Every MBA aspirant who understands percentages and the principles of arithmetic and geometric progressions will be able to handle problems of simple interest (SI) and compound interest (CI) easily. In CAT, XAT or other MBA exams, the problems on simple and compound interest are asked as a part of percentages only. Let me start this lesson with a simple problem:
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?
We can see from the table that the 2nd plant outgrows the 1st plant on the 8th 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 2nd plant everyday?
4. What will be the difference in heights of the plants on the 10th day?
You can see that the 1st 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 1st plant is increasing with a fixed percentage of its original value. In other words, the first plant is growing with a simple interest (SI) rate.
You can also see that the 2nd 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 1st 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 10th day we use the formula for the Nth terms of the arithmetic and geometric progressions.
For the 1st plant
Tn = a + (n - 1)d --> T10 = 10 + 9 × 10 = 100
For the 2nd plant
Tn = arn – 1 --> T10 = 1 × 29 = 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, … (P + n × 0.05P), …and so on. Let the amount be double of the original amount after n years.
Therefore, P + n × 0.05P = 2P --> n = 20 years.
Alternative Answer: The amount is growing by 5% every year or 1/20th every year. Since 1/20th of the sum is added every year, the whole sum will be added in 20 years.