TOPIC: SIMPLE INTEREST
FORMULAE:
i. SI = (P * N * R)/100
Where,
SI – Simple Interest
P – Principal
R – Rate percent per annum
N – Number of years
A – Amount
ii. Amount (A) – Principal (P) + Interest (SI)
iii. Principal (P) – (100 * A) / (100 + RT)
iv. (ART) / (100 + RT)
EXAMPLE PROBLEMS
EXAMPLE 1:
In what time 1200 will become 1450. When annual rate of interest is 20%?
Solution:
P (Principle) = 1200, A (Amount) = 1450, R (Rate of Interest) =20%
A = P + SI
1450 = 1200 + SI
SI = 1450 – 1200
SI = 250
SI = (PRT) / 100
250 = (1200* 20 * T) / 100
250 = 240T
T= 25 / 24
Answer => T = 1 (1 /24) yrs
EXAMPLE 2 :
A Sum at simple interest of 4% per annum amounts to 3120 in 5 yrs. Find the sum?
Solution :
T (No. of Years) = 5 yrs, R (Rate of Interest) = 4%, A (Amount) = 3120
Method 1 :
P= (100 * A) / (100 + RT)
= (100 * 3120) / (100 + 4 * 5)
= (100 * 3120) / 120
Answer => P = 2600
Method 2:
SI = PRT / 100
= (P * 4 * 5) / 100
SI = P / 5
Amount = P + SI
= P + (P / 5) = 6P / 5
According to the question,
Amount = 3120 = 6P / 5 = 3120
P = (3120 * 5) / 6 = 2600
Answer => P = 2600
Note 1 :
If a sum of money becomes n times in T yr at simple interest, then formula for calculating rate of interest will be given as
R = [ 100 (n – 1)] / T %
Example :
A sum of money becomes 4 times in 20 yrs at simple interest. Find the rate of interest?
Solution :
T (No. of Years) = 20 yrs, n = 4
R = [100 (n – 1) / T]
= [100 (4 – 1) / 20]
= (100 * 3) / 20
Answer => R = 15 %
Note 2 :
If a sum of money at a certain rate of interest becomes “n” times in T1 yr and “m” times in T2 yr, then formula for T2 will be given as
T2 = [(m – 1) / (n – 1)] * T1
Example :
If a sum of money at simple interest becomes 4 times of itself in 15 yrs, it will become 3 times in,
Solution :
T1 = 15 yrs, n = 4, m = 3,
T2 = [(m -1) / (n – 1)] * T1
= [(3 – 1) / (4 – 1)] * 15
= (2 / 3) * 15
Answer => T2 = 10 yrs
Note 3 :
If a sum of money in a certain time becomes “n” times at R1 rate of interest and “m” times at R2 rate of interest, then formula for R2 will be given as,
R2 = [(m – 1) / (n – 1)] * R1
Example :
In a certain time, a sum becomes 3 times at the rate of 5 % per annum. At what rate of interest, the sum becomes 6 times in same duration?
Solution :
n = 3, m = 6, R1 = 5%, R2 = ?
R2 = [(m – 1) / (n – 1)] * R1
= [(6 – 1) / (3 – 1)] * 5
= [5 / 2] * 5
= 25 / 2
Answer => R2 = 12.5 %
Exercise Problems for Practice
1 ) In what time, does a sum of money become fourfold at the simple interest rate of 10 % per annum?
a) 10 years b) 20 years c) 30 years d) 40 years
N = 4, R = 10 %
R = [100 * (n – 1)] / T
10 = [100 * (4 – 1)] / T
10 = [100 * 3] / T
T = 30 yrs
Answer => c) 30 years
2 ) A certain sum becomes 3 fold at 4 % annual rate of interest. At what rate, it will become 6 fold?
a) 10 % b) 12 % c) 15 % d) 20 %
M = 6, n = 3, R1 = 4
R2 = [(m – 1) / (n – 1)] * R1
= [(6 – 1) / (3 – 1)] * 4
= [5 / 2] * 4
R2 = 10 %
Answer => a) 10 %
3 ) A sum of 2668 amounts to 4669 in 5 yrs at the rate of simple interest. Find the rate percent
a) 10 % b) 12 % c) 15 % d) 20 %
N = 5, P = 2668, A = 4669
SI = A – P
= 4669 – 2668 = 2001
SI = 2001
SI = PNR / 100
2001 = (2668 * 5 * R) / 100
R = 15 %
Answer => c) 15%
4 ) Find the difference in amount and principal for ₹ 4000 at the rate of 5 % amount interest in 4 yr?
