[Mission 2024] Insta–DART (Daily Aptitude and Reasoning Test) 2 April 2024

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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 Two men set out at the same time to walk towards each other from points X and Y; 108 km apart. The first man walks at the speed of 6 kmph while the second one walks 4km in the first hour, 4.5km in second hour, 5 km in the third hour, and so on. Two men will meet in how many hours? a) 6 b) 8 c) 9 d) 12 Correct Solution: c) 9 Explanation: Total distance covered by both in first hour = 6 + 4 = 10km Total distance covered by both in second hour = 6 + 4.5 = 10.5 km Total distance covered by both in third hour = 6 + 5 = 11km By observing above calculations we can conclude that the distance covered by the two is in Series with difference of 0.5km. To get the sum of the series as 108km, we have to apply the concept of sum of series. i.e. 108 = (n/2) × (2a+(n-1)d) (d = 0.5, a = 10, n = hours) 108 = (n/2) × (2×10 + (n – 1) × 0.5) 108 = (n/2) × (20 + (n – 1) × 0.5) 108 × 2 = n (20 + (n-1)/2) 216 = (20n + n (n-1)/2) 216 – 20n = (n2 – n)/2 432 – 40n = n2 – n n2 + 39n – 432 = 0 n2 + 48n – 9n – 432 = 0 n (n + 48) – 9 (n + 48) = 0 (n – 9) (n + 48) = 0 n – 9 = 0 or n + 48 = 0 n = 9 Incorrect Solution: c) 9 Explanation: Total distance covered by both in first hour = 6 + 4 = 10km Total distance covered by both in second hour = 6 + 4.5 = 10.5 km Total distance covered by both in third hour = 6 + 5 = 11km By observing above calculations we can conclude that the distance covered by the two is in Series with difference of 0.5km. To get the sum of the series as 108km, we have to apply the concept of sum of series. i.e. 108 = (n/2) × (2a+(n-1)d) (d = 0.5, a = 10, n = hours) 108 = (n/2) × (2×10 + (n – 1) × 0.5) 108 = (n/2) × (20 + (n – 1) × 0.5) 108 × 2 = n (20 + (n-1)/2) 216 = (20n + n (n-1)/2) 216 – 20n = (n2 – n)/2 432 – 40n = n2 – n n2 + 39n – 432 = 0 n2 + 48n – 9n – 432 = 0 n (n + 48) – 9 (n + 48) = 0 (n – 9) (n + 48) = 0 n – 9 = 0 or n + 48 = 0 n = 9

#### 1. Question

Two men set out at the same time to walk towards each other from points X and Y; 108 km apart. The first man walks at the speed of 6 kmph while the second one walks 4km in the first hour, 4.5km in second hour, 5 km in the third hour, and so on. Two men will meet in how many hours?

Solution: c) 9

Explanation:

Total distance covered by both in first hour = 6 + 4 = 10km

Total distance covered by both in second hour = 6 + 4.5 = 10.5 km

Total distance covered by both in third hour = 6 + 5 = 11km

By observing above calculations we can conclude that the distance covered by the two is in Series with difference of 0.5km.

To get the sum of the series as 108km, we have to apply the concept of sum of series.

i.e. 108 = (n/2) × (2a+(n-1)d) (d = 0.5, a = 10, n = hours)

108 = (n/2) × (2×10 + (n – 1) × 0.5)

108 = (n/2) × (20 + (n – 1) × 0.5)

108 × 2 = n (20 + (n-1)/2)

216 = (20n + n (n-1)/2)

216 – 20n = (n2 – n)/2

432 – 40n = n2 – n

n2 + 39n – 432 = 0

n2 + 48n – 9n – 432 = 0

n (n + 48) – 9 (n + 48) = 0

(n – 9) (n + 48) = 0

n – 9 = 0 or n + 48 = 0

Solution: c) 9

Explanation:

Total distance covered by both in first hour = 6 + 4 = 10km

Total distance covered by both in second hour = 6 + 4.5 = 10.5 km

Total distance covered by both in third hour = 6 + 5 = 11km

By observing above calculations we can conclude that the distance covered by the two is in Series with difference of 0.5km.

