simple interest - fdaytalk4 a sum of money will become double in 3 years at a certain rate of c.i....

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Author : J. Maha Laxamaiah Mailid: [email protected]

SIMPLE INTEREST

S. I = 𝑷𝑻𝑹

𝟏𝟎𝟎

S. I = Simple interest

P = principal

T = time in years

R = rate of interest

A = P + S. I

A = total amount

COMPOUND INTEREST

C. I = P × (1 + 𝑹

𝟏𝟎𝟎)𝑻 − 𝑷

C.I = Compound interest

P = principal

T = time in years

R = rate of interest

A = P + C. I

A = Total amount

Compound interest for half yearly

C. I = P × (1 + 𝑹

𝟐∗𝟏𝟎𝟎)𝟐𝑻 − 𝑷

Compound interest for quarterly

C. I = P × (1 + 𝑹

𝟒∗𝟏𝟎𝟎)𝟒𝑻 − 𝑷

When T = 2 years, R = any rate of interest

C. I – S. I = 𝑷𝑹𝟐

𝟏𝟎𝟎𝟐

When T = 3 years, R = any rate of interest

C. I – S. I = 𝑷𝑹𝟐 (𝟑𝟎𝟎+𝑹)

𝟏𝟎𝟎𝟑

EXAMPLE:

The difference between Compound and Simple rates of interest on 10, 000/-

for 3 – years at 5% per annum is ………..

SOLUTION :

SHORT METHOD for Difference between C. I & S. I for 3 – years C. I –

S. I = 𝑷𝑹𝟐 (𝟑𝟎𝟎+𝑹)

𝟏𝟎𝟎𝟑

= 10,000 ∗ 5 ∗ 5 ∗ (300 +5)

𝟏𝟎𝟎𝟑

= 76. 25 (answer)

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SHORT METHODS FOR COMPOUND INTEREST

When T = 2 years R = any rate of interest

For example R = 6%

C. I = Compound interest

P = principal

For example R = 8%

When T = 3 years, R = 3%

C. I = P × 𝟗.𝟐𝟕𝟐𝟕

𝟏𝟎𝟎

When = 3 years, R = 4%

C. I = P × 𝟏𝟐.𝟒𝟖𝟔𝟒

𝟏𝟎𝟎


C. I = P × 𝟏𝟓.𝟕𝟔𝟐𝟓

𝟏𝟎𝟎


C. I = P × 𝟏𝟗.𝟏𝟎𝟏𝟔

𝟏𝟎𝟎


C. I = P × 𝟐𝟐.𝟓𝟎𝟒𝟑

𝟏𝟎𝟎


C. I = P × 𝟐𝟓.𝟗𝟕𝟏𝟐

𝟏𝟎𝟎


C. I = P × 𝟐𝟗. 𝟓𝟎𝟐𝟗

𝟏𝟎𝟎


C. I = P × 𝟑𝟑.𝟏

𝟏𝟎𝟎


C. I = P × 𝟑𝟔.𝟕𝟔𝟑𝟏

𝟏𝟎𝟎

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C. I = P × 𝟒𝟎.𝟒𝟗𝟐𝟖

𝟏𝟎𝟎


C. I = P × 𝟒𝟒.𝟐𝟖𝟗𝟕

𝟏𝟎𝟎


C. I = P × 𝟒𝟖.𝟏𝟓𝟒𝟒

𝟏𝟎𝟎


C. I = P × 𝟓𝟐.𝟎𝟖𝟕𝟓

𝟏𝟎𝟎

1 On simple interest any amount becomes 812/- in two years and 924/- in

four years. What is the rate of interest on that amount ?

SOLUTION:

S.I = PTR/100 & A = S.I + P

Simple interest (S.I) => 924 - 812 = 112/-

Principal (p) => 812 - 112

S.I = PTR/100

=> R= 𝟏𝟎𝟎 × 𝑺.𝑰

𝑷𝑻

= 𝟏𝟎𝟎 × 𝟏𝟏𝟐

𝟕𝟎𝟎 × 𝟐

R = 8% (answer)

2 A man deposits am amount of 54600/- at a S.I rate of 12% for 3 years.

What total amount will he get at the end of 3 years ?

SOLUTION:

Short method:

12 × 3 = 36%

If 100% money …………….. 54600

136% (100 + 36) ………….. ?

Total amount (A) = 𝟏𝟑𝟔∗𝟓𝟒𝟔𝟎𝟎

𝟏𝟎𝟎

= 74256/- (answer)

3 A sum of money will become 108800/- in four years at 7% per annum on

S.I. Find the sum ………..

