[Mission 2024] Insta–DART (Daily Aptitude and Reasoning Test) 16 May 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 What is the sum of all natural numbers between 100 and 200 which are multiples of 3? (a) 4950 (b) 4300 (c) 5000 (d) 5200 Correct Solution: A) 4950 Multiples of 3 between 100 and 200 are 102, 105, 108,.. 198. Here, the first term = 102 last term = 198 Let the number of Multiples of 3 between 100 and 200 = n Arithmetic Progression Formula: an = a1 + (n – 1)d Where, an = last term = 198 a1 = first term = 102 d = common difference = 105 – 102 = 3 —> 198 = 102 + (n – 1) 3 —> 198 – 102 = (n – 1) 3 —> 96 = (n – 1) 3 —> (n – 1) = 96/3 = 32 —> n = 32 + 1 —> n = 33 Formula: Sum of n terms = Sn = (n/2) (a1 + an ) where n = number of elements = 33 a1 = first term = 102 an= last term = 198 Thus, using the above formula, Sum of all natural numbers between 100 and 200 which are multiples of 3 = (33/2) * (102 + 198) = (33/2) * 300 = 33 * 150 = 4950 Incorrect Solution: A) 4950 Multiples of 3 between 100 and 200 are 102, 105, 108,.. 198. Here, the first term = 102 last term = 198 Let the number of Multiples of 3 between 100 and 200 = n Arithmetic Progression Formula: an = a1 + (n – 1)d Where, an = last term = 198 a1 = first term = 102 d = common difference = 105 – 102 = 3 —> 198 = 102 + (n – 1) * 3 —> 198 – 102 = (n – 1) * 3 —> 96 = (n – 1) * 3 —> (n – 1) = 96/3 = 32 —> n = 32 + 1 —> n = 33 Formula: Sum of n terms = Sn = (n/2) (a1 + an ) where n = number of elements = 33 a1 = first term = 102 an= last term = 198 Thus, using the above formula, Sum of all natural numbers between 100 and 200 which are multiples of 3 = (33/2) (102 + 198) = (33/2) 300 = 33 150 = 4950

#### 1. Question

What is the sum of all natural numbers between 100 and 200 which are multiples of 3?

Solution: A) 4950

Multiples of 3 between 100 and 200 are 102, 105, 108,.. 198. Here, the first term = 102 last term = 198 Let the number of Multiples of 3 between 100 and 200 = n

Arithmetic Progression Formula: an = a1 + (n – 1)d Where, an = last term = 198 a1 = first term = 102 d = common difference = 105 – 102 = 3 —> 198 = 102 + (n – 1) 3 —> 198 – 102 = (n – 1) 3 —> 96 = (n – 1) 3 —> (n – 1) = 96/3 = 32 —> n = 32 + 1 —> n = 33*

Formula: Sum of n terms = Sn = (n/2) * (a1 + an ) where n = number of elements = 33 a1 = first term = 102 an= last term = 198 Thus, using the above formula, Sum of all natural numbers between 100 and 200 which are multiples of 3 = (33/2) (102 + 198) = (33/2) 300 = 33 150 = 4950*

Solution: A) 4950

Multiples of 3 between 100 and 200 are 102, 105, 108,.. 198. Here, the first term = 102 last term = 198 Let the number of Multiples of 3 between 100 and 200 = n

• Question 2 of 5 2. Question Ramya invested an amount of Rs.60000 to start a business . After 8 months, Bharathi joined her with an amount of Rs . 30000 . They earned profit RS . 24500 after one year. What is Bharathi profit ? (a) 2500 (b) 1500 (c) 3500 (d) 4000 Correct Solution: C) 3500 Ratio of capital invested by each partner. i.e Ramya and Bharathi for one year in order to get same profit. Ramya invested for 12 months whereas Bharathi invested for 4 months. Therefore, = 12 60000 : 4 30000 = 720000 : 120000 = 6 : 1 Share of Bharathi out of Rs .24500 = 1/7 24500 = Rs. 3500 Incorrect Solution: C) 3500 Ratio of capital invested by each partner. i.e Ramya and Bharathi for one year in order to get same profit. Ramya invested for 12 months whereas Bharathi invested for 4 months. Therefore, = 12 60000 : 4 30000 = 720000 : 120000 = 6 : 1 Share of Bharathi out of Rs .24500 = 1/7 24500 = Rs. 3500

#### 2. Question

Ramya invested an amount of Rs.60000 to start a business . After 8 months, Bharathi joined her with an amount of Rs . 30000 . They earned profit RS . 24500 after one year. What is Bharathi profit ?

Solution: C) 3500

Ratio of capital invested by each partner. i.e Ramya and Bharathi for one year in order to get same profit.

Ramya invested for 12 months whereas Bharathi invested for 4 months. Therefore, = 12 60000 : 4 30000 = 720000 : 120000 = 6 : 1 Share of Bharathi out of Rs .24500 = 1/7 24500 = Rs. 3500*

Solution: C) 3500

Ratio of capital invested by each partner. i.e Ramya and Bharathi for one year in order to get same profit.

Ramya invested for 12 months whereas Bharathi invested for 4 months. Therefore, = 12 60000 : 4 30000 = 720000 : 120000 = 6 : 1 Share of Bharathi out of Rs .24500 = 1/7 24500 = Rs. 3500*

• Question 3 of 5 3. Question Dheeraj sold an article for Rs. 6,750 after giving a discount of 10% on the labelled price. He would have earned a profit of 50%, had there been no discount. What was the actual percentage of profit earned? (a) 26% (b) 30% (c) 35% (d) 45% Correct Solution: C) 35% Let the marked price of the article be Rs. x Discount = 10% Then, the selling price = – 10% of = (9x/10) Now,= (9x/10)= 6750 Or x = Rs. 7,500 As per the question, if the selling price is Rs. 7,500, then the profit earned by the Dheeraj is 50%. So, the cost price = Rs. 5,000 Therefore, the actual percentage of profit earned by selling the article for Rs. 6,750 = (Profit/CP) × 100 =(1750/5000)× 100 = 35% Hence, option (c) is the correct. Incorrect Solution: C) 35% Let the marked price of the article be Rs. x Discount = 10% Then, the selling price = – 10% of = (9x/10) Now,= (9x/10)= 6750 Or x = Rs. 7,500 As per the question, if the selling price is Rs. 7,500, then the profit earned by the Dheeraj is 50%. So, the cost price = Rs. 5,000 Therefore, the actual percentage of profit earned by selling the article for Rs. 6,750 = (Profit/CP) × 100 =(1750/5000)× 100 = 35% Hence, option (c) is the correct.

#### 3. Question

Dheeraj sold an article for Rs. 6,750 after giving a discount of 10% on the labelled price. He would have earned a profit of 50%, had there been no discount. What was the actual percentage of profit earned?

Solution: C) 35% Let the marked price of the article be Rs. x

Discount = 10% Then, the selling price = – 10% of = (9x/10) Now,= (9x/10)= 6750 Or x = Rs. 7,500 As per the question, if the selling price is Rs. 7,500, then the profit earned by the Dheeraj is 50%. So, the cost price = Rs. 5,000 Therefore, the actual percentage of profit earned by selling the article for Rs. 6,750 = (Profit/CP) × 100 =(1750/5000)× 100 = 35% Hence, option (c) is the correct.

Solution: C) 35% Let the marked price of the article be Rs. x

• Question 4 of 5 4. Question The length of a rectangle is decreased by 20% and breadth increased by 20%. What will be the net percentage increase/decrease in the area of the rectangle? (a) Increase by 44% (b) Remain the same (c) Increase by 4% (d) Decrease by 4% Correct Solution: D) Decrease by 4% Let the original length of the rectangle is x units Let the original length of the rectangle be y units. Thus, original area of rectangle = xy units^2 . If the length of a rectangle is decreased by 20%, then new length will be 0.8 x units. If breadth of rectangle is increased by 20%, then new breadth will be 1.2 y units. Area of new rectangle will be 0.8x 1.2y units^2 Area of new rectangle will be = 0.96 xy units^2 Hence, Area of new rectangle will be 4 % less than the area of original rectangle. Hence, option (d) is correct. Incorrect Solution: D) Decrease by 4% Let the original length of the rectangle is x units Let the original length of the rectangle be y units. Thus, original area of rectangle = xy units^2 . If the length of a rectangle is decreased by 20%, then new length will be 0.8 x units. If breadth of rectangle is increased by 20%, then new breadth will be 1.2 y units. Area of new rectangle will be 0.8x 1.2y units^2 Area of new rectangle will be = 0.96 xy units^2 Hence, Area of new rectangle will be 4 % less than the area of original rectangle. Hence, option (d) is correct.

#### 4. Question

The length of a rectangle is decreased by 20% and breadth increased by 20%. What will be the net percentage increase/decrease in the area of the rectangle?

• (a) Increase by 44%

• (b) Remain the same

• (c) Increase by 4%

• (d) Decrease by 4%

Solution: D) Decrease by 4%

Let the original length of the rectangle is x units

Let the original length of the rectangle be y units.

Thus, original area of rectangle = xy units^2 .

If the length of a rectangle is decreased by 20%, then new length will be 0.8 x units.

If breadth of rectangle is increased by 20%, then new breadth will be 1.2 y units.

Area of new rectangle will be 0.8x * 1.2y units^2

• Area of new rectangle will be = 0.96 xy units^2

Hence, Area of new rectangle will be 4 % less than the area of original rectangle.

Hence, option (d) is correct.

Solution: D) Decrease by 4%

Let the original length of the rectangle is x units

Let the original length of the rectangle be y units.

Thus, original area of rectangle = xy units^2 .

If the length of a rectangle is decreased by 20%, then new length will be 0.8 x units.

If breadth of rectangle is increased by 20%, then new breadth will be 1.2 y units.

Area of new rectangle will be 0.8x * 1.2y units^2

• Area of new rectangle will be = 0.96 xy units^2

Hence, Area of new rectangle will be 4 % less than the area of original rectangle.

Hence, option (d) is correct.

• Question 5 of 5 5. Question A tap can fill a water tank in 8 minutes. After one fourth of the tank is filled, two more taps with same capacity are opened. What is the total time taken to fill the tank completely? (a) 6 minutes (b) 4 minutes (c) 3 minutes (d) 5 minutes Correct Solution: B) 4 minutes Since 1 part is filled in 8 minutes. Therefore, 1/4 part is filled in 8×1/4 = 2 minutes. Remaining part = 1 – ¼ = 3/4. Since 1 part is filled in 8 minute, therefore, in 1 minute 1/8 part is filled. Hence, in 1 minute, part filled by 3 (=1+2) taps = 3/8. Since 3/8 part is filled in 1 minute. Therefore, 1 part is filled in 1/(3/8) = 8/3 minutes. Therefore, 3/4 part is filled in (8/3)×(3/4) = 2 minutes. So, the total time taken = 2 + 2 = 4 minutes. Hence, option (b) is correct. Incorrect Solution: B) 4 minutes Since 1 part is filled in 8 minutes. Therefore, 1/4 part is filled in 8×1/4 = 2 minutes. Remaining part = 1 – ¼ = 3/4. Since 1 part is filled in 8 minute, therefore, in 1 minute 1/8 part is filled. Hence, in 1 minute, part filled by 3 (=1+2) taps = 3/8. Since 3/8 part is filled in 1 minute. Therefore, 1 part is filled in 1/(3/8) = 8/3 minutes. Therefore, 3/4 part is filled in (8/3)×(3/4) = 2 minutes. So, the total time taken = 2 + 2 = 4 minutes. Hence, option (b) is correct.

#### 5. Question

A tap can fill a water tank in 8 minutes. After one fourth of the tank is filled, two more taps with same capacity are opened. What is the total time taken to fill the tank completely?

• (a) 6 minutes

• (b) 4 minutes

• (c) 3 minutes

• (d) 5 minutes

Solution: B) 4 minutes

Since 1 part is filled in 8 minutes. Therefore, 1/4 part is filled in 8×1/4 = 2 minutes. Remaining part = 1 – ¼ = 3/4. Since 1 part is filled in 8 minute, therefore, in 1 minute 1/8 part is filled. Hence, in 1 minute, part filled by 3 (=1+2) taps = 3/8.

Since 3/8 part is filled in 1 minute. Therefore, 1 part is filled in 1/(3/8) = 8/3 minutes. Therefore, 3/4 part is filled in (8/3)×(3/4) = 2 minutes. So, the total time taken = 2 + 2 = 4 minutes.

Hence, option (b) is correct.

Solution: B) 4 minutes

Since 3/8 part is filled in 1 minute. Therefore, 1 part is filled in 1/(3/8) = 8/3 minutes. Therefore, 3/4 part is filled in (8/3)×(3/4) = 2 minutes. So, the total time taken = 2 + 2 = 4 minutes.

Hence, option (b) is correct.

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