Mosquitoes are growing at a rate of 10%
a year. If there were 200 mosquitoes in
the begining. Write down the number of
mosquitoes after (i) 3 years (ii) 10 years
(iii) n years.
Answers
Answer:
10%/100of 200 = 20
200 + 20= 220
that means 20 mosquitoes are increasing per year
Step-by-step explanation:
1) 3 years = 200 + ( 20 × 3 )
= 260 mosquitoes
2) 10 years= 200 +(20×10)
=400
3) n years = 200 + ( 20× n )
= 200+20n
Answer:
Given :
Number of mosquitoes = 200
Growth rate = 10%
★ To find :
I) number of mosquitoes in 3 years.
ii) Number of mosquitoes in 10 years
iii) Number of mosquitoes in n years..
★ Solution :
We know that,
A = P\left(1 + \frac{R}{100}\right)^tA=P(1+
100
R
)
t
Here,
P = 200 , R = 10 and t = 3 years
by putting values
= 200 \times \left(1 + \frac{10}{100}\right)^{3}=200×(1+
100
10
)
3
= 200 \times \left(1 + \frac{1}{10}\right)^{3}=200×(1+
10
1
)
3
= 200 \times \frac{11}{10} \times \frac{11}{10} \times \frac{11}{10}=200×
10
11
×
10
11
×
10
11
= \dfrac{266200}{1000}=
1000
266200
\purple{\leadsto 266.2}⇝266.2
ii) Here,
P = 200 , R = 10 , t = 10
Applying the same formula
= 200 \times \left(1 + \frac{10}{100}\right)^{10}=200×(1+
100
10
)
10
= 200 \times \left(1 + \frac{1}{10}\right)^{10}=200×(1+
10
1
)
10
= 200 \times \left(\frac{11}{10}\right)^{10}=200×(
10
11
)
10
= \frac{200 \times 1331 \times 1331 \times 1331 \times 11}{10^{10}}=
10
10
200×1331×1331×1331×11
\purple{\leadsto 518. 748 = 519(approx)}⇝518.748=519(approx)
iii) for n years
= \purple{P \left(1 + \frac{R}{100}\right)^{n}}P(1+
100
R
)
n
Where,
P = original population
R = rate of growth
n = number of years..