UPSC Insta–DART (Daily Aptitude and Reasoning Test) 16 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 Find the number of ways of selecting 9 balls from 6 red balls, 5 white balls and 5 blue balls if each selection consists of 3 balls of each colour. (a) 1500 (b) 2000 (c) 2500 (d) 1200 Correct Solution: Option (b) Explanation 3 red balls out of 6 can be selected in 6C3 ways. 3 white balls out of 5 can be selected in 5C3 ways. 3 blue balls out of 5 can be selected in 5C3 ways. ∴ By multiplication principle, 9 balls can be selected in 6C3 × 5C3 × 5C3 = 6C3 × 5C2 × 5C2 [because, nCr = nCn – r] = 2000 ways. Incorrect Solution: Option (b) Explanation 3 red balls out of 6 can be selected in 6C3 ways. 3 white balls out of 5 can be selected in 5C3 ways. 3 blue balls out of 5 can be selected in 5C3 ways. ∴ By multiplication principle, 9 balls can be selected in 6C3 × 5C3 × 5C3 = 6C3 × 5C2 × 5C2 [because, nCr = nCn – r] = 2000 ways.
#### 1. Question
Find the number of ways of selecting 9 balls from 6 red balls, 5 white balls and 5 blue balls if each selection consists of 3 balls of each colour.
Solution: Option (b)
Explanation
3 red balls out of 6 can be selected in 6C3 ways.
3 white balls out of 5 can be selected in 5C3 ways.
3 blue balls out of 5 can be selected in 5C3 ways.
∴ By multiplication principle, 9 balls can be selected in
6C3 × 5C3 × 5C3 = 6C3 × 5C2 × 5C2 [because, nCr = nCn – r]
= 2000 ways.
Solution: Option (b)
Explanation
3 red balls out of 6 can be selected in 6C3 ways.
3 white balls out of 5 can be selected in 5C3 ways.
3 blue balls out of 5 can be selected in 5C3 ways.
∴ By multiplication principle, 9 balls can be selected in
6C3 × 5C3 × 5C3 = 6C3 × 5C2 × 5C2 [because, nCr = nCn – r]
= 2000 ways.
• Question 2 of 5 2. Question It is required to seat 5 men and 4 women in a row so that the women occupy the even places. How many such arrangements are possible? (a) 2800 (b) 2870 (c) 2880 (d) 2850 Correct Solution: Option (c) Explanation We know that in the row of 9 places, the second, fourth, sixth and the eighth places are the even places. Four women can be arranged in four even places in 4P4 ways. Five men can be arranged in the remaining five odd places in 5P5 ways. By the Fundamental Principle of Counting (Multiplication), the required number of seating arrangements is 4P4 × 5P5 = 4! × 5! = (4 × 3 × 2 × 1) × (5 × 4 × 3 × 2 × 1) = 24 × 120 = 2880 Incorrect Solution: Option (c) Explanation We know that in the row of 9 places, the second, fourth, sixth and the eighth places are the even places. Four women can be arranged in four even places in 4P4 ways. Five men can be arranged in the remaining five odd places in 5P5 ways. By the Fundamental Principle of Counting (Multiplication), the required number of seating arrangements is 4P4 × 5P5 = 4! × 5! = (4 × 3 × 2 × 1) × (5 × 4 × 3 × 2 × 1) = 24 × 120 = 2880
#### 2. Question
It is required to seat 5 men and 4 women in a row so that the women occupy the even places. How many such arrangements are possible?
Solution: Option (c)
Explanation
We know that in the row of 9 places, the second, fourth, sixth and the eighth places are the even places.
Four women can be arranged in four even places in 4P4 ways.
Five men can be arranged in the remaining five odd places in 5P5 ways.
By the Fundamental Principle of Counting (Multiplication), the required number of seating arrangements is
4P4 × 5P5 = 4! × 5! = (4 × 3 × 2 × 1) × (5 × 4 × 3 × 2 × 1) = 24 × 120 = 2880
Solution: Option (c)
Explanation
We know that in the row of 9 places, the second, fourth, sixth and the eighth places are the even places.
Four women can be arranged in four even places in 4P4 ways.
Five men can be arranged in the remaining five odd places in 5P5 ways.
