- 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
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- Mensuration
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- Trigonometric Ratios and Standard Angles
- Coordinate Geometry
- Statistics
ML Aggarwal Solutions Class 9 Mathematics Solutions for Compound Interest Exercise 2.3 in Chapter 2 - Compound Interest
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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
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
\begin{aligned} &\mathrm{A} / \mathrm{P}=(1+\mathrm{r} / 100)^{\mathrm{n}}\\ &\text { Substituting the values }\\ &92610 / 80000=(1+\mathrm{r} / 100)^{3} \end{aligned}
\begin{aligned} &\text { By further calculation }\\ &(21 / 20)^{3}=(1+\mathrm{r} / 100)^{3} \end{aligned}
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 depreciation of cars is 5% p.a.
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