Aptitude - Compound Interest - Discussion
Discussion Forum : Compound Interest - General Questions (Q.No. 13)
13.
The difference between simple interest and compound on Rs. 1200 for one year at 10% per annum reckoned half-yearly is:
Answer: Option
Explanation:
| S.I. = Rs |  |
1200 x 10 x 1 |  |
= Rs. 120. |
| 100 |
| C.I. = Rs. |  |
1200 x | |
1 + | 5 | |
2 | - 1200 |  |
= Rs. 123. |
| 100 |
Difference = Rs. (123 - 120) = Rs. 3.
Discussion:
53 comments Page 3 of 6.
Simte said:
8 years ago
Can you give me the formulae to find out rate percent in compound interests?
Name said:
7 years ago
Can anyone explain me the Last step?
Name said:
7 years ago
Can anyone explain me the Last step?
Rohit said:
7 years ago
It is mentioned in the question SI-CI so they have just subtracted.
K.BHARGAVI said:
9 years ago
for one year S.I is ptr/100,
S.I = 120.
C.I : FOR HALF YEAR.
a = p(1 + (R/2)/100)^2 = 1200.
We know that,
A = P + C.I.
C.I = A - P,
= 1323 - 1200,
= 123.
Difference = C.I - S.I
= 123 - 120.
Difference = 3.
S.I = 120.
C.I : FOR HALF YEAR.
a = p(1 + (R/2)/100)^2 = 1200.
We know that,
A = P + C.I.
C.I = A - P,
= 1323 - 1200,
= 123.
Difference = C.I - S.I
= 123 - 120.
Difference = 3.
Ami said:
6 years ago
Why 5/100? Please explain that.
Naina said:
6 years ago
Because as r=10% for 1 yr so for the half year it would be 5%=5/100.
Skywalker said:
1 month ago
S.I = 2 (PRT)/100 ,
T = 0.5,
= 120.
A = P(1+ R/2/100)^2
= 1323.
Then C.I = 1323 - 1200,
= 123
And we get, S.I - C.I = 3.
T = 0.5,
= 120.
A = P(1+ R/2/100)^2
= 1323.
Then C.I = 1323 - 1200,
= 123
And we get, S.I - C.I = 3.
Priya said:
1 decade ago
On the account of half-yearly, the value of n in S.I would be 1/2 only na?
Abi said:
1 decade ago
Hello, how the 5/100 came? can anyone tell me.
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