Aptitude - Compound Interest - Discussion
Discussion Forum : Compound Interest - General Questions (Q.No. 15)
15.
The compound interest on a certain sum for 2 years at 10% per annum is Rs. 525. The simple interest on the same sum for double the time at half the rate percent per annum is:
Answer: Option
Explanation:
Let the sum be Rs. P.
| Then, |  |
P |  |
1 + | 10 |  |
2 | - P |  |
= 525 |
| 100 |
 |
P | |
|
11 | |
2 | - 1 | |
= 525 |
| 10 |
| P = |
|
525 x 100 | |
= 2500. |
| 21 |
Sum = Rs . 2500.
| So, S.I. = Rs. | |
2500 x 5 x 4 | |
= Rs. 500 |
| 100 |
Discussion:
37 comments Page 1 of 4.
Asif said:
5 years ago
In the simplest way, we can slove this by;
Suppose principle amount is 100.
For two years CI 100*10%=10 again 110*10%=21 (total interest).
Total increased 21%, we can take it as 525 is 21% then 100% will be the principle amount which 2500.
Suppose principle amount is 100.
For two years CI 100*10%=10 again 110*10%=21 (total interest).
Total increased 21%, we can take it as 525 is 21% then 100% will be the principle amount which 2500.
(8)
Lokesh said:
1 year ago
Hello guys.
The simplest way is;
Here the sum will be the same.
C. I for 2yrs@10%p.a is;
21% = 525.
For S.I 4yrs @5 % p.a is;
20% = 500.
The simplest way is;
Here the sum will be the same.
C. I for 2yrs@10%p.a is;
21% = 525.
For S.I 4yrs @5 % p.a is;
20% = 500.
(2)
Rohan said:
7 years ago
A farmer takes a loan of Rs:45000 at 15% per annum to instal a pump in his field. He pays back 15000 to the bank along with the interest due at the end of the first year. At the end of second year he pays back the balance to the bank. How much does the farmer pay the bank at the end of second year?
Can anyone solve this?
Can anyone solve this?
(2)
Tanisha said:
5 years ago
@Nisha.
Actually they take the common factor so when we take p common there it is replaced by 1.
Actually they take the common factor so when we take p common there it is replaced by 1.
(2)
P Kumar said:
1 decade ago
CI = Amount - Principle.
525 = Principle(1+10/100)^2 - Principle.
Rate = 10, Time = 2.
525 = Principle(110/100)^2 - Principle.
525 = Principle((22/20)^2-1)).
525 = Principle(121/100-1).
525 = Principle (21/100).
Principle = (525*100)/21.
Principle = 2500.
Now to calculate S.I principle = 2500 T=4 (double the time taken for C.I i.e 2*2 = 4).
R = 5 (Half of the rate taken for C.I i.e 10/2 = 5).
S.I = (P*T*R)/100.
S.I = (2500*4*5)/100.
S.I = 500.
525 = Principle(1+10/100)^2 - Principle.
Rate = 10, Time = 2.
525 = Principle(110/100)^2 - Principle.
525 = Principle((22/20)^2-1)).
525 = Principle(121/100-1).
525 = Principle (21/100).
Principle = (525*100)/21.
Principle = 2500.
Now to calculate S.I principle = 2500 T=4 (double the time taken for C.I i.e 2*2 = 4).
R = 5 (Half of the rate taken for C.I i.e 10/2 = 5).
S.I = (P*T*R)/100.
S.I = (2500*4*5)/100.
S.I = 500.
(1)
Vinayak said:
7 years ago
How get 21 in the denominator? I am not getting. Please explain me.
(1)
Babai said:
7 years ago
1+ 10/100 = 11?
(1)
Mujahid Aziz said:
7 years ago
Given that ;
p*21/100=525,
so P= 525*100/21,
p=2500.
Now S.I= PTR/100.
where time is doubled So T=4 and Rate is halved So R=5.
Now S.I= 2500*4*5/100.
S.I =500.
p*21/100=525,
so P= 525*100/21,
p=2500.
Now S.I= PTR/100.
where time is doubled So T=4 and Rate is halved So R=5.
Now S.I= 2500*4*5/100.
S.I =500.
(1)
Sagar said:
5 years ago
How p becomes 1, We don't know the value of p right? Explain please.
(1)
Sandy said:
1 decade ago
Can anyone elaborate why there is a -P ? I'm unable to understand. !
(1)
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