Question

the income of a man increase by 10% every year.If his present annual income is Rs 8,80,000.

(a) what will be his income 2 year later?

(b) what was his income 1 year earlier

Answer 1

1.principal=8,80,000
rate = 10%/yr
period=2yrs
so we can use compound interest formula which is p(1+r/100)*n
so
=8,80,000(1+10/100)2
=8,80,000(110/100*110/100)
=88*110*110
=88*12100
=105600rs
2.let his income was 100%
and after 1yr means now his income is 110%
now,
110%= 8,80,000
100%=?
so
=8,80,000*100/110
=8,00,000rs

here are Ur ans ☺

Answer 2

Answer: a) Rs 1,064,800

b) Rs 800,000

Step-by-step explanation:

The exponential growth function is given by :-

$y=A(1+r)^x$ , where A is initial amount , r is rate of growth and x is time period.

Given : Present annual income : A= Rs 8,80,000.

Rate : r= 10%=0.10

a) Time : x= 2

$y=880000(1+0.1)^2\\\\=880000(1.1)^2=1,064,800$

Hence, his income 2 year later = Rs 1,064,800

b) Let m be his income 1 year earlier.

Take y= 8,80,000, then

$880000=m(1+0.1)^1\\\\\Rightarrow\ 1.1m=880000\\\\\Rightarrow\ m=\dfrac{880000}{1.1}=800,000$

Hence,  his income 1 year earlier=  Rs 800,000