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 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 10\times 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 20}{100}=44%

Scheme A=15% S.I

Assume principal is 100

S.I=\frac{PTR}{100}=\frac{100\times 2\times 15}{100}=130%

44%——–>2860

100%——–>?

\frac{2860\times 100}{44}=6500

\frac{6500\times 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 \frac{9}{10}    :        3\times \frac{6}{5}

1              :            2

\frac{1}{1}\times 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 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 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 100}{121}=2000

121———->2420

21————–>?

\frac{2420\times 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 4}{1}= 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 49}{1}=1960 =Principle

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

14———>?\frac{40\times 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 400}{9}=8000 is principal