SOLUTION:

Short method

7 × 4 = 28%

If 128% (100 + 28) ………………. 108800

100% …………………?

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Principal (P) = 𝟏𝟎𝟎 × 𝟏𝟎𝟖𝟖𝟎𝟎

𝟏𝟐𝟖

= 85000/- (answer)

4 A sum of money will become double in 3 years at a certain rate of C.I. In

what time will it becomes four times itself ?

SOLUTION:

Short method

In 3 years becomes ………… double

In 6 years (3+3) becomes ……… 4 times (double double)

Therefore, Answer = 6 years

5 What will be the amount on 25000/- in 2 years at C.I if the rates for the

successive years be 4% & 5% per year ……………

SOLUTION:

Short method:

Total amount (A) =

25000 × 𝟏𝟎𝟒

𝟏𝟎𝟎×

𝟏𝟎𝟓

𝟏𝟎𝟎

A = 27300/- (answer)

6 The difference of S.I on 5000/- from two banks in 2 years is 100/-. Find the

difference of their rate …

SOLUTION:

Given that, P = 5000/- T = 2 years S.I = 100/-

Formula

S. I = 𝑷 × 𝑻(𝑹𝟏−𝑹𝟐)

𝟏𝟎𝟎

100 = 𝟓𝟎𝟎𝟎 × 𝟐(𝑹𝟏−𝑹𝟐)

𝟏𝟎𝟎

R1 – R2 = 1% (answer)

7 A sum becomes 3 times in 10 years. In how many years will amount to 5

times at the same rate of S.I ……….

SOLUTION:

Formula,

T2 = (𝒀−𝟏) × 𝑻𝟏

𝑿−𝟏

Given that, X = 3 times, Y = 5 times, T1 = 10 years & T2 = ?

𝑻𝟐 =(𝟓 − 𝟏) × 𝟏𝟎

𝟑 − 𝟏

T2 = 20 years (answer)

8 Find the sum which amount to 1125/- in 5 years and becomes 1200/- in 8

years at the rate of S.I ………

SOLUTION:

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Formula

P = [A2 – (𝑇1

𝑇2−𝑇1) × (𝐴2 − 𝐴1)]

Given that, A1 = 1125/-, A2 = 1200/-, T1 = 5 years , T2 = 8 years & P = ?

P = [1125 – (5

8−5) × (1200 − 1125)]

P = 1000/- (answer)

9 If 5, 000/- becomes 5, 700/- in one year, what will 7, 000/- become at the

end of 5- years at the same rate of simple interest?

SOLUTION:

S. I = 5, 700- 5, 000

= 700/-

S. I = 𝑷𝑻𝑹

𝟏𝟎𝟎

700 = 𝟓𝟎𝟎𝟎 × 𝟏 × 𝑹

𝟏𝟎𝟎

R = 14%

NOW

S. I = 𝑷𝑻𝑹

𝟏𝟎𝟎

S. I = 𝟕𝟎𝟎𝟎 × 𝟓 × 𝟏𝟒

𝟏𝟎𝟎

= 4, 900/-

Total money=> 7, 000+ 4, 900

= 11, 900/- (answer)

10 A sum of money 12, 000/- deposited at compound interest became double

after 5- years. How much will it be after 20 years ?

SOLUTION:

SHORT METHOD

5 (If in 5- years) ……………….. 2 times

10 ……………………………………. 4 times

15 ……………………………………. 8 times

20 ……………………………………. 16 times

Required answer, 12, 000 × 16

= 1, 92, 000/- (answer)

11 If the difference between the C. I, compounded every 6 months and the

S. I on certain sum of money at the rate of 12% p. a for one year is 36/-, the

sum is

SOLUTION:

Formula, D = 𝑷𝑹𝟐

𝟏𝟎𝟎𝟐

Here, R = 𝑹

𝟐 =>

𝟏𝟐

𝟐 = 6%

36 = 𝑷 × 𝟔 × 𝟔

𝟏𝟎𝟎 × 𝟏𝟎𝟎

P = 10, 000/- (answer)

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NOTE: As the interest was compounded half- yearly, we changed R to 𝑹

𝟐 and T to

𝑻

𝟐

12 Arun lends 20, 000/- to two of his friends. He gives 12, 000/- to the first

one at 8% per annum at simple interest. Arun wants to make a profit of 10%

on the whole. The simple interest rate at which he should lend the remaining

sum of money to the second friend is…

SOLUTION:

