Jump to

- Rational and Irrational Numbers
- Compound Interest
- Expansions
- Factorization
- Simultaneous Linear Equations
- Problems on Simultaneous Linear Equations
- Quadratic Equations
- Indices
- Logarithms
- Triangles
- Mid Point Theorem
- Pythagoras Theorem
- Rectilinear Figures
- Theorems on Area
- Circle
- Mensuration
- Trigonometric Ratios
- Trigonometric Ratios and Standard Angles
- Coordinate Geometry
- Statistics

A man invests ₹ 8000 for three years at the rate of 10% per annum compound interest. Find the interest

for the second year. Also, find the sum due at the end of the third year.

Answer:

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 the 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

Related Questions

Was This helpful?

Exercises

Chapters

Lido

Courses

Quick Links

Terms & Policies

Terms & Policies

2022 © Quality Tutorials Pvt Ltd All rights reserved