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Question 1 of 10
1. Question
Equal quantities of three mixtures of milk and water are mixed in the ratio of 1:2, 2:3, and 3:4. The ratio of milk and water in the final mixture is:
Correct
Solution: b
Justification:
Since all the mixtures are of equal quantity, we have to find the LCM of (1+2), (2+3) and (3+4).
LCM of 3, 5 and 7 = 105.
Let the quantity of each mixture be 105x litres.
Quantities of milk and water in the first mixture are (1/3 of 105x litres = 35x litres) and (2/3 of 105x litres = 70x litres) respectively.
Quantities of milk and water in the second mixture are (2/5 of 105x litres = 42x litres) and (2/3 of 105x litres = 63x litres) respectively.
Quantities of milk and water in the third mixture are (3/7 of 105x litres = 45x litres) and (4/7 of 105x litres = 60x litres) respectively.
The ratio of milk and water in the mixture = (35x+42x+45x) : (70x+63x+60x)
= 122x : 193x
= 122 : 193
Incorrect
Solution: b
Justification:
Since all the mixtures are of equal quantity, we have to find the LCM of (1+2), (2+3) and (3+4).
LCM of 3, 5 and 7 = 105.
Let the quantity of each mixture be 105x litres.
Quantities of milk and water in the first mixture are (1/3 of 105x litres = 35x litres) and (2/3 of 105x litres = 70x litres) respectively.
Quantities of milk and water in the second mixture are (2/5 of 105x litres = 42x litres) and (2/3 of 105x litres = 63x litres) respectively.
Quantities of milk and water in the third mixture are (3/7 of 105x litres = 45x litres) and (4/7 of 105x litres = 60x litres) respectively.
The ratio of milk and water in the mixture = (35x+42x+45x) : (70x+63x+60x)
= 122x : 193x
= 122 : 193

Question 2 of 10
2. Question
Tap A can fill the empty tank in 12 hours, but due to a leak in the bottom it is filled in 15 hours. If the tank is full and then tap A is closed then in how many hours the leak can empty it?
Correct
Solution: d) 60
Justification:
Tap A can fill 1/12 of the tank in 1 hour.
Along with the leak, Tap A can fill 1/15 of the tank in 1 hour.
So the fraction of the tank that can leak in 1 hour = (1/12) – (1/15) = (1/60)
Hence, the time taken by the leak to empty the full tank = 60 hours.
Incorrect
Solution: d) 60
Justification:
Tap A can fill 1/12 of the tank in 1 hour.
Along with the leak, Tap A can fill 1/15 of the tank in 1 hour.
So the fraction of the tank that can leak in 1 hour = (1/12) – (1/15) = (1/60)
Hence, the time taken by the leak to empty the full tank = 60 hours.

Question 3 of 10
3. Question
The average temperature on Monday, Tuesday and Wednesday is 38˚C. The average temperature on Tuesday, Wednesday and Thursday is 43˚C. If the average temperature on Monday and Thursday is 18.5˚C. The average temperature on Monday is:
Correct
Solution: a) 11˚C
Justification:
Total of the temperatures on Monday, Tuesday and Wednesday = 38 × 3 = 114˚C …….Eq1
Total of the temperatures on Tuesday, Wednesday and Thursday = 43 × 3 = 129˚C …….Eq2
Subtracting Eq1 from Eq2, we get
Temperature on Thursday – Temperature on Monday = T – M = 129 – 114 = 15˚C ……Eq3
But average of the temperatures on Monday and Thursday = (M + T)/2 = 18.5˚C
M + T = 37˚C …….Eq4
Solving Eq3 and Eq4, we get
M = 11˚C
T = 26˚C
The average temperature on Monday is 11˚C.
Incorrect
Solution: a) 11˚C
Justification:
Total of the temperatures on Monday, Tuesday and Wednesday = 38 × 3 = 114˚C …….Eq1
Total of the temperatures on Tuesday, Wednesday and Thursday = 43 × 3 = 129˚C …….Eq2
Subtracting Eq1 from Eq2, we get
Temperature on Thursday – Temperature on Monday = T – M = 129 – 114 = 15˚C ……Eq3
But average of the temperatures on Monday and Thursday = (M + T)/2 = 18.5˚C
M + T = 37˚C …….Eq4
Solving Eq3 and Eq4, we get
M = 11˚C
T = 26˚C
The average temperature on Monday is 11˚C.

Question 4 of 10
4. Question
Eighteen years ago, a father was thrice as old as his son. Now, the son’s age is half of that of father. Then the sum of the present ages (in years) of father and son is:
Correct
Solution: c) 108
Justification:
Let the present age of father be ‘f’ years and son be ‘s’ years.
So, f = 2s
18 years ago, father’s age was (f – 18) and son’s age was (s – 18).
So, f – 18 = 3(s – 18)
Substituting the value of f in above equation, we get
2s – 18 = 3 (s – 18)
2s – 18 = 3s – 54
s = 36
Hence, f = 2s = 72 years
Hence, the sum of the present ages = 72 + 36 = 108 years.
Incorrect
Solution: c) 108
Justification:
Let the present age of father be ‘f’ years and son be ‘s’ years.
So, f = 2s
18 years ago, father’s age was (f – 18) and son’s age was (s – 18).
So, f – 18 = 3(s – 18)
Substituting the value of f in above equation, we get
2s – 18 = 3 (s – 18)
2s – 18 = 3s – 54
s = 36
Hence, f = 2s = 72 years
Hence, the sum of the present ages = 72 + 36 = 108 years.

