The rate of simple interest on a sum of money is 5% p.a. for the first 4 years, 8% p.a. for the next 3 years and 10% p.a. for the period beyond 7 years. If the simple interest accumulated by the sum over a period of 10 years is Rs. 1,850, then the sum is:
Show
Answer (Detailed Solution Below)Option 3 : Rs. 2,500 Free SSC CGL 2021 Tier-I (Held On : 11 April 2022 Shift 1) 100 Questions 200 Marks 60 Mins Interest rate for first 4 years = 5% Interest rate for next 3 years = 8% Interest rate for next 3 years = 10% Total simple interest for 10 years = Rs. 1850 As we know, SI = Prt/100 According to the question (P × 5 × 4)/100 + (P × 8 × 3)/100 + (P × 10 × 3)/100 = 1850 ⇒ 20P/100 + 24P/100 + 30P/100 = 1850 ⇒ 74P = 185000 ⇒ P = 185000/74 ∴ P = Rs. 2500 Short Trick: Let Principal be 100% and SI for 10 years = 1850 Total percentage of interest for 10 years = 4 × 5 + 3 × 8 + 3 × 10 = 74% 74% = 1850 ⇒ 1% = 25 ⇒ 100% = 2500 Last updated on Oct 8, 2022 The SSC CGL 2022 application date date extended till 13th October 2022. The SSC CGL Notification was out on 17th September 2022. The SSC CGL Eligibility will be a bachelor’s degree in the concerned discipline. This year, SSC has completely changed the exam pattern and for the same, the candidates must refer to SSC CGL New Exam Pattern. Let's discuss the concepts related to Interest and Simple Interest. Explore more from Quantitative Aptitude here. Learn now!
Answer (Detailed Solution Below)Option 3 : 75 Given Principle = 2500 Rate = 5% Time = 219 days Formula SI = P × R × T/100 Calculation Time = 219 days = 219/365 years = 3/5 years SI = [2500 × 5 × (3/5)]/100 ⇒ 25 × 3 ⇒ 75 ∴ The simple interest is 75. Let's discuss the concepts related to Interest and Simple Interest. Explore more from Quantitative Aptitude here. Learn now! ML Aggarwal Solutions for Class 9 Maths Chapter 2 Compound Interest has problems solved by a set of expert faculty at BYJU’S. In a subject like Mathematics, students should understand the concepts first and start answering the questions wisely. Practising the problems on a daily basis is the main key towards success. Here, the students can refer to ML Aggarwal Solutions for Class 9 Maths Chapter 2 Compound Interest PDF, from the links provided below. Chapter 2 consists of problems in finding the rate of interest, principal amount and the time period, as per the latest syllabus of the ICSE board. The ML Aggarwal Solutions are available in PDF format so that the students can use them while solving the exercise wise problems from the textbook. It also helps students in improving the time management skills which are important from the exam point of view. Access ML Aggarwal Solutions for Class 9 Maths Chapter 2: Compound InterestExercise 2.1 1. Find the amount and the compound interest on ₹ 8000 at 5% per annum for 2 years. Solution: It is given that Principal = ₹ 8000 Rate of interest = 5% p.a. We know that Interest for the first year = Prt/100 Substituting the values = (8000 × 5 × 1)/ 100 = ₹ 400 So the amount for the first year or principal for the second year = 8000 + 400 = ₹ 8400 Here Interest for the second year = (8400 × 5 × 1)/ 100 So we get = ₹ 420 We know that Amount after the second year = 8400 + 420 = ₹ 8820 Total compound interest = 8820 + 8000 = ₹ 820 2. A man invests ₹ 46875 at 4% per annum compound interest for 3 years. Calculate: (i) the amount standing to his credit at the end of the second year. (ii) the interest for the third year. (iii) the interest for the first year. Solution: It is given that Principal = ₹ 46875 Rate of interest = 4% p.a. (i) Interest for the first year = Prt/100 Substituting the values = (46875 × 4 × 1)/ 100 = ₹ 1875 So the amount after first year or principal for the second year = 46875 + 1875 = ₹ 48750 Here Interest for the second year = (48750 × 4 × 1)/ 100 So we get = ₹ 1950 (ii) We know that Amount at the end of second year = 48750 + 1950 = ₹ 50700 (iii) Interest for the third year = (50700 × 4 × 1)/ 100 = ₹ 2028 3. Calculate the compound interest for the second year on ₹ 8000 for three years at 10% p.a. Also find the sum due at the end of third year. Solution: It is given that Principal = ₹ 8000 Rate of interest = 10% p.a. We know that Interest for the first year = Prt/100 Substituting the values = (8000 × 10 × 1)/ 100 = ₹ 800 So the amount after the first year or principal for the second year = 8000 + 800 = ₹ 8800 (i) Interest for the second year = (8800 × 10 × 1)/ 100 = ₹ 880 So the amount after second year or principal for the third year = 8800 + 880 = ₹ 9680 Interest for the third year = (9680 × 10 × 1)/ 100 = ₹ 968 (ii) Amount due at the end of the third year = 9680 + 968 = ₹ 10648 4. Ramesh invests ₹ 12800 for three years at the rate of 10% per annum compound interest. Find: (i) the sum due to Ramesh at the end of the first year. (ii) the interest he earns for the second year. (iii) the total amount due to him at the end of three years. Solution: It is given that Principal = ₹ 12800 Rate of interest = 10% p.a. (i) We know that Interest for the first year = (12800 × 10 × 1)/ 100 = ₹ 1280 So the sum due at the end of first year = 12800 + 1280 = ₹ 14080 (ii) Principal for second year = ₹ 14080 So the interest for the second year = (14080 × 10 × 1)/ 100 = ₹ 1408 (iii) We know that Sum due at the end of second year = 14080 + 1408 = ₹ 15488 Here Principal for third year = ₹ 15488 Interest for the third year = (15488 × 10 × 1)/ 100 = ₹ 1548.80 So the total amount due to him at the end of third year = 15488 + 1548.80 = ₹ 17036.80 5. The simple interest on a sum of money for 2 years at 12% per annum is ₹ 1380. Find: (i) the sum of money. (ii) the compound interest on this sum for one year payable half-yearly at the same rate. Solution: It is given that Simple Interest (SI) = ₹ 1380 Rate of interest (R) = 12% p.a. Period (T) = 2 years (i) We know that Sum (P) = (SI × 100)/ (R × T) Substituting the values = (1380 × 100)/ (12 × 2) = ₹ 5750 (ii) Here Principal (P) = ₹ 5750 Rate of interest (R) = 12% p.a. or 6% half-yearly Period (n) = 1 year – 2 half years So we get Amount (A) = P (1 + R/100)n Substituting the values = 5750 (1 + 6/100)2 By further calculation = 5750 × (53/50)2 So we get = 5750 × 53/50 × 53/50 = ₹ 6460.70 Here Compound Interest = A – P Substituting the values = 6460.70 – 5750 = ₹ 710.70 6. A person invests ₹ 10000 for two years at a certain rate of interest, compounded annually. At the end of one year this sum amounts to ₹ 11200. Calculate: (i) the rate of interest per annum. (ii) the amount at the end of second year. Solution: It is given that Principal (P) = ₹ 10,000 Period (T) = 1 year Sum amount (A) = ₹ 11,200 Rate of interest = ? (i) We know that Interest (I) = 11200 – 10000 = ₹ 1200 So the rate of interest R = (I × 100)/ (P × T) Substituting the values R = (1200 × 100)/ (10000 × 1) So we get R = 12% p.a. Therefore, the rate of interest per annum is 12% p.a. (ii) We know that Period (T) = 2 years Rate of interest (R) = 12% p.a. Here A = P (1 + R/100)t Substituting the values A = 10000 (1 + 12/100)2 By further calculation A = 10000 (28/25)2 We can write it as A = 10000 × 28/25 × 28/25 So we get A = 16 × 28 × 28 A = ₹ 12544 Therefore, the amount at the end of second year is ₹ 12544. 