UPSC Insta–DART (Daily Aptitude and Reasoning Test) 27 Dec 2025
Kartavya Desk Staff
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).
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• Question 1 of 5 1. Question The monthly incomes of Amit and Bharat are in the ratio of 5:4. Their monthly expenses are in the ratio 4:3. If each saves ₹5,000 per month, what are their monthly incomes? (a) ₹25,000 and ₹20,000 (b) ₹30,000 and ₹24,000 (c) ₹35,000 and ₹28,000 (d) ₹40,000 and ₹32,000 Correct Answer: (a) Solution: Let incomes be: Amit = 5x, Bharat = 4x Expenses: Amit = 4y, Bharat = 3y Savings = Income − Expense = ₹5000 for both So: 5x − 4y = 5000 …(i) 4x − 3y = 5000 …(ii) Multiply (i) by 3: 15x − 12y = 15000 Multiply (ii) by 4: 16x − 12y = 20000 Subtract: (16x − 15x) = (20000 − 15000) x = 5000 Then y = (5x − 5000) / 4 = (25000 − 5000)/4 = 5000 So incomes: Amit = 5x = ₹25,000 Bharat = 4x = ₹20,000 Hence, option (A) is correct. Incorrect Answer: (a) Solution: Let incomes be: Amit = 5x, Bharat = 4x Expenses: Amit = 4y, Bharat = 3y Savings = Income − Expense = ₹5000 for both So: 5x − 4y = 5000 …(i) 4x − 3y = 5000 …(ii) Multiply (i) by 3: 15x − 12y = 15000 Multiply (ii) by 4: 16x − 12y = 20000 Subtract: (16x − 15x) = (20000 − 15000) x = 5000 Then y = (5x − 5000) / 4 = (25000 − 5000)/4 = 5000 So incomes: Amit = 5x = ₹25,000 Bharat = 4x = ₹20,000 Hence, option (A) is correct.
#### 1. Question
The monthly incomes of Amit and Bharat are in the ratio of 5:4. Their monthly expenses are in the ratio 4:3. If each saves ₹5,000 per month, what are their monthly incomes?
• (a) ₹25,000 and ₹20,000
• (b) ₹30,000 and ₹24,000
• (c) ₹35,000 and ₹28,000
• (d) ₹40,000 and ₹32,000
Answer: (a)
Solution: Let incomes be: Amit = 5x, Bharat = 4x Expenses: Amit = 4y, Bharat = 3y Savings = Income − Expense = ₹5000 for both
So: 5x − 4y = 5000 …(i) 4x − 3y = 5000 …(ii)
Multiply (i) by 3: 15x − 12y = 15000 Multiply (ii) by 4: 16x − 12y = 20000 Subtract: (16x − 15x) = (20000 − 15000) x = 5000 Then y = (5x − 5000) / 4 = (25000 − 5000)/4 = 5000
So incomes: Amit = 5x = ₹25,000 Bharat = 4x = ₹20,000
Hence, option (A) is correct.
Answer: (a)
Solution: Let incomes be: Amit = 5x, Bharat = 4x Expenses: Amit = 4y, Bharat = 3y Savings = Income − Expense = ₹5000 for both
So: 5x − 4y = 5000 …(i) 4x − 3y = 5000 …(ii)
Multiply (i) by 3: 15x − 12y = 15000 Multiply (ii) by 4: 16x − 12y = 20000 Subtract: (16x − 15x) = (20000 − 15000) x = 5000 Then y = (5x − 5000) / 4 = (25000 − 5000)/4 = 5000
So incomes: Amit = 5x = ₹25,000 Bharat = 4x = ₹20,000
Hence, option (A) is correct.
