A sum of ₹x amounts to ₹83,853 in 2 $$\frac{1}{2}$$ years at 12% p.a., interest compounded 10 – monthly. What is the value of x?

Correct option is D

Given:

Amount (A) = 83,853

Rate (R) = 12% p.a.

Time = $$2 \frac{1}{2 }$$​years = $$\frac{5}{2}$$​ years

Interest is compounded every 10 months

Formula Used:

​$$A = P \times (1 + \frac{R}{100})^n$$​​

Solution:

Let P = x

Since interest is compounded every 10 months, number of compounding periods in 2.5 years =($$\frac{30 \ months }{10 \ months}$$​) = 3

Effective rate for 10 months = $$12 × ( \frac{10}{12}) = 10$$​​

So, 

​$$A = x × (1 + \frac{10}{100})^3 \\\ \\ 83853 = x × (\frac{110}{100})^3\\\ \\83853 = x × (\frac{1331}{1000})\\\ \\x = 83853 × (\frac{1000}{1331})\\\ \\$$
x = 63,000

Alternate Solution:

$$10 \% = \frac{1}{10}$$

     p       :    A  

   10     :    11 

   10     :   11 

   10    :  11 

~~~~~~~~~~~~~ 

1000   :   1331  

According to the question 

1331 unit = 83853

1 unit   = $$\frac{83853}{1331}$$  = 63

Then principal = 1000 unit = 63000