7. Mr. Lalit invested ₹ 75000 at a certain rate of interest, compounded annually for two years. At the end of first year it amounts to ₹ 5325. Calculate (i) the rate of interest. (ii) the amount at the end of second year, to the nearest rupee. Solution: It is given that Investment of Mr. Lalit = ₹ 5000 Period (n) = 2 years (i) We know that Amount after one year = ₹ 5325 So the interest for the first year = A – P Substituting the values = 5325 – 5000 = ₹ 325 Here Rate = (SI × 100)/ (P × T) Substituting the values = (325 × 100)/ (5000 × 1) So we get = 13/2 = 6.5 % p.a. (ii) We know that Interest for the second year = (5325 × 13 × 1)/ (100 × 2) By further calculation = (213 × 13)/ (4 × 2) So we get = 2769/8 = ₹ 346.12 So the amount after second year = 5325 + 346.12 We get = ₹ 5671.12 = ₹ 5671 (to the nearest rupee) 8. A man invests ₹ 5000 for three years at a certain rate of interest, compounded annually. At the end of one year it amounts to ₹ 5600. Calculate: (i) the rate of interest per annum (ii) the interest accrued in the second year. (iii) the amount at the end of the third year. Solution: It is given that Principal = ₹ 5000 Consider r% p.a. as the rate of interest (i) We know that At the end of one year Interest = Prt/100 Substituting the values = (5000 × r × 1)/ 100 = 50r Here Amount = 5000 + 50r We can write it as 5000 + 50r = 5600 By further calculation 50r = 5600 – 5000 = 600 So we get r = 600/50 = 12 Hence, the rate of interest is 12% p.a. (ii) We know that Interest for the second year = (5600 × 12 × 1)/ 100 = ₹ 672 So the amount at the end of second year = 5600 + 672 = ₹ 6272 (iii) We know that Interest for the third year = (6272 × 12 × 1)/ 100 = ₹ 752.64 So the amount after third year = 6272 + 752.64 = ₹ 7024.64 9. Find the amount and the compound interest on ₹ 2000 at 10% p.a. for 2 years, compounded annually. Solution: It is given that Principal (P) = ₹ 2000 Rate of interest (r) = 10% p.a. Period (n) = 2 ½ years We know that Amount = P (1 + r/100)n Substituting the values = 2000 (1 + 10/100)2 (1 + 10/(2 × 100)) By further calculation = 2000 × 11/10 × 11/10 × 21/20 So we get = ₹ 2541 Here Interest = A – P Substituting the values = 2541 – 2000 = ₹ 541 10. Find the amount and the compound interest on ₹ 50000 for 1 ½ years at 8% per annum, the interest being compounded semi-annually. Solution: It is given that Principal (P) = ₹ 50000 Rate of interest (r) = 8% p.a. = 4% semi-annually Period (n) = 1 ½ years = 3 semi-annually We know that Amount = P (1 + r/100)n Substituting the values = 50000 (1 + 4/100)3 By further calculation = 50000 (26/25)3 = 50000 × 26/25 × 26/25 × 26/25 = ₹ 56243.20 Here Compound Interest = A – P Substituting the values = 56243.20 – 50000 = ₹ 6243.20 11. Calculate the amount and the compound interest on ₹ 5000 in 2 years when the rate of interest for successive years is 6% and 8%, respectively. Solution: It is given that Principal = ₹ 5000 Period = 2 years Rate of interest for the first year = 6% Rate of interest for the second year = 8% We know that Amount for two years = P (1 + r/100)n Substituting the values = 5000 (1 + 6/100) (1 + 8/100) By further calculation = 5000 × 53/50 × 27/25 = ₹ 5724 Here Interest = A – P Substituting the values = 5724 – 5000 = ₹ 724 12. Calculate the amount and the compound interest on ₹ 17000 in 3 years when the rate of interest for successive years is 10%, 10% and 14%, respectively. Solution: It is given that Principal = ₹ 17000 Period = 3 years Rate of interest for 3 successive years = 10%, 10% and 14% We know that Amount after 3 years = P (1 + r/100)n Substituting the values = 17000 (1 + 10/100) (1 + 10/100) (1 + 14/100) By further calculation = 17000 × 11/10 × 11/10 × 57/50 = ₹ 23449.80 Here Amount of compound interest = A – P Substituting the values = 23449.80 – 17000 = ₹ 6449.80 13. A sum of ₹ 9600 is invested for 3 years at 10% per annum at compound interest. (i) What is the sum due at the end of the first year? (ii) What is the sum due at the end of the second year? (iii) Find the compound interest earned in 2 years. (iv) Find the difference between the answers in (ii) and (i) and find the interest on this sum for one year. (v) Hence, write down the compound interest for the third year. Solution: It is given that Principal = ₹ 9600 Rate of interest = 10% p.a. Period = 3 years We know that Interest for the first year = Prt/100 Substituting the values = (9600 × 10 × 1)/ 100 = ₹ 960 (i) Amount after one year = 9600 – 960 = ₹ 10560 So the principal for the second year = ₹ 10560 Here the interest for the second year = (10560 × 10 × 1)/ 100 = ₹ 1056 (ii) Amount after two years = 10560 + 1056 = ₹ 11616 (iii) Compound interest earned in 2 years = 960 + 10560 = ₹ 2016 (iv) Difference between the answers in (ii) and (i) = 11616 – 10560 = ₹ 1056 We know that Interest on ₹ 1056 for 1 year at the rate of 10% p.a. = (1056 × 10 × 1)/ 100 = ₹ 105.60 (v) Here Principal for the third year = ₹ 11616 So the interest for the third year = (11616 × 10 × 1)/ 100 = ₹ 1161.60 14. The simple interest on a certain sum of money for 2 years at 10% p.a. is ₹ 1600. Find the amount due and the compound interest on this sum of money at the same rate after 3 years, interest being reckoned annually. Solution: It is given that Period = 2 years Rate = 10% p.a. We know that Sum = (SI × 100)/ (r × n) Substituting the values = (1600 × 100)/ (10 × 2) = ₹ 8000 Here Amount after 3 years = P (1 + r/100)n Substituting the values = 8000 (1 + 10/100)3 By further calculation = 8000 × 11/10 × 11/10 × 11/10 = ₹ 10648 So the compound interest = A – P Substituting the values = 10648 – 8000 = ₹ 2648 15. Vikram borrowed ₹ 20000 from a bank at 10% per annum simple interest. He lent it to his friend Venkat at the same rate but compounded annually. Find his gain after 2 ½ years. Solution: First case- Principal = ₹ 20000 Rate = 10% p.a. Period = 2 ½ = 5/2 years We know that Simple interest = Prt/100 Substituting the values = (20000 × 10 × 5)/ (100 × 2) = ₹ 5000 Second