RD Sharma Solutions Class 7 Chapter 13 Simple Interest

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Question :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 1:
(i) Principal = Rs 2000, Rate of Interest = 5% per annum and Time = 5 years.
Formula of Simple interest = ((P × T × R))/100
By putting the values in formula, 
SI = ((2000 × 5 × 5))/100 = Rs 500
 
(ii) Principal = Rs 500, Rate of Interest = 12.5% per annum and Time = 4 years.
Formula of simple interest = ((P × T × R))/100
By putting the values in formula,
SI = ((500 × 4 × 12.5))/100 = Rs 250
 
(iii) Principal = Rs 4500, Rate of Interest = 4% per annum and Time = 6 months = ½ years
Formula simple interest = ((P × T × R))/100
By putting the values in formula,
SI = ((4500 × 1/2  × 4))/100
SI = ((4500 × 1 × 4))/(100 ×2) = Rs 90
 
(iv) Principal = Rs 12000, Rate of Interest = 18% per annum and Time= 4 months 
4 months = (4/12) = (1/3) years
Formula of simple interest = ((P × T × R))/100
By putting the values in formula,
SI = ((12000 × 1/3  × 18))/100
SI = ((12000 × 1 × 18))/(100 ×  3) = Rs 720
 
(v) Principal = Rs 1000, Rate of Interest = 10% per annum and
Time = 73 days = (73/365) days
Formula of simple interest = ((P × T × R))/100
By putting the values in formula,
SI = ((1000 × 73/365  × 10))/100
SI = ((1000 × 73 × 10))/(100×365) = Rs 20
 
Question :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  2:
Principal amount P = Rs 500; Time period T = 4 years; Rate of interest R = 8% p.a.
Formula of Simple interest = ((P × T × R))/100
By putting the values in formula
SI = ((500 × 4 × 8))/100 = Rs 160
Amount = Principal amount + Interest
Amount = Rs 500 + 160
Amount = Rs 660
 
Question :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  3:
Principal amount P = Rs 400; Time period T = 2 years; Rate of interest R = 5% p.a.
Formula of simple interest = ((P × T × R))/100
By putting the values in formula,
SI = ((400 × 2 × 5))/100
SI = Rs 40
 
Question :4. A sum of Rs 400 is lent for 3 years at the rate of 6% per annum. Find the interest. 
Solution   4:
Principal amount P = Rs 400; Time period T = 3 years; Rate of interest R = 6% p.a.
Formula of simple interest = ((P × T × R))/100
By putting the values in formula,
SI = ((400 × 3 × 6))/100
SI = Rs 72
 
Question :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   5:
Principal amount P = Rs 25000; Time period T = 1 year; Rate of interest R = 20% p.a.
Formula of simple interest = ((P × T × R))/100
By putting the values in formula
SI = ((25000 × 1 × 20))/100
SI = Rs 5000
 
Question :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 6:
Principal amount P = Rs 8000; Time period T = 4 ½ years = 9/2 years; Rate of interest R = 8% p.a.
Formula of simple interest = ((P × T × R))/100
By putting the values in formula 
SI = ((8000 × 9/2  × 8))/100
SI = Rs, 2880
Amount = Principal amount + Interest
Amount = Rs 8000 + 2880
Amount = Rs 10880
 
Question :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  7:
Principal amount P = Rs 8000; Time period T = 5 years; Rate of interest R = 15% p.a.
Formula of simple interest = ((P × T × R))/100
By putting the values in formula,
SI = ((8000 × 5 × 15))/100
SI = Rs 6000
 
Principal amount P = Rs 6000; Time period T = 3 years; Rate of interest R = 12% p.a.
Formula of simple interest = ((P × T × R))/100
By putting the values in formula,
SI = ((6000 × 3 × 12))/100
SI = Rs 2160
Amount gained by Rakesh = Rs 6000 − Rs 2160
 = Rs 3840
 
Question :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  8:
Principal amount P = Rs 1000; Time period T = 1 year; Rate of interest R = 5% p.a.
Formula of simple interest = ((P × T × R))/100
By putting the values in Formula
SI = ((1000 × 1 × 5))/100
SI = Rs 50
Total amount paid after 1 year = Principal amount + Interest
= Rs 1000 + Rs 50
= Rs 1050
 
Question :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  9:
Principal amount P = Rs 550; Time period T = ½ year; Rate of interest R = 8% p.a.
Formula of simple interest = ((P × T × R))/100
On putting the values in formula
SI = ((550 × 1/2  × 8))/100
SI = Rs 22
Total amount paid after ½ year = Principal amount + Interest
= Rs 550 + Rs 22
= Rs 572
 
