The problems on simple and compound interest are asked as a
part of percentages only. Let me start this lesson with a simple problem:

We can see from the table that the 2

Tommy and his dad planted two plants on the
same day. The first plant is 10 ft high and grows by 10 ft every day. The
second plant is 1 ft high and doubles its height everyday. On which day will
the 1 ft high plant overtake the 10 ft high plant?

Answer: Let's make a simple table to answer this question.
We note down the heights of the plants on each day, as shown below:We can see from the table that the 2

^{nd}plant outgrows the 1^{st}plant on the 8^{th}day (colored orange in the table). Now, answer the following questions:
1. Which
plant has a fixed increase in height everyday? How much is this increase?

2. Which plant is growing at a faster rate?

3. What is the percentage increase in the height of 2

4. What will be the difference in heights of the plants on the 10

2. Which plant is growing at a faster rate?

3. What is the percentage increase in the height of 2

^{nd}plant everyday?4. What will be the difference in heights of the plants on the 10

^{th}day?
You can see that the 1

^{st}plant has a fixed increase of 10 ft in its height everyday. If you observe harder, this increase of 10 ft is nothing but 100 % of its original height. Therefore,**the 1**. In other words, the first plant is growing with a simple interest (SI) rate.^{st}plant is increasing with a fixed percentage of its original value
You can also see that the 2

^{nd}plant is increasing at a faster rate and**increases by 100% on its present value**everyday. In other words, the second plant is increasing at a compound interest (CI) rate.
Therefore,

**both the plants are increasing by the same percentage, first on the initial value and the second on the current value. Hence, the first is in simple interest and the second is in the compound interest.**
Note that the heights of the 1

^{st}plant form an arithmetic progression with a common difference of 10 ft and the heights of the second plant form a geometric progression with a common ratio of 2.
Therefore, to calculate the heights on the 10

^{th}day we use the formula for the N^{th}terms of the arithmetic and geometric progressions.
For the 1

^{st}plant
T

_{n}= a + (n - 1)d --> T_{10}= 10 + 9 x 10 = 100
For the 2

^{nd}plant
T

_{n}= ar^{n - 1 }--> T_{10}= 1 x 2^{9}= 512.
Therefore difference in heights = 512 - 100 = 412 ft

This time, we invest the two amounts of Rs100 each, both
at the annual rate of 10%, one at simple interest and the other at compound
interest. Let's see their growth:

Notice a few things:

**SOME SOLVED EXAMPLES:**

1. In how
many years will an amount invested in simple interest at the rate of 5% get
doubled?

Answer: we know that yearly
amounts in a simple interest scheme are in arithmetic progression with common
difference of Pr/100. In this case, the amounts will be in arithmetic
progression with a common difference of 0.05P.

The series is P, 1.05P, 1.1P,
1.15P, â€¦ (P + n x 0.05P), ...and so on. Let
the amount be double of the original amount after n years.

Therefore, P + n x 0.05P = 2P --> n = 20 years.

Therefore, P + n x 0.05P = 2P --> n = 20 years.

Alternative Answer: The amount
is growing by 5% every year or 1/20

^{th}every year. Since 1/20^{th}of the sum is added every year, the whole sum will be added in 20 years.
Courtesy:Totalgadha

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