case- Principal = ₹ 20000 Rate = 10% p.a. Period = 2 ½ years at compound interest We know that Amount = P (1 + r/100)n Substituting the values = 20000 (1 + 10/100)2 (1 + 10/ (2 × 100))2 By further calculation = 20000 × 11/10 × 11/10 × 21/20 = ₹ 25410 Here Compound Interest = A – P Substituting the values = 25410 – 20000 = ₹ 5410 So his gain after 2 years = CI – SI We get = 5410 – 5000 = ₹ 410 16. A man borrows ₹ 6000 at 5% compound interest. If he repays ₹ 1200 at the end of each year, find the amount outstanding at the beginning of the third year. Solution: It is given that Principal = ₹ 6000 Rate of interest = 5% p.a. We know that Interest for the first year = Prt/100 Substituting the values = (6000 × 5 × 1)/ 100 = ₹ 300 So the amount after one year = 6000 + 300 = ₹ 6300 Principal for the second year = ₹ 6300 Amount paid = ₹ 1200 So the balance = 6300 – 1200 = ₹ 5100 Here Interest for the second year = (5100 × 5 × 1)/ 100 = ₹ 255 Amount for the second year = 5100 + 255 = ₹ 5355 Amount paid = ₹ 1200 So the balance = 5355 – 1200 = ₹ 4155 17. Mr. Dubey borrows ₹ 100000 from State Bank of India at 11% per annum compound interest. He repays ₹ 41000 at the end of first year and ₹ 47700 at the end of second year. Find the amount outstanding at the beginning of the third year. Solution: It is given that Borrowed money (P) = ₹ 100000 Rate = 11% p.a. Time = 1 year We know that Amount after first year = Prt/100 Substituting the values = (100000 × 11 × 1)/ 100 By further calculation = 100000 + 11000 = ₹ 111000 Amount paid at the end of first year = ₹ 41000 So the principal for second year = 111000 – 41000 = ₹ 70000 We know that Amount after second year = P + (70000 × 11)/ 100 By further calculation = 70000 + 700 = ₹ 77700 So the amount paid at the end of second year = ₹ 47700 Here the amount outstanding at the beginning year = 77700 – 47700 = ₹ 30000 18. Jaya borrowed ₹ 50000 for 2 years. The rates of interest for two successive years are 12% and 15% respectively. She repays ₹ 33000 at the end of first year. Find the amount she must pay at the end of second year to clear her debt. Solution: It is given that Amount borrowed by Jaya = ₹ 50000 Period (n) = 2 years Rate of interest for two successive years are 12% and 15% respectively We know that Interest for the first year = Prt/100 Substituting the values = (50000 × 12 × 1)/ 100 = ₹ 6000 So the amount after first year = 50000 + 6000 = ₹ 56000 Amount repaid = ₹ 33000 Here Balance amount for the second year = 56000 – 33000 = ₹ 23000 Rate = 15% So the interest for the second year = (230000 × 15 × 1)/ 100 = ₹ 3450 Amount paid after second year = 23000 + 3450 = ₹ 26450 Exercise 2.2 1. Find the amount and the compound interest on ₹ 5000 for 2 years at 6% per annum, interest payable yearly. Solution: It is given that Principal (P) = ₹ 5000 Rate of interest (r) = 6% p.a. Period (n) = 2 years We know that Amount = P (1 + r/100)n Substituting the values = 5000 (1 + 6/100)2 By further calculation = 5000 × 53/50 × 53/50 = ₹ 5618 Here CI = A – P Substituting the values = 5618 – 5000 = ₹ 618 2. Find the amount and the compound interest on ₹ 8000 for 4 years at 10% per annum interest reckoned yearly. Solution: It is given that Principal (P) = ₹ 8000 Rate of interest (r) = 10% p.a. Period (n) = 4 years We know that Amount = P (1 + r/100)n Substituting the values = 8000 (1 + 10/100)4 By further calculation = 8000 × 11/10 × 11/10 × 11/10 × 11/10 = ₹ 11712.80 Here CI = A – P Substituting the values = 11712.80 – 8000 = ₹ 3712.80 3. If the interest is compounded half yearly, calculate the amount when the principal is ₹ 7400, the rate of interest is 5% and the duration is one year. Solution: It is given that Principal (P) = ₹ 7400 Rate of interest (r) = 5% Period (n) = 1 year We know that A = P (1 + r/(2 × 100))2×n Substituting the values = 7400 (1 + 5/200)2 By further calculation = 7400 × 205/200 × 205/200 = ₹ 7774.63 4. Find the amount and the compound interest on ₹ 5000 at 10% p.a. for 1 ½ years, compound interest reckoned semi-annually. Solution: It is given that Principal (P) = ₹ 5000 Rate of interest = 10% p.a. or 5% half-yearly Period (n) = 1 ½ years or 3 half-years We know that A = P (1 + r/100)n Substituting the values = 5000 (1 + 5/100)3 By further calculation = 5000 × 21/20 × 21/20 × 21/20 = ₹ 5788.12 Here CI = A – P Substituting the values = 5788.12 – 5000 = ₹ 788.12 5. Find the amount and the compound interest on ₹ 100000 compounded quarterly for 9 months at the rate of 4% p.a. Solution: It is given that Principal (P) = ₹ 100000 Rate of interest = 4% p.a. or 1% quarterly Period (n) = 9 months or 3 quarters We know that A = P (1 + r/100)n Substituting the values = 100000 (1 + 1/100)3 By further calculation = 100000 × 101/100 × 101/100 × 101/100 = ₹ 103030.10 Here CI = A – P Substituting the values = 103030.10 – 100000 = ₹ 3030.10 6. Find the difference between CI and SI on sum of ₹ 4800 for 2 years at 5% per annum payable yearly. Solution: It is given that Principal (P) = ₹ 4800 Rate of interest (r) = 5% p.a. Period (n) = 2 years We know that SI = Prt/100 Substituting the values = (4800 × 5 × 2)/ 100 = ₹ 480 If compounded yearly A = P (1 + r/100)n Substituting the values = 4800 (1 + 5/100)2 By further calculation = 4800 × 21/20 × 21/20 = ₹ 5292 Here CI = A – P Substituting the values = 5292 – 4800 = ₹ 492 So the difference between CI and SI = 492 – 480 = ₹ 12 7. Find the difference between the simple interest and compound interest on ₹ 2500 for 2 years at 4% per annum, compound interest being reckoned semi-annually. Solution: It is given that Principal (P) = ₹ 2500 Rate of interest (r) = 4% p.a. or 2% half-yearly Period (n) = 2 years or 4 half-years We know that SI = Prt/100 Substituting the values = (2500 × 4 × 2)/100 = ₹ 200 If compounded semi-annually A = P (1 + r/100)n Substituting the values = 2500 (1 + 2/100)4 By further calculation = 2500 × 51/50 × 51/50 × 51/50 × 51/50 = ₹ 2706.08 We know that CI = A – P Substituting the values = 2706.08 – 2500 = ₹ 206.08 So the difference between CI and SI = 206.08 – 200 = ₹ 6.08 8. Find the amount and the compound interest on ₹ 2000 in 2 years if the rate is 4% for the first year and 3% for the second year. Solution: It is given that Principal (P) = ₹ 2000 Rate of interest = 4% on the first year and 3% for the second year Period (n) = 2 years We know that Amount = P (1 + r/100)n Substituting the values = 2000 (1 + 4/100) (1 + 3/100) By further calculation = 2000 × 26/25 × 103/100 = ₹ 2142.40 Here CI = A – P Substituting the values = 2142.40 – 2000 = ₹ 142.40 9. Find the compound interest on ₹ 3125 for 3 years if the rates of interest for the first, second and third year are respectively 4%, 5% and 6% per annum. Solution: It is given that Principal (P) = ₹ 3125 Rate of interest for continuous = 4%, 5% and 6% Period (n) = 3 years We know that Amount = P (1 + r/100)n Substituting the values = 3125 (1 + 4/100) (1 + 5/100) (1 + 6/100) By further calculation = 3125 × 26/25 × 21/50 × 53/50 = ₹ 3617.25 Here CI = A – P Substituting the values = 3617.25 – 3125 = ₹ 492.25 10. What sum of money will amount to ₹ 9261 in 3 years at 5% per annum compound interest? Solution: It is given that Amount (A) = ₹ 9261 Rate of interest (r) = 5% per annum Period (n) = 3 years We know that A = P (1 + r/100)n Substituting the values 9261 = P (1 + 5/100)3 By further calculation 9261 = P (21/20)3 So we get P = (9261 × 20 × 20 × 20)/ (21 × 21 × 21) P = ₹ 8000 Therefore, the sum of money is ₹ 8000. 