Aptitude - Compound Interest - Discussion
Discussion Forum : Compound Interest - General Questions (Q.No. 10)
10.
The effective annual rate of interest corresponding to a nominal rate of 6% per annum payable half-yearly is:
Answer: Option
Explanation:
| Amount of Rs. 100 for 1 year when compounded half-yearly |
 |
= Rs. |  |
100 x |  |
1 + | 3 |  |
2 |  |
= Rs. 106.09 |
| 100 |
Effective rate = (106.09 - 100)% = 6.09%
Discussion:
43 comments Page 4 of 5.
Povizha said:
7 years ago
What is effective rate?
Zaini said:
6 years ago
How to solve if there is a whole year?
Please explain.
Please explain.
Datta said:
6 years ago
@Durga.
106.09-106 = ?
10690/100 -106 = 6.09.
106.09-106 = ?
10690/100 -106 = 6.09.
Summer said:
6 years ago
Please explain, , how come 3 in (3/100) ?
(3)
Syamu said:
6 years ago
The given interest is 6%.
According to formulae of compound interest for half-yearly [p (1 + (R/2) /100) ^2].
According to formulae of compound interest for half-yearly [p (1 + (R/2) /100) ^2].
Padminiyadav said:
5 years ago
@All.
Since,it says rate to be 6% per annumn and we need to find the effective rate for half yearly
So, rate=6/100=0.06% (mandatory step).
E.F(effective rate)=(1+r/n)^n-1 (n=time; i.e-if time is 1 year ,annually=1,half .yearly=2,quarterly=4).
And then,
(E.F * 100 ) will give you an actual effective rate.
Since,it says rate to be 6% per annumn and we need to find the effective rate for half yearly
So, rate=6/100=0.06% (mandatory step).
E.F(effective rate)=(1+r/n)^n-1 (n=time; i.e-if time is 1 year ,annually=1,half .yearly=2,quarterly=4).
And then,
(E.F * 100 ) will give you an actual effective rate.
(3)
Ravi Kumar said:
5 years ago
The amount for a sum of 'p' for 'n' years with effective rate 'r' compounded yearly should be equal to the amount for the same sum 'p' for 'n' years with rate 6% p.a compounded half-yearly.
p(1+r/100)^n = p(1+(6/2)/100)^2n
'p' and 'n' on both sides gets cancelled and remaining EQ becomes;
(1+r/100) = (1+3/100)^2.
1+r/100 = 1.0609 (on simplication),
r/100 = 0.0609 ==> r= 6.09,
Therefore, 'r' becomes 6.09% which is the required answer.
p(1+r/100)^n = p(1+(6/2)/100)^2n
'p' and 'n' on both sides gets cancelled and remaining EQ becomes;
(1+r/100) = (1+3/100)^2.
1+r/100 = 1.0609 (on simplication),
r/100 = 0.0609 ==> r= 6.09,
Therefore, 'r' becomes 6.09% which is the required answer.
(3)
Nitin said:
5 years ago
How can we know that we have to find the compound interest in this question? Anyone, please explain to me to get it.
(3)
Gojo said:
4 years ago
r=6.
t=1.
Let sum=100.
CI = (p(1+r/2*100)^t)-p.
=100(1+6/200)^2*1-100 //as we have to solve for 1 year
=100(1+3/100)^2-100,
=100(103/100)^2-100,
=100*103/100*103/100-100,
=10609/100-100,
=106.09-100,
=6.09 Ans.
t=1.
Let sum=100.
CI = (p(1+r/2*100)^t)-p.
=100(1+6/200)^2*1-100 //as we have to solve for 1 year
=100(1+3/100)^2-100,
=100(103/100)^2-100,
=100*103/100*103/100-100,
=10609/100-100,
=106.09-100,
=6.09 Ans.
(3)
Luke said:
4 years ago
Where R=3 from? Please explain.
(2)
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