Class 7: Maths Chapter 13 solutions. Complete Class 7 Maths Chapter 13 Notes.<\/p>\n\n\n\n
RD Sharma 7th Maths Chapter 13, Class 7 Maths Chapter 13 solutions<\/p>\n\n\n\n
1. Find the simple interest, when: (iv) Principal = Rs 12000, Rate of Interest = 18% per annum and Time = 4 months. Solution:<\/strong><\/p>\n\n\n\n (i) Given Principal = Rs 2000, Rate of Interest = 5% per annum and Time = 5 years.<\/p>\n\n\n\n We know that simple interest = (P \u00d7 T \u00d7 R)\/100<\/p>\n\n\n\n On substituting these values in above equation we get<\/p>\n\n\n\n SI = (2000 \u00d7 5 \u00d7 5)\/100<\/p>\n\n\n\n = Rs 500<\/p>\n\n\n\n (ii) Given Principal = Rs 500, Rate of Interest = 12.5% per annum and Time = 4 years.<\/p>\n\n\n\n We know that simple interest = (P \u00d7 T \u00d7 R)\/100<\/p>\n\n\n\n On substituting these values in above equation we get<\/p>\n\n\n\n SI = (500 \u00d7 4 \u00d7 12.5)\/100<\/p>\n\n\n\n = Rs 250<\/p>\n\n\n\n (iii) Given Principal = Rs 4500, Rate of Interest = 4% per annum and Time = 6 months = \u00bd years<\/p>\n\n\n\n We know that simple interest = (P \u00d7 T \u00d7 R)\/100<\/p>\n\n\n\n On substituting these values in above equation we get<\/p>\n\n\n\n SI = (4500 \u00d7 \u00bd \u00d7 4)\/100<\/p>\n\n\n\n SI = (4500 \u00d7 1 \u00d7 4)\/100 \u00d7 2<\/p>\n\n\n\n = Rs 90<\/p>\n\n\n\n (iv) Given Principal = Rs 12000, Rate of Interest = 18% per annum and Time = 4 months = (4\/12) = (1\/3) years<\/p>\n\n\n\n We know that simple interest = (P \u00d7 T \u00d7 R)\/100<\/p>\n\n\n\n On substituting these values in above equation we get<\/p>\n\n\n\n SI = (12000 \u00d7 (1\/3) \u00d7 18)\/100<\/p>\n\n\n\n SI = (12000 \u00d7 1 \u00d7 18)\/100 \u00d7 3<\/p>\n\n\n\n = Rs 720<\/p>\n\n\n\n (v) Given Principal = Rs 1000, Rate of Interest = 10% per annum and<\/p>\n\n\n\n Time = 73 days = (73\/365) days<\/p>\n\n\n\n We know that simple interest = (P \u00d7 T \u00d7 R)\/100<\/p>\n\n\n\n On substituting these values in above equation we get<\/p>\n\n\n\n SI = (1000 \u00d7 (73\/365) \u00d7 10)\/100<\/p>\n\n\n\n SI = (1000 \u00d7 73 \u00d7 10)\/100 \u00d7 365<\/p>\n\n\n\n = Rs 20<\/p>\n\n\n\n 2. Find the interest on Rs 500 for a period of 4 years at the rate of 8% per annum. Also, find the amount to be paid at the end of the period.<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n Given Principal amount P = Rs 500<\/p>\n\n\n\n Time period T = 4 years<\/p>\n\n\n\n Rate of interest R = 8% p.a.<\/p>\n\n\n\n We know that simple interest = (P \u00d7 T \u00d7 R)\/100<\/p>\n\n\n\n On substituting these values in above equation we get<\/p>\n\n\n\n SI = (500 \u00d7 4 \u00d7 8)\/100<\/p>\n\n\n\n = Rs 160<\/p>\n\n\n\n Amount = Principal amount + Interest<\/p>\n\n\n\n = Rs 500 + 160<\/p>\n\n\n\n = Rs 660<\/p>\n\n\n\n 3. A sum of Rs 400 is lent at the rate of 5% per annum. Find the interest at the end of 2 years.<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n Given Principal amount P = Rs 400<\/p>\n\n\n\n Time period T = 2 years<\/p>\n\n\n\n Rate of interest R = 5% p.a.