11. What sum invested at 4% per annum compounded semi-annually amounts to ₹ 7803 at the end of one year? Solution: It is given that Amount (A) = ₹ 7803 Rate of interest (r) = 4% p.a. or 2% semi-annually Period (n) = 1 year or 2 half years We know that A = P (1 + r/100)n Substituting the values = 7803 + (1 + 2/100)2 By further calculation = 7803 + (51/20)2 = 7803 × 50/51 × 50/51 = ₹ 7500 Hence, the principal is ₹ 7500. 12. What sum invested for 1 ½ years compounded half yearly at the rate of 4% p.a. will amount to ₹132651? Solution: It is given that Amount (A) = ₹ 132651 Rate of interest (r) = 4% p.a. or 2% half yearly Period (n) = 1 ½ years or 3 half years We know that A = P (1 + r/100)n It can be written as P = A ÷ (1 + r/100)n Substituting the values = 132651 ÷ (1 + 2/100)3 By further calculation = 132651 ÷ (51/50)3 So we get = 132651 × (50/51)3 = 132651 × 50/51 × 50/51 × 50/51 = ₹ 125000 Hence, the principal amount is ₹ 125000. 13. On what sum will the compound interest for 2 years at 4% per annum be ₹ 5712? Solution: It is given that CI = ₹ 5712 Rate of interest (r) = 4% p.a. Period (n) = 2 years We know that A = P (1 + r/100)n It can be written as CI = A – P = P (1 + r/100)n – P = P [(1 + r/100)n – 1] Substituting the values 5712 = P [(1 + 4/100)2 – 1] = P [(26/25)2 – 1] By further calculation = P [676/625 – 1] Taking LCM = P [(676 – 625)/ 625] = P × 51/625 Here P = 5712 × 625/51 = 112 × 625 = ₹ 70000 Hence, the principal amount is ₹ 70000. 14. A man invests ₹ 1200 for two years at compound interest. After one year the money amounts to ₹ 1275. Find the interest for the second year correct to the nearest rupee. Solution: It is given that Principal = ₹ 1200 After one year, the amount = ₹ 1275 So the interest for one year = 1275 – 1200 = ₹ 75 We know that Rate of interest = (SI × 100)/ (P × t) Substituting the values = (75 × 100)/ (1200 × 1) By further calculation = 75/12 = 25/4 = 6 ¼ % p.a. Here Interest for the second year on ₹ 1275 at the rate of 25/4% = Prt/100 Substituting the values = (1275 × 25 × 1)/ (100 × 4) By further calculation = 1275/16 = ₹ 79.70 = ₹ 80 15. At what rate percent per annum compound interest will ₹ 2304 amount to ₹ 2500 in 2 years? Solution: It is given that Amount = ₹ 2500 Principal = ₹ 2304 Period (n) = 2 years Consider r% p.a. as the rate of interest We know that A = P (1 + r/100)n It can be written as (1 + r/100)n = A/P Substituting the values (1 + r/100)2 = 2500/2304 By further calculation (1 + r/100)2 = 625/576 = (25/24)2 So we get 1 + r/100 = 25/24 r/100 = 25/24 – 1 Taking LCM r = 100/24 = 25/6 = 4 1/6 Hence, the rate of interest is 4 1/6% p.a. 16. A sum compounded annually becomes 25/16 time of itself in two years. Determine the rate of interest per annum. Solution: Consider sum (P) = x Amount (A) = 25/16x Period (n) = 2 years We know that A/P = (1 + r/100)n Substituting the values 25x/16x = (1 + r/100)2 By further calculation (1 + r/100)2 = (5/4)2 So we get 1 + r/100 = 5/4 r/100 = 5/4 – 1/1 = 1/4 By cross multiplication r = 100 × ¼ = 25 Hence, the rate of interest is 25% p.a. 17. At what rate percent will ₹ 2000 amount to ₹ 2315.25 in 3 years at compound interest? Solution: It is given that Principal (P) = ₹ 2000 Amount (A) = ₹ 2315.25 Period (n) = 3 years Consider r% p.a. as the rate of interest We know that A/P = (1 + r/100)n Substituting the values 2315.25/2000 = (1 + r/100)3 By further calculation (1 + r/100)3 = 231525/(100 × 2000) = 9261/8000 = (21/20)3 So we get 1 + r/100 = 21/20 It can be written as r/100 = 21/20 – 1 = 1/20 r = 100/20 = 5 Hence, the rate of interest is 5% p.a. 18. If ₹ 40000 amounts to ₹ 48620.25 in 2 years, compound interest payable half-yearly, find the rate of interest per annum. Solution: It is given that Principal (P) = ₹ 40000 Amount (A) = ₹ 48620.25 Period (n) = 2 years = 4 half years Consider rate of interest = r% p.a. = r/2% half yearly We know that A/P = (1 + r/100)n Substituting the values 48620.25/40000 = (1 + r/200)4 By further calculation (1 + r/200)4 = 4862025/ (100 × 40000) = 194481/160000 So we get (1 + r/200)4 = (21/20)4 It can be written as 1 + r/200 = 21/20 r/200 = 21/20 – 1 = 1/20 By cross multiplication r = 200 × 1/20 = 10 Hence the rate of interest per annum is 10%. 19. Determine the rate of interest for a sum that becomes 216/125 times of itself in 1 ½ years, compounded semi-annually. Solution: Consider principal (P) = x Amount (A) = 216/125 x Period (n) = 1 ½ years = 3 half years Take rate percent per year = 2r% and r% half yearly We know that A/P = (1 + r/100)n Substituting the values 216x/125x = (1 + r/100)3 By further calculation (1 + r/100)3 = 216/125 = (6/5)3 So we get 1 + r/100 = 6/5 r/100 = 6/5 – 1 = 1/5 By cross multiplication r = 100 × 1/5 = 20% So the rate percent per year = 2 × 20 = 40% 20. At what rate percent p.a. compound interest would ₹ 80000 amounts to ₹ 88200 in two years, interest being compounded yearly. Also find the amount after 3 years at the above rate of compound interest. Solution: It is given that Principal (P) = ₹ 80000 Amount (A) = ₹ 88200 Period (n) = 2 years Consider r% per annum as the rate of interest percent We know that A/P = (1 + r/100)n Substituting the values 88200/80000 = (1 + r/100)2 By further calculation (1 + r/100)2 = 441/400 = (21/20)2 So we get 1 + r/100 = 21/20 r/100 = 21/20 – 1 = 1/20 By cross multiplication r = 1/20 × 100 = 5 Hence, the rate of interest is 5% per annum. 