Class 7: Maths Chapter 13 solutions. Complete Class 7 Maths Chapter 13 Notes.
RD Sharma Solutions for Class 7 Maths Chapter 13–Simple Interest
RD Sharma 7th Maths Chapter 13, Class 7 Maths Chapter 13 solutions
1. Find the simple interest, when:
(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.
(iv) Principal = Rs 12000, Rate of Interest = 18% per annum and Time = 4 months.
(v) Principal = Rs 1000, Rate of Interest = 10% per annum and Time = 73 days.
Solution:
(i) Given Principal = Rs 2000, Rate of Interest = 5% per annum and Time = 5 years.
We know that simple interest = (P × T × R)/100
On substituting these values in above equation we get
SI = (2000 × 5 × 5)/100
= Rs 500
(ii) Given Principal = Rs 500, Rate of Interest = 12.5% per annum and Time = 4 years.
We know that simple interest = (P × T × R)/100
On substituting these values in above equation we get
SI = (500 × 4 × 12.5)/100
= Rs 250
(iii) Given Principal = Rs 4500, Rate of Interest = 4% per annum and Time = 6 months = ½ years
We know that simple interest = (P × T × R)/100
On substituting these values in above equation we get
SI = (4500 × ½ × 4)/100
SI = (4500 × 1 × 4)/100 × 2
= Rs 90
(iv) Given Principal = Rs 12000, Rate of Interest = 18% per annum and Time = 4 months = (4/12) = (1/3) years
We know that simple interest = (P × T × R)/100
On substituting these values in above equation we get
SI = (12000 × (1/3) × 18)/100
SI = (12000 × 1 × 18)/100 × 3
= Rs 720
(v) Given Principal = Rs 1000, Rate of Interest = 10% per annum and
Time = 73 days = (73/365) days
We know that simple interest = (P × T × R)/100
On substituting these values in above equation we get
SI = (1000 × (73/365) × 10)/100
SI = (1000 × 73 × 10)/100 × 365
= Rs 20
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.
Solution:
Given Principal amount P = Rs 500
Time period T = 4 years
Rate of interest R = 8% p.a.
We know that simple interest = (P × T × R)/100
On substituting these values in above equation we get
SI = (500 × 4 × 8)/100
= Rs 160
Amount = Principal amount + Interest
= Rs 500 + 160
= Rs 660
3. A sum of Rs 400 is lent at the rate of 5% per annum. Find the interest at the end of 2 years.
Solution:
Given Principal amount P = Rs 400
Time period T = 2 years
Rate of interest R = 5% p.a.
We know that simple interest = (P × T × R)/100
On substituting these values in above equation we get
SI = (400 × 2 × 5)/100
= Rs 40
4. A sum of Rs 400 is lent for 3 years at the rate of 6% per annum. Find the interest.
Solution:
Principal amount P = Rs 400
Time period T = 3 years
Rate of interest R = 6% p.a.
We know that simple interest = (P × T × R)/100
On substituting these values in above equation we get
SI = (400 × 3 × 6)/100
= Rs 72
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.
Solution:
Given Principal amount P = Rs 25000
Time period T = 1 year
Rate of interest R = 20% p.a.
We know that simple interest = (P × T × R)/100
On substituting these values in above equation we get
SI = (25000 × 1 × 20)/100
= Rs 5000
6. A man borrowed Rs 8000 from a bank at 8% per annum. Find the amount he has to pay after 4 ½ years.
Solution:
Given Principal amount P = Rs 8000
Time period T = 4 ½ years = 9/2 years
Rate of interest R = 8% p.a.
We know that simple interest = (P × T × R)/100
On substituting these values in above equation we get
SI = (8000 × (9/2) × 8)/100
= Rs 2880
Amount = Principal amount + Interest
= Rs 8000 + 2880
= Rs 10880
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?
Solution:
Given Principal amount P = Rs 8000
Time period T = 5 years
Rate of interest R = 15% p.a.
We know that simple interest = (P × T × R)/100
On substituting these values in above equation we get
SI = (8000 × 5 × 15)/100
= Rs 6000
Principal amount P = Rs 6000
Time period T = 3 years
Rate of interest R = 12% p.a.
We know that simple interest = (P × T × R)/100
On substituting these values in above equation we get
SI = (6000 × 3 × 12)/100
= Rs 2160
Amount gained by Rakesh = Rs 6000 − Rs 2160
= Rs 3840
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?
Solution:
Given Principal amount P = Rs 1000
Time period T = 1 year
Rate of interest R = 5% p.a.
We know that simple interest = (P × T × R)/100
On substituting these values in above equation we get
SI = (1000 × 1 × 5)/100
= Rs 50
Total amount paid after 1 year = Principal amount + Interest
= Rs 1000 + Rs 50
= Rs 1050
9. Nalini borrowed Rs 550 from her friend at 8% per annum. She returned the amount after 6 months. How much did she pay?
Solution:
Given Principal amount P = Rs 550
Time period T = ½ year
Rate of interest R = 8% p.a.
We know that simple interest = (P × T × R)/100
On substituting these values in above equation we get
SI = (550 × ½ × 8)/100
= Rs 22
Total amount paid after ½ year = Principal amount + Interest
= Rs 550 + Rs 22
= Rs 572
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?
Solution:
Given Principal amount P = Rs 60000
Time period T = 2 years
Rate of interest R = 10% p.a.
