Aptitude - Compound Interest - Discussion
Discussion Forum : Compound Interest - General Questions (Q.No. 3)
3.
There is 60% increase in an amount in 6 years at simple interest. What will be the compound interest of Rs. 12,000 after 3 years at the same rate?
Answer: Option
Explanation:
Let P = Rs. 100. Then, S.I. Rs. 60 and T = 6 years.
 R = |
 |
100 x 60 |  |
= 10% p.a. |
| 100 x 6 |
Now, P = Rs. 12000. T = 3 years and R = 10% p.a.
| C.I. |
|
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|
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| = 3972. |
Discussion:
94 comments Page 4 of 10.
Rajeev rathor said:
7 years ago
Using pascal traingle.
= 3*10%of12000+3*10%of1200+1*10%of120,
= 3600+360+12,
= 3972.
= 3*10%of12000+3*10%of1200+1*10%of120,
= 3600+360+12,
= 3972.
Aashi said:
7 years ago
Please tell me how 331 is coming? Please explain me.
MONI SHANKAR said:
8 years ago
1200*(11/10)*(11/10)*(11/10)=15972.
CI=15972-12000=3972,
P.S : (HERE 11/10=110/100 WHICH MEANS 10% RATE AND CAME FROM STATEMENT 60% INCREMENT IN 6 YEARS.)
CI=15972-12000=3972,
P.S : (HERE 11/10=110/100 WHICH MEANS 10% RATE AND CAME FROM STATEMENT 60% INCREMENT IN 6 YEARS.)
Lokesh said:
8 years ago
How Will find P & S.I?
Mohammad Gazali said:
8 years ago
Please tell me how R=100*60/100*6?
(1)
Riya said:
8 years ago
helpful
Saleem said:
8 years ago
60%increase in 6 yrs.
i. e 10%every yr.
i. e R=10%.
Now A=P(1+10/100)^3=1200(1+1/10)^3=15972.
CI=15972-12000=3972.
i. e 10%every yr.
i. e R=10%.
Now A=P(1+10/100)^3=1200(1+1/10)^3=15972.
CI=15972-12000=3972.
(3)
Dron said:
8 years ago
How to find p=100?
Manisha said:
8 years ago
Thanks @Sanjay.
Sudarsan said:
8 years ago
Thanks @Neha.
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