A sum of money is borrowed at compound interest payable annually. The interests for the 2nd and 3rd year
are 330 and 363 respectively. Find the sum.
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Answer:
It is given that
Interest for the first year = 225
Interest for the second year = 240
So the difference = 240 - 225 = 15
Here 15 is the interest on 225 for 1 year
(i) Rate = (SI×100)/(P×t)
Substituting the values
=(15×100)/(225×1)
So we get
=
3
20
=6
3
2
% p.a.
(ii) We know that
Sum=(SI×100)/(R×t)
Substituting the values
=(225×100)/(20/3×1)
It can be written as
=(225×100×3)/(20×1)
So we get
= 225×15
=3375
(iii) Here
AMount after second year = 225 + 240 + 3375 = 3840
So the interest for the third year = Prt/100
Substituting the values
=(3840×20×1)/(100×3)
=256
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