a) 780 b) 800 c) 810 d) 820
R = 5 %, N = 4, P = 4000
SI = PNR / 100
= (4000 * 5 * 4) / 100
SI = 800
A = SI + P
= 800 + 4000
A = 4800
Difference in amount and principal,
SI = A – P => 4800 – 4000
SI = 800
Answer => b) 800
5 ) A sum of money triple itself at SI in 12 yrs. In how many years will it become 9 times?
a) 36 years b) 40 years c) 42 years d) 48 years
M = 9, n = 3, T1 = 12
T2 = [(m – 1) / (n – 1)] * T1
= [8 / 2] * 12
T2 = 48 yrs
Answer => d) 48 years
6 ) Amit takes some loan from Akash for 2 yr at the rate of 5 % per annum and after 2 yr he gave back ₹ 6600 to Akash and completed the payment of his loan. Find the interest paid by Amit?
a) 330 b) 450 c) 600 d) 820
T = 2 yr, R = 5%, A = 6600
SI = ART / (100 + RT)TOPIC: SIMPLE INTEREST
FORMULAE:
i. SI = (P * N * R)/100
Where,
SI – Simple Interest
P – Principal
R – Rate percent per annum
N – Number of years
A – Amount
ii. Amount (A) – Principal (P) + Interest (SI)
iii. Principal (P) – (100 * A) / (100 + RT)
iv. (ART) / (100 + RT)
EXAMPLE PROBLEMS
EXAMPLE 1:
In what time 1200 will become 1450. When annual rate of interest is 20%?
Solution:
P (Principle) = 1200, A (Amount) = 1450, R (Rate of Interest) =20%
A = P + SI
1450 = 1200 + SI
SI = 1450 – 1200
SI = 250
SI = (PRT) / 100
250 = (1200* 20 * T) / 100
250 = 240T
T= 25 / 24
Answer => T = 1 (1 /24) yrs
EXAMPLE 2 :
A Sum at simple interest of 4% per annum amounts to 3120 in 5 yrs. Find the sum?
Solution :
T (No. of Years) = 5 yrs, R (Rate of Interest) = 4%, A (Amount) = 3120
Method 1 :
P= (100 * A) / (100 + RT)
= (100 * 3120) / (100 + 4 * 5)
= (100 * 3120) / 120
Answer => P = 2600
Method 2:
SI = PRT / 100
= (P * 4 * 5) / 100
SI = P / 5
Amount = P + SI
= P + (P / 5) = 6P / 5
According to the question,
Amount = 3120 = 6P / 5 = 3120
P = (3120 * 5) / 6 = 2600
Answer => P = 2600
Note 1 :
If a sum of money becomes n times in T yr at simple interest, then formula for calculating rate of interest will be given as
R = [ 100 (n – 1)] / T %
Example :
A sum of money becomes 4 times in 20 yrs at simple interest. Find the rate of interest?
Solution :
T (No. of Years) = 20 yrs, n = 4
R = [100 (n – 1) / T]
= [100 (4 – 1) / 20]
= (100 * 3) / 20
Answer => R = 15 %
Note 2 :
If a sum of money at a certain rate of interest becomes “n” times in T1 yr and “m” times in T2 yr, then formula for T2 will be given as
T2 = [(m – 1) / (n – 1)] * T1
Example :
If a sum of money at simple interest becomes 4 times of itself in 15 yrs, it will become 3 times in,
Solution :
T1 = 15 yrs, n = 4, m = 3,
T2 = [(m -1) / (n – 1)] * T1
= [(3 – 1) / (4 – 1)] * 15
= (2 / 3) * 15
Answer => T2 = 10 yrs
Note 3 :
If a sum of money in a certain time becomes “n” times at R1 rate of interest and “m” times at R2 rate of interest, then formula for R2 will be given as,
R2 = [(m – 1) / (n – 1)] * R1
Example :
In a certain time, a sum becomes 3 times at the rate of 5 % per annum. At what rate of interest, the sum becomes 6 times in same duration?
Solution :
n = 3, m = 6, R1 = 5%, R2 = ?