<\/p>\n\n\n\n We know that simple interest = (P \u00d7 T \u00d7 R)\/100<\/p>\n\n\n\n On substituting these values in above equation we get<\/p>\n\n\n\n SI = (400 \u00d7 2 \u00d7 5)\/100<\/p>\n\n\n\n = Rs 40<\/p>\n\n\n\n 4. A sum of Rs 400 is lent for 3 years at the rate of 6% per annum. Find the interest.<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n Principal amount P = Rs 400<\/p>\n\n\n\n Time period T = 3 years<\/p>\n\n\n\n Rate of interest R = 6% p.a.<\/p>\n\n\n\n We know that simple interest = (P \u00d7 T \u00d7 R)\/100<\/p>\n\n\n\n On substituting these values in above equation we get<\/p>\n\n\n\n SI = (400 \u00d7 3 \u00d7 6)\/100<\/p>\n\n\n\n = Rs 72<\/p>\n\n\n\n 5. A person deposits Rs 25000 in a firm who pays an interest at the rate of 20% per annum. Calculate the income he gets from it annually.<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n Given Principal amount P = Rs 25000<\/p>\n\n\n\n Time period T = 1 year<\/p>\n\n\n\n Rate of interest R = 20% p.a.<\/p>\n\n\n\n We know that simple interest = (P \u00d7 T \u00d7 R)\/100<\/p>\n\n\n\n On substituting these values in above equation we get<\/p>\n\n\n\n SI = (25000 \u00d7 1 \u00d7 20)\/100<\/p>\n\n\n\n = Rs 5000<\/p>\n\n\n\n 6. A man borrowed Rs 8000 from a bank at 8% per annum. Find the amount he has to pay after 4 \u00bd years.<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n Given Principal amount P = Rs 8000<\/p>\n\n\n\n Time period T = 4 \u00bd years = 9\/2 years<\/p>\n\n\n\n Rate of interest R = 8% p.a.<\/p>\n\n\n\n We know that simple interest = (P \u00d7 T \u00d7 R)\/100<\/p>\n\n\n\n On substituting these values in above equation we get<\/p>\n\n\n\n SI = (8000 \u00d7 (9\/2) \u00d7 8)\/100<\/p>\n\n\n\n = Rs 2880<\/p>\n\n\n\n Amount = Principal amount + Interest<\/p>\n\n\n\n = Rs 8000 + 2880<\/p>\n\n\n\n = Rs 10880<\/p>\n\n\n\n 7. Rakesh lent out Rs 8000 for 5 years at 15% per annum and borrowed Rs 6000 for 3 years at 12% per annum. How much did he gain or lose?<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n Given Principal amount P = Rs 8000<\/p>\n\n\n\n Time period T = 5 years<\/p>\n\n\n\n Rate of interest R = 15% p.a.<\/p>\n\n\n\n We know that simple interest = (P \u00d7 T \u00d7 R)\/100<\/p>\n\n\n\n On substituting these values in above equation we get<\/p>\n\n\n\n SI = (8000 \u00d7 5 \u00d7 15)\/100<\/p>\n\n\n\n = Rs 6000<\/p>\n\n\n\n Principal amount P = Rs 6000<\/p>\n\n\n\n Time period T = 3 years<\/p>\n\n\n\n Rate of interest R = 12% p.a.<\/p>\n\n\n\n We know that simple interest = (P \u00d7 T \u00d7 R)\/100<\/p>\n\n\n\n On substituting these values in above equation we get<\/p>\n\n\n\n SI = (6000 \u00d7 3 \u00d7 12)\/100<\/p>\n\n\n\n = Rs 2160<\/p>\n\n\n\n Amount gained by Rakesh = Rs 6000 \u2212 Rs 2160<\/p>\n\n\n\n = Rs 3840<\/p>\n\n\n\n 8. Anita deposits Rs 1000 in a savings bank account. The bank pays interest at the rate of 5% per annum. What amount can Anita get after one year?