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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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Author : J. Maha Laxamaiah Mailid: [email protected]
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
When T = 3 years, R = 5%
C. I = P × 𝟏𝟓.𝟕𝟔𝟐𝟓
𝟏𝟎𝟎
When T = 3 years, R = 6%
C. I = P × 𝟏𝟗.𝟏𝟎𝟏𝟔
𝟏𝟎𝟎
When T = 3 years, R = 7%
C. I = P × 𝟐𝟐.𝟓𝟎𝟒𝟑
𝟏𝟎𝟎
When T = 3 years, R = 8%
C. I = P × 𝟐𝟓.𝟗𝟕𝟏𝟐
𝟏𝟎𝟎
When T = 3 years, R = 9%
C. I = P × 𝟐𝟗. 𝟓𝟎𝟐𝟗
𝟏𝟎𝟎
When T = 3 years, R = 10%
C. I = P × 𝟑𝟑.𝟏
𝟏𝟎𝟎
When T = 3 years, R = 11%
C. I = P × 𝟑𝟔.𝟕𝟔𝟑𝟏
𝟏𝟎𝟎
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When T = 3 years, R = 12%
C. I = P × 𝟒𝟎.𝟒𝟗𝟐𝟖
𝟏𝟎𝟎
When T = 3 years, R = 13%
C. I = P × 𝟒𝟒.𝟐𝟖𝟗𝟕
𝟏𝟎𝟎
When T = 3 years, R = 14%
C. I = P × 𝟒𝟖.𝟏𝟓𝟒𝟒
𝟏𝟎𝟎
When T = 3 years, R = 15%
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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Author : J. Maha Laxamaiah Mailid: [email protected]
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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Author : J. Maha Laxamaiah Mailid: [email protected]
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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Author : J. Maha Laxamaiah Mailid: [email protected]
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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Author : J. Maha Laxamaiah Mailid: [email protected]
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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Author : J. Maha Laxamaiah Mailid: [email protected]
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)