Hola friends !!
Solve it please
The interest on a sum of ₹ 12800 is being compounded annually at the rate of 7.5% per annum.Find the period for which the compound interest is ₹ 1992.
Answers
Answered by
96
Solutions :-
Some of the abbreviations used in the solutions are :-
Principle = p
Time = t
Rate = r
Compound Interest = C.I
Amount = A
Given :
p = ₹ 12800
r = 7.5% p.a
C.I = ₹ 1992
Find the Amount :-
A = p + C.I
= ₹ (12800 + 1992)
= ₹ 14792
Find the Time :-
A = p (1 + r)^t
14792 = 12800 (1 + 7.5%)^t
14792 = 12800 (1 + 0.075)^t
14792 = 12800 (1.075)^t
14792/12800 = (1075/1000)^t
7396/6400 = (86/80)^t
(86/80)^2 = (86/80)^t
2 = t
Hence,
Time = 2 years
Some of the abbreviations used in the solutions are :-
Principle = p
Time = t
Rate = r
Compound Interest = C.I
Amount = A
Given :
p = ₹ 12800
r = 7.5% p.a
C.I = ₹ 1992
Find the Amount :-
A = p + C.I
= ₹ (12800 + 1992)
= ₹ 14792
Find the Time :-
A = p (1 + r)^t
14792 = 12800 (1 + 7.5%)^t
14792 = 12800 (1 + 0.075)^t
14792 = 12800 (1.075)^t
14792/12800 = (1075/1000)^t
7396/6400 = (86/80)^t
(86/80)^2 = (86/80)^t
2 = t
Hence,
Time = 2 years
VickyskYy:
Bahut achhe bihari ji.
Answered by
79
Solutions :-
We have,
Principle = p = ₹ 12800
Rate = r = 7.5% p.a
Compound Interest = C.I = ₹ 1992
Amount = A = ?
Period = t = ?
Find the Amount :-
A = p + C.I
= ₹ (12800 + 1992)
= ₹ 14792
Find the Time :-
By using Compound Interest formula
A = p (1 + r)^t
=> 14792 = 12800 (1 + 7.5%)^t
=> 14792 = 12800 (1 + 0.075)^t
=> 14792 = 12800 (1.075)^t
=> 14792/12800 = (1075/1000)^t
=> 7396/6400 = (86/80)^t
=> (86/80)^2 = (86/80)^t
=> 2 = t
Answer : Period = 2 years
We have,
Principle = p = ₹ 12800
Rate = r = 7.5% p.a
Compound Interest = C.I = ₹ 1992
Amount = A = ?
Period = t = ?
Find the Amount :-
A = p + C.I
= ₹ (12800 + 1992)
= ₹ 14792
Find the Time :-
By using Compound Interest formula
A = p (1 + r)^t
=> 14792 = 12800 (1 + 7.5%)^t
=> 14792 = 12800 (1 + 0.075)^t
=> 14792 = 12800 (1.075)^t
=> 14792/12800 = (1075/1000)^t
=> 7396/6400 = (86/80)^t
=> (86/80)^2 = (86/80)^t
=> 2 = t
Answer : Period = 2 years
Similar questions
Math,
7 months ago
History,
7 months ago
English,
7 months ago
History,
1 year ago
Environmental Sciences,
1 year ago
English,
1 year ago
Environmental Sciences,
1 year ago