• Question 2 of 5 2. Question A person buys three articles A, B, and C for ₹3960. If A costs 20% more than C, and C costs 25% more than B, then what is the cost of A? (a) ₹1320 (b) ₹1440 (c) ₹1584 (d) ₹1600 Correct Answer: C Solution: Let the cost of B be ₹x. Then, C = 125% of B = (5/4)x A = 120% of C = (6/5) × (5/4)x = (6/4)x = (3/2)x Now, total cost = A + B + C = (3/2)x + x + (5/4)x = (6x + 4x + 5x)/4 = 15x/4 Given total cost = ₹3960 ⇒ 15x/4 = 3960 ⇒ x = (3960 × 4)/15 = ₹1056 Therefore, Cost of A = (3/2) × 1056 = ₹1584 Incorrect Answer: C Solution: Let the cost of B be ₹x. Then, C = 125% of B = (5/4)x A = 120% of C = (6/5) × (5/4)x = (6/4)x = (3/2)x Now, total cost = A + B + C = (3/2)x + x + (5/4)x = (6x + 4x + 5x)/4 = 15x/4 Given total cost = ₹3960 ⇒ 15x/4 = 3960 ⇒ x = (3960 × 4)/15 = ₹1056 Therefore, Cost of A = (3/2) × 1056 = ₹1584
#### 2. Question
A person buys three articles A, B, and C for ₹3960. If A costs 20% more than C, and C costs 25% more than B, then what is the cost of A?
Answer: C
Solution: Let the cost of B be ₹x. Then, C = 125% of B = (5/4)x A = 120% of C = (6/5) × (5/4)x = (6/4)x = (3/2)x
Now, total cost = A + B + C = (3/2)x + x + (5/4)x = (6x + 4x + 5x)/4 = 15x/4
Given total cost = ₹3960 ⇒ 15x/4 = 3960 ⇒ x = (3960 × 4)/15 = ₹1056
Therefore, Cost of A = (3/2) × 1056 = ₹1584
Answer: C
Solution: Let the cost of B be ₹x. Then, C = 125% of B = (5/4)x A = 120% of C = (6/5) × (5/4)x = (6/4)x = (3/2)x
Now, total cost = A + B + C = (3/2)x + x + (5/4)x = (6x + 4x + 5x)/4 = 15x/4
Given total cost = ₹3960 ⇒ 15x/4 = 3960 ⇒ x = (3960 × 4)/15 = ₹1056
Therefore, Cost of A = (3/2) × 1056 = ₹1584
• Question 3 of 5 3. Question A man borrowed a sum from a bank to be repaid in equal monthly installments without interest. After paying 12 installments, he found that 40% of the total loan was cleared. How many installments were there in the agreement? (a) 20 (b) 25 (c) 28 (d) 30 Correct Answer: (d) Solution: Let total loan amount = T Let installment amount = A After 12 installments: 12 × A = 40% of T = (40/100) × T ⇒ 12A = 0.4T ⇒ T/A = 12 / 0.4 = 30 Total number of installments = T / A = 30 Incorrect Answer: (d) Solution: Let total loan amount = T Let installment amount = A After 12 installments: 12 × A = 40% of T = (40/100) × T ⇒ 12A = 0.4T ⇒ T/A = 12 / 0.4 = 30 Total number of installments = T / A = 30
#### 3. Question
A man borrowed a sum from a bank to be repaid in equal monthly installments without interest. After paying 12 installments, he found that 40% of the total loan was cleared. How many installments were there in the agreement?