<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n Given Principal amount P = Rs 1000<\/p>\n\n\n\n Time period T = 1 year<\/p>\n\n\n\n Rate of interest R = 5% p.a.<\/p>\n\n\n\n We know that simple interest = (P \u00d7 T \u00d7 R)\/100<\/p>\n\n\n\n On substituting these values in above equation we get<\/p>\n\n\n\n SI = (1000 \u00d7 1 \u00d7 5)\/100<\/p>\n\n\n\n = Rs 50<\/p>\n\n\n\n Total amount paid after 1 year = Principal amount + Interest<\/p>\n\n\n\n = Rs 1000 + Rs 50<\/p>\n\n\n\n = Rs 1050<\/p>\n\n\n\n 9. Nalini borrowed Rs 550 from her friend at 8% per annum. She returned the amount after 6 months. How much did she pay?<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n Given Principal amount P = Rs 550<\/p>\n\n\n\n Time period T = \u00bd year<\/p>\n\n\n\n Rate of interest R = 8% p.a.<\/p>\n\n\n\n We know that simple interest = (P \u00d7 T \u00d7 R)\/100<\/p>\n\n\n\n On substituting these values in above equation we get<\/p>\n\n\n\n SI = (550 \u00d7 \u00bd \u00d7 8)\/100<\/p>\n\n\n\n = Rs 22<\/p>\n\n\n\n Total amount paid after \u00bd year = Principal amount + Interest<\/p>\n\n\n\n = Rs 550 + Rs 22<\/p>\n\n\n\n = Rs 572<\/p>\n\n\n\n 10. Rohit borrowed Rs 60000 from a bank at 9% per annum for 2 years. He lent this sum of money to Rohan at 10% per annum for 2 years. How much did Rohit earn from this transaction?<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n Given Principal amount P = Rs 60000<\/p>\n\n\n\n Time period T = 2 years<\/p>\n\n\n\n Rate of interest R = 10% p.a.<\/p>\n\n\n\n We know that simple interest = (P \u00d7 T \u00d7 R)\/100<\/p>\n\n\n\n On substituting these values in above equation we get<\/p>\n\n\n\n SI = (60000 \u00d7 2 \u00d7 10)\/100<\/p>\n\n\n\n = Rs 12000<\/p>\n\n\n\n Principal amount P = Rs 60000<\/p>\n\n\n\n Time period T = 2 years<\/p>\n\n\n\n Rate of interest R = 9% p.a.<\/p>\n\n\n\n We know that simple interest = (P \u00d7 T \u00d7 R)\/100<\/p>\n\n\n\n On substituting these values in above equation we get<\/p>\n\n\n\n SI = (60000 \u00d7 2 \u00d7 9)\/100<\/p>\n\n\n\n = Rs 10800<\/p>\n\n\n\n Amount gained by Rohit = Rs 12000 \u2212 Rs 10800<\/p>\n\n\n\n = Rs 1200<\/p>\n\n\n\n 11. Romesh borrowed Rs 2000 at 2% per annum and Rs 1000 at 5% per annum. He cleared his debt after 2 years by giving Rs 2800 and a watch. What is the cost of the watch?<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n Given Principal amount P = Rs 2000<\/p>\n\n\n\n Time period T = 2 years<\/p>\n\n\n\n Rate of interest R = 2% p.a.<\/p>\n\n\n\n We know that simple interest = (P \u00d7 T \u00d7 R)\/100<\/p>\n\n\n\n On substituting these values in above equation we get<\/p>\n\n\n\n SI = (2000 \u00d7 2 \u00d7 2)\/100<\/p>\n\n\n\n = Rs 80<\/p>\n\n\n\n Principal amount P = Rs 1000<\/p>\n\n\n\n Time period T = 2 years<\/p>\n\n\n\n Rate of interest R = 5% p.a.<\/p>\n\n\n\n We know that simple interest = (P \u00d7 T \u00d7 R)\/100<\/p>\n\n\n\n On substituting these values in above equation we get<\/p>\n\n\n\n SI = (1000 \u00d7 2 \u00d7 5)\/100<\/p>\n\n\n\n = Rs 100<\/p>\n\n\n\n Total amount that he will have to return = Rs. 2000 + 1000 + 80 + 100 = Rs. 3180<\/p>\n\n\n\n Amount repaid = Rs. 2800<\/p>\n\n\n\n Value of the watch = Rs. 3180 \u2013 2800 = Rs. 380<\/p>\n\n\n\n 12. Mr Garg lent Rs 15000 to his friend. He charged 15% per annum on Rs 12500 and 18% on the rest. How much interest does he earn in 3 years?<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n Given Principal amount P = Rs 15000<\/p>\n\n\n\n Time period T = 3 years<\/p>\n\n\n\n Rate of interest R = 15% p.a.<\/p>\n\n\n\n We know that simple interest = (P \u00d7 T \u00d7 R)\/100<\/p>\n\n\n\n On substituting these values in above equation we get<\/p>\n\n\n\n SI = (15000 \u00d7 3 \u00d7 15)\/100<\/p>\n\n\n\n = Rs 6750<\/p>\n\n\n\n Rest of the amount lent = Rs 15000 \u2212 Rs 12500 = Rs 2500<\/p>\n\n\n\n Rate of interest = 18 % p.a.<\/p>\n\n\n\n Time period = 3 years<\/p>\n\n\n\n We know that simple interest = (P \u00d7 T \u00d7 R)\/100<\/p>\n\n\n\n On substituting these values in above equation we get<\/p>\n\n\n\n SI = (2500 \u00d7 3 \u00d7 18)\/100<\/p>\n\n\n\n = Rs 1350<\/p>\n\n\n\n Total interest earned = Rs 6750 + Rs 1350 = Rs 8100<\/p>\n\n\n\n 13. Shikha deposited Rs 2000 in a bank which pays 6% simple interest. She withdrew Rs 700 at the end of first year. What will be her balance after 3 years?<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n Given Principal amount P = Rs 2000<\/p>\n\n\n\n Time period T = 1 year<\/p>\n\n\n\n Rate of interest R = 6% p.a.<\/p>\n\n\n\n We know that simple interest = (P \u00d7 T \u00d7 R)\/100<\/p>\n\n\n\n On substituting these values in above equation we get<\/p>\n\n\n\n SI = (2000 \u00d7 1 \u00d7 6)\/100<\/p>\n\n\n\n = Rs 120<\/p>\n\n\n\n So amount after 1 year = Principal amount + Interest = 2000 + 120 = Rs 2120<\/p>\n\n\n\n after 1 year, amount withdrawn = Rs 700<\/p>\n\n\n\n Principal amount left = Rs 2120 \u2212 Rs 700 = Rs 1420<\/p>\n\n\n\n Time period = 2 years<\/p>\n\n\n\n Rate of interest = 6% p.a.<\/p>\n\n\n\n We know that simple interest = (P \u00d7 T \u00d7 R)\/100<\/p>\n\n\n\n On substituting these values in above equation we get<\/p>\n\n\n\n SI = (1420 \u00d7 2 \u00d7 6)\/100<\/p>\n\n\n\n Interest after two years = Rs 170.40<\/p>\n\n\n\n Total amount after 3 years = Rs 1420 + Rs 170.40 = Rs 1590.40<\/p>\n\n\n\n 14. Reema took a loan of Rs 8000 from a money lender, who charged interest at the rate of 18% per annum. After 2 years, Reema paid him Rs 10400 and wrist watch to clear the debt. What is the price of the watch?<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n Given Principal amount P = Rs 8000<\/p>\n\n\n\n Time period T = 2 years<\/p>\n\n\n\n Rate of interest R = 18% p.a.