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
A tap having diameter ‘d can empty a tank in 40 min. How long another tap having diameter ‘2d’ take to empty the same tank?
Correct
Answer : c
Area of tap a Work done by pipe.
When diameter is doubled, area will be
four times. So, it will work four times
faster.
Hence, required time taken to empty the
Tank = 40 * (1/4) = 10 min.
Incorrect
Answer : c
Area of tap a Work done by pipe.
When diameter is doubled, area will be
four times. So, it will work four times
faster.
Hence, required time taken to empty the
Tank = 40 * (1/4) = 10 min.
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Question 2 of 5
2. Question
Two pipes can fill a cistern in 14 and 16 h, respectively. The pipes are opened simultaneously and it is found that due to leakage in the bottom, it took 92 min more to fill the cistern. When the cistern is full, in what time will the leak empty it?
Correct
Answer : a
Part filled by 1st pipe in 1 h = (1/14)
Part filled by 2nde pipe in 1 h = (1/16)
Part filled by the two pipes in 1 h = [(1/14)+(1/16)]= [(8+7)/112]=(15/112)
∴Time taken by these two pipes to fill the cistern = (112/15)h = 7 h 28 min
Due to leakage, the time taken
7 h 28 min + 92 min = 9h
∴Work done by (two pipes + leak) in 1 h = (1/9)
Work done by the leak in 1 h = (1/9)-(15/112)= [(112-135)/1008]
= – (23/1008)
∴Time taken by leak to empty the full cistern = [(1008)/23] = 43 (19/23) h
Incorrect
Answer : a
Part filled by 1st pipe in 1 h = (1/14)
Part filled by 2nde pipe in 1 h = (1/16)
Part filled by the two pipes in 1 h = [(1/14)+(1/16)]= [(8+7)/112]=(15/112)
∴Time taken by these two pipes to fill the cistern = (112/15)h = 7 h 28 min
Due to leakage, the time taken
7 h 28 min + 92 min = 9h
∴Work done by (two pipes + leak) in 1 h = (1/9)
Work done by the leak in 1 h = (1/9)-(15/112)= [(112-135)/1008]
= – (23/1008)
∴Time taken by leak to empty the full cistern = [(1008)/23] = 43 (19/23) h
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Question 3 of 5
3. Question
Two pipes A and B can fill a tank in 24 and 32 min, respectively. If both the pipes are opened together, after how much time pipe B should be closed so that the tank is full in 9 min?
Correct
Answer : d
Part filled by A in 1 min = (1/24)
Part filled by B in 1 min = (1/32)
Let B is closed after x min. Then, [Part filled by (A+B) in x min] + [Part filled by A in (9-x) min] = 1
∴x [(1/24)+(1/32)] + (9-x) ´ (1/24)=1
X [(4+3)/96]+[(9-x)/24]=1
[(7x+4(9 – x))/96] = 1
7x + 4(9-x) = 96
7x + 36 – 4x = 96-36
3x=60
X = (60/3) = 20
Incorrect
Answer : d
Part filled by A in 1 min = (1/24)
Part filled by B in 1 min = (1/32)
Let B is closed after x min. Then, [Part filled by (A+B) in x min] + [Part filled by A in (9-x) min] = 1
∴x [(1/24)+(1/32)] + (9-x) ´ (1/24)=1
X [(4+3)/96]+[(9-x)/24]=1
[(7x+4(9 – x))/96] = 1
7x + 4(9-x) = 96
7x + 36 – 4x = 96-36
3x=60
X = (60/3) = 20
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Question 4 of 5
4. Question
Two pipes A and B can fill a cistern in 15 and 20 min, respectively. Both the pipes are opened together, but after 2 min, pipe A is turned off. What is the total time required to fill the tank?
Correct
Answer : b
Part filled by both in 2 min=2 * [(1/15)+(1/20)] = 2*[(4+3)/60]
= 2 * (7/60)=(7/30)
Part unfilled = 1 – (7/30) = (30 – 7)/30 = (23/30)
Now, B fills (1/20) part in 1 min
∴(23/30) part will be filled by B in [(20 *(23/30)] min or in (46/3) min.
∴Required time taken to fill the tank = [(2+(46/3)]=(52/3) min.
Incorrect
Answer : b
Part filled by both in 2 min=2 * [(1/15)+(1/20)] = 2*[(4+3)/60]
= 2 * (7/60)=(7/30)
Part unfilled = 1 – (7/30) = (30 – 7)/30 = (23/30)
Now, B fills (1/20) part in 1 min
∴(23/30) part will be filled by B in [(20 *(23/30)] min or in (46/3) min.
∴Required time taken to fill the tank = [(2+(46/3)]=(52/3) min.
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Question 5 of 5
5. Question
A pipe can fill a cistern in 12 min and another pipe can fill it in 15 min, but a third pipe can empty it in 6 min, The first two pipes are kept open for 5 min in the beginning and then the third pipe is also opened. Time taken to empty the cistern is
Correct
Answer : d
Let the number of minutes taken to empty the cistern be x min.
According to the question ,
(x/6) – [(x+5)/12] – [(x +5)/15] = 0
(x/6) – (x/12)-(5/12)-(x/15)=0
(x/6)-(x/12)-(x/15)=(5/12)+(5/15)
[(10x-5x-4x)/60]=[(25+20)/60]
(x/60) = (45/60)
X = 45 min.
Incorrect
Answer : d
Let the number of minutes taken to empty the cistern be x min.
According to the question ,
(x/6) – [(x+5)/12] – [(x +5)/15] = 0
(x/6) – (x/12)-(5/12)-(x/15)=0
(x/6)-(x/12)-(x/15)=(5/12)+(5/15)
[(10x-5x-4x)/60]=[(25+20)/60]
(x/60) = (45/60)
X = 45 min.
