SIMPLE INTEREST - WORD PROBLEM -
Example.1) Mr. Goldy took loan of $ 60,000 to construct a house at 8 % per annum. He repaid the loan after 10 years. How much interest did he pay?
Ans.) Principal (P) amount of loan taken by Mr. Goldy = $ 60,000 Rate of Interest (R) on loan paid by Mr. Goldy = 8 %
Time (T) taken by Mr. Gupta to repay back the loan = 10 years
60000 × 8 × 10
Interest (S.I.) paid by Mr. Goldy on the loan = $ -----------------
100
[60000 and 100 have the common factor 100]
= $ (600 × 8 × 10)
= $ 48000 (Ans.)
2) Mr. Michel borrows $ 2500 from a bank at 10 % per annum. What
1
interest must he pay to the bank if he repays the loan in (i) 1 -----
2
years, (ii) 6 months (iii) 3 years 9 months?
Ans.)
i) Principal (P) amount borrowed by Mr. Michel = $ 2500 Rate of Interest
(R) paid by Mr. Michel = 10 %
1
Time (T) taken by Mr. Madan to repay back the loan = 1 ------ years
2
[Convert the mixed number into improper fraction]
(1 × 2) + 1
= ------------- years
2
1
= --------- years
2
2500 × 10 × 3
Interest (S.I.) paid by Mr. Michel on the loan = $ ------------------
100 × 2
[25000 and 100 have the common factor 100 & 10 and 2 have the common factor 2]
25 × 5 × 3
= $ --------------
1 × 1
= $ 375 (Ans.)
ii) Principal (P) amount borrowed by Mr.Michel = $ 2500 Rate of Interest (R) paid by Mr.Michel = 10 %
Time (T) taken by Mr.Michel to repay back the loan = 6 months
6
= ------ years [Since, 1 year = 12 months]
12
1
= ------
2 [6 and 12 have the common factor 6]
2500 × 10 × 1
Interest (S.I.) paid by Mr.Michel on the loan = $ ----------------
100 × 2
[25000 and 100 have the common factor 100 & 10 and 2 have the common factor 2]
25 × 5 × 1
= $ --------------
1 × 1 × 1
= $ 125 (Ans.)
iii) Principal (P) amount borrowed by Mr.Michel = $ 2500 Rate of Interest (R) paid by Mr.Michel = 10 %
Time (T) taken by Mr.Michel to repay back the loan = 3 years 9 months
9
= 3 ------ years [Since, 1 year = 12 months]
12
3
= 3 ----- years
4 [9 and 12 have the common factor 3]
(3 × 4) + 3
= ------------- years
4
[Convert the mixed number into improper fraction]
15
= -------
4
2500 × 10 × 15
Interest (S.I.) paid by Mr.Michel on the loan = $ -----------------
100 × 4
[2500 and 100 have the common factor 100, 10 and 4 have the common factor 2]
25 × 5 × 15
= $ --------------- 1 × 2 × 1
= $ 937.5 (Ans.)