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Arithmetic I

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Interest & Growth
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CAT 2025 Lesson : Interest & Growth - Common Types

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5. Common Types

5.1 Simple Interest as a multiple of Principal

Example 16

If a certain deposit, yielding simple interest, triples itself every 55 years, then in how many years does the deposit become 77 times the amount invested?

Solution

Let p be the principal and r% be the interest rate per annum.
Where n =5= 5, amount is 3p3p.
∴ Simple Interest (SI) =3p−p== 3p - p = 2p.

SI =pnr100= \dfrac{pnr}{100}

⇒ 2p=p×5×r100 2p = \dfrac{p \times 5 \times r}{100} ⇒ r=40 r = 40

For the amount to become 7p7p, Simple Interest =7p−p== 7p - p = 6p
Let n be the number of years for simple interest to become 6p6p.

6p=p×n×401006p = \dfrac{p \times n \times 40}{100} ⇒ n=15n = 15 years

Alternatively

Let the initial deposit (or Principal) placed be p. As the amount invested triples itself in 55 years, the deposit becomes 3p.

∴ Every 55 years, the interest earned is 2p.

To become 77 times it's original amount, the deposit should earn an interest of 6p.

As it earns 2p2p of interest every 55 years, the deposit will earn 6p6p in 3×5=153 \times 5 = 15 years.

Answer: 1515 years

Example 17

If a certain deposit, yielding simple interest, doubles itself every 44 years, then in 1010 years it becomes xx times itself. x=x = ?

Solution

Let the principal be pp. If the deposit amounts to 2p2p in 44 years, then the simple interest is 2p−p=2p - p = p every 4 years.

In 2.52.5 times the 44-year time period, i.e. 1010 years, the interest is 2.5×p=2.5 \times p = 2.5p

Amount at end of 1010 years = Principal + Interest =p+2.5p== p + 2.5p = 3.5p

So, the deposit becomes 3.53.5 times itself in 1010 years.

Answer: 3.53.5

5.2 Compound Interest as a multiple of Principal

Example 18

A certain deposit, yielding compound interest, triples itself every 33 years. In 1212 years, the overall percentage growth in the deposit is _____%\%.

Solution

Let the principal be pp. Amount at end of 33 years =3p= 3p.

3p=p(1+r100)33p = p \left(1 + \dfrac{r}{100} \right)^{3} ⇒ (1+r100)3=3⟶(1) \left(1 + \dfrac{r}{100} \right)^{3} = 3 \longrightarrow (1)

In 1212 years, the amount becomes

A=p(1+r100)12=p((1+r100)3)4A = p \left(1 + \dfrac{r}{100} \right)^{12} = p\left( \left(1 + \dfrac{r}{100} \right)^{3} \right)^{4}

Substituting (1)(1),

⇒ A=p(3)4=81p A = p(3)^{4} = 81p

Deposit has grown from pp to 81p81p.

%\% growth =81p−pp×100%=8000%= \dfrac{81p - p}{p} \times 100 \% = 8000 \%

Answer: 8000%8000 \%

5.3 Questions with Simple and Compound Interest for 2 years

If the principal amount and rate of interest are the same and the time period is 22 years, the only difference between Simple Interest and Compound Interest is the interest on the interest component.

Example 19

On a 22-year deposit, John receives Rs. 240240 as Simple Interest. If the interest would have been compounded annually, he would have received Rs. 258258 as interest. What is the initial value of John's deposit?

Solution

As John receives Rs. 240240 as SI for 22 years, SI for each year is Rs. 120120.

Simple Interest and Compound Interest are the same in the first year. Whereas, in the second year, the difference between them is the interest on the first year's interest.

If interest would have been compounded, John would have received interest on his first year's interest, which is 258−240=Rs.18.258 - 240 = \text{Rs}. 18.

Interest Rate =18120×100%=15%= \dfrac{18}{120} \times 100 \% = 15 \%

As he earns Simple interest of Rs. 120120 in the first year,

120=p×1×15100120 = \dfrac{p \times 1 \times 15}{100} ⇒ p=Rs.800p = \text{Rs.} 800

Answer: Rs. 800800

5.4 Compound Interest in nth\bm{n^{\text{th}}} year and (n+1)th\bm{(n + 1)^{th}} year given

The growth in the annual Compound Interest equals the Compound Interest Rate.

Example 20

The compound interest earned on a deposit in the 8th8^{\text{th}} and 9th9^{\text{th}} years are Rs. 400400 and Rs. 420420 respectively. What is the annual rate of interest?

Solution

The annual interest grows at the rate of the applicable compound interest rate. Interest in 9th9^{\text{th}} year is higher than that in the 8th8^{\text{th}} year by Rs. 20. This is on account of the interest on interest of the previous year.

∴ Annual Growth in compound interest =420−400400×100%=5%= \dfrac{420 - 400}{400} \times 100 \% = 5 \%

The rate of interest is 5% per annum.

Answer: 5%5 \%

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