#simpleandcompoundinterest

Que: Rohan borrows Rs 3600 for 6 months at 15% per annum simple interest and pays Rs 100, 200, 400, 800, and 1600 at the end of the first to fifth month respectively. Find the amount of his last installment.

(a) 712.25 ₹

(b) 698.75 ₹

(c) 674.25 ₹

(d) 756.75 ₹

Solution: 15% Annual= (15/12)%= (5/4)% monthly

(5/4)%= (5/4)×(1/100)= 1/80

It means 1 rs monthly interest on a principal of 80 rs.

In 6 months, the amount of 3600 rs becomes= 3600+3600×(1/80)×6=3870 rs

In five months, 100 rs becomes=100+100×(1/80)×5=(100+6.25) rs

In four months, 200 rs becomes=200+200×(1/80)×4=(200+10) rs

In three months, 400 rs becomes=400+400×(1/80)×3=(400+15) rs

In two months, 800 rs becomes=800+800×(1/80)×2=(800+20) rs

In one month, 1600 rs becomes=1600+1600×(1/80)×1=(1600+20) rs

Total= (100+200+400+800+1600)+(6.25+10+15+20+20)

=3171.25 rs

The amount of the last installment= 3870–3171.25= 698.75 rs

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Que: A sum of Rs 100 is lent to be returned in 10 equal monthly installments of Rs 12 each. Find the annual rate of simple interest.

(a) 52(4/23)%

(b) 51(12/23)%

(c) 49(17/23)%

(d) 50(1/23)%

Solution: Let the amount of the monthly installment on the principal of Rs.12 be ‘r’

So, the amount of monthly installment on the principal of rs 100= (r/12)×100=(25r)/3

Total interest paid on 100 rs in 10 months= (250r)/3

Total amount after 10 months= 100+ (250r)/3 ……….(i)

The first installment of rs 12 becomes 12+9r at the end if not paid

The second installment of rs 12 becomes 12+8r at the end if not paid

The third installment of rs 12 becomes 12+7r at the end if not paid

…….

The last installment is rs 12

Total Sum= (12+9r)+(12+8r)+(12+7r)+……+(12+r)+12

=12×10+ (0+1+2+3+4+……+9)×r

=120+45×r ……….(ii)

120+45×r =100+ (250r)/3

r=12/23

Monthly rate percentage= (r/12)×100= (100/23)%

Annual rate percentage= (100/23)% ×12=(1200/23)%=52(4/23)%

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