## Insta–DART (Daily Aptitude and Reasoning Test) 2020 - 21

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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

All members of a club went to Mumbai and stayed in a hotel. On the first day, 80% went for shopping and 50% went for sight-seeing, whereas 10% took rest in the hotel. Which of the following conclusion(s) can be drawn from the above data?

- 40% members went for shopping as well as sight-seeing.
- 20% members went for only shopping.

Select the correct answer using the codes given below.

CorrectSol. (a)

Let the total members be 100.

Members who take rest = 10

Number of members left = 100 – 10 = 90

Now, number of members who went for both shopping and sight-seeing = (80 + 50) – 90 = 130

-90 = 40. Hence, following Venn diagram may be drawn.

It is clear that 40% members went for shopping only, 10% members went for sight-seeing only, and 40% members went for shopping as well as sight-seeing. Hence, only conclusion 1 is correct.

IncorrectSol. (a)

Let the total members be 100.

Members who take rest = 10

Number of members left = 100 – 10 = 90

Now, number of members who went for both shopping and sight-seeing = (80 + 50) – 90 = 130

-90 = 40. Hence, following Venn diagram may be drawn.

It is clear that 40% members went for shopping only, 10% members went for sight-seeing only, and 40% members went for shopping as well as sight-seeing. Hence, only conclusion 1 is correct.

- Question 2 of 5
##### 2. Question

In a group of 15 people, 7 can read French, 8 can read English while 3 of them can read neither of these two languages. The number of people who can read exactly one language is

CorrectSol. (b)

According to above Venn diagram, those who can read exactly one language is 4 + 5 = 9.

IncorrectSol. (b)

According to above Venn diagram, those who can read exactly one language is 4 + 5 = 9.

- Question 3 of 5
##### 3. Question

Out of 130 students appearing in an examination, 62 failed in English, 52 failed in Mathematics, whereas 24 failed in both English and Mathematics. The number of students who passed finally is

CorrectSol. (a)

Number of students who failed in only Mathematics = 52 – 24 = 28

Number of students who failed in only English = 62 – 24 = 38

Here,

Total number of students who failed in one or both = 28 +24 +38 = 90

So, total number of students who passed final. = 130- 90 = 40

IncorrectSol. (a)

Number of students who failed in only Mathematics = 52 – 24 = 28

Number of students who failed in only English = 62 – 24 = 38

Here,

Total number of students who failed in one or both = 28 +24 +38 = 90

So, total number of students who passed final. = 130- 90 = 40

- Question 4 of 5
##### 4. Question

The number of persons who read magazine X only is thrice the number of persons who read magazine Y. The number of persons who read magazine Y only is thrice the number of persons who read magazine X. Then, which of the following conclusions can be drawn?

- The number of persons who read both the magazines is twice the number of persons who read only magazine X.
- The total number of persons who read either one magazine or both the magazines is twice the number of persons who read both the magazines.

Select the correct answer using the codes given below.

CorrectSol. (d)

According to the question, following Venn diagram may be drawn.

Number of people who read both the magazines = a.

Number of people who read Y magazine only = c.

Number of people who read X magazine only = b.

According to the question,

b = 3(c + a)

b = 3c + 3a

a= (b – 3c)/3 ……. (i)

and c = 3(b + a)

c = 3b + 3a

a = (c – 3b)/3 ……. (ii)

According to statement 1,

a = 2b

Which is not met by equations (i) and (ii).

According to the statement 2,

a +b + c = 2a

a = b + c

Which is not met by equations (i) and (ii).

Hence, neither of the statements is true.

IncorrectSol. (d)

According to the question, following Venn diagram may be drawn.

Number of people who read both the magazines = a.

Number of people who read Y magazine only = c.

Number of people who read X magazine only = b.

According to the question,

b = 3(c + a)

b = 3c + 3a

a= (b – 3c)/3 ……. (i)

and c = 3(b + a)

c = 3b + 3a

a = (c – 3b)/3 ……. (ii)

According to statement 1,

a = 2b

Which is not met by equations (i) and (ii).

According to the statement 2,

a +b + c = 2a

a = b + c

Which is not met by equations (i) and (ii).

Hence, neither of the statements is true.

- Question 5 of 5
##### 5. Question

There are 50 students admitted to a nursery class. Some students can speak only English and some can speak only Hindi. 10 students can speak both English and Hindi. If the number of students who can speak English is 21, then how many students can speak Hindi, how many can speak only Hindi and how many can speak only English?

CorrectSol. (d)

Let the sets E and H represent the number of students who can speak English and Hindi, respectively.

Now, n(E Ս H) = 50

n(E Ո H) = 10

n(E) = 21

We need to determine n(H), n(H Ո E^C) and n(E Ո H^C)

Since, n(E U H) = n(E) + n(H) – n(E Ո H)

= 50 = 21 + n(H) – 10

n(H)= 39

and n(H Ո E^C) = n(H) – n(E Ո H)

= 39 – 10 = 29

and n(E Ո H^C) = n(E) – n(E Ո H)

= 21 – 10 = 11

Hence, option (d) is correct.IncorrectSol. (d)

Let the sets E and H represent the number of students who can speak English and Hindi, respectively.

Now, n(E Ս H) = 50

n(E Ո H) = 10

n(E) = 21

We need to determine n(H), n(H Ո E^C) and n(E Ո H^C)

Since, n(E U H) = n(E) + n(H) – n(E Ո H)

= 50 = 21 + n(H) – 10

n(H)= 39

and n(H Ո E^C) = n(H) – n(E Ո H)

= 39 – 10 = 29

and n(E Ո H^C) = n(E) – n(E Ո H)

= 21 – 10 = 11

Hence, option (d) is correct.

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