Question

A man borrows 15000 at 12% per annum, compounded annually. If he repays 4400 at the
end of each year, find the amount outstanding against him at the beginning of the third year.​

Answer 1

$\large\bf\underline{Given:-}$

• Principal = 15000
• Rate = 12%
• Repays amount = 4400

$\large\bf\underline {To \: find:-}$

• find the amount outstanding against him at the beginning of the third year.

$\huge\bf\underline{Solution:-}$

$\bf \: s imple\: intrest = \frac{principal \times rate \times timr}{100}$

Intrest for the 1st year :-

$: \implies\rm\: \frac{15000 \times 12 \times 1}{100} \\ \\ : \implies\rm\: \frac{150 \cancel{00 }\times 12}{ \cancel{100}} \\ \\ : \implies\rm\: 150 \times 12 \\ \\ : \implies\bf\:1800$

Amount at the end of first year = P + I

$: \implies\rm\:15000 + 1800 \\ \\ : \implies\rm\:16800$

Amount left after repaying = 16800-4400

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀=12400

Principal for the second year :-

$:\implies\rm\: \frac{12400 \times 12 \times 1}{100} \\ \\:\implies\rm\: \frac{124 \cancel{00 }\times 12}{ \cancel{100}} \\ \\ :\implies\rm\:124 \times 12 \\ \\ :\implies\bf\:1488$

Amount with intrest = 12400+1488

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀= 13888

Amount after repaying = 13888 - 4400

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ = 9488

The amount outstanding against him at the beginning of the third year is 9488

Answer 2

Given that,

•Principal (p) = Rs.15000

•rate of interest = 12% per annum.

SOLUTION :-

=>Interest for the first year = 12% of 15000.

=12/100 × 15000

=12 × 150

=1800

=>Interest for the first year =Rs. 1800.

=>Amount after 1 year = 15000 + 1800 = Rs.16800.

=>He repay Rs. 4400.

=> So, the amount at the beginning of the second year = Rs.(16800 - 4400) = Rs.12400

=> Interest for the second year = 12% of 12400

= 12/100 × 12400

= 12 × 124

=1448

=>Interest for the second year = Rs.1448

=>Amount after second year = Rs.(12400 + 1448) = Rs.13888 .

=> He repay Rs. 4400.

=> Therefore, the outstanding amount at the beginning of third year = Rs. (13888 - 4400) = Rs. 9488.