Question :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  10:
Principal amount P = Rs 60000; Time period T = 2 years; Rate of interest R = 10% p.a.
Formula of simple interest = ((P × T × R))/100
By putting the values in formula,
SI = ((60000 × 2 × 10))/100
SI = Rs 12000
 
Principal amount P = Rs 60000; Time period T = 2 years; Rate of interest R = 9% p.a.
Formula of simple interest = ((P × T × R))/100
On putting the values in formula
SI = ((60000 × 2 × 9))/100
SI = Rs 10800
Amount gained by Rohit = Rs 12000 − Rs 10800
= Rs 1200
 
Question :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  11:
Principal amount P = Rs 2000; Time period T = 2 years; Rate of interest R = 2% p.a.
Formula of simple interest = ((P × T × R))/100
By putting the values in above formula,
SI = ((2000 × 2 × 2))/100
SI = Rs 80
 
Principal amount P = Rs 1000; Time period T = 2 years; Rate of interest R = 5% p.a.
Formula of simple interest = ((P × T × R))/100
On putting the values in formula
SI = ((1000 × 2 × 5))/100
SI = 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
 
Question :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  12:
Principal amount P = Rs 15000; Time period T = 3 years; Rate of interest R = 15% p.a.
Formula of simple interest = ((P × T × R))/100
On putting the values in above formula 
SI = ((15000 × 3 × 15))/100 
SI = Rs 6750
Rest of the amount lent = Rs 15000 − Rs 12500 = Rs 2500
Rate of interest = 18 % p.a.
Time period = 3 years
SI = ((2500 × 3 × 18))/100
SI= Rs 1350
Total interest earned = Rs 6750 + Rs 1350 = Rs 8100
 
Question :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  13:
Principal amount P = Rs 2000
Time period T = 1 year
Rate of interest R = 6% p.a.
Formula of simple interest = ((P × T × R))/100
On putting the values in above formula
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.; Simple interest = ((P × T × R))/100
BY putting these values in formula 
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
 
Question :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  14:
Principal amount P = Rs 8000; Time period T = 2 years; Rate of interest R = 18% p.a.
Formula of simple interest = ((P × T × R))/100
By putting the values in above formula 
SI = ((8000 × 2 × 18))/100
SI = 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
 