R2 = [(m – 1) / (n – 1)] * R1
= [(6 – 1) / (3 – 1)] * 5
= [5 / 2] * 5
= 25 / 2
Answer => R2 = 12.5 %
Exercise Problems for Practice
1 ) In what time, does a sum of money become fourfold at the simple interest rate of 10 % per annum?
a) 10 years b) 20 years c) 30 years d) 40 years
N = 4, R = 10 %
R = [100 * (n – 1)] / T
10 = [100 * (4 – 1)] / T
10 = [100 * 3] / T
T = 30 yrs
Answer => c) 30 years
2 ) A certain sum becomes 3 fold at 4 % annual rate of interest. At what rate, it will become 6 fold?
a) 10 % b) 12 % c) 15 % d) 20 %
M = 6, n = 3, R1 = 4
R2 = [(m – 1) / (n – 1)] * R1
= [(6 – 1) / (3 – 1)] * 4
= [5 / 2] * 4
R2 = 10 %
Answer => a) 10 %
3 ) A sum of 2668 amounts to 4669 in 5 yrs at the rate of simple interest. Find the rate percent
a) 10 % b) 12 % c) 15 % d) 20 %
N = 5, P = 2668, A = 4669
SI = A – P
= 4669 – 2668 = 2001
SI = 2001
SI = PNR / 100
2001 = (2668 * 5 * R) / 100
R = 15 %
Answer => c) 15%
4 ) Find the difference in amount and principal for ₹ 4000 at the rate of 5 % amount interest in 4 yr?
a) 780 b) 800 c) 810 d) 820
R = 5 %, N = 4, P = 4000
SI = PNR / 100
= (4000 * 5 * 4) / 100
SI = 800
A = SI + P
= 800 + 4000
A = 4800
Difference in amount and principal,
SI = A – P => 4800 – 4000
SI = 800
Answer => b) 800
5 ) A sum of money triple itself at SI in 12 yrs. In how many years will it become 9 times?
a) 36 years b) 40 years c) 42 years d) 48 years
M = 9, n = 3, T1 = 12
T2 = [(m – 1) / (n – 1)] * T1
= [8 / 2] * 12
T2 = 48 yrs
Answer => d) 48 years
6 ) Amit takes some loan from Akash for 2 yr at the rate of 5 % per annum and after 2 yr he gave back ₹ 6600 to Akash and completed the payment of his loan. Find the interest paid by Amit?
a) 330 b) 450 c) 600 d) 820
T = 2 yr, R = 5%, A = 6600
SI = ART / (100 + RT)
SI = (6600 * 5 * 2) / (100 + 5 * 2)
= (6600 * 10) / 110
SI = 600
Answer => c) 600
SI = (6600 * 5 * 2) / (100 + 5 * 2)
= (6600 * 10) / 110
SI = 600
Answer => c) 600
FORMULAE:
i. SI = (P * N * R)/100
Where,
SI – Simple Interest
P – Principal
R – Rate percent per annum
N – Number of years
A – Amount
ii. Amount (A) – Principal (P) + Interest (SI)
iii. Principal (P) – (100 * A) / (100 + RT)
iv. (ART) / (100 + RT)
EXAMPLE PROBLEMS
EXAMPLE 1:
In what time 1200 will become 1450. When annual rate of interest is 20%?
Solution:
P (Principle) = 1200, A (Amount) = 1450, R (Rate of Interest) =20%
A = P + SI
1450 = 1200 + SI
SI = 1450 – 1200
SI = 250
SI = (PRT) / 100
250 = (1200* 20 * T) / 100
250 = 240T
T= 25 / 24
Answer => T = 1 (1 /24) yrs
EXAMPLE 2 :
A Sum at simple interest of 4% per annum amounts to 3120 in 5 yrs. Find the sum?
Solution :
T (No. of Years) = 5 yrs, R (Rate of Interest) = 4%, A (Amount) = 3120
Method 1 :
P= (100 * A) / (100 + RT)
= (100 * 3120) / (100 + 4 * 5)
= (100 * 3120) / 120
Answer => P = 2600
Method 2:
SI = PRT / 100
= (P * 4 * 5) / 100
SI = P / 5
Amount = P + SI
= P + (P / 5) = 6P / 5
According to the question,
Amount = 3120 = 6P / 5 = 3120
P = (3120 * 5) / 6 = 2600
Answer => P = 2600
Note 1 :
If a sum of money becomes n times in T yr at simple interest, then formula for calculating rate of interest will be given as
R = [ 100 (n – 1)] / T %
Example :
A sum of money becomes 4 times in 20 yrs at simple interest. Find the rate of interest?
