Question

Rohan borrow ₹22500 at 10% compound. If he repay ₹11250 at the end of one year and ₹12550 at the end of second year. Find the amount of loan outstanding against him at the end of third year.

Answer 1

Answer:

2530

Step-by-step explanation:

For 1st year

P=22500

R=10%

T=1year

I=(P*R*T)/100

=(22500*10*1)/100

=2250

A=P+I

=22500+2250

=24750

He repaid=11250

A left=24750-11250

=13500

For 2nd year

P=13500

R=10%

T=1year

I=(13500*10*1)/100

=1350

A=13500+1350

=14850

He repaid=12550

A left=14850-12550

=2300

For 3rd year

P=2300

R=10%

T=1

I=(2300*10*1)/100

=230

A=2300+230

=2530