Question

The current population of a city is
11,02,500. If it has been increasing at
the rate of 5% per annum, what was its
population 2 years ago?
A 10,00,000
B 9,92,500
C 12,51,506
D 9,95,006​

Answer 1

Answer :-

✧ Population 2 years ago of the city = 1,000,000.

Given :-

◍ The current population of the city = 11,02,500

◍ Rate at which it is increasing (R) = 5%

◍ Time (n) = 2 years ago

To find :-

• Population of the city 2 years ago

Solution :-

Let the population 2 years ago be P.

So,

$\tt{Current \: \: population \: = P(1 + \dfrac{r}{100} )^{n} }$

Substituting the values we have in the formula

$\mapsto \: \tt{1,102,500\: = P(1 + \dfrac{5}{100} )^{2} } \\ \\ \mapsto \: \tt{1,102,500\: = P(1 + \dfrac{1}{20} )^{2} } \\ \\ \mapsto \: \tt{1,102,500\: = P(\dfrac{20 + 1}{20} )^{2} }\\ \\ \mapsto \: \tt{1,102,500\: = P(\dfrac{21}{20} )^{2} }\\ \\ \mapsto \: \tt{1,102,500\: = P \times (\dfrac{441}{400} ) }\\ \\ \mapsto \: \tt{(\dfrac{400 \times 1,102,500}{441} ) \: = P}\\ \\ \mapsto \: \tt{(\dfrac{441,000,000}{441} ) \: = P}\\ \\ \mapsto \: \tt{1,000,000 \: = P}$

Therefore, population 2 years ago = 1,000,000.

Hence, option A. is the correct option.

More Information :-

• $\tt{A\: = P(1 + \dfrac{r}{100} )^{n} }$

where,

A = Population after n years

P = Current population

r = rate at which population is increasing or decreasing

n = time in years.

Answer 2

Answer:

let the current population of a city be p

ATQ:

current population =p(1+r/100)^n

1102500=p(1+5/100)^2

1102500×20×20/21×21=p

1000000=p