Solution :
T (No. of Years) = 20 yrs, n = 4
R = [100 (n – 1) / T]
= [100 (4 – 1) / 20]
= (100 * 3) / 20
Answer => R = 15 %
Note 2 :
If a sum of money at a certain rate of interest becomes “n” times in T1 yr and “m” times in T2 yr, then formula for T2 will be given as
T2 = [(m – 1) / (n – 1)] * T1
Example :
If a sum of money at simple interest becomes 4 times of itself in 15 yrs, it will become 3 times in,
Solution :
T1 = 15 yrs, n = 4, m = 3,
T2 = [(m -1) / (n – 1)] * T1
= [(3 – 1) / (4 – 1)] * 15
= (2 / 3) * 15
Answer => T2 = 10 yrs
Note 3 :
If a sum of money in a certain time becomes “n” times at R1 rate of interest and “m” times at R2 rate of interest, then formula for R2 will be given as,
R2 = [(m – 1) / (n – 1)] * R1
Example :
In a certain time, a sum becomes 3 times at the rate of 5 % per annum. At what rate of interest, the sum becomes 6 times in same duration?
Solution :
n = 3, m = 6, R1 = 5%, R2 = ?
R2 = [(m – 1) / (n – 1)] * R1
= [(6 – 1) / (3 – 1)] * 5
= [5 / 2] * 5
= 25 / 2
Answer => R2 = 12.5 %
Exercise Problems for Practice
1 ) In what time, does a sum of money become fourfold at the simple interest rate of 10 % per annum?
a) 10 years b) 20 years c) 30 years d) 40 years
N = 4, R = 10 %
R = [100 * (n – 1)] / T
10 = [100 * (4 – 1)] / T
10 = [100 * 3] / T
T = 30 yrs
Answer => c) 30 years
2 ) A certain sum becomes 3 fold at 4 % annual rate of interest. At what rate, it will become 6 fold?
a) 10 % b) 12 % c) 15 % d) 20 %
M = 6, n = 3, R1 = 4
R2 = [(m – 1) / (n – 1)] * R1
= [(6 – 1) / (3 – 1)] * 4
= [5 / 2] * 4
R2 = 10 %
Answer => a) 10 %
3 ) A sum of 2668 amounts to 4669 in 5 yrs at the rate of simple interest. Find the rate percent
a) 10 % b) 12 % c) 15 % d) 20 %
N = 5, P = 2668, A = 4669
SI = A – P
= 4669 – 2668 = 2001
SI = 2001
SI = PNR / 100
2001 = (2668 * 5 * R) / 100
R = 15 %
Answer => c) 15%
4 ) Find the difference in amount and principal for ₹ 4000 at the rate of 5 % amount interest in 4 yr?
a) 780 b) 800 c) 810 d) 820
R = 5 %, N = 4, P = 4000
SI = PNR / 100
= (4000 * 5 * 4) / 100
SI = 800
A = SI + P
= 800 + 4000
A = 4800
Difference in amount and principal,
SI = A – P => 4800 – 4000
SI = 800
Answer => b) 800
5 ) A sum of money triple itself at SI in 12 yrs. In how many years will it become 9 times?
a) 36 years b) 40 years c) 42 years d) 48 years
M = 9, n = 3, T1 = 12
T2 = [(m – 1) / (n – 1)] * T1
= [8 / 2] * 12
T2 = 48 yrs
Answer => d) 48 years
6 ) Amit takes some loan from Akash for 2 yr at the rate of 5 % per annum and after 2 yr he gave back ₹ 6600 to Akash and completed the payment of his loan. Find the interest paid by Amit?
a) 330 b) 450 c) 600 d) 820
T = 2 yr, R = 5%, A = 6600
SI = ART / (100 + RT)TOPIC: SIMPLE INTEREST
FORMULAE:
i. SI = (P * N * R)/100
Where,
SI – Simple Interest
P – Principal
R – Rate percent per annum
N – Number of years
A – Amount
ii. Amount (A) – Principal (P) + Interest (SI)
iii. Principal (P) – (100 * A) / (100 + RT)
iv. (ART) / (100 + RT)
EXAMPLE PROBLEMS
EXAMPLE 1:
In what time 1200 will become 1450. When annual rate of interest is 20%?
Solution:
P (Principle) = 1200, A (Amount) = 1450, R (Rate of Interest) =20%
A = P + SI
1450 = 1200 + SI
SI = 1450 – 1200
SI = 250
SI = (PRT) / 100
250 = (1200* 20 * T) / 100
250 = 240T
T= 25 / 24
Answer => T = 1 (1 /24) yrs
EXAMPLE 2 :
A Sum at simple interest of 4% per annum amounts to 3120 in 5 yrs. Find the sum?