Question 5 of 10
5. Question
A starts business with Rs.3500 and after 5 months, B joins with A as his partner. After a year, the profit is divided in the ratio 2:3. What is B’s contribution in the capital?
Correct
Solution: d) 9000rs
Justification:
The ratio shares in the profits = 2 : 3
(A’s investment × A’s period of investment) : (B’s investment × B’s period of investment)= 2:3
(3500 × 12) : (B’s investment × 7) = 2 : 3
42000 : (B’s investment × 7) = 2 : 3
(B’s investment × 7) × 2 = 126000
B’s investment = 126000/14
B’s investment = Rs.9000
Incorrect
Solution: d) 9000rs
Justification:
The ratio shares in the profits = 2 : 3
(A’s investment × A’s period of investment) : (B’s investment × B’s period of investment)= 2:3
(3500 × 12) : (B’s investment × 7) = 2 : 3
42000 : (B’s investment × 7) = 2 : 3
(B’s investment × 7) × 2 = 126000
B’s investment = 126000/14
B’s investment = Rs.9000

Question 6 of 10
6. Question
A sum of money is invested in Fund A at 7.5% per annum for SI. After 4 years an amount is received and that amount is invested in Fund B at 20% per annum for CI. If the interest received at the end of 2 years is 1430, find the initial invested amount?
Correct
Correct Answer: B) 2500rs
Explanation:
P[(1 + 20/100)^2 – 1] = 1430
P = 130 * 25 = Amount of Fund A
P[RT/100 + 1] = 130 * 25
P[30/100 + 1] = 130 * 25
P = 2500Incorrect
Correct Answer: B) 2500rs
Explanation:
P[(1 + 20/100)^2 – 1] = 1430
P = 130 * 25 = Amount of Fund A
P[RT/100 + 1] = 130 * 25
P[30/100 + 1] = 130 * 25
P = 2500 
Question 7 of 10
7. Question
A vessel contains milk and water in the ratio of 4:3. If 14 litres of the mixture is drawn and filled with water, the ratio changes to 3:4. How much milk was there in the vessel initially?
Correct
Correct Answer: B) 32
Explanation:
milk = 4x and water = 3x
milk = 4x – 14*4/7 and water = 3x – 14*3/7 + 14
4x – 8: 3x + 8 = 3:4
X = 8, so milk = 8*4 = 32 litres
Incorrect
Correct Answer: B) 32
Explanation:
milk = 4x and water = 3x
milk = 4x – 14*4/7 and water = 3x – 14*3/7 + 14
4x – 8: 3x + 8 = 3:4
X = 8, so milk = 8*4 = 32 litres

Question 8 of 10
8. Question
A bag contains 25p coins, 50p coins and 1 rupee coins whose values are in the ratio of 8:4:2. The total values of coins are 840. Then find the total amount in rupees.
Correct
Correct Answer: D) 280
Explanation:
Value is given in the ratio 8:4:2.
(8x/0.25) + (4x/0.5) + (2x/1) = 840.
X = 20.
Total amount = 14*20 = 280
Incorrect
Correct Answer: D) 280
Explanation:
Value is given in the ratio 8:4:2.
(8x/0.25) + (4x/0.5) + (2x/1) = 840.
X = 20.
Total amount = 14*20 = 280

Question 9 of 10
9. Question
Vikas lends Rs 30,000 to two of his friends for 1 year. He gives Rs 15,000 to the first friend Praveen at 6% p.a. simple interest. He wants to make a profit of 20% on the whole. The simple interest rate at which he should lend the remaining sum of money to the second friend, Rahul is
Correct
Correct Answer: D) 34%
Explanation:
S.I. on Rs 15000 = (15000×6×1)/100 = Rs. 900
Profit to made on Rs 30000 = 30000×20/100=Rs 6000
S.I. on Rs.15000 = 6000900 = Rs.5100
Rate=(S.I.* 100)/(P * T) = (5100×100)/15000 = 34% per annumIncorrect
Correct Answer: D) 34%
Explanation:
S.I. on Rs 15000 = (15000×6×1)/100 = Rs. 900
Profit to made on Rs 30000 = 30000×20/100=Rs 6000
S.I. on Rs.15000 = 6000900 = Rs.5100
Rate=(S.I.* 100)/(P * T) = (5100×100)/15000 = 34% per annum 
Question 10 of 10
10. Question
The average expenditure of Sharma for the January to June is Rs. 4200 and he spent Rs. 1200 in January and Rs.1500 in July. The average expenditure for the months of February to July is:
Correct
Correct Answer: C) 4250
Explanation:
Total Expenditure (Jan – June) = 4200 * 6 = 25200
Total Expenditure (Feb – June) = 25200 – 1200 = 24000
Total Expenditure (Feb – July) = 24000 + 1500 = 25500/6 = 4250
Incorrect
Correct Answer: C) 4250
Explanation:
Total Expenditure (Jan – June) = 4200 * 6 = 25200
Total Expenditure (Feb – June) = 25200 – 1200 = 24000
Total Expenditure (Feb – July) = 24000 + 1500 = 25500/6 = 4250