To get the sum of the series as 108km, we have to apply the concept of sum of series.

i.e. 108 = (n/2) × (2a+(n-1)d) (d = 0.5, a = 10, n = hours)

108 = (n/2) × (2×10 + (n – 1) × 0.5)

108 = (n/2) × (20 + (n – 1) × 0.5)

108 × 2 = n (20 + (n-1)/2)

216 = (20n + n (n-1)/2)

216 – 20n = (n2 – n)/2

432 – 40n = n2 – n

n2 + 39n – 432 = 0

n2 + 48n – 9n – 432 = 0

n (n + 48) – 9 (n + 48) = 0

(n – 9) (n + 48) = 0

n – 9 = 0 or n + 48 = 0

• Question 2 of 5 2. Question Ajay decides to sell milk at cost price. In what ratio should he mix water and milk so that he can earn 20% profit? (a) 1:9 (b) 1:11 (c) 1:5 (d) 1:10 Correct Solution: C) 1:5 Let cost price of pure milk = Rs 1 per litre. Now he adds some water in this milk. Then selling price of mixture of milk and water is also = Rs 1 per litre Suppose he had only 1 litre milk costing Rs 1. To earn profit of 20 % he has to earn Rs 1.2 As mixture also costs Rs 1 per litre means to earn Rs 1.2 he has sold total 1.2 litre of mixture . That means he added 0.2 litre of milk in 1 litre. Thus, ratio is 0.2: 1 = 1:5 . Hence, option (c) is correct. Incorrect Solution: C) 1:5 Let cost price of pure milk = Rs 1 per litre. Now he adds some water in this milk. Then selling price of mixture of milk and water is also = Rs 1 per litre Suppose he had only 1 litre milk costing Rs 1. To earn profit of 20 % he has to earn Rs 1.2 As mixture also costs Rs 1 per litre means to earn Rs 1.2 he has sold total 1.2 litre of mixture . That means he added 0.2 litre of milk in 1 litre. Thus, ratio is 0.2: 1 = 1:5 . Hence, option (c) is correct.

#### 2. Question

Ajay decides to sell milk at cost price. In what ratio should he mix water and milk so that he can earn 20% profit?

Solution: C) 1:5

Let cost price of pure milk = Rs 1 per litre. Now he adds some water in this milk. Then selling price of mixture of milk and water is also = Rs 1 per litre Suppose he had only 1 litre milk costing Rs 1. To earn profit of 20 % he has to earn Rs 1.2 As mixture also costs Rs 1 per litre means to earn Rs 1.2 he has sold total 1.2 litre of mixture . That means he added 0.2 litre of milk in 1 litre. Thus, ratio is 0.2: 1 = 1:5 .

Hence, option (c) is correct.

Solution: C) 1:5

Hence, option (c) is correct.

• Question 3 of 5 3. Question Three bells chime at an interval of 18, 24 and 32 minutes respectively. At a certain time they begin to chime together. What length of time will elapse before they chime together again? (a) 2 hours 24 minutes (b) 4 hours 48 minutes (c) 1 hour 36 minutes (d) 5 hours Correct Solution: B) 4 hours 48 minutes L.C.M of 18, 24 and 32 = 288 Hence, they would chime after every 288 min. or 4 hrs 48 min. Therefore, option (b) is correct. Incorrect Solution: B) 4 hours 48 minutes L.C.M of 18, 24 and 32 = 288 Hence, they would chime after every 288 min. or 4 hrs 48 min. Therefore, option (b) is correct.

#### 3. Question

Three bells chime at an interval of 18, 24 and 32 minutes respectively. At a certain time they begin to chime together. What length of time will elapse before they chime together again?

• (a) 2 hours 24 minutes

• (b) 4 hours 48 minutes

• (c) 1 hour 36 minutes

• (d) 5 hours

Solution: B) 4 hours 48 minutes

L.C.M of 18, 24 and 32 = 288 Hence, they would chime after every 288 min. or 4 hrs 48 min.

Therefore, option (b) is correct.

Solution: B) 4 hours 48 minutes

L.C.M of 18, 24 and 32 = 288 Hence, they would chime after every 288 min. or 4 hrs 48 min.

Therefore, option (b) is correct.

• Question 4 of 5 4. Question A report consists of 10 sheets each of 110 lines and each such line consists of 65 characters. This report is retyped into sheets each of 65 lines such that each line consists of 22 characters. The approximate percentage change in number of sheets are (a) 200 (b) 800 (c) 400 (d) 350 Correct Solution: C) 400 Total volume of first report = (110 × 10 × 65) = 71500 Let n be the number of pages. Therefore, the new volume will be (65 × 22 × n) = 1430n As the volume remains same, we can say that 71500 = 1430n è n = 50 sheets. Clearly, the sheets increased by (50 –10 = 40) Therefore, percentage increase in number of sheets = (40/10) 100 = 400 % Hence, option (c) is correct. Incorrect Solution: C) 400 Total volume of first report = (110 × 10 × 65) = 71500 Let n be the number of pages. Therefore, the new volume will be (65 × 22 × n) = 1430n As the volume remains same, we can say that 71500 = 1430n è n = 50 sheets. Clearly, the sheets increased by (50 –10 = 40) Therefore, percentage increase in number of sheets = (40/10) 100 = 400 % Hence, option (c) is correct.

#### 4. Question

A report consists of 10 sheets each of 110 lines and each such line consists of 65 characters. This report is retyped into sheets each of 65 lines such that each line consists of 22 characters. The approximate percentage change in number of sheets are

Solution: C) 400

Total volume of first report = (110 × 10 × 65) = 71500 Let n be the number of pages. Therefore, the new volume will be (65 × 22 × n) = 1430n As the volume remains same, we can say that 71500 = 1430n è n = 50 sheets. Clearly, the sheets increased by (50 –10 = 40)

Therefore, percentage increase in number of sheets = (40/10) * 100 = 400 %

Hence, option (c) is correct.

Solution: C) 400

Therefore, percentage increase in number of sheets = (40/10) * 100 = 400 %

Hence, option (c) is correct.

• Question 5 of 5 5. Question Carpenter A can make a chair in 8 hours, carpenter B in 7 hours and carpenter C in 6 hours. If each carpenter works for 8 hours per day, how many chairs will be made in 21 days? (a) 73 (b) 63 (c) 83 (d) 75 Correct Solution: A) 73 It is given that: Carpenter A can make a chair in 8 hours. Carpenter B can make a chair in 7 hours. Carpenter C can make a chair in 6 hours Now, if each carpenter works for 8 hours per day for 21 days, then total number of hours worked is 21 8 or 168 hours by each carpenter. Now , in 168 hours Carpenter A can make (168/8) chairs = 21 chairs. In 168 hours Carpenter B can make (168/7) chairs = 24 chairs. In 168 hours Carpenter C can make (168/6) chairs = 28 chairs. Therefore, total number of chairs done together = 21+24+28 = 73 chairs Hence, option (a) is correct. Incorrect Solution: A) 73 It is given that: Carpenter A can make a chair in 8 hours. Carpenter B can make a chair in 7 hours. Carpenter C can make a chair in 6 hours Now, if each carpenter works for 8 hours per day for 21 days, then total number of hours worked is 21 8 or 168 hours by each carpenter. Now , in 168 hours Carpenter A can make (168/8) chairs = 21 chairs. In 168 hours Carpenter B can make (168/7) chairs = 24 chairs. In 168 hours Carpenter C can make (168/6) chairs = 28 chairs. Therefore, total number of chairs done together = 21+24+28 = 73 chairs Hence, option (a) is correct.

#### 5. Question

Carpenter A can make a chair in 8 hours, carpenter B in 7 hours and carpenter C in 6 hours. If each carpenter works for 8 hours per day, how many chairs will be made in 21 days?

Solution: A) 73

It is given that:

Carpenter A can make a chair in 8 hours.

Carpenter B can make a chair in 7 hours.

Carpenter C can make a chair in 6 hours

Now, if each carpenter works for 8 hours per day for 21 days, then total number of hours worked is 21 * 8 or 168 hours by each carpenter.

Now , in 168 hours Carpenter A can make (168/8) chairs = 21 chairs.

In 168 hours Carpenter B can make (168/7) chairs = 24 chairs.

In 168 hours Carpenter C can make (168/6) chairs = 28 chairs.

Therefore, total number of chairs done together = 21+24+28 = 73 chairs

Hence, option (a) is correct.