) Explanation: Here, S = {1, 2, 3, 4, …. , 19, 20}. Let E = event of getting a multiple of 4 or 6 = {4, 6, 8, 12, 16, 18, 20}. P(E) = n(E)/n(S) = 7/20. Hence Option (a) is correct answer
#### 5. Question
Tickets numbered 1 to 20 are mixed up and then a ticket is drawn at random. What is the probability that the ticket drawn has a number which is a multiple of 4 or 6?
Solution: Option (a)
Explanation:
Here, S = {1, 2, 3, 4, …. , 19, 20}.
Let E = event of getting a multiple of 4 or 6 = {4, 6, 8, 12, 16, 18, 20}.
P(E) = n(E)/n(S) = 7/20. Hence Option (a) is correct answer
Solution: Option (a)
Explanation:
Here, S = {1, 2, 3, 4, …. , 19, 20}.
Let E = event of getting a multiple of 4 or 6 = {4, 6, 8, 12, 16, 18, 20}.
P(E) = n(E)/n(S) = 7/20. Hence Option (a) is correct answer
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