[Mission 2024] Insta–DART (Daily Aptitude and Reasoning Test) 18- 19 March 2024
Kartavya Desk Staff
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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 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: a) 2 : 3 b) 122 : 193 c) 61 : 97 d) 137 : 178 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
#### 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:
• b) 122 : 193
• c) 61 : 97
• d) 137 : 178
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
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? a) 40 b) 45 c) 48 d) 60 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.
#### 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?
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.
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: a) 11˚C b) 21˚C c) 26˚C d) 27˚C 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. Gulmarg is located at a distance of 31 km (19 mi) from Baramulla and 49 km (30 mi) from Srinagar. Kargil, Nubra Valley and Hemis are part of Ladakh Refer: https://www.insightsonindia.com/2024/03/23/ladakhs-statehood-demand/
#### 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:
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
The average temperature on Monday is 11˚C.
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
The average temperature on Monday is 11˚C.
• Gulmarg is located at a distance of 31 km (19 mi) from Baramulla and 49 km (30 mi) from Srinagar.
• Kargil, Nubra Valley and Hemis are part of Ladakh
Refer: https://www.insightsonindia.com/2024/03/23/ladakhs-statehood-demand/
• 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: a) 54 b) 72 c) 108 d) 144 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.
#### 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:
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
Hence, f = 2s = 72 years
Hence, the sum of the present ages = 72 + 36 = 108 years.
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
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? a) Rs.7500 b) Rs.8000 c) Rs.8500 d) Rs.9000 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
#### 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?
• a) Rs.7500
• b) Rs.8000
• c) Rs.8500
• d) Rs.9000
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
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 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 (a) 28% (b) 36% (c) 32% (d) 34% 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 = 6000-900 = Rs.5100 Rate=(S.I. 100)/(P T) = (5100×100)/15000 = 34% per annum Incorrect 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 = 6000-900 = Rs.5100 Rate=(S.I. 100)/(P T) = (5100×100)/15000 = 34% per annum
#### 6. 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 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 = 6000-900 = Rs.5100 Rate=(S.I. 100)/(P T) = (5100×100)/15000 = 34% per annum
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 = 6000-900 = Rs.5100 Rate=(S.I. 100)/(P T) = (5100×100)/15000 = 34% per annum
• Question 7 of 10 7. 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? (a) 1500 rs (b) 2500 rs (c) 3500 rs (d) 4500 rs 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 = 2500 Incorrect 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
#### 7. 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?
• (a) 1500 rs
• (b) 2500 rs
• (c) 3500 rs
• (d) 4500 rs
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
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 8 of 10 8. 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? (a) 24 (b) 32 (c) 40 (d) 48 Correct Correct Answer: B) 32 Explanation: milk = 4x and water = 3x milk = 4x – 144/7 and water = 3x – 143/7 + 14 4x – 8: 3x + 8 = 3:4 X = 8, so milk = 84 = 32 litres Incorrect Correct Answer: B) 32 Explanation: milk = 4x and water = 3x milk = 4x – 144/7 and water = 3x – 143/7 + 14 4x – 8: 3x + 8 = 3:4 X = 8, so milk = 84 = 32 litres
#### 8. 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 Answer: B) 32
Explanation:
milk = 4x and water = 3x
milk = 4x – 144/7 and water = 3x – 143/7 + 14
4x – 8: 3x + 8 = 3:4
X = 8, so milk = 8*4 = 32 litres
Correct Answer: B) 32
Explanation:
milk = 4x and water = 3x
milk = 4x – 144/7 and water = 3x – 143/7 + 14
4x – 8: 3x + 8 = 3:4
X = 8, so milk = 8*4 = 32 litres
• Question 9 of 10 9. 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. (a) 220 (b) 240 (c) 260 (d) 280 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 = 1420 = 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 = 1420 = 280
#### 9. 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 Answer: D) 280
Explanation:
Value is given in the ratio 8:4:2.
(8x/0.25) + (4x/0.5) + (2x/1) = 840.
Total amount = 14*20 = 280
Correct Answer: D) 280
Explanation:
Value is given in the ratio 8:4:2.
(8x/0.25) + (4x/0.5) + (2x/1) = 840.
Total amount = 14*20 = 280
• 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: (a) 2750 (b) 3250 (c) 4250 (d) 4500 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
#### 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 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
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
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