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
Three taps A, B and C together can fill an empty cistern in 10 min. The tap A alone can fill it in 30 min and the tap B alone can fill it in 40 min. How long will the tap C alone take to fill it?
Correct
Answer : b
Part filled by (A+B+C) in min = (1/10)
Part filled by A in 1 min = (1/30)
Part filled by B in 1 min
Part filled by (A+B) in 1 min=(1/30) +(1/40)= (4+3)/120 = (7/120)
∴Part filled by C in 1 min= (1/10)-(7/120) = (12-7)/120 = (5/120) = 1/24
∴Tap C will fill the cistern in 24 min.
Incorrect
Answer : b
Part filled by (A+B+C) in min = (1/10)
Part filled by A in 1 min = (1/30)
Part filled by B in 1 min
Part filled by (A+B) in 1 min=(1/30) +(1/40)= (4+3)/120 = (7/120)
∴Part filled by C in 1 min= (1/10)-(7/120) = (12-7)/120 = (5/120) = 1/24
∴Tap C will fill the cistern in 24 min.
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Question 2 of 5
2. Question
Two pipes A and B can fill a tank in 1 h and 75 min, respectively. There is also an outlet C. If all the three pipes are opened together. The tank is full in 50 min. How much time will be taken by C to empty the full tank?
Correct
Answer : a
Work done by C in 1 min=[(1/60)+(1/75)-(1/50)]
=[(5+4-6)/300] = (3/300)=(1/100)
Hence, C can empty the full tank in
100 min.
Incorrect
Answer : a
Work done by C in 1 min=[(1/60)+(1/75)-(1/50)]
=[(5+4-6)/300] = (3/300)=(1/100)
Hence, C can empty the full tank in
100 min.
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Question 3 of 5
3. Question
A tank has a leak which would empty it in 8 h. A tap is turned on which admits 3L a min into the tank and it is now emptied in 12 h. How many litres does the tank hold?
Correct
Answer : a
Work done by the inlet in 1 h = [(1/8)-(1/12)]=(1/24)
Work done by the inlet in 1 min = (1/24) ´ (1/60) = (1/1440)
∴Volume of (1/(1440) part = 3 L
∴Volume of the whole = 3 * 1440 = 4320 L
Incorrect
Answer : a
Work done by the inlet in 1 h = [(1/8)-(1/12)]=(1/24)
Work done by the inlet in 1 min = (1/24) ´ (1/60) = (1/1440)
∴Volume of (1/(1440) part = 3 L
∴Volume of the whole = 3 * 1440 = 4320 L
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Question 4 of 5
4. Question
A, B and C are three pipes connected to a tank. A and B together fill the tank in 6h, B and C together fill the tank in 10 h and A and C together fill the tank in 12 h. In how much time A, B and C fill up the tank together?
Correct
Answer : d
Part filled by (A + B) in h = 1/6
Part filled by (B + C ) in 1 h = 1/10
Part filled by (A + C ) in 1 h = 1/12
Part filled by 2(A+B+C) in 1 hr
=1/6+1/10+1/12
=10/60+6/60+5/60
=21/60
∴Part filled by (A+B+C) in 1 h=(7/(2 * 20) = 7/40
∴Required time = (40/7) = 5 (5/7) h
Incorrect
Answer : d
Part filled by (A + B) in h = 1/6
Part filled by (B + C ) in 1 h = 1/10
Part filled by (A + C ) in 1 h = 1/12
Part filled by 2(A+B+C) in 1 hr
=1/6+1/10+1/12
=10/60+6/60+5/60
=21/60
∴Part filled by (A+B+C) in 1 h=(7/(2 * 20) = 7/40
∴Required time = (40/7) = 5 (5/7) h
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Question 5 of 5
5. Question
Two pipes P and Q can fill a cistern in 12 and 15 min, respectively. If both are opened together and at the end of 3 min, the first is closed. How much longer will the cistern take to fill?
Correct
Answer : a
Part filled by pipe P in 1 min = (1/12)
Part filled by pipe Q in 1 min=1/15
Part filled by pipe P in 1 min = (1/12) + (1/15) = (5+4)/60 = (9/60)
Now, part filled by both pipes in 3 min = (3 *9)/60 =(27/60) = (9/20)
\Remaining part = 1 –(9/20)=11/20
Let the remaining part is filled by pipe Q in x min.
Then, x* (1/15) = (11/20)
X = (15 * 11)/20 = (3*11)/4 = 33/4 = 8 (1/4)min
Incorrect
Answer : a
Part filled by pipe P in 1 min = (1/12)
Part filled by pipe Q in 1 min=1/15
Part filled by pipe P in 1 min = (1/12) + (1/15) = (5+4)/60 = (9/60)
Now, part filled by both pipes in 3 min = (3 *9)/60 =(27/60) = (9/20)
\Remaining part = 1 –(9/20)=11/20
Let the remaining part is filled by pipe Q in x min.
Then, x* (1/15) = (11/20)
X = (15 * 11)/20 = (3*11)/4 = 33/4 = 8 (1/4)min