## Compound Interest Part 9

Problem 1: When a certain principal is invested at 16.66% compound interest instead of 12.5% per annum for 2 years the interest is 550 more. Find the principal?

Solution:

12.5% ——–>$\frac{1}{8}$——–>64

16.66%——–>$\frac{1}{6}$——–>36

LCM Of 64 and 36 is 576 is principal

72+72+9=153                                             96+96+16=208

208-153=55

55———->550

576——–>?

$\frac{550\times&space;576}{55}=5760$

Problem 2: Raju took Rs.38400 from Rama at 10% rate of SI and give it to Surya who gives 12.5% rate of C.I compounded annually. If Raju gives the money to Surya for 3 years and returns to Rama immediately. Find the profit of Raju?

Solution:

S.I=$\frac{PTR}{100}=\frac{38400\times&space;10\times&space;3}{100}=11520$

4800*3=14400

600*3=1800

14400+1800+75=16275

16275-11520=4755

Problem 3: A man invested a sum of money in scheme A, at the rate of 15% per annum for S.I, at the end of two years the amount received by him is invested in scheme B at 20% per annum for C.I. If the interest received by him from scheme B at the end of the second year is RS. 2860, then find the sum invested by a man in the beginning?

Solution:

Scheme B=20% C.I

20+20+$\frac{20\times&space;20}{100}=44$%

Scheme A=15% S.I

Assume principal is 100

$S.I=\frac{PTR}{100}=\frac{100\times&space;2\times&space;15}{100}=130$%

44%——–>2860

100%——–>?

$\frac{2860\times&space;100}{44}=6500$

$\frac{6500\times&space;100}{130}=50,000$

Problem 4: A and B have an amount in 2:3 if A buy a car from his money whose price is depreciated by 10% whereas B invested that money in the bank which gives C.I at the rate of 20% per annum. Find the total percentage change in the total amount?

Solution:

10%=$\frac{1}{10}$–  ——>  Depreciates —–>  $\frac{9}{10}$

20%=$\frac{1}{5}$    ———> Increases  ——-> $\frac{6}{5}$

A             :                B

2              :                3

$2\times&space;\frac{9}{10}$    :        $3\times&space;\frac{6}{5}$

1              :            2

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

## 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