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 7 of 10.
Lokesh said:
8 years ago
How Will find P & S.I?
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.)
Aashi said:
7 years ago
Please tell me how 331 is coