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
In a race of 600 m, A can beat B by 30 m and in a race of 500 m, B can beat C by 25 m. By how many metres will A beat C in a race of 400 m?
Correct
Answer : a
If A runs 600 m, then B runs 570 m
If A runs 400 m, B then runs
[(570/600)*400]m = 380 m
When B runs 500 m, then C runs 475 m
When B runs 380 m, then C runs
[(475/500) * 380] = 361 m
∴A beats C by (400 – 361) = 39 m
Incorrect
Answer : a
If A runs 600 m, then B runs 570 m
If A runs 400 m, B then runs
[(570/600)*400]m = 380 m
When B runs 500 m, then C runs 475 m
When B runs 380 m, then C runs
[(475/500) * 380] = 361 m
∴A beats C by (400 – 361) = 39 m
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Question 2 of 5
2. Question
A motor car does a journey in 17.5 hours, covering the first half at 30km/h and the second half at 40 km/h. Find the distance of the journey.
Correct
Ans-b
If the car does half the journey @30 kmph and the other half at 40kmph its average speed can be estimated using weighted averages.
Since ,the distance travelled in each part of the journey is equal ,the ratio of time for which the car would travel would be inverse to the ratio of speeds .Since ,the speed ratio 3:4,the time ratio for the two halves of the journey would be 4:3.the average speed of the carwould be-
(30*4+40 *3)/7=240/7kmph.
It is further known that the car traveled for 17.5 hours(which is also equal to 35/2 hours)
Thus ,total distance=average speed *total time
=(240*35)/(2*7)
=120*5
=600km.
Incorrect
Ans-b
If the car does half the journey @30 kmph and the other half at 40kmph its average speed can be estimated using weighted averages.
Since ,the distance travelled in each part of the journey is equal ,the ratio of time for which the car would travel would be inverse to the ratio of speeds .Since ,the speed ratio 3:4,the time ratio for the two halves of the journey would be 4:3.the average speed of the carwould be-
(30*4+40 *3)/7=240/7kmph.
It is further known that the car traveled for 17.5 hours(which is also equal to 35/2 hours)
Thus ,total distance=average speed *total time
=(240*35)/(2*7)
=120*5
=600km.
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Question 3 of 5
3. Question
From a group of 12 dancers, five have to be taken for a stage show. Among them Radha and Mohan decide either both of them would join or none of them would join. In how many ways can the 5 dancers be chosen?
Correct
Ans-d
We have to select 5 out of 12.
If Radha and Mohan join- then we have to select only 5 – 2 = 3 dancers out of 12 – 2 = 10 which can be done in ‘°C3 = 120 ways.If Radha and Mohan do not join, then we have to
select 5 out of 12 – 2 = 10-> 10C5, = 252 ways.
Total number of ways = 120 + 252 = 372.
Hence, Option (d) is correct.
Incorrect
Ans-d
We have to select 5 out of 12.
If Radha and Mohan join- then we have to select only 5 – 2 = 3 dancers out of 12 – 2 = 10 which can be done in ‘°C3 = 120 ways.If Radha and Mohan do not join, then we have to
select 5 out of 12 – 2 = 10-> 10C5, = 252 ways.
Total number of ways = 120 + 252 = 372.
Hence, Option (d) is correct.
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Question 4 of 5
4. Question
Find the number of 6-digit numbers that can be found using the digits 1, 2, 3, 4, 5, 6 once such that the 6-digit number is divisible by its unit digit. (The unit digit is not 1.)
Correct
Ans-d
The unit digit can either be 2, 3, 4, 5 or 6. When the unit digit is 2, the number would be even and hence will be divisible by 2. Hence all numbers with unit digit 2 will be included which is equal to 5! Or 120.When the unit digit is 3, then in every case the sum of the digits of the number would be 21 ,which is a multiple of 3 ,hence all numbers with unit digit 3 will be divisible by 3 and hence will be included ,total no of such numbers is !5 or 120.Similarly for unit digit is 4 ,then the number would be divisible by 4 only if the tens digit is 2 or 6,total no of such numbers is 2*!4 or 48. Hence total no of required nos is (4*120)+48=528. Hence, Option (d) is correct.
Incorrect
Ans-d
The unit digit can either be 2, 3, 4, 5 or 6. When the unit digit is 2, the number would be even and hence will be divisible by 2. Hence all numbers with unit digit 2 will be included which is equal to 5! Or 120.When the unit digit is 3, then in every case the sum of the digits of the number would be 21 ,which is a multiple of 3 ,hence all numbers with unit digit 3 will be divisible by 3 and hence will be included ,total no of such numbers is !5 or 120.Similarly for unit digit is 4 ,then the number would be divisible by 4 only if the tens digit is 2 or 6,total no of such numbers is 2*!4 or 48. Hence total no of required nos is (4*120)+48=528. Hence, Option (d) is correct.
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Question 5 of 5
5. Question
1 L of water is added to 5 L of alcohol and water solution containing 40% alcohol strength. The strength of alcohol in the new solution will be
Correct
Answer : b
Quantity of alcohol in 5 L of solution
= (40/100 )*5 = 2L
Quantity of alcohol in 6 L of solution = 2 L
∴ Strength of alcohol in new solution
=[(2/6)*100)% = 33(1/3)%
Incorrect
Answer : b
Quantity of alcohol in 5 L of solution
= (40/100 )*5 = 2L
Quantity of alcohol in 6 L of solution = 2 L
∴ Strength of alcohol in new solution
=[(2/6)*100)% = 33(1/3)%
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