## Compound Interest Part 1

Problem 1: Find the successive discount of  $37\tfrac{1}{2}$% and 40%

Solution:

Fractional value of  $37\tfrac{1}{2}$% is $\frac{3}{8}$

Fractional value of 40% is  $\frac{2}{5}$

8        5

5         3

8*5=40

5*3=15

$\frac{40}{15}=\frac{8}{3}$

The result is 8 and 3

8 is the marked price and 3 is the S.P

8-3=5

$\frac{5}{8}\times&space;100=62\frac{1}{2}$%

There is a successive discount of  $62\frac{1}{2}$%

Problem 2: Find the successive increase of $11\frac{1}{9}$% and $12\frac{1}{2}$%

Solution:

Fraction for $11\frac{1}{9}$% is$\frac{1}{9}$

Fraction for $12\frac{1}{2}$% is $\frac{1}{8}$

9         10

8           9

Cancel common terms and the result is 8 and 10

$\frac{8}{10}=\frac{4}{5}$

4-5=1

$\frac{1}{5}\times&space;100=25$%

There is a successive increase of 25%

Note: when there is a successive decrease it is discount, When there is a successive increase it is compound interest.

## Compound Interest Part 2

Based on part 1 we will solve some problems in compound interest.

Problem 1: Find the principle and compound interest if the rate of interest is 10% and time is 2years and amount is 2420?

Solution:

R=10%

Time =2 years

Amount=2420

Principle=?

C.I=?

The fraction of 10% is $\frac{1}{10}$

10      11

10      11

10*10=100

11*11-121

Let 100 as a principle

121 as amount

121-100=21 is C.I

121———>2420

100———->?

$\frac{2420\times&space;100}{121}=2000$

121———->2420

21————–>?

$\frac{2420\times&space;21}{121}=420$

Principle =2000 and C.I=420

Problem 2: P=?, T=2 years,$R_{1}$=$11\frac{1}{9}$%, $R_{2}=12\tfrac{1}{2}$%,CI=480

Solution :

Convert $11\frac{1}{9}$% into fraction i.e $\frac{1}{9}$

Convert $12\tfrac{1}{2}$% into fraction i.e $\frac{1}{8}$

9      10

8       9

Cancel the common term 9 and the result is 8 and 10$\frac{8}{10}=\frac{4}{5}$

Here 4 is the principle and 5 is the amount

4-5=1 is C.I

1———->480

4———–> ?

$\frac{480\times&space;4}{1}=&space;1920$  is the principle.

## Compound Interest Part 3

Problem 1: P=?, R=$14\tfrac{2}{3}$%, T=2years, C.I-S.I=40,C.I=?

Solution: When such type of problems are given we have to follow one method as shown below.

Fraction for $14\tfrac{2}{7}$% is $\frac{1}{7}$

7*7=49

14-15=1 is difference between C.I and S.I

1———>40

49———>?   $\frac{40\times&space;49}{1}=1960$ =Principle

15———>?  $\frac{15\times&space;40}{1}=600$ = C.I

14———>?$\frac{40\times&space;14}{1}=560$ = S.I

Problem 2: P=?, R=15%,T=2 years,C.I-S.I=180

Solution:

Fraction for 15% is $\frac{3}{20}$

20*20=400

129-120=9

9———>180

400——->?

$\frac{180\times&space;400}{9}=8000$ is principal

## Compound Interest Part 4

Problem 1: P=?, R=$8\tfrac{1}{3}$%, T=2 years, 2nd year C.I=9.10

Solution:

Fraction for $8\tfrac{1}{3}$% is $\frac{1}{12}$

2nd year C.I=12+1=13

13———>9.10

144——–>?

$\frac{9.10\times&space;144}{13}=100.8$ is principle.

Problem 2: P=?, R=$6\frac{2}{3}$%, T=2years, 2nd year C.I=8

Solution:

Fraction for $6\frac{2}{3}$% is$\frac{1}{15}$

2nd year C.I 15+1=16

16——->8

225—–>?

$\frac{8\times&space;225}{16}=112.5$ is principle

## Compound Interest Part 5

Problem 1: P=?, R=20%, T=1 year 73 days, C.I=1240

Solution:

Fraction for 20%=$\frac{1}{5}$

$\frac{6}{365}\times&space;73=1.2$

5+1.2 =6.2 is C.I

6.2———–>1240

25————->?

$\frac{1240\times&space;25}{6.2}=5000$ is principle.

Problem 2: P=?,R=$12\tfrac{1}{2}$%, T=1 year 4 months, C.I=1782

Solution:

Fraction for $12\tfrac{1}{2}$% is $\frac{1}{8}$

8+1=9

$\frac{9}{12}\times&space;4=3$

8+3=11

11———->1782

64————>?

$\frac{1782\times&space;64}{11}=10468$ is principal.