We know that simple interest = (P × T × R)/100
On substituting these values in above equation we get
SI = (60000 × 2 × 10)/100
= Rs 12000
Principal amount P = Rs 60000
Time period T = 2 years
Rate of interest R = 9% p.a.
We know that simple interest = (P × T × R)/100
On substituting these values in above equation we get
SI = (60000 × 2 × 9)/100
= Rs 10800
Amount gained by Rohit = Rs 12000 − Rs 10800
= Rs 1200
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?
Solution:
Given Principal amount P = Rs 2000
Time period T = 2 years
Rate of interest R = 2% p.a.
We know that simple interest = (P × T × R)/100
On substituting these values in above equation we get
SI = (2000 × 2 × 2)/100
= Rs 80
Principal amount P = Rs 1000
Time period T = 2 years
Rate of interest R = 5% p.a.
We know that simple interest = (P × T × R)/100
On substituting these values in above equation we get
SI = (1000 × 2 × 5)/100
= Rs 100
Total amount that he will have to return = Rs. 2000 + 1000 + 80 + 100 = Rs. 3180
Amount repaid = Rs. 2800
Value of the watch = Rs. 3180 – 2800 = Rs. 380
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?
Solution:
Given Principal amount P = Rs 15000
Time period T = 3 years
Rate of interest R = 15% p.a.
We know that simple interest = (P × T × R)/100
On substituting these values in above equation we get
SI = (15000 × 3 × 15)/100
= Rs 6750
Rest of the amount lent = Rs 15000 − Rs 12500 = Rs 2500
Rate of interest = 18 % p.a.
Time period = 3 years
We know that simple interest = (P × T × R)/100
On substituting these values in above equation we get
SI = (2500 × 3 × 18)/100
= Rs 1350
Total interest earned = Rs 6750 + Rs 1350 = Rs 8100
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?
Solution:
Given Principal amount P = Rs 2000
Time period T = 1 year
Rate of interest R = 6% p.a.
We know that simple interest = (P × T × R)/100
On substituting these values in above equation we get
SI = (2000 × 1 × 6)/100
= Rs 120
So amount after 1 year = Principal amount + Interest = 2000 + 120 = Rs 2120
after 1 year, amount withdrawn = Rs 700
Principal amount left = Rs 2120 − Rs 700 = Rs 1420
Time period = 2 years
Rate of interest = 6% p.a.
We know that simple interest = (P × T × R)/100
On substituting these values in above equation we get
SI = (1420 × 2 × 6)/100
Interest after two years = Rs 170.40
Total amount after 3 years = Rs 1420 + Rs 170.40 = Rs 1590.40
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?
Solution:
Given Principal amount P = Rs 8000
Time period T = 2 years
Rate of interest R = 18% p.a.
We know that simple interest = (P × T × R)/100
On substituting these values in above equation we get
SI = (8000 × 2 × 18)/100
= Rs 2880
Total amount payable by Reema after 2 years = Rs 8,000 + Rs 2,880
= Rs 10,880
Amount paid = Rs 10,400
Value of the watch = Rs 10,880 − Rs 10,400 = Rs 480
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.
Solution:
Given Principal amount P = Rs 20000
Time period T = 1 year
Rate of interest R = 10% p.a.
We know that simple interest = (P × T × R)/100
On substituting these values in above equation we get
SI = (20000 × 1 × 10)/100
= Rs 2000
Amount deducted as income tax = 30% of 2000 = (30 × 2000)/100
= Rs 600
Annual interest after tax deduction = Rs 2,000 − Rs 600 = Rs 1,400
RD Sharma Solutions for Class 7 Maths Chapter 13: Download PDF
RD Sharma Solutions for Class 7 Maths Chapter 13–Simple Interest
Download PDF: RD Sharma Solutions for Class 7 Maths Chapter 13–Simple Interest PDF
Chapterwise RD Sharma Solutions for Class 7 Maths :
- Chapter 1–Integers
- Chapter 2–Fractions
- Chapter 3–Decimals
- Chapter 4–Rational Numbers
- Chapter 5–Operations On Rational Numbers
- Chapter 6–Exponents
- Chapter 7–Algebraic Expressions
- Chapter 8–Linear Equations in One Variable
- Chapter 9–Ratio And Proportion
- Chapter 10–Unitary Method
- Chapter 11–Percentage
- Chapter 12–Profit And Loss
- Chapter 13–Simple Interest
- Chapter 14–Lines And Angles
- Chapter 15–Properties of Triangles
- Chapter 16–Congruence
- Chapter 17–Constructions
- Chapter 18–Symmetry
- Chapter 19–Visualising Solid Shapes
- Chapter 20–Mensuration – I (Perimeter and area of rectilinear figures)
- Chapter 21–Mensuration – II (Area of Circle)
- Chapter 22–Data Handling – I (Collection and Organisation of Data)
- Chapter 23–Data Handling – II Central Values
- Chapter 24–Data Handling – III (Constructions of Bar Graphs)
- Chapter 25–Data Handling – IV (Probability)
About RD Sharma
RD Sharma isn't the kind of author you'd bump into at lit fests. But his bestselling books have helped many CBSE students lose their dread of maths. Sunday Times profiles the tutor turned internet star
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 — mathematics teacher and author of 25 reference books — 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.
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.