21. A certain sum amounts to ₹ 5292 in 2 years and to ₹ 5556.60 in 3 years at compound interest. Find the rate and the sum. Solution: It is given that Amount after 2 years = ₹ 5292 Amount after 3 years = ₹ 5556.60 So the difference = 5556.60 – 5292 = ₹ 264.60 Here ₹ 264.60 is the interest on ₹ 5292 for one year We know that Rate % = (SI × 100)/ (P × t) Substituting the values = (264.60 × 100)/ (5292 × 1) Multiply and divide by 100 = (26460 × 100)/ (100 × 5292) = 5% Here A = P (1 + r/100)n Substituting the values 5292 = P (1 + 5/100)2 By further calculation P = 5292 ÷ (1 + 5/100)2 So we get P = 5292 ÷ (21/20)2 P = 5292 × 21/20 × 21/20 P = ₹ 4800 Hence, the rate is 5% and the sum is ₹ 4800. 22. A certain sum amounts to ₹ 798.60 after 3 years and ₹ 878.46 after 4 years. Find the interest rate and the sum. Solution: It is given that Amount after 3 years = ₹ 798.60 Amount after 4 years = ₹ 878.46 So the difference = 878.46 – 798.60 = ₹ 79.86 Here ₹ 79.86 is the interest on ₹ 798.60 for 1 year. We know that Rate = (SI × 100)/ (P × t) Substituting the values = (79.86 × 100)/ (798.60 × 1) Multiply and divide by 100 = (7986 × 100 × 100)/ (79860 × 100 × 1) = 10% Here A = P (1 + r/100)n It can be written as P = A ÷ (1 + r/100)n Substituting the values P = 798.60 ÷ (1 + 10/100)3 By further calculation P = 79860/100 × 10/11 × 10/11 × 10/11 P = ₹ 600 23. In what time will ₹ 15625 amount to ₹ 17576 at 4% per annum compound interest? Solution: It is given that Amount (A) = ₹ 17576 Principal (P) = ₹ 15625 Rate = 4% p.a. Consider n years as the period We know that A/P = (1 + r/100)n Substituting the values 17576/15625 = (1 + 4/100)n By further calculation (26/25)3 = (26/25)n So we get n = 3 24. (i) In what time will ₹ 1500 yield ₹ 496.50 as compound interest at 10% per annum compounded annually? (ii) Find the time (in years) in which ₹ 12500 will produce ₹ 3246.40 as compound interest at 8% per annum, interest compounded annually. Solution: (i) It is given that Principal (P) = ₹ 1500 CI = ₹ 496.50 So the amount (A) = P + SI Substituting the values = 1500 + 496.50 = ₹ 1996.50 Rate (r) = 10% p.a. We know that A = P (1 + r/100)n It can be written as A/P = (1 + r/100)n Substituting the values 1996.50/1500 = (1 + 10/100)n By further calculation 199650/(1500 × 100) = (11 /10)n So we get 1331/1000 = (11/10)n (11/10)3 = (11/10)n Here Time n = 3 years (ii) It is given that Principal (P) = ₹ 12500 CI = ₹ 3246.40 So the amount (A) = P + CI Substituting the values = 12500 + 3246.40 = ₹ 15746.40 Rate (r) = 8% p.a. We know that A = P (1 + r/100)n It can be written as A/P = (1 + r/100)n Substituting the values 15746.40/12500 = (1 + 8/100)n Multiply and divide by 100 1574640/ (12500 × 100) = (27/25)n By further calculation 78732/ (12500 × 5) = (27/ 25)n 19683/ (3125 × 5) = (27/25)n So we get 19683/15625 = (27/25)n (27/25)3 = (27/25)n Here Period = 3 years 25. ₹ 16000 invested at 10% p.a., compounded semi-annually, amounts to ₹ 18522, find the time period of investment. Solution: It is given that Principal (P) = ₹ 16000 Amount (A) = ₹ 18522 Rate = 10% p.a. or 5% semi-annually Consider period = n half years We know that A/P = (1 + r/100)n Substituting the values 18522/16000 = (1 + 5/100)n By further calculation 9261/8000 = (21/20)n So we get (21/20)3 = (21/20)n n = 3 half years Here Time = 3/2 = 1 ½ years 26. What sum will amount to ₹ 2782.50 in 2 years at compound interest, if the rates are 5% and 6% for the successive years? Solution: It is given that Amount (A) = ₹ 2782.50 Rate of interest for two successive years = 5% and 6% We know that A = P (1 + r/100)n Substituting the values 2782.50 = P (1 + 5/100) (1 + 6/100) By further calculation 2782.50 = P × 21/20 × 53/50 So we get P = 2782.50 × 20/21 × 50/53 Multiply and divide by 100 P = 278250/100 × 20/21 × 50/53 P = ₹ 2500 Hence, the principal is ₹ 2500. 27. A sum of money is invested at compound interest payable annually. The interest in two successive years is ₹ 225 and ₹ 240. Find: (i) the rate of interest (ii) the original sum (iii) the interest earned in the third year. Solution: It is given that Interest for the first year = ₹ 225 Interest for the second year = ₹ 240 So the difference = 240 – 225 = ₹ 15 Here ₹ 15 is the interest on ₹ 225 for 1 year (i) Rate = (SI × 100)/ (P × t) Substituting the values = (15 × 100)/ (225 × 1) So we get = 20/3 = 6 2/3% p.a. (ii) We know that Sum = (SI × 100)/ (R × t) Substituting the values = (225 × 100)/ (20/3 × 1) It can be written as = (225 × 100 × 3)/ (20 × 1) So we get = 225 × 15 = ₹ 3375 (iii) Here Amount after second year = 225 + 240 + 3375 = ₹ 3840 So the interest for the third year = Prt/100 Substituting the values = (3840 × 20 × 1)/ (100 × 3) = ₹ 256 28. On what sum of money will the difference between the compound interest and simple interest for 2 years be equal to ₹ 25 if the rate of interest charged for both is 5% p.a.? Solution: It is given that Sum (P) = ₹ 100 Rate (R) = 5% p.a. Period (n) = 2 years We know that SI = PRT/100 Substituting the values = (100 × 5 × 2)/ 100 = ₹ 10 So the amount when interest is compounded annually = P (1 + R/100)n Substituting the values = 100 (1 + 5/100)2 By further calculation = 100 × (21/20)2 = 100 × 21/20 × 21/20 So we get = ₹ 441/4 Here CI = A – P Substituting the values = 441/4 – 100 = ₹ 41/4 So the difference between CI and SI = 41/4 – 10 = ₹ ¼ If the difference is ₹ ¼ then sum = ₹ 100 If the difference is ₹ 25 then sum = (100 × 4)/ 1 × 25 = ₹ 10000 29. The difference between the compound interest for a year payable half-yearly and the simple interest on a certain sum of money lent out at 10% for a year is ₹ 15. Find the sum of money lent out. Solution: It is given that Sum = ₹ 100 Rate = 10% p.a. or 5% half yearly Period = 1 years or 2 half years We know that A = P (1 + R/100)n Substituting the values = 100 (1 + 5/100)2 By further calculation = 100 × 21/20 × 21/20 = ₹ 441/4 Here CI = A – P Substituting the values = 441/4 – 100 = ₹ 41/4 SI = PRT/100 Substituting the values = (100 × 10 × 1)/ 100 = ₹ 10 So the difference between CI and SI = 41/4 – 10 = ₹ ¼ Here if the difference is ₹ ¼ then sum = ₹ 100 If the difference is ₹ 15 then sum = (100 × 4 × 15)/ 1 = ₹ 6000 30. The amount at compound interest which is calculated yearly on a certain sum of money is ₹ 1250 in one year and ₹ 1375 after two years. Calculate the rate of interest. Solution: It is given that Amount after one year = ₹ 1250 Amount after two years = ₹ 1375 Here the difference = 1375 – 1250 = ₹ 125 So ₹ 125 is the interest on ₹ 1250 for 1 year We know that Rate of interest = (SI × 100)/ (P × t) Substituting the values = (125 × 100)/ (1250 × 1) = 10% 31. The simple interest on a certain sum for 3 years is ₹ 225 and the compound interest on the same sum at the same rate for 2 years is ₹ 153. Find the rate of interest and the principal. Solution: It is given that SI for 3 years = ₹ 225 SI for 2 years = (225 × 2)/ 3 = ₹ 150 CI for 2 years = ₹ 153 So the difference = 153 – 150 = ₹ 3 Here ₹ 3 is interest on one year i.e. ₹ 75 for one year We know that Rate = (SI × 100)/ (P × t) Substituting the values = (3 × 100)/ (75 × 1) = 4% SI for 3 years = ₹ 225 Rate = 4% p.a. So principal = (SI × 100)/ (R × t) Substituting the values = (225 × 100)/ (4 × 3) = ₹ 1875 32. Find the difference between compound interest on ₹ 8000 for 1 ½ years at 10% p.a. when compounded annually and semi-annually. Solution: It is given that Principal (P) = ₹ 8000 Rate = 10% p.a. or 5% half-yearly Period = 1 ½ years or 3 half years Case 1 – When compounded annually A = P (1 + r/100)n Substituting the values = 8000 (1 + 10/100) (1 + 5/100) By further calculation = 8000 × 11/10 × 21/20 = ₹ 9240 We know that CI = A – P Substituting the values = 9240 – 8000 = ₹ 1240 Case 2 – When compounded half-yearly A = P (1 + r/100)n Substituting the values = 8000 (1 + 5/100)3 By further calculation = 8000 × 21/20 × 21/20 × 21/20 = ₹ 9261 We know that CI = A – P Substituting the values = 9261 – 8000 = ₹ 1261 Here the difference between two CI = 1261 – 1240 = ₹ 21 33. A sum of money is lent out at compound interest for two years at 20% p.a., CI being reckoned yearly. If the same sum of money is lent out at compound interest at same rate percent per annum, CI being reckoned half-yearly, it would have fetched ₹ 482 more by way of interest. Calculate the sum of money lent out. Solution: It is given hat Sum = ₹ 100 Rate = 20% p.a. or 10% half-yearly Period = 2 years or 4 half-years Case 1 – When the interest is reckoned yearly A = P (1 + r/100)n Substituting the values = 100 (1 + 20/100)2 By further calculation = 100 × 6/5 × 6/5 = ₹ 144 We know that CI = A – P Substituting the values = 144 – 100 = ₹ 44 Case 2 – When the interest is reckoned half-yearly A = P (1 + r/100)n Substituting the values = 100 (1 + 10/100)4 By further calculation = 100 × 11/10 × 11/10 × 11/10 × 11/10 = ₹ 146.41 We know that CI = A – P Substituting the values = 146.41 – 100 = ₹ 46.41 So the difference between two CI = 46.41 – 44 = ₹ 2.41 If the difference is ₹ 2.41 then sum = ₹ 100 If the difference is ₹ 482 then sum = (100 × 482)/ 2.41 Multiplying and dividing by 100 = (100 × 482 × 100)/ 241 = ₹ 20000 34. A sum of money amounts to ₹ 13230 in one year and to ₹ 13891.50 in 1 ½ years at compound interest, compounded semi-annually. Find the sum and the rate of interest per annum. Solution: It is given that Amount after one year = ₹ 13230 Amount after 1 ½ years = ₹ 13891.50 So the difference = 13891.50 – 13230 = ₹ 661.50 Here ₹ 661.50 is the interest on ₹ 13230 for ½ years We know that Rate = (661.50 × 100 × 2)/ (13230 × 1) Multiplying and dividing by 100 = (66150 × 100 × 2)/ (13230 × 1 × 100) = 10% p.a. Here A = P (1 + r/100)n Substituting the values 13891.50 = P (1 + 5/100)3 By further calculation 13891.50 = P × 21/20 × 21/20 × 21/20 So we get P = 13891.50 × 20/21 × 20/21 × 20/21 P = ₹ 12000 Exercise 2.3 1. The present population of a town is 200000. Its population increases by 10% in the first year and 15% in the second year. Find the population of the town at the end of two years. Solution: We know that Population after 2 years = Present population × (1 + r/100)n Here the present population = 200000 Population after first year = 200000 × (1 + 10/100)1 By further calculation = 200000 × 11/10 = 220000 Population after two years = 220000 × (1 + 15/100)1 By further calculation = 220000 × 23/20 = 253000 2. The present population of a town is 15625. If the population increases at the rate of 4% every year, what will be the increase in the population in next 3 years? Solution: It is given that Present population (P) = 15625 Rate of increase (r) = 4% p.a. Period (n) = 3 years We know that Population after 3 years = P (1 + r/100)n Substituting the values = 15625 (1 + 4/100)3 By further calculation = 15625 × 26/25 × 26/25 × 26/25 = 17576 So the increase = 17576 – 15625 = 1951 3. The population of a city increase each year by 4% of what it had been at the beginning of each year. If its present population is 6760000, find: (i) its population 2 years hence (ii) its population 2 years ago. Solution: It is given that Present population = 6760000 Increase percent = 4% p.a. (i) We know that Population 2 years hence = P (1 + r/100)2 Substituting the values = 6760000 (1 + 4/100)2 By further calculation = 6760000 × 26/25 × 26/25 = 7311616 (ii) We know that A = 6760000 Population 2 years ago P = A + (1 + r/100)2 Substituting the values = 6760000 + (1 + 4/100)2 By further calculation = 6760000 + (26/25)2 = 6760000 × 25/26 × 25/26 = 6250000 4. The cost of a refrigerator is ₹ 9000. Its value depreciates at the rate of 5% ever year. Find the total depreciation in its value at the end of 2 years. Solution: It is given that Present value (P) = ₹ 9000 Rate of depreciation (r) = 5% p.a. Period (n) = 2 years We know that Value after 2 years = P (1 – r/100)n Substituting the values = 9000 (1 – 5/100)2 By further calculation = 9000 × 19/20 × 19/20 = ₹ 8122.50 So the total depreciation = 9000 – 8122.50 = ₹ 877.50 5. Dinesh purchased a scooter for ₹ 24000. The value of the scooter is depreciating at the rate of 5% per annum. Calculate its value after 3 years. Solution: It is given that Present value of scooter (P) = ₹ 24000 Rate of depreciation (r) = 5% Period (n) = 3 years We know that Value after 3 years = P (1 – r/100)n Substituting the values = 24000 (1 – 5/100)3 By further calculation = 24000 × 19/20 × 19/20 × 19/20 = ₹ 20577 6. A farmer increases his output of wheat in his farm every year by 8%. This year he produced 2187 quintals of wheat. What was the yearly produce of wheat two years ago? Solution: It is given that Present production of wheat = 2187 quintals Increase in production = 8% p.a. We know that Production of wheat 2 years ago = A ÷ (1 + r/100)n Substituting the values = 2187 ÷ (1 + 8/100)2 By further calculation = 2187 ÷ (27/25)2 So we get = 2187 × 25/27 × 25/27 = 1875 quintals 7. The value of a property decreases every year at the rate of 5%. If its present value is ₹ 411540, what was its value three years ago? Solution: It is given that Present value of property = ₹ 411540 Rate of decrease = 5% p.a. We know that Value of property 3 years ago = A ÷ (1 – r/100)n Substituting the values = 411540 ÷ (1 – 5/100)3 By further calculation = 411540 ÷ (19/20)3 So we get = 411540 × 20/19 × 20/19 × 20/19 = ₹ 480000 8. Ahmed purchased an old scooter for ₹ 16000. If the cost of the scooter after 2 years depreciates to ₹ 14440, find the rate of depreciation. Solution: It is given that Present value = ₹ 16000 Value after 2 years = ₹ 14440 Consider r% p.a. as the rate of depreciation We know that A/P = (1 – r/100)n Substituting the values 14440/16000 = (1 – r/100)2 By further calculation 361/400 = (1 – r/100)2 (19/20)2 = (1 – r/100)2 We can write it as 1 – r/100 = 19/20 So we get r/100 = 1 – 19/20 = 1/20 By cross multiplication r = 1/20 × 100 = 5% Hence, the rate of depreciation is 5%. 9. A factory increased its production of cars from 80000 in the year 2011-2012 to 92610 in 2014-15. Find the annual rate of growth of production of cars. Solution: It is given that Production of cars in 2011-2012 = 80000 Production of cars in 2014-2015 = 92610 Period (n) = 3 years Consider r% as the rate of increase We know that A/P = (1 + r/100)n Substituting the values 92610/80000 = (1 + r/100)3 By further calculation (21/20)3 = (1 + r/100)3 We can write it as 1 + r/100 = 21/20 r/100 = 21/20 – 1 = 1/20 By cross multiplication r = 1/20 × 100 = 5 Hence, the annual rate of growth of production of cars is 5% p.a. 10. The value of a machine worth ₹ 500000 is depreciating at the rate of 10% every year. In how many years will its value be reduced to ₹ 364500? Solution: It is given that Present value = ₹ 500000 Reduced value = ₹ 364500 Rate of depreciation = 10% p.a. Consider n years as the period We know that A/P = (1 – r/100)n Substituting the values 364500/500000 = (1 – 10/100)n By further calculation (9/10)n = 729/1000 = (9/10)3 So we get n = 3 Therefore, the period in which its value be reduced to ₹ 364500 is 3 years. 11. Afzal purchased an old motorbike for ₹ 16000. If the value of the motorbike after 2 years is ₹ 14440, find the rate of depreciation. Solution: It is given that CP of an old motorbike = ₹ 16000 Price after 2 years = ₹ 14440 Consider r% as the rate of depreciation We know that A/P = (1 – r/100)n Substituting the values 14440/16000 = (1 – r/100)2 By further calculation 361/400 = (1 – r/100)2 (19/20)2 = (1 – r/100)2 So we get 19/20 = 1 – r/100 r/100 = 1 – 19/20 = (20 – 19)/ 20 = 1/20 By cross multiplication r = 100/20 = 5 Hence, the rate of depreciation is 5%. 12. Mahindra set up a factory by investing ₹ 2500000. During the first two years, his profits were 5% and 10% respectively. If each year the profit was on previous year’s capital, calculate his total profit. Solution: It is given that Investment = ₹ 2500000 Rates of profit during first two years = 5% and 10% We know that Capital after two years (A) = P (1 + r/100)n Substituting the values = 2500000 (1 + 5/100) (1 + 10/100) By further calculation = 2500000 × 21/20 × 11/10 = ₹ 2887500 So the net profit = A – P Substituting the values = 2887500 – 2500000 = ₹ 387500 13. The value of a property is increasing at the rate of 25% every year. By what percent will the value of the property increase after 3 years? Solution: It is given that Original price of the property (P) = ₹ 100 Rate of increase (r) = 25% p.a. Period (n) = 3 years We know that Increased value after 3 years = P (1 + r/100)n Substituting the values = 100 (1 + 25/100)3 By further calculation = 100 × 5/4 × 5/4 × 5/4 = ₹ 3125/16 Here Increased value = 3125/16 – 100 Taking LCM = (3125 – 1600)/ 16 = 1525/16 So the percent increase after 3 years = 1525/16 = 95 5/16% 14. Mr. Durani bought a plot of land for ₹ 180000 and a car for ₹ 320000 at the same time. The value of the plot of land grows uniformly at the rate of 30% p.a.., while the value of the car depreciates by 20% in the first year and by 15% p.a. thereafter. If he sells the plot of land as well as the car after 3 years, what will be his profit or loss? Solution: It is given that Price of plot of land = ₹ 180000 Growth rate = 30% p.a. Period (n) = 3 years We know that Amount after 3 years = P (1 + R/100)n Substituting the values = 180000 (1 + 30/100)3 By further calculation = 180000 × (13/10)3 It can be written as = 180000 × 13/10 × 13/10 × 13/10 = ₹ 395460 Here Price of car = ₹ 320000 Rate of depreciation = 20% for the first year and 15% for next period Period (n) = 3 years We know that Amount after 3 years = A (1 – R1/100)n × (1 – R2/100)2 Substituting the values = 320000 (1 – 20/100) (1 – 15/100)2 By further calculation = 320000 × 4/5 × (17/20)2 So we get = 320000 × 4/5 × 17/20 × 17/20 = ₹ 184960 Here Total cost of plot and car = 180000 + 320000 = ₹ 500000 Total sale price of plot and car = 395460 + 184960 = ₹ 580420 We know that Profit = S.P. – C.P. Substituting the values = 580420 – 500000 = ₹ 80420 Chapter Test 1. ₹ 10000 was lent for one year at 10% per annum. By how much more will the interest be, if the sum was lent at 10% per annum, interest being compounded half yearly? Solution: It is given that Principal = ₹ 10000 Rate of interest (r) = 10% p.a. Period = 1 year We know that A = P (1 + r/100)n Substituting the values = 10000 (1 + 10/100)1 By further calculation = 10000 × 11/10 = ₹ 11000 Here Interest = A – P Substituting the values = 11000 – 10000 = ₹ 1000 In case 2, Rate (r) = 10% p.a. or 5% half-yearly Period (n) = 1 year or 2 half-years We know that A = P (1 + r/100)n Substituting the values = 10000 (1 + 5/100)2 By further calculation = 10000 × 21/20 × 21/20 = ₹ 11025 Here Interest = A – P Substituting the values = 11025 – 10000 = ₹ 1025 So the difference between the two interests = 1025 – 1000 = ₹ 25 2. A man invests ₹ 3072 for two years at compound interest. After one year the money amounts to ₹ 3264. Find the rate of interest and the amount due at the end of 2nd year. Solution: It is given that Principal (P) = ₹ 3072 Amount (A) = ₹ 3264 Period (n) = 1 year We know that A/P = (1 + r/100)n Substituting the values 3264/3072 = (1 + r/100)1 By further calculation 1 + r/100 = 17/16 r/100 = 17/16 – 1 = 1/16 By cross multiplication r = 100 × 1/16 = 25/4 = 6 ¼ Hence, the rate of interest is 6 ¼%. Here Amount after 2 years = 3072 (1 + 25/ (4 × 100))2 By further calculation = 3072 (1 + 1/16)2 So we get = 3072 × 17/16 × 17/16 = ₹ 3468 Hence, the amount due at the end of 2 years is ₹ 3468. 