Question

In what period of time will Rs 12000 yield Rs 3972 as C.I. at 10% p.a. if compounded on yearly basis *​

Answer 1

Answer:

3 years

Step-by-step explanation:

P= 12000 r=10 c.i = 3972

c.i = p[(1+r/100)^n-1]

3972= 12000[(1+10/100)^n-1]

3972/12000= (11/10)^n - 1

(3972/12000) + 1 =  (11/10)^n

1331/1000 =  (11/10)^n

(11/10)^3 =  (11/10)^n

n=3

Answer 2

$\sf\large\underline{Given:}$

• $\rm{Principal=Rs.12000}$
• $\rm{CI=Rs.3972}$
• $\rm{Rate=10\%}$

$\sf\large\underline{To\: Find:}$

• $\rm{The\: period\:of\:time=n}$
• Here,"n" is the period of time in compound interest the period of time is represented by n so we take time=n

$\sf\large\underline{Solution:}$

• [By applying formula to calculate A:]

$\rm{\implies A=P+CI}$

$\rm{\implies A=12000+3972}$

$\rm{\implies A=Rs.15972}$

• [Now calculate time by using formula:]

$\rm{\implies A=P(1+\frac{r}{100})^n}$

$\rm{\implies 15972=12000(1+\frac{10}{100})^n}$

$\rm{\implies \frac{15972}{12000}=(\frac{110}{100})^n}$

$\rm{\implies \frac{3991}{3000}=(\frac{110}{100})^n}$

$\rm{\implies \frac{1331}{1000}=(\frac{110}{100})^n}$

$\rm{\implies (\frac{11}{10})^3=(\frac{11}{10})^n}$

$\rm\purple{\implies n=3}$

Hence,

• [Rs.12000 yield Rs.3972 as CI is in 3 years:]