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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 vessels having volumes in the ratio of 2: 3. 5 are full of a mixture of water and milk. In the first vessel ratio of water and milk is 1 : 3, in second 2 : 3 and in third vessel ,2:5 If all the three vessels were poured out in a large container, what is the resulting ratio of milk and water?
Correct
Answer : D
Explanation : Ratio of W1 M1 1 : 3
W2 M2 2 : 3
W3 M3 2 : 5
Proportion of water : : 1/4:2/ 5 :2/7
(1/4)* (5×7/5*7) : 2/5 * (4×7/4*7) : 2 /7 * (4×5 /4*5)
=35/140 : 56/140 : 40/140
Now since all these three mixtures are mixed in the ratio of 2: 3 5
Therefore new ratio = —35/140 x 2/2 : 56 x 3 /140 x 3 : 40 x5/140* 5
=70 /280 , 168/420 , 200/700
Now, the amount of water = 70 + 168 + 200 = 438
The amount of milk = (280 + 420 + 700) – 438 = 962 Ratio of milk to water = 962: 438
Incorrect
Answer : D
Explanation : Ratio of W1 M1 1 : 3
W2 M2 2 : 3
W3 M3 2 : 5
Proportion of water : : 1/4:2/ 5 :2/7
(1/4)* (5×7/5*7) : 2/5 * (4×7/4*7) : 2 /7 * (4×5 /4*5)
=35/140 : 56/140 : 40/140
Now since all these three mixtures are mixed in the ratio of 2: 3 5
Therefore new ratio = —35/140 x 2/2 : 56 x 3 /140 x 3 : 40 x5/140* 5
=70 /280 , 168/420 , 200/700
Now, the amount of water = 70 + 168 + 200 = 438
The amount of milk = (280 + 420 + 700) – 438 = 962 Ratio of milk to water = 962: 438
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Question 2 of 5
2. Question
After measuring 120 metres of a rope, it was discovered that the metre rod was 3 cm longer. The true length of the rope measured is :
Correct
Answer : D
Explanation – Actual length has to be 120m + (120 × 3) cm = 120 m + 360 cm = 123 m 60 cm.
Incorrect
Answer : D
Explanation – Actual length has to be 120m + (120 × 3) cm = 120 m + 360 cm = 123 m 60 cm.
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Question 3 of 5
3. Question
There are two clocks, both set to show the correct time at 10 a.m. One clock gains two minutes in one hour while the other gain one minute in one hour. If the clock which gains 2 minute shows the time as 22 minutes past 9 p. m., then what time does the other watch show?
Correct
Answer : C
Explanation – Difference in minute between the two clocks in one hour = 1 minute.
Number of hours = 11 hours.
In 11 hours, one of the clock gains 22 minutes and shows the time as 9:22 p.m. The other clock which gains 1 minute per hour shows the time as 9:11 p.m.
Incorrect
Answer : C
Explanation – Difference in minute between the two clocks in one hour = 1 minute.
Number of hours = 11 hours.
In 11 hours, one of the clock gains 22 minutes and shows the time as 9:22 p.m. The other clock which gains 1 minute per hour shows the time as 9:11 p.m.
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Question 4 of 5
4. Question
In a question on division with zero remainders, a candidate took 12 as divisor instead of 21. The quotient obtained by him was 35. The correct quotient is:
Correct
Answer : D
Explanation: Number = (35 × 12) = 420 Correct quotient = 420/21 = 20
Incorrect
Answer : D
Explanation: Number = (35 × 12) = 420 Correct quotient = 420/21 = 20
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Question 5 of 5
5. Question
The sum of ages of 5 children born at the intervals of 3 years each is 50 years. What is the age of the youngest child?
Correct
Answer: A
Explanation:
Let the ages of children be x, (x + 3), (x + 6), (x + 9) and (x + 12) years.
Then, x + (x + 3) + (x + 6) + (x + 9) + (x + 12) = 50
5x = 20
x = 4.
Age of the youngest child = x = 4 years
Incorrect
Answer: A
Explanation:
Let the ages of children be x, (x + 3), (x + 6), (x + 9) and (x + 12) years.
Then, x + (x + 3) + (x + 6) + (x + 9) + (x + 12) = 50
5x = 20
x = 4.
Age of the youngest child = x = 4 years