S. I on 12, 000/- => 𝟏𝟐𝟎𝟎𝟎 × 𝟏 × 𝟖

𝟏𝟎𝟎

= 960/-

Desired gain on 20, 000/-

= 20, 000 × 𝟏𝟎

𝟏𝟎𝟎

= 2, 000/-

S. I on 8, 000 => 2, 000- 960

= 1040/-

Therefore, S. I = 𝑷𝑻𝑹

𝟏𝟎𝟎

1040 = 𝟖𝟎𝟎𝟎 × 𝟏 × 𝑹

𝟏𝟎𝟎

R = 13% (answer)

13 An amount of money at C. I grow up to 3840/- in 4- years and up to

3956/- in 5- years. Find the rate of interest

SOLUTION:

FORMULA

R = [ ( A2

A1)

1

T2− T1 − 1 ] ∗ 100

R = [ ( 3936

3840 )

1

5−4 − 1 ] ∗ 100

R = 2. 5% (answer)

14 A man borrows 21, 000/- at 10% compound interest. How much he has

to pay equally at the end of each year, to settle his loan in two years is

SOLUTION:

Let annual payment = X x

(1 + 10

100)2

+ x

(1 + 10

100)2

= 21, 000

By solving, we get x = 12, 100/- (answer)

15 1, 500/- is invested at a rate of 10% S. I and interest is added to the

principal after every 5- years. In how many years will it amount to 2, 500/-?

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SOLUTION:

First we will find the S. I for 5- years

S. I = 𝑷𝑻𝑹

𝟏𝟎𝟎

S. I = 𝟏𝟓𝟎𝟎 × 𝟓 × 𝟏𝟎

𝟏𝟎𝟎

= 750/-

Now principal after 5- years => 1, 500 + 750

= 2250/-

Final amount = 2500/- (given)

S. I= 2500 - 2250

= 250/-

S. I = 𝑷𝑻𝑹

𝟏𝟎𝟎

250 = 𝟐𝟐𝟓𝟎 × 𝑻 × 𝟏𝟎

𝟏𝟎𝟎

T = 𝟏𝟎

𝟗 years

Therefore, total time => 5 + 𝟏𝟎

𝟗

= 𝟔𝟗𝟏 (answer)

16 What annual payment will discharge a debt of 6450/- due in 4- years at

5% per annum simple interest?

SOLUTION:

Let the annual installment = X

X + (X + 𝑋 × 3 × 5

100) + (𝑋 +

𝑋 × 2 × 5

100) + (𝑋 +

𝑋 × 1 × 5

100) = 6450

X = 1500/- (answer)

17 A person invested in three banks in the ratio of 1: 2: 3 and the rates were

in the ratio of 2: 3: 4 at S.I, also he got the income in the ratio of 3: 4: 5, then

find the ratio of time periods

SOLUTION:

S.I = 𝑷𝑻𝑹

𝟏𝟎𝟎

T = 𝑺.𝑰 ×𝟏𝟎𝟎

𝑷𝑹

Therefore, the ratio of time periods

T1 : T2 : T3 = 𝟏𝟎𝟎 ×𝑺𝟏

𝑷𝟏 × 𝑹𝟏∶

𝟏𝟎𝟎 × 𝑺𝟐

𝑷𝟐 × 𝑹𝟐∶

𝟏𝟎𝟎 × 𝑺𝟑

𝑷𝟑 × 𝑹𝟑

T1 : T2 : T3 = 𝟏𝟎𝟎 × 𝟑

𝟏 × 𝟐∶

𝟏𝟎𝟎 × 𝟒

𝟐 × 𝟑∶

𝟏𝟎𝟎 × 𝟓

𝟑 × 𝟒

= 18: 8: 5 (answer)

18 A person invested in all 2, 600/- at 4%, 6% and 8% per annum S.I. At

the end of the year, he got the same interest in all three cases. The money

invested at 4% is

SOLUTION:

Let the amount interested at 4%, 6% and 8% per annum S.I are X, Y and Z

respectively

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S.I = 𝑷𝑻𝑹

𝟏𝟎𝟎

S. I = x × 1 × 4

100 S. I =

y × 1 × 6

100 S. I =

z × 1 × 8

100

X: Y: Z = 𝟏

𝟒 :

𝟏

𝟔 :

𝟏

𝟖

= 6: 4: 3

Therefore, X = 2600 × 𝟔

𝟏𝟑

= 1200/- (answer)

19 A person has 5, 000/-. He invests a part of it at 3% per annum and the

remainder at 8% per annum S.I. His total income I 3- years is 750/-. Find the

sum invested at different rate of interest.

