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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).
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Question 1 of 5
1. Question
Apurva can do a piece of work in 12 days. Apurva and Amit complete the work together and were paid Rs. 54 and Rs. 81 respectively. How many days must they have taken to complete the work together?
Correct
Answer: c
Since the ratio of money given to Apurva and Amit is 2:3, their work done would also be in the same ratio. Thus, their time ratio would be 3:2 (inverse of 2:3). So, if Apurva takes 12 days, Amith would take 8 days and total number of days required(t) would be given by the equation:
(1/12+ 1/8)=1 →1= 24/5 = 4.8 days
Incorrect
Answer: c
Since the ratio of money given to Apurva and Amit is 2:3, their work done would also be in the same ratio. Thus, their time ratio would be 3:2 (inverse of 2:3). So, if Apurva takes 12 days, Amith would take 8 days and total number of days required(t) would be given by the equation:
(1/12+ 1/8)=1 →1= 24/5 = 4.8 days
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Question 2 of 5
2. Question
Rishikant,during his journey ,travel for 20 minutes at a speed of 30km/h,another 30 min at a speed of 50km/h,and 1 hour at a speed of 50km/h and 1 hour at a speed of 60km/h.what is the average velocity.
Correct
Ans-a
The distance covered in the various phases of his travel would be
10km +25km+50km+60km.
Thus the total distance covered =145 km in 2 hours 50 minutes
145km in 2.8333 hours
51.18kmph
Incorrect
Ans-a
The distance covered in the various phases of his travel would be
10km +25km+50km+60km.
Thus the total distance covered =145 km in 2 hours 50 minutes
145km in 2.8333 hours
51.18kmph
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Question 3 of 5
3. Question
The simple interest on a certain sum of money for 2(1/2) yr at 12% per annum is Rs. 20 less than the simple interest on the same sum for 3(1/2) yr at 10% per annum. Find the sum.
Correct
Answer : d
Given, T1 = (5/2), R1 = 12%, T2 = (7/2) yr
And R2 = 10%
Let the sum be P.
Then, [(Px10x7)/(100´2) ] – [(Px12´5)/(100×2)] = 20
(7P/20) – (3P/10 ) = 20
∴P = 20 ´20 = Rs. 400
Incorrect
Answer : d
Given, T1 = (5/2), R1 = 12%, T2 = (7/2) yr
And R2 = 10%
Let the sum be P.
Then, [(Px10x7)/(100´2) ] – [(Px12´5)/(100×2)] = 20
(7P/20) – (3P/10 ) = 20
∴P = 20 ´20 = Rs. 400
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Question 4 of 5
4. Question
In a game of 160 points, A can give 10 points to B and 30 points to C. How many points B can give C in a game of 60?
Correct
Answer : d
A:B=160:150, A: C = 160 : 130
(B/C) = [(B/A) x (A/C)] = [ (150/160) x (160/130)]
= (15/13) = (60/52) = 60 : 52
∴ In a game of 60, B can give C = (60 – 52) = 8 points
Incorrect
Answer : d
A:B=160:150, A: C = 160 : 130
(B/C) = [(B/A) x (A/C)] = [ (150/160) x (160/130)]
= (15/13) = (60/52) = 60 : 52
∴ In a game of 60, B can give C = (60 – 52) = 8 points
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Question 5 of 5
5. 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 x (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)x 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 x (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)x 10 = 2 (1/2) times ≈ 2 times
