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 game of billiards, A can give B 20 points in the game of 120 points and he can give C 30 points in the game of 120 points. How many points can B give C in a game of 90?
Correct
Answer : a
If A scores 120 points, then B scores 100 points and C scores 90 points.
When B scores 100 points, then C scores 90 points.
When B Scores 90 points, then
C scores [(90/100)*90] points = 81 points
B can give ,9 points in game of 90.
Incorrect
Answer : a
If A scores 120 points, then B scores 100 points and C scores 90 points.
When B scores 100 points, then C scores 90 points.
When B Scores 90 points, then
C scores [(90/100)*90] points = 81 points
B can give ,9 points in game of 90.
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Question 2 of 5
2. Question
Arun and Bhaskar start from place P at 6:00 am and 7: 30 am, respectively and run in the same direction. Arun and Bhaskar run at 8 km/h and 12 km/h, respectively. Bhaskar overtakes Arun
Correct
Answer : a
Distance between Arun and Bhasker at 7 : 30 am = 8 * [1(1/2)]=12 km
Time taken by Bhasker in covering a distance of 12 km = [ 12/(12-8)] = 3 h
∴Required time = 10 : 30 am
Incorrect
Answer : a
Distance between Arun and Bhasker at 7 : 30 am = 8 * [1(1/2)]=12 km
Time taken by Bhasker in covering a distance of 12 km = [ 12/(12-8)] = 3 h
∴Required time = 10 : 30 am
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Question 3 of 5
3. Question
A runs 7/4 times as fast as B a start of 30 m, how far must the winning post be, so that A and B reach in at the same time?
Correct
Answer : d
A is 7/4 faster than B.
Then ratio of the rates of A and B=7:4
3 m are gained in a race of 7 m
30 m are gained in a race of
(7/3)*30 = 70 m
Incorrect
Answer : d
A is 7/4 faster than B.
Then ratio of the rates of A and B=7:4
3 m are gained in a race of 7 m
30 m are gained in a race of
(7/3)*30 = 70 m
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Question 4 of 5
4. Question
A 10 km race is organised at 800 m circular race course. P and Q are the contestants of this race. If the ratio Of the speeds of P and Q is 5:4, how many times will the winner overtake the loser?
Correct
Answer : c
Speed of P : Speed of Q = 5: 4
Time taken by P to cover 5 rounds
Distance covered by Pin 5 rounds
= 5* (800/1000) = 4 km
Distance covered by Q in 4 rounds = 4 *(9800/1000) = (16/5) km
In 5 rounds, P will overtake Q every time.
It means that after covering 4 km, P will overtake one time.
After covering 10 km P will overtake
Q, (1/4) * 10 = 2 (1/2) times » 2 times
Incorrect
Answer : c
Speed of P : Speed of Q = 5: 4
Time taken by P to cover 5 rounds
Distance covered by Pin 5 rounds
= 5* (800/1000) = 4 km
Distance covered by Q in 4 rounds = 4 *(9800/1000) = (16/5) km
In 5 rounds, P will overtake Q every time.
It means that after covering 4 km, P will overtake one time.
After covering 10 km P will overtake
Q, (1/4) * 10 = 2 (1/2) times » 2 times
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Question 5 of 5
5. Question
A runs 5/3 times as fast as B. If A gives B a start of 40 m, how far must the Winning post be, so that A and B might reach it at the same time?
Correct
Answer : c
Ratio of the speeds of A and B = 5 : 3
Thus, in a race of 5m, A gains 2 m over B
2 m are gained by A in a race ot5 m.
40 m will be qained by A in a, race of [(5/2) * 40 ] m = 100 m
∴winning post is 100 m away from the Starting point.
Incorrect
Answer : c
Ratio of the speeds of A and B = 5 : 3
Thus, in a race of 5m, A gains 2 m over B
2 m are gained by A in a race ot5 m.
40 m will be qained by A in a, race of [(5/2) * 40 ] m = 100 m
∴winning post is 100 m away from the Starting point.