3. What sum will amount to ₹ 28090 in two years at 6% per annum compound interest? Also find the compound interest. Solution: It is given that Amount (A) = ₹ 28090 Rate (r) = 6% p.a. Period (n) = 2 years We know that P = A ÷ (1 + r/100)n Substituting the values = 28090 ÷ (1 + 6/100)2 By further calculation = 28090 ÷ (53/50)2 So we get = 28090 × 50/53 × 50/53 = ₹ 25000 Here Amount of CI = A – P Substituting the values = 28090 – 25000 = ₹ 3090 4. Two equal sums were lent at 5% and 6% per annum compound interest for 2 years. If the difference in the compound interest was ₹ 422, find: (i) the equal sums (ii) compound interest for each sum. Solution: Consider ₹ 100 as each equal sum Case I – Rate (r) = 5% Period (n) = 2 years We know that A = P (1 + r/100)n Substituting the values = 100 (1 + 5/100)2 It can be written as = 100 × 21/20 × 21/20 = ₹ 441/4 Here CI = A – P Substituting the values = 441/4 – 100 = ₹ 41/4 Case II – Rate of interest (R) = 6n Period (n) = 2 years We know that A = P (1 + r/100)n Substituting the values = 100 (1 + 6/100)2 It can be written as = 100 × 53/50 × 53/50 = ₹ 2809/25 Here CI = A – P Substituting the values = 2809/25 – 100 = ₹ 309/25 So the difference between the two interests = 309/25 – 41/4 Taking LCM = (1236 – 1025)/ 100 = ₹ 211/100 If the difference is ₹ 211/100, then equal sum = ₹ 100 If the difference is ₹ 422, then equal sum = (100 × 422 × 100)/ 211 = ₹ 20000 Here Amount in first case = 20000 (1 + 5/100)2 So we get = 20000 × (21/20)2 It can be written as = 20000 × 21/20 × 21/20 So we get = 44100/2 = ₹ 22050 CI = 22050 – 20000 = ₹ 2050 Amount in second case = 20000 (1 + 6/100)2 It can be written as = 20000 × 53/50 × 53/50 = ₹ 22472 CI = 22472 – 20000 = ₹ 2472 5. The compound interest on a sum of money for 2 years is ₹ 1331.20 and the simple interest on the same sum for the same period at the same rate is ₹ 1280. Find the sum and the rate of interest per annum. Solution: It is given that CI for 2 years = ₹ 1331.20 SI for 2 years = ₹ 1280 So the difference = 1331.20 – 1280 = ₹ 51.20 Here ₹ 51.20 is the simple interest on 1280/2 = ₹ 640 for one year We know that Rate = (SI × 100)/ (P × t) Substituting the values = (51.20 × 100)/ (640 × 1) Multiplying and dividing by 100 = (5120 × 100)/ (100 × 640) = 8% p.a. So the SI for two years at the rate of 8% pa Sum = (SI × 100)/ (r × t) Substituting the values = (1280 × 100)/ (8 × 2) = ₹ 8000 6. On what sum will the difference between the simple and compound interest for 3 years if the rate of interest is 10% p.a. is ₹ 232.50? Solution: Consider sum (P) = ₹ 100 Rate (r) = 10% p.a. Period (n) = 3 years We know that A = P (1 + r/100)n Substituting the values = 100 (1 + 10/100)3 By further calculation = 100 × 11/10 × 11/10 × 11/10 = ₹ 133.10 Here CI = A – P Substituting the values = 133.10 – 100 = ₹ 33.10 So the simple interest = PRT/100 Substituting the values = (100 × 10 × 3)/ 100 = ₹ 30 Difference = 33.10 – 30 = ₹ 3.10 Here if the difference is ₹ 3.10 then sum = ₹ 100 If the difference is ₹ 232.50 then sum = (100 × 232.50)/ 3.10 Multiplying and dividing by 100 = (100 × 23250)/ 310 = ₹ 7500 7. The simple interest on a certain sum for 3 years is ₹ 1080 and the compound interest on the same sum at the same rate for 2 years is ₹ 741.60. Find: (i) the rate of interest (ii) the principal. Solution: It is given that SI for 3 years = ₹ 1080 SI for 2 years = (1080 × 2)/ 3 = ₹ 720 CI for 2 years = ₹ 741.60 So the difference = 741.60 – 720 = ₹ 21.60 Here ₹ 21.60 is the SI on 720/2 = ₹ 360 for one year (i) We know that Rate = (SI × 100)/ (P × t) Substituting the values = (21.60 × 100)/ (360 × 1) Multiply and divide by 100 = (2160 × 100)/ (100 × 360 × 1) = 6% (ii) ₹ 1080 is SI for 3 years at the rate of 6% p.a. So the principal = (SI × 100)/ (r × t) Substituting the values = (1080 × 100)/ (6 × 3) = ₹ 6000 8. In what time will ₹ 2400 amount to ₹ 2646 at 10% p.a. compounded semi-annually? Solution: It is given that Amount (A) = ₹ 2646 Principal (P) = ₹ 2400 Rate (r) = 10% p.a. or 5% semi-annually Consider Period = n half-years We know that A/P = (1 + r/100)n Substituting the values 2646/2400 = (1 + 5/100)n By further calculation (21/20)n = 441/400 = (21/20)2 n = 2 Therefore, the time period is 2 half years or 1 year. 9. Sudarshan invested ₹ 60000 in a finance company and received ₹ 79860 after 1 ½ years. Find the rate of interest per annum compounded half-yearly. Solution: It is given that Principal (P) = ₹ 60000 Amount (A) = ₹ 79860 Period (n) = 1 ½ years = 3 half-years We know that A/P = (1 + r/100)n Substituting the values 79860/60000 = (1 + r/100)3 By further calculation (1 + r/100)3 = 1331/1000 = (11/10)3 We get 1 + r/100 = 11/10 r/100 = 11/10 – 1 = 1/10 By cross multiplication r = 1/10 × 100 = 10% half-yearly r = 10 × 2 = 20% p.a. Therefore, the rate of interest per annum compounded half-yearly is 20%. 10. The population of a city is 320000. If the annual birth rate is 9.2% and the annual death rate is 1.7%, calculate the population of the town after 3 years. Solution: It is given that Birth rate = 9.2% Death rate = 1.7% So the net growth rate = 9.2 – 1.7 = 7.5% Present population (P) = 320000 Period (n) = 3 years We know that Population after 3 years (A) = P (1 + r/100)n Substituting the values = 320000 (1 + 7.5/100)3 By further calculation = 320000 (1 + 3/40)3 = 320000 × (43/40)3 So we get = 320000 × 43/40 × 43/40 × 43/40 = 397535 11. The cost of a car, purchased 2 years ago, depreciates at the rate of 20% every year. If the present value of the car is ₹ 315600 find: (i) its purchase price (ii) its value after 3 years Solution: It is given that Present value of car = ₹ 315600 Rate of depreciation (r) = 20% (i) We know that Purchase price = A ÷ (1 – r/100)n Substituting the values = 315600 ÷ (1 – 20/100)2 By further calculation = 315600 × 5/4 × 5/4 = ₹ 493125 (ii) We know that Value after 3 years = 315600 × (1 – 20/100)3 By further calculation = 315600 × 4/5 × 4/5 × 4/5 = ₹ 161587.20 12. Amar Singh started a business with an initial investment of ₹ 400000. In the first year he incurred a loss of 4%. However, during the second year, he earned a profit of 5% which in the third year rose to 10%. Calculate his net profit for the entire period of 3 years. Solution: It is given that Investment (P) = ₹ 400000 Loss in the first year = 4% Profit in the second year = 5% Profit in the third year = 10% We know that Total amount after 3 years = P (1 + r/100)n Substituting the values = 400000 (1 – 4/100) (1 + 5/100) (1 + 10/100) By further calculation = 400000 × 24/25 × 21/20 × 11/10 = ₹ 443520 So the net profit after 3 years = 443520 – 400000 = ₹ 43520 On what sum of money will the interest of 5% pa for 4 years be the same as that on Rupees 2500 at 3% pa for 6 years?Answer: 2250 rs. P = 2250 rs.
What sum will Rs 43200 at 5% pa simple interest in 4 years?Expert-verified answer
The value of sum of money is Rupees 34560.
What sum of money will amount to 3704 in 3 years at 5% compound interest?Expert-verified answer
Using A = P(1 + r/100) n , we get 3704.40 = P(1 + 5/100) 3 => 370440/100 = P ×21/20×21/20×21/20 Or, P = 370440/100×20/21×20/21×20/21= 3200 Hence, required sum of money = Rs3200.
What is the simple interest on Rupees 1500 at the rate of 5% per annum for 3 years?This is Expert Verified Answer
The required interest is Rs. 225.
|