<\/p>\n\n\n\n We know that simple interest = (P \u00d7 T \u00d7 R)\/100<\/p>\n\n\n\n On substituting these values in above equation we get<\/p>\n\n\n\n SI = (8000 \u00d7 2 \u00d7 18)\/100<\/p>\n\n\n\n = Rs 2880<\/p>\n\n\n\n Total amount payable by Reema after 2 years = Rs 8,000 + Rs 2,880<\/p>\n\n\n\n = Rs 10,880<\/p>\n\n\n\n Amount paid = Rs 10,400<\/p>\n\n\n\n Value of the watch = Rs 10,880 \u2212 Rs 10,400 = Rs 480<\/p>\n\n\n\n 15. Mr Sharma deposited Rs 20000 as a fixed deposit in a bank at 10% per annual. If 30% is deducted as income tax on the interest earned, find his annual income.<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n Given Principal amount P = Rs 20000<\/p>\n\n\n\n Time period T = 1 year<\/p>\n\n\n\n Rate of interest R = 10% p.a.<\/p>\n\n\n\n We know that simple interest = (P \u00d7 T \u00d7 R)\/100<\/p>\n\n\n\n On substituting these values in above equation we get<\/p>\n\n\n\n SI = (20000 \u00d7 1 \u00d7 10)\/100<\/p>\n\n\n\n = Rs 2000<\/p>\n\n\n\n Amount deducted as income tax = 30% of 2000 = (30 \u00d7 2000)\/100<\/p>\n\n\n\n = Rs 600<\/p>\n\n\n\n Annual interest after tax deduction = Rs 2,000 \u2212 Rs 600 = Rs 1,400<\/p>\n\n\n\n RD Sharma Solutions for Class 7 Maths Chapter 13\u2013Simple Interest<\/p>\n\n\n\n Download PDF<\/strong>: RD Sharma Solutions for Class 7 Maths Chapter 13\u2013Simple Interest PDF<\/a><\/p>\n\n\n\n RD Sharma isn’t the kind of author you’d bump into at lit fests. But his bestselling books have helped many <\/em>CBSE students lose their dread of <\/em>maths. Sunday Times profiles the tutor turned internet star<\/em> R D Sharma had a good laugh but said he shared little with his on-screen persona except for the love for maths. “I like to spend all my time thinking and writing about maths problems. I find it relaxing,” he says. When he is not writing books explaining mathematical concepts for classes 6 to 12 and engineering students, Sharma is busy dispensing his duty as vice-principal and head of department of science and humanities at Delhi government’s Guru Nanak Dev Institute of Technology.<\/p>\n\n\n\n
(i) Principal = Rs 2000, Rate of Interest = 5% per annum and Time = 5 years.
(ii) Principal = Rs 500, Rate of Interest = 12.5% per annum and Time = 4 years.
(iii) Principal = Rs 4500, Rate of Interest = 4% per annum and Time = 6 months.<\/strong><\/p>\n\n\n\n
(v) Principal = Rs 1000, Rate of Interest = 10% per annum and Time = 73 days.<\/strong><\/p>\n\n\n\nRD Sharma Solutions for Class 7 Maths Chapter 13: Download PDF<\/strong><\/h2>\n\n\n\n
Chapterwise RD Sharma Solutions for Class 7 Maths :<\/strong><\/h2>\n\n\n\n
About RD Sharma<\/h2>\n\n\n\n
He dreams of algorithms that would give most people nightmares. And, spends every waking hour thinking of ways to explain concepts like ‘series solution of linear differential equations’. Meet Dr Ravi Dutt Sharma \u2014 mathematics teacher and author of 25 reference books \u2014 whose name evokes as much awe as the subject he teaches. And though students have used his thick tomes for the last 31 years to ace the dreaded maths exam, it’s only recently that a spoof video turned the tutor into a YouTube star.<\/p>\n\n\n\nRead More<\/h2>\n\n\n\n