By the Fundamental Principle of Counting (Multiplication), the required number of seating arrangements is
4P4 × 5P5 = 4! × 5! = (4 × 3 × 2 × 1) × (5 × 4 × 3 × 2 × 1) = 24 × 120 = 2880
• Question 3 of 5 3. Question The English alphabet has 5 vowels and 21 consonants. How many words with two different vowels and 2 different consonants can be formed from the alphabet? (a) 50500 (b) 50300 (c) 50450 (d) 50400 Correct Solution: Option (d) Explanation 2 vowels out of 5 can be selected in 5C2 = 10 ways 2 consonants out of 21 can be selected in 21C2 = 210 ways ∴ The number of selections of 2 vowels and 2 consonants is 10 × 210 = 2100. Now, each of these 2100 selections has 4 letters which can be arranged among themselves in 4P4 = 4 ! = 1 × 2 × 3 × 4 = 24 ways. Therefore, the required number of different words = 2100 × 24 = 50400. Incorrect Solution: Option (d) Explanation 2 vowels out of 5 can be selected in 5C2 = 10 ways 2 consonants out of 21 can be selected in 21C2 = 210 ways ∴ The number of selections of 2 vowels and 2 consonants is 10 × 210 = 2100. Now, each of these 2100 selections has 4 letters which can be arranged among themselves in 4P4 = 4 ! = 1 × 2 × 3 × 4 = 24 ways. Therefore, the required number of different words = 2100 × 24 = 50400.
#### 3. Question
The English alphabet has 5 vowels and 21 consonants. How many words with two different vowels and 2 different consonants can be formed from the alphabet?
Solution: Option (d)
Explanation
2 vowels out of 5 can be selected in 5C2 = 10 ways
2 consonants out of 21 can be selected in 21C2 = 210 ways
∴ The number of selections of 2 vowels and 2 consonants is 10 × 210 = 2100.
Now, each of these 2100 selections has 4 letters which can be arranged among themselves in
4P4 = 4 ! = 1 × 2 × 3 × 4 = 24 ways.
Therefore, the required number of different words = 2100 × 24 = 50400.
Solution: Option (d)
Explanation
2 vowels out of 5 can be selected in 5C2 = 10 ways
2 consonants out of 21 can be selected in 21C2 = 210 ways
∴ The number of selections of 2 vowels and 2 consonants is 10 × 210 = 2100.
Now, each of these 2100 selections has 4 letters which can be arranged among themselves in
4P4 = 4 ! = 1 × 2 × 3 × 4 = 24 ways.
Therefore, the required number of different words = 2100 × 24 = 50400.
• Question 4 of 5 4. Question Seven people are sitting around a round table. The probability that two distinguished persons will be next to each other is (a) 1/3 (b) 1/2 (c) 1/4 (d) 2/3 Correct Solution: Option (a) Explanation: Seven people can sit around a round table in 6! ways. The number of ways in which two distinguished persons will be next to each other = 2 (5!) Hence, the required probability = 2(5)!/6! = 1/3 Incorrect Solution: Option (a) Explanation: Seven people can sit around a round table in 6! ways. The number of ways in which two distinguished persons will be next to each other = 2 (5!) Hence, the required probability = 2(5)!/6! = 1/3
#### 4. Question
Seven people are sitting around a round table. The probability that two distinguished persons will be next to each other is
Solution: Option (a)
Explanation:
Seven people can sit around a round table in 6! ways.
The number of ways in which two distinguished persons will be next to each other = 2 (5!)
Hence, the required probability = 2(5)!/6! = 1/3
Solution: Option (a)
Explanation:
Seven people can sit around a round table in 6! ways.
The number of ways in which two distinguished persons will be next to each other = 2 (5!)
Hence, the required probability = 2(5)!/6! = 1/3
• Question 5 of 5 5. Question How many 6-digit numbers can be formed from the digits 0, 1, 3, 5, 7 and 9 which are divisible by 10 and no digit is repeated? (a) 115 (b) 120 (c) 130 (d) 135 Correct Solution: Option (b) Explanation: Numbers divisible by 10 must have ‘0’ in the unit’s place. _ _ _ _ _ 0 The remaining 5 digits can be arranged in the remaining 5 vacant places in 5P5 = 5 ! ways. ∴ The required number of 6-digit numbers = 5 ! = 120. Incorrect Solution: Option (b) Explanation: Numbers divisible by 10 must have ‘0’ in the unit’s place. _ _ _ _ _ 0 The remaining 5 digits can be arranged in the remaining 5 vacant places in 5P5 = 5 ! ways. ∴ The required number of 6-digit numbers = 5 ! = 120.
#### 5. Question
How many 6-digit numbers can be formed from the digits 0, 1, 3, 5, 7 and 9 which are divisible by 10 and no digit is repeated?
Solution: Option (b)
Explanation:
Numbers divisible by 10 must have ‘0’ in the unit’s place.
_ _ _ _ _ 0
The remaining 5 digits can be arranged in the remaining 5 vacant places in 5P5 = 5 ! ways.
∴ The required number of 6-digit numbers = 5 ! = 120.
Solution: Option (b)
Explanation:
Numbers divisible by 10 must have ‘0’ in the unit’s place.
_ _ _ _ _ 0
The remaining 5 digits can be arranged in the remaining 5 vacant places in 5P5 = 5 ! ways.
∴ The required number of 6-digit numbers = 5 ! = 120.
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