[Mission 2023] Insta–DART (Daily Aptitude and Reasoning Test) 20 March 2023

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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).

Wish you all the best ! 🙂

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Question 1 of 5

1. Question

A basket contains 6 black balls and 5 white balls. In how many ways 5 balls can be selected randomly, so that at least 2 black balls must be included in every pick?

Correct

Solution: D

When question is about arrangement it is combination.

Different possibilities of selecting 5 balls with at least 2 black balls are included in it are: (2 black balls AND 3 white balls) OR

(3 black balls AND 2 white balls) OR

(4 black balls AND 1 white ball) OR

(all 5 black balls).

Thus, it can be written through formula as (6c2*5c3) + (6c3*5c2) + (6c4*5c1) + 6c5. [For AND we should use multiplication and for OR we should use addition

Standard formula for nCr = n!/( r! * (n-r)!).

Therefore, solving the above equation we get the answer as 1526.

Hence, option (d) is correct.

Incorrect

Solution: D

When question is about arrangement it is combination.

Different possibilities of selecting 5 balls with at least 2 black balls are included in it are: (2 black balls AND 3 white balls) OR

(3 black balls AND 2 white balls) OR

(4 black balls AND 1 white ball) OR

(all 5 black balls).

Thus, it can be written through formula as (6c2*5c3) + (6c3*5c2) + (6c4*5c1) + 6c5. [For AND we should use multiplication and for OR we should use addition

Standard formula for nCr = n!/( r! * (n-r)!).

Therefore, solving the above equation we get the answer as 1526.

Hence, option (d) is correct.

Question 2 of 5

2. Question

15 men take 20 days to complete a work working 8 hours a day. How many number of hours a day should 20 men take to complete the work in 12 days?

Correct

Solution: B

Formula: If M1 persons can do a piece of work in D1 days for h1 hours and M2 persons can do the same piece of work in D2 days for h2 hours,

then: M1D1h1 = M2D2h2.

In the above given problem M1=15 men, D1=20 days, h1=8 hours, M2=20 men, D2= 12 days.

Substituting the above values in the formula, we get

15 * 20 * 8 = 20 * 12 * h2

2400 = 240 * h2

h2 = 10 hours

Therefore, 10 hrs a day should 20 men take to complete the same work in 12 days.

Hence, option (b) is correct.

Incorrect

Solution: B

Formula: If M1 persons can do a piece of work in D1 days for h1 hours and M2 persons can do the same piece of work in D2 days for h2 hours,

then: M1D1h1 = M2D2h2.

In the above given problem M1=15 men, D1=20 days, h1=8 hours, M2=20 men, D2= 12 days.

Substituting the above values in the formula, we get

15 * 20 * 8 = 20 * 12 * h2

2400 = 240 * h2

h2 = 10 hours

Therefore, 10 hrs a day should 20 men take to complete the same work in 12 days.

Hence, option (b) is correct.

Question 3 of 5

3. Question

A card is drawn from a pack of 52 cards. The card is drawn at random, find the probability that it is neither club nor queen?

Correct

Solution: D

Probability of getting club = 13/52.

Probability of getting queen = 4/52

Probability of getting a club card which is queen = 1/52.

Therefore, the probability that it is neither club nor queen

= 1- (13/52 + 4/52 – 1/52)

= 1- 16/52 = 36/52 = 9/13.

Hence, option (d) is correct.

Incorrect

Solution: D

Probability of getting club = 13/52.

Probability of getting queen = 4/52

Probability of getting a club card which is queen = 1/52.

Therefore, the probability that it is neither club nor queen

= 1- (13/52 + 4/52 – 1/52)

= 1- 16/52 = 36/52 = 9/13.

Hence, option (d) is correct.

Question 4 of 5

4. Question

Direction for the following 5 (five) items: Consider the given information and answer the five items that follow:

Study the table carefully to answer the questions that follow: Candidates who appeared and passed in the test from four schools in six different years

What was the total number of failed candidates from school C in the year 2008 and the number of candidates who appeared in the exam from school D in the year 2006?

Correct

Solution: B

Total failed in school C in 2008 = 354 – 258 = 96

Total appeared in school D in 2006 = 235.

Therefore, total = 235 + 96 = 331 Hence, option (b) is correct.

Incorrect

Solution: B

Total failed in school C in 2008 = 354 – 258 = 96

Total appeared in school D in 2006 = 235.

Therefore, total = 235 + 96 = 331 Hence, option (b) is correct.

Question 5 of 5

5. Question

Direction for the following 5 (five) items: Consider the given information and answer the five items that follow:

Study the table carefully to answer the questions that follow: Candidates who appeared and passed in the test from four schools in six different years

In which year was the difference between the number of candidates who appeared and passed in the exam from school B is the second lowest?

Correct

Solution: B

The difference between the number of candidates who appeared and passed in the exam from school B is lowest in 2004 with the difference of 91.

The second lowest difference is year 2005 with the difference of 110.

Hence, the option (b) is correct.

Incorrect

Solution: B

The difference between the number of candidates who appeared and passed in the exam from school B is lowest in 2004 with the difference of 91.

The second lowest difference is year 2005 with the difference of 110.