SOLUTION:

Average rate of interest => S.I = 𝑷𝑻𝑹

𝟏𝟎𝟎

750 = 𝟓𝟎𝟎𝟎 × 𝟑 × 𝑹

𝟏𝟎𝟎 R = 5%

By using Allegations and mixture method

Required ratio = 3: 2

Investment at 3% per annum

= (3

3 + 2) × 5000

= 3, 000/- (answer)

Investment at 8% per annum

= (2

3+2) × 5000

= 2, 000/- (answer)

20 The S.I and C.I on a sum of money for are 8, 400/- and 8652/-

respectively. The rate of interest per annum is

SOLUTION:

SHORT METHOD for 2- years

S. I

C. I=

200 × R

200 + R

8400

8652=

200 × R

200 + R

R = 6% (answer)

21 A sum of money invested at C.I amounts to 800/- in 2- years and 880/- in

3- years. Find the rate of interest per annum and the sum

SOLUTION:

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SHORT METHOD

Here, A1 = 800/-

A2 = 880/-

For rate of interest

= A2 − A1

A1 × 100

=880 − 800

800× 100

= 10% (answer)

And Sum

= A1( A1

A2 )T

= 800 ( 800

880 )2

= 661. 15/- (answer)

22 A man borrows 3000/- at 10% compound rate of interest. At the end of

each year he pays back 1000/-, how much amount should he pay at the end of

the third year to clear all his dues.

SOLUTION:

C. I for 1st year = S. I for 1st year

= 10% of 3000

= 300/-

Principle for 2nd year = (3000 + 300) – 1000

= 2300/-

C. I for 2nd year = S. I of 2300 at 10%

= 230/-

Principle for 3rd year = (2300 + 230) – 1000

= 1530/-

C. I for 3rd year = 10% of 1530

= 153/-

Therefore, total amount pay at the end of 3rd year = 1530 + 153

= 1683/- (answer)

23 A certain sum is invested for certain time. It amounts to 450/- at 7% per

annum. But, when invested at 5% per annum, it amounts to 350/-. Find out

the sum and time.

SOLUTION:

Using the formula, 𝒑 = 𝑨𝟐𝑹𝟏− 𝑨𝟏𝑹𝟐

𝑹𝟏− 𝑹𝟐

Here A1 = 450/-, R1 = 7%, A2 = 350/-, R2 = 5%

P = 350 × 7 − 450 × 5

7 − 5

P = 100/-

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Also using the formula, T = [A1− A2

A2R1− A1R2] ∗ 100

T = [450 − 350

350 × 7 − 450 × 5] × 100

T = 50 years (answer)

24 A sum 2/- is lent to be paid back in 3 equal monthly instalments of 1/-

each. Find the rate percent ….

SOLUTION:

Using the formula, Z = na − Ra

100∗b∗

n(n−1)

2

Here, Z = debt to be paid = 2/-, a = 1/-, n = 3

b = number of installments per year = 12

R = ?

2 = 3 × 1 − R × 1

100 × 12 ×

3 × 2

2

R = 400% (answer)

25 A man borrows 3, 000/- at 10% compound interest per annum. At the

end of each year he pays back 1, 000/-. How much amount he pay at the end of

3rd year to clear all his dues?

SOLUTION:

Amount due at the end of 1st year

= (3000 × 110

100) − 1000

= 2, 300/-

Amount due at the end of the 2nd year

= (2300 × 110

100) − 1000

= 1, 350/-

Amount due at the end of the 3rd year

= 1530 × 110

100

= 1683/- (answer)

26 Divide 15, 494/- A and B so that A’s share at the end of 9 – years may be

equals to B’s share at the end of 11 – years, Compound Interest being 20%

per annum. Then A’s share is ………..

SOLUTION:

Let A’s share = x & B’s share = 15, 494 – x

According to question

𝑥 × (1 + 20

100)9 = (15, 494 − 𝑥) × (1 +

20

100)11 X = 9, 144/- (answer)

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27 The Simple Interest for 4 –years is one fourth of the Principal. The sum

that will be amount to 450/- in two years at the same rate is ………..

SOLUTION:

Let the Principal be P/-

According to question

S. I = 𝑃 × 𝑇 × 𝑅

100 [Formula]

𝑃

4=

𝑃 × 4 × 𝑅

100

R = 25

4% Per annum

P = 𝑆.𝐼 × 100

𝑅 × 𝑇 [Formula]

P = (450−𝑃) × 100

25

4 × 2

[From question]

By solving, we get

P = 400/- (answer)

28 Kartik lost 9% by selling pencils at the rate of 15 a rupee. How many for

a rupee must Kartik sell them to gain 5% ……….

SOLUTION:

PERCENTAGE METHOD:

15 × 91

100= 𝑥 ×

105

100

x = 13 (answer)

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simple interest - fdaytalk4 a sum of money will become double in 3 years at a certain rate of c.i....

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