Answer: (d)
Solution: Let total loan amount = T Let installment amount = A
After 12 installments: 12 × A = 40% of T = (40/100) × T ⇒ 12A = 0.4T ⇒ T/A = 12 / 0.4 = 30
Total number of installments = T / A = 30
Answer: (d)
Solution: Let total loan amount = T Let installment amount = A
After 12 installments: 12 × A = 40% of T = (40/100) × T ⇒ 12A = 0.4T ⇒ T/A = 12 / 0.4 = 30
Total number of installments = T / A = 30
• Question 4 of 5 4. Question A person took a loan of ₹20,000 at 10% compound interest per annum. He paid ₹7,000 at the end of the first year and ₹7,700 at the end of the second year. How much should he pay at the end of the third year to clear all dues? (a) ₹7,500 (b) ₹9,680 (c) ₹8,310 (d) ₹8,580 Correct Answer: (b) Solution: Principal = ₹20,000 Interest rate = 10% compounded yearly After 1st year: Interest = 10% of 20,000 = ₹2,000 Total = 20,000 + 2,000 = ₹22,000 Paid = ₹7,000 Remaining = ₹15,000 After 2nd year: Interest = 10% of 15,000 = ₹1,500 Total = 15,000 + 1,500 = ₹16,500 Paid = ₹7,700 Remaining = ₹8,800 After 3rd year: Interest = 10% of 8,800 = ₹880 Total = 8,800 + 880 = ₹9,680 Incorrect Answer: (b) Solution: Principal = ₹20,000 Interest rate = 10% compounded yearly After 1st year: Interest = 10% of 20,000 = ₹2,000 Total = 20,000 + 2,000 = ₹22,000 Paid = ₹7,000 Remaining = ₹15,000 After 2nd year: Interest = 10% of 15,000 = ₹1,500 Total = 15,000 + 1,500 = ₹16,500 Paid = ₹7,700 Remaining = ₹8,800 After 3rd year: Interest = 10% of 8,800 = ₹880 Total = 8,800 + 880 = ₹9,680
#### 4. Question
A person took a loan of ₹20,000 at 10% compound interest per annum. He paid ₹7,000 at the end of the first year and ₹7,700 at the end of the second year. How much should he pay at the end of the third year to clear all dues?
• (a) ₹7,500
• (b) ₹9,680
• (c) ₹8,310
• (d) ₹8,580
Answer: (b)
Solution: Principal = ₹20,000 Interest rate = 10% compounded yearly
After 1st year: Interest = 10% of 20,000 = ₹2,000 Total = 20,000 + 2,000 = ₹22,000 Paid = ₹7,000 Remaining = ₹15,000
After 2nd year: Interest = 10% of 15,000 = ₹1,500 Total = 15,000 + 1,500 = ₹16,500 Paid = ₹7,700 Remaining = ₹8,800
After 3rd year: Interest = 10% of 8,800 = ₹880 Total = 8,800 + 880 = ₹9,680
Answer: (b)
Solution: Principal = ₹20,000 Interest rate = 10% compounded yearly
After 1st year: Interest = 10% of 20,000 = ₹2,000 Total = 20,000 + 2,000 = ₹22,000 Paid = ₹7,000 Remaining = ₹15,000
After 2nd year: Interest = 10% of 15,000 = ₹1,500 Total = 15,000 + 1,500 = ₹16,500 Paid = ₹7,700 Remaining = ₹8,800
After 3rd year: Interest = 10% of 8,800 = ₹880 Total = 8,800 + 880 = ₹9,680
• Question 5 of 5 5. Question A principal amount becomes in one year when compounded quarterly at an annual interest rate of . If the same amount becomes in one year when compounded annually at , which one of the following is correct? (a) R=S (b) R>S (c) R<S (d) Cannot be determined Correct Answer: C Solution: Let’s assume , and test with a 12% annual rate: If compounded quarterly: Each quarter rate = 12% / 4 = 3% Amount after 1 year = If compounded annually at 12%: Amount = Thus, for the same amount , the quarterly compounding gives more than annual. So to reach the same , S > R Incorrect Answer: C Solution: Let’s assume , and test with a 12% annual rate: If compounded quarterly: Each quarter rate = 12% / 4 = 3% Amount after 1 year = If compounded annually at 12%: Amount = Thus, for the same amount , the quarterly compounding gives more than annual. So to reach the same , S > R
#### 5. Question
A principal amount becomes in one year when compounded quarterly at an annual interest rate of . If the same amount becomes in one year when compounded annually at , which one of the following is correct?
• (d) Cannot be determined
Answer: C
Solution: Let’s assume , and test with a 12% annual rate:
If compounded quarterly: Each quarter rate = 12% / 4 = 3% Amount after 1 year =
If compounded annually at 12%: Amount =
Thus, for the same amount , the quarterly compounding gives more than annual. So to reach the same , S > R
Answer: C
Solution: Let’s assume , and test with a 12% annual rate:
If compounded quarterly: Each quarter rate = 12% / 4 = 3% Amount after 1 year =
If compounded annually at 12%: Amount =
Thus, for the same amount , the quarterly compounding gives more than annual. So to reach the same , S > R
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