Insta–DART (Daily Aptitude and Reasoning Test) 4 January 2021

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

If x varies as y then ( x^{2} + y^{2} ) varies as

Correct

Answer : d

x varies as y, means x = ky. This does not have any relation to the variance of ( x^{2} + y^{2})

Incorrect

Answer : d

x varies as y, means x = ky. This does not have any relation to the variance of ( x^{2} + y^{2})

Question 2 of 5

2. Question

If f(x) = (x+1)/(x-1), then the ratio of x to f(y) where y = f(x) is

Correct

Answer : c

Let x = 5

Then f(x) = 6/4 = 1.5 = y

And f(y) = 2.5 / 0.5 = 5

Thus, the ratio of x : f(y) = 1:1

Note: Even if you take some other value of y, you would still get the same answer.

Incorrect

Answer : c

Let x = 5

Then f(x) = 6/4 = 1.5 = y

And f(y) = 2.5 / 0.5 = 5

Thus, the ratio of x : f(y) = 1:1

Note: Even if you take some other value of y, you would still get the same answer.

Question 3 of 5

3. Question

A contractor employs 200 men to build a bund. They finish 5/6 of the work in 10 weeks. Then rain sets in and not only does the work remain suspended for 4 weeks but also half of the work already done is washed away. After the rain, when the work is resumed, only 140 men turn up. The total time in which the contractor is able to complete the work assuming that there are no further disruptions in the schedule is

Correct

Answer : c

2000 man weeks before the rain, 5/6th of the work is completed. Hence, 2400 men weeks will be the total amount of work. However, due to the rain half the work gets washed off → This means that 1000 man weeks worth of work must have got washed of This leaves 1400 men weeks of work to be completed by the 140 men. They will take 10 more weeks and hence the total time required is 24 weeks.

Incorrect

Answer : c

2000 man weeks before the rain, 5/6th of the work is completed. Hence, 2400 men weeks will be the total amount of work. However, due to the rain half the work gets washed off → This means that 1000 man weeks worth of work must have got washed of This leaves 1400 men weeks of work to be completed by the 140 men. They will take 10 more weeks and hence the total time required is 24 weeks.

Question 4 of 5

4. Question

In a journey of 48 km performed by tonga, rickshaw and cycle in that order, the distance covered by the three ways in that order are in the ratio of 8:1:3 and charges per kilometre in that order are in the ratio of 8:1:4. If the tonga charges being 24 paise per kilometre, the total cost of the journey is

Correct

Answer : a

Total distances covered under each mode 32, 4 and 12 km respectively.

Total charges = 32 × 24 + 4 x 3 + 12 × 12 = 924 paise = Rs. 9.24

Incorrect

Answer : a

Total distances covered under each mode 32, 4 and 12 km respectively.

Total charges = 32 × 24 + 4 x 3 + 12 × 12 = 924 paise = Rs. 9.24

Question 5 of 5

5. Question

A bag contains 25 paise, 50 paise and 1 Re. coins. There are 220 coins in all and the total amount in the bag is ₹ 160. If there are thrice as many 1 Re. coins as there are 25 paise coins, then what is the number of 50 paise coins?

Correct

Answer : a

The no. of coins of 1 Re =

3x and 25p = x.

Conventionally, we can solve this using equations as

follows.

A + B + C = 220 …….. (1)

A = 3C ……….(2)

A + 0.5B + 0.25C = 160 ……(3)

We have a situation with 3 equations and 3 un

knowns. and we can solve for

A (no. of 1 Re coins),

B (no. of 50 paise coins)

C (no. of 25 paise coins)

However, a much smarter approach would be to go through the options. If we check option (a)- number of 50 paise coins = 60 we would get the number of 1 Re coins as 120 and the number of 25 paise coins as 40.

120*1+60*0.5+40*0.25=160

This fits the conditions perfectly and is hence the correct answer,

Incorrect

Answer : a

The no. of coins of 1 Re =

3x and 25p = x.

Conventionally, we can solve this using equations as

follows.

A + B + C = 220 …….. (1)

A = 3C ……….(2)

A + 0.5B + 0.25C = 160 ……(3)

We have a situation with 3 equations and 3 un

knowns. and we can solve for

A (no. of 1 Re coins),

B (no. of 50 paise coins)

C (no. of 25 paise coins)

However, a much smarter approach would be to go through the options. If we check option (a)- number of 50 paise coins = 60 we would get the number of 1 Re coins as 120 and the number of 25 paise coins as 40.

120*1+60*0.5+40*0.25=160

This fits the conditions perfectly and is hence the correct answer,

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