Question

In a town the population grows at a simple rate of 10% in a decade and compounds from decade to decade. find the population at the beginning of the 1970s if the population at the beginning of the 1990s is 3,63,000 people.

Answer 1

This question applies compound interest formula. The number of the population compounds after every decade.

It is 2 decades between 1970 and 1990. That means it compounded twice - we use 2.

Let us assume that the population at the beginning of 1970 was x number of people.

Find the sum that will compound after 1st decade and 2nd decade:

1st decade --->   10/100 × x = 0.1x

In the 1980's the population was x + 0.1 x = 1.1x

2nd decade-----> 10/100 × 1.1x = 0.11x

 The population at the beginning of the 1990's was 1.1x + 0.11x =  1.21x

1.21x  should be equal to 363000

1.21x = 363,000
       x = 36300/1.21 
       = 300,000

Therefore the population at the beginning of the 1970's was 300,000 people

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You can as well use compound interest formula to find the answer

A = P (1 + r/a) ∧ (ab)

Where:
 
A is final population
P initial population
a is time compounded per decade
r interest
t is number of decades

363000 = P (1 + 0.1)²
363000 = P(1.1)²
363000 = 1.21P
P = 363000/1.21
P = 300,000

The population at the beginning of the 1970's was 300,000 people

Answer 2

Answer:

300000

Step-by-step explanation:

1 decade =10 years

so in this problem there are 2 decades

i.e., 1970-1980 1st decade

1980-1990 2nd decade

for every decade there is a increase of 10%

1990s population --> 363000

Lets take the population in 1970 --> 100

+10% increass. (1980) -->10= 110

+10% increass. (1990)--> 11 = 121

cross multiplication

100. 121

? 363000

Ans : 300,000