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Interest & Growth
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CAT 2025 Lesson : Interest & Growth - Growth & CAGR

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4. Growth & Depreciation

Successive growth rates follow the same concept as Compound Interest. Growth rate of an item is applied on the value of the item at the end of the prior period.

Example 13

The price of potatoes was Rs. 8080 per kg in 20112011. Inflation over the next three years was 20%20 \%, 25%25 \% and 3313%33 \dfrac{1}{3} \% respectively. What is the price of potatoes in 20142014?

Solution

Price per kg in 2012=80×(1+20100)2012 = 80 \times \left( 1 + \dfrac{20}{100} \right)

Price per kg in 2013=80×(1+20100)×(1+25100)2013 = 80 \times \left(1 + \dfrac{20}{100} \right) \times \left(1 + \dfrac{25}{100} \right)

Price per kg in 2014=80×(1+20100)×(1+25100)×(1+100300)2014 = 80 \times \left(1 + \dfrac{20}{100} \right) \times \left(1 + \dfrac{25}{100} \right) \times \left(1 + \dfrac{100}{300} \right)

=80×(1+15)×(1+14)×(1+13)= 80 \times \left(1 + \dfrac{1}{5} \right) \times \left(1 + \dfrac{1}{4} \right) \times \left(1 + \dfrac{1}{3} \right)

=80×65×54×43== 80 \times \dfrac{6}{5} \times \dfrac{5}{4} \times \dfrac{4}{3} = Rs. 160160

Answer: Rs. 160160 per kg

In the case of depreciation, the rate has to be subtracted instead of added.

Example 14

The price of petrol in 20282028 was Rs. 200200 per litre. It fell by 20%20 \% a year for 33 successive years. What was the price of petrol in 20312031?

Solution

p=200p = 200 ; r=−20%r = -20 \% ;  n = 33
Price of a litre of petrol in 2031=p(1+r100)n=200(1−20100)32031 = p\left(1 + \dfrac{r}{100} \right)^{n} = 200 \left(1 - \dfrac{20}{100} \right)^{3}

=200×0.83=200×0.512=102.4= 200 \times 0.8^{3} = 200 \times 0.512 = 102.4

Answer: Rs. 102.4102.4

4.1 Compound Annual Growth Rate

Where a certain initial investment has grown by different rates at the end of every year, Compound Annual Growth Rate (CAGR) is the geometric mean of the actual annual growth rates.

In other words, CAGR provides a constant rate at which the initial investment could have been compounded over nn years to yield its final value. The Compound Interest formula applies.

A=p(1+r100)nA = p\left(1 + \dfrac{r}{100} \right)^{n}

When a certain amount pp has grown to an amount AA over nn years, then Compound Annual Growth Rate (CAGR) is the variable rr.

Example 15

An investment grew from Rs. 1,0001,000 to Rs. 2,5002,500 over 55 years. What was the Compound Annual Growth Rate?

(1) 20%20 \%            (2) 25%25 \%            (3) 30%30 \%            (4) 35%35 \%           

Solution

p=1,000;p = 1,000; n=5n = 5; A=2,500A = 2,500

A=p(1+r100)nA = p\left(1 + \dfrac{r}{100} \right)^{n}

⇒ 2500=1000(1+r100)5 2500 = 1000 \left(1 + \dfrac{r}{100} \right)^{5}

⇒ (1+r100)5=2.5 \left(1 + \dfrac{r}{100} \right)^{5} = 2.5

Using the calculator in the CAT exam and substituting the options, we find that option (11) satisfies.

(1+20100)5=(1.2)5=2.488∼2.5\left(1 + \dfrac{20}{100} \right)^{5} = (1.2)^{5} = 2.488 \thicksim 2.5

∴ r=20% r = 20 \%

Answer: 20%20 \%

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