## Installments Part 1

Problem 1: What annual installment will discharge a dept of 6450/- due in 4 years at 5% S.I?

Solution: $S.I=\frac{100\times&space;A}{100\times&space;Time+\frac{R\times&space;t(T-1)}{2}}$

$\frac{100\times&space;6450}{400+\frac{5\times&space;4\times&space;3}{2}}=\frac{100\times&space;6450}{430}=1500$

Method:

100 |105 |110|115

430——->6450

100——->?

$\frac{6450\times&space;100}{430}=1500$

Problem 2: What annual payment will discharge a debt of RS 944 in 4 annual equal installments at the rate of 12% on S.I?

Solution:

100| 112| 124 | 136

472——->944

100———>?

$\frac{944\times&space;100}{472}=200$

Problem 3: The annual payment of Rs.700 in 5 years at the rate of 10% S.I will discharge a debt of what amount?

Solution:

1st year——>700—->10%$\frac{700\times&space;10}{100}=70$

2nd year—–>770+70

3rd year—–>840+70

4th year—–>910+70

5th year——>980

Total amount=4200

## Installments Part 2

Problem 1: A man borrows RS.2100 and undertakes to pay back with C.I at the rate of 10% in 2 equal yearly installments at the end of 1st and 2nd year. What is the amount of each installment?

Solution:

Formula:$C.I=A=\frac{X}{(1+\frac{R}{100})}+\frac{X}{(1+\frac{R}{100})^{2}}+\frac{X}{(1+\frac{R}{100})^{2}}$

2100=$\frac{X}{(1+\frac{10}{100})}+\frac{X}{(1+\frac{10}{100})^{2}}$

2100=$\frac{X}{(\frac{110}{100})}+\frac{X}{(\frac{110}{100})^{2}}$

2100=$X\times&space;\frac{10}{11}+X\times&space;\frac{100}{121}$

2100=$\frac{110X+100X}{121}=\frac{210X}{121}$

2100=$\frac{210X}{121}=1210$

Shortcut: (1+r%)=>$(1+\frac{r}{100})=1+\frac{10}{100}=\frac{11}{10}$

$I=A\times&space;\frac{Big&space;number}{sum}\times&space;1+r$%

=$2100\times&space;\frac{11}{21}\times&space;\frac{11}{10}=1210$

Problem 2: A=1820  R=20% Time=3 C.I=?

Solution: $(1+\frac{r}{100})=1+\frac{20}{100}=\frac{120}{100}=\frac{6}{5}$

$1:\frac{5}{6}:(\frac{5}{6})^{2}$

$1:\frac{5}{6}:\frac{25}{36}$

36:30:25

$I=1820\times&space;\frac{36}{91}\times&space;\frac{6}{5}=216\times&space;4=864$

Problem 3: A man buys a scooter on making a cash draw payment of 16224 and promises to pay two more yearly installment of equivalent amount in the next two years. If the rate of the installment is 4% per annum then the cash value of the scooter is?

Solution:$X=A\times&space;(1+\frac{R}{100})$

$16224=P_{1}(1+\frac{R}{100})$

$16224=P_{1}(1+\frac{4}{100})$

$P_{1}=\frac{1622\times&space;25}{26}=15600$

$16224=P_{2}(1+\frac{4}{100})^{2}$

$P_{2}=15000$

Total value = 16244+15600+15000 = 46,844