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Compound Interest | Compound Interest Exercise 2.3

Question 7

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

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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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subject-cta

Question 7

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

Looking to do well in your science exam ? Learn from an expert tutor. Book a free class!

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.

Our top 5% students will be awarded a special scholarship to Lido.

subject-cta
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