8. A man borrows 5800 at 12% per annum
compound interest. He repays 1800 at the
end of every six months. Calculate the amount
standing at the end of third payment.
Give your answer to the nearest rupee.
Answers
Answer:For the first year
P=Rs5,800
N=
2
1
year
R=12 %
We have S.I.=
100
PNR
=
100
5,800×
2
1
×12
=Rs348
And Amount at the end of first 6 months P+S.I.=Rs5,800+Rs348=Rs6,148
Now, for the second half year
P=Rs6,148−Rs1,800=Rs4,348
N=
2
1
year
R=12 %
We have S.I.=
100
PNR
=
100
4,348×
2
1
×12
=Rs260.88
And Amount at the end of second half - year P+S.I.=Rs4,348+Rs260.88=Rs4,608.88
Also, for the third half year
P=Rs4,608.88−Rs1,800=Rs2,808.88
N=
2
1
year
R=12 %
We have S.I.=
100
PNR
=
100
2,808.88×
2
1
×12
=Rs168.5328
And Amount at the end of third half - year P+S.I.=Rs2,808.88+Rs168.5328=Rs2977.4128 or approximately Rs2977
And after the third payment amount outstanding =2977.4128−1,800=Rs1177.4128 or approximately Rs1177
Step-by-step explanation: