The population of a village increase by 5% annually. If its present population is 4410, then its population 2 years ago was:
A. 4500
B. 4000
C. 3800
D. 3500
Answer: Option B
Solution(By Examveda Team)
Using compounding formula$${\text{a}} + {\text{b}} + \frac{{{\text{ab}}}}{{100}}$$
we have
$$\eqalign{ & 5 + 5 + \frac{{25}}{{100}} \cr & = 10.25\% \cr} $$
so if the population 2 yrs ago be x
then
$$\eqalign{ & {\text{x}} + \frac{{10.25x}}{{100}} = 4410 \cr & {\text{or}},\,110.25{\text{x}} = 441000 \cr & \therefore {\text{x}} = 4000 \cr} $$
This Question Belongs to Arithmetic Ability >> Percentage
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Comments ( 4 )
Related Questions on Percentage
A. $$\frac{1}{4}$$
B. $$\frac{1}{3}$$
C. $$\frac{1}{2}$$
D. $$\frac{2}{3}$$
Ans is 4300
Let it be 100,
For 1st yr is 105
2nd yr is 110.25
So, 4410-110.25 is approx 4300
B
for the 1st year = (5/100)*4410 = 220.5
population in the last year = 4410-220.5 = 4189.5
for the 2nd year = (5/100)*4189.5 = 209.475
population in the 2nd last year = 4189.5-209.475 = 3980.025
according to my calculation using the concept of increase.
coming from the options B. 4000 is answer after 1st year 5% increase i.e.4000+200=4200.
again 2nd year 5% increase 4200+210=4410.
its like a compound interest.