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.
Siva said:
1 decade ago
@Pavitra.
It is "106.09-100 = 6.09"
It is "106.09-100 = 6.09"
Pavitra said:
1 decade ago
How could it be possible?
106.09-106 = 6.09?
106.09-106 = 6.09?
Parvathy said:
1 decade ago
Still got no clue. Somebody please explain in simple context. What this effective rate and annual rate anyway?
Gaurav said:
1 decade ago
Effective rate is actual rate for full year corresponding to rate for a year half yearly.
Eg. Rs.100 A = 100(1+6/100) = 106 for full year.
A = 100(1+3/100)^2 = 106.09 for a year half yearly. So effective rate is 106.09 = 100(1+R/100) is equal to 6.09. That's it.
Eg. Rs.100 A = 100(1+6/100) = 106 for full year.
A = 100(1+3/100)^2 = 106.09 for a year half yearly. So effective rate is 106.09 = 100(1+R/100) is equal to 6.09. That's it.
RASHMI PODDAR said:
1 decade ago
@Sathya can you please explain me.
What will be the value of n according to this question in the following formula?
Effective rate = ((1+i/n)^n)-1.
What will be the value of n according to this question in the following formula?
Effective rate = ((1+i/n)^n)-1.
Xyz said:
1 decade ago
@Arun.
For half yearly, is in't the formula:
C.I=(P(1+(r/2)/100)^2n-P)? where n=1/2. why have you missed n in the formula? Please explain.
For half yearly, is in't the formula:
C.I=(P(1+(r/2)/100)^2n-P)? where n=1/2. why have you missed n in the formula? Please explain.
Arun said:
1 decade ago
Effective rate of Interest means,
the simple interest rate that produces the same amount of interest as the compound interest rate.
i.e assume the principal amount be 'x'
Then according to definition,
p*r*t = (p(1+r/100)^n - p),
For S.I., as per question, t= 1 year;
For C.I., as per question, we use formula C.I= (P(1+(r/2)/100)^2 - P);
Therefore,
x*r*1 = ( x(1+(6/2)/100)^2 - x),
r = 0.0609;
or, rate percent = 6.09.
the simple interest rate that produces the same amount of interest as the compound interest rate.
i.e assume the principal amount be 'x'
Then according to definition,
p*r*t = (p(1+r/100)^n - p),
For S.I., as per question, t= 1 year;
For C.I., as per question, we use formula C.I= (P(1+(r/2)/100)^2 - P);
Therefore,
x*r*1 = ( x(1+(6/2)/100)^2 - x),
r = 0.0609;
or, rate percent = 6.09.
Shyamlata said:
1 decade ago
@Chandrashekhar Ji, How could u solve the Effective rate?
= 106.09-106 = 6.09%.
Please explain it.
= 106.09-106 = 6.09%.
Please explain it.
Vimal said:
1 decade ago
Thanks chandrashekhar.
Chadrashekhar said:
1 decade ago
Let p=rs 100
As we know that
for yearly:
A=P(1+R/100)^n
=100(1+6/100)
=Rs.106
for half yearly:
A=p(1+R/2*100)^2n
=100(1+6/2*100)^2*1
=100(103/100*103/100)
=Rs.106.09
Effective rate=(106.09-106)=6.09%
As we know that
for yearly:
A=P(1+R/100)^n
=100(1+6/100)
=Rs.106
for half yearly:
A=p(1+R/2*100)^2n
=100(1+6/2*100)^2*1
=100(103/100*103/100)
=Rs.106.09
Effective rate=(106.09-106)=6.09%
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