Question

5. The population of a town in the year 2002 was 22,050. The population grows at the rate of 5%
p.a. What was the population in the year
(a) 2000
(6) 2004?​

Answer 1

Answer:

(a)

→20000

(6)

→24310{Approximately}

Step-by-step explanation:

(a)2000

Population in 2002: 22,050

Rate of Growth of population p.a.: 5%

Time: 2002-2000

→2 years.

Let the population in 2000 be: x.

Amount: P×[1+(R/100)]^n

x[1+(5/100)]^2=22050

→x[1+(1/20)]^2=22050

→x[21/20]^2=22050

→x(21/20)(21/20)=22050

→441x/400=22050

{Transposing 441/400 yo RHS...}

→x = 22050(400/441)

→x = 20000

Hence, the population in 2000 was 20,000.

(6)

Population in 2002: 22050.

Rate of Growth of population p.a.: 5%.

Time: 2004-2002

→2 years.

Amount: P[1+(R/100)]^n

→22050×[1+(5/100)]^2

→22050×[1+(1/20)]^2

→22050×[21/20]^2

→22050×(21/20)×(21/20)

→24310.125

→24310{Approximately}

Hence, the population in 2004 would be Approximately 24,310.

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Answer 2

Step-by-step explanation:

this is your answer. please mark as brainlist answer