Question

A man borrows rupees 6000 at 5% compound interest . If he repays rupees 1200 at the end of each year,find the amount outstanding at the beginning of the third year.​

Answer 1

It is given that

Principal = ₹ 6000

Rate of interest = 5% p.a.

We know that

Interest for the first year = Prt/100

Substituting the values

= (6000 × 5 × 1)/ 100

= ₹ 300

So the amount after one year = 6000 + 300 = ₹ 6300

Principal for the second year = ₹ 6300

Amount paid = ₹ 1200

So the balance = 6300 – 1200 = ₹ 5100

Here

Interest for the second year = (5100 × 5 × 1)/ 100 = ₹ 255

Amount for the second year = 5100 + 255 = ₹ 5355

Amount paid = ₹ 1200

So the balance = 5355 – 1200 = ₹ 4155

${\fcolorbox{blue}{black}{\orange{\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: DecentMortal\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:}}}$

Answer 2

Step-by-step explanation:

It is given that

Principal = ₹ 6000

Rate of interest = 5% p.a.

We know that

Interest for the first year = Prt/100

Substituting the values

= (6000 × 5 × 1)/ 100

= ₹ 300

So the amount after one year = 6000 + 300 = ₹ 6300

Principal for the second year = ₹ 6300

Amount paid = ₹ 1200

So the balance = 6300 – 1200 = ₹ 5100

Here

Interest for the second year = (5100 × 5 × 1)/ 100 = ₹ 255

Amount for the second year = 5100 + 255 = ₹ 5355

Amount paid = ₹ 1200

So the balance = 5355 – 1200 = ₹ 4155