Question :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  15:
Principal amount P = Rs 20000; Time period T = 1 year; Rate of interest R = 10% p.a.
Formula of simple interest = ((P × T × R))/100
By putting the values in above formula
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
NCERT Exemplar Solutions Class 7 Maths Algebraic Expressions
NCERT Exemplar Solutions Class 7 Maths Comparing Quantities
NCERT Exemplar Solutions Class 7 Maths Data Handling
NCERT Exemplar Solutions Class 7 Maths Exponents and Powers
NCERT Exemplar Solutions Class 7 Maths Fractions and Decimals
NCERT Exemplar Solutions Class 7 Maths Integers
NCERT Exemplar Solutions Class 7 Maths Lines and Angles
NCERT Exemplar Solutions Class 7 Maths Perimeter and Area
NCERT Exemplar Solutions Class 7 Maths Practical Geometry
NCERT Exemplar Solutions Class 7 Maths Rational Numbers
NCERT Exemplar Solutions Class 7 Maths Simple Equation
NCERT Exemplar Solutions Class 7 Maths Triangles and Its Properties
RS Aggarwal Class 7 Mathematics Solutions Chapter 1 Integers
RS Aggarwal Class 7 Mathematics Solutions Chapter 2 Fractions
RS Aggarwal Class 7 Mathematics Solutions Chapter 3 Decimals
RS Aggarwal Class 7 Mathematics Solutions Chapter 4 Rational Numbers
RS Aggarwal Class 7 Mathematics Solutions Chapter 5 Exponents
RS Aggarwal Class 7 Mathematics Solutions Chapter 6 Algebraic Expressions
RS Aggarwal Class 7 Mathematics Solutions Chapter 7 Linear Equations in One Variable
RS Aggarwal Class 7 Mathematics Solutions Chapter 8 Ratio and Proportion
RS Aggarwal Class 7 Mathematics Solutions Chapter 9 Unitary Method
RS Aggarwal Class 7 Mathematics Solutions Chapter 10 Percentage
RS Aggarwal Class 7 Mathematics Solutions Chapter 11 Profit and Loss
RS Aggarwal Class 7 Mathematics Solutions Chapter 12 Simple Interest
RS Aggarwal Class 7 Mathematics Solutions Chapter 13 Lines and Angles
RS Aggarwal Class 7 Mathematics Solutions Chapter 14 Properties of Parallel Lines
RS Aggarwal Class 7 Mathematics Solutions Chapter 15 Properties of Triangles
RS Aggarwal Class 7 Mathematics Solutions Chapter 16 Congruence
RS Aggarwal Class 7 Mathematics Solutions Chapter 17 Constructions
RS Aggarwal Class 7 Mathematics Solutions Chapter 18 Reflection and Rotational Symmetry
RS Aggarwal Class 7 Mathematics Solutions Chapter 19 Three-Dimensional Shapes
RS Aggarwal Class 7 Mathematics Solutions Chapter 20 Mensuration
RS Aggarwal Class 7 Mathematics Solutions Chapter 21 Collection and Organisation of Data
RS Aggarwal Class 7 Mathematics Solutions Chapter 22 Bar Graph
RS Aggarwal Class 7 Mathematics Solutions Chapter 23 Probability
RD Sharma Solutions Class 7 Maths
RD Sharma Solutions Class 7 Chapter 1 Integers
RD Sharma Solutions Class 7 Chapter 2 Fraction
RD Sharma Solutions Class 7 Chapter 3 Decimals
RD Sharma Solutions Class 7 Chapter 4 Rational Numbers
RD Sharma Solutions Class 7 Chapter 5 Operation on Decimal Numbers
RD Sharma Solutions Class 7 Chapter 6 Exponents
RD Sharma Solutions Class 7 Chapter 7 Algebraic Expressions
RD Sharma Solutions Class 7 Chapter 8 Linear Equations in One Variable
RD Sharma Solutions Class 7 Chapter 9 Ratio and Proportion
RD Sharma Solutions Class 7 Chapter 10 Unitary Method
RD Sharma Solutions Class 7 Chapter 11 Percentage
RD Sharma Solutions Class 7 Chapter 12 Profit and Loss
RD Sharma Solutions Class 7 Chapter 13 Simple Interest
RD Sharma Solutions Class 7 Chapter 14 Line and Angles
RD Sharma Solutions Class 7 Chapter 15 Properties of Triangle
RD Sharma Solutions Class 7 Chapter 16 Congruence
RD Sharma Solutions Class 7 Chapter 17 Constructions
RD Sharma Solutions Class 7 Chapter 18 Symmetry
RD Sharma Solutions Class 7 Chapter 19 Visualising Solid Shapes
RD Sharma Solutions Class 7 Chapter 20 Mensuration Perimeter and Area of Rectilinear Figures
RD Sharma Solutions Class 7 Chapter 21 Mensuration Area of Circle
RD Sharma Solutions Class 7 Chapter 22 Data Handling Collection and Organisation of Data
RD Sharma Solutions Class 7 Chapter 23 Data Handling Central Values
RD Sharma Solutions Class 7 Chapter 24 Data Handling Constructions of Bar Graph
RD Sharma Solutions Class 7 Chapter 25 Data Handling Probability
RS Aggarwal Class 7 Mathematics Solutions
RS Aggarwal Class 7 Mathematics Solutions Chapter 1 Integers
RS Aggarwal Class 7 Mathematics Solutions Chapter 2 Fractions
RS Aggarwal Class 7 Mathematics Solutions Chapter 3 Decimals
RS Aggarwal Class 7 Mathematics Solutions Chapter 4 Rational Numbers
RS Aggarwal Class 7 Mathematics Solutions Chapter 5 Exponents
RS Aggarwal Class 7 Mathematics Solutions Chapter 6 Algebraic Expressions
RS Aggarwal Class 7 Mathematics Solutions Chapter 7 Linear Equations in One Variable
RS Aggarwal Class 7 Mathematics Solutions Chapter 8 Ratio and Proportion
RS Aggarwal Class 7 Mathematics Solutions Chapter 9 Unitary Method
RS Aggarwal Class 7 Mathematics Solutions Chapter 10 Percentage
RS Aggarwal Class 7 Mathematics Solutions Chapter 11 Profit and Loss
RS Aggarwal Class 7 Mathematics Solutions Chapter 12 Simple Interest
RS Aggarwal Class 7 Mathematics Solutions Chapter 13 Lines and Angles
RS Aggarwal Class 7 Mathematics Solutions Chapter 14 Properties of Parallel Lines
RS Aggarwal Class 7 Mathematics Solutions Chapter 15 Properties of Triangles
RS Aggarwal Class 7 Mathematics Solutions Chapter 16 Congruence
RS Aggarwal Class 7 Mathematics Solutions Chapter 17 Constructions
RS Aggarwal Class 7 Mathematics Solutions Chapter 18 Reflection and Rotational Symmetry
RS Aggarwal Class 7 Mathematics Solutions Chapter 19 Three-Dimensional Shapes
RS Aggarwal Class 7 Mathematics Solutions Chapter 20 Mensuration
RS Aggarwal Class 7 Mathematics Solutions Chapter 21 Collection and Organisation of Data
RS Aggarwal Class 7 Mathematics Solutions Chapter 22 Bar Graph
RS Aggarwal Class 7 Mathematics Solutions Chapter 23 Probability