Solution :
T (No. of Years) = 5 yrs, R (Rate of Interest) = 4%, A (Amount) = 3120
Method 1 :
P= (100 * A) / (100 + RT)
= (100 * 3120) / (100 + 4 * 5)
= (100 * 3120) / 120
Answer => P = 2600
Method 2:
SI = PRT / 100
= (P * 4 * 5) / 100
SI = P / 5
Amount = P + SI
= P + (P / 5) = 6P / 5
According to the question,
Amount = 3120 = 6P / 5 = 3120
P = (3120 * 5) / 6 = 2600
Answer => P = 2600
Note 1 :
If a sum of money becomes n times in T yr at simple interest, then formula for calculating rate of interest will be given as
R = [ 100 (n – 1)] / T %
Example :
A sum of money becomes 4 times in 20 yrs at simple interest. Find the rate of interest?
Solution :
T (No. of Years) = 20 yrs, n = 4
R = [100 (n – 1) / T]
= [100 (4 – 1) / 20]
= (100 * 3) / 20
Answer => R = 15 %
Note 2 :
If a sum of money at a certain rate of interest becomes “n” times in T1 yr and “m” times in T2 yr, then formula for T2 will be given as
T2 = [(m – 1) / (n – 1)] * T1
Example :
If a sum of money at simple interest becomes 4 times of itself in 15 yrs, it will become 3 times in,
Solution :
T1 = 15 yrs, n = 4, m = 3,
T2 = [(m -1) / (n – 1)] * T1
= [(3 – 1) / (4 – 1)] * 15
= (2 / 3) * 15
Answer => T2 = 10 yrs
Note 3 :
If a sum of money in a certain time becomes “n” times at R1 rate of interest and “m” times at R2 rate of interest, then formula for R2 will be given as,
R2 = [(m – 1) / (n – 1)] * R1
Example :
In a certain time, a sum becomes 3 times at the rate of 5 % per annum. At what rate of interest, the sum becomes 6 times in same duration?
Solution :
n = 3, m = 6, R1 = 5%, R2 = ?
R2 = [(m – 1) / (n – 1)] * R1
= [(6 – 1) / (3 – 1)] * 5
= [5 / 2] * 5
= 25 / 2
Answer => R2 = 12.5 %
Exercise Problems for Practice
1 ) In what time, does a sum of money become fourfold at the simple interest rate of 10 % per annum?
a) 10 years b) 20 years c) 30 years d) 40 years
N = 4, R = 10 %
R = [100 * (n – 1)] / T
10 = [100 * (4 – 1)] / T
10 = [100 * 3] / T
T = 30 yrs
Answer => c) 30 years
2 ) A certain sum becomes 3 fold at 4 % annual rate of interest. At what rate, it will become 6 fold?
a) 10 % b) 12 % c) 15 % d) 20 %
M = 6, n = 3, R1 = 4
R2 = [(m – 1) / (n – 1)] * R1
= [(6 – 1) / (3 – 1)] * 4
= [5 / 2] * 4
R2 = 10 %
Answer => a) 10 %
3 ) A sum of 2668 amounts to 4669 in 5 yrs at the rate of simple interest. Find the rate percent
a) 10 % b) 12 % c) 15 % d) 20 %
N = 5, P = 2668, A = 4669
SI = A – P
= 4669 – 2668 = 2001
SI = 2001
SI = PNR / 100
2001 = (2668 * 5 * R) / 100
R = 15 %
Answer => c) 15%
4 ) Find the difference in amount and principal for ₹ 4000 at the rate of 5 % amount interest in 4 yr?
a) 780 b) 800 c) 810 d) 820
R = 5 %, N = 4, P = 4000
SI = PNR / 100
= (4000 * 5 * 4) / 100
SI = 800
A = SI + P
= 800 + 4000
A = 4800
Difference in amount and principal,
SI = A – P => 4800 – 4000
SI = 800
Answer => b) 800
5 ) A sum of money triple itself at SI in 12 yrs. In how many years will it become 9 times?
a) 36 years b) 40 years c) 42 years d) 48 years
M = 9, n = 3, T1 = 12
T2 = [(m – 1) / (n – 1)] * T1
= [8 / 2] * 12
T2 = 48 yrs
Answer => d) 48 years
6 ) Amit takes some loan from Akash for 2 yr at the rate of 5 % per annum and after 2 yr he gave back ₹ 6600 to Akash and completed the payment of his loan. Find the interest paid by Amit?
a) 330 b) 450 c) 600 d) 820
T = 2 yr, R = 5%, A = 6600
SI = ART / (100 + RT)
SI = (6600 * 5 * 2) / (100 + 5 * 2)
= (6600 * 10) / 110
SI = 600
Answer => c) 600
SI = (6600 * 5 * 2) / (100 + 5 * 2)
= (6600 * 10) / 110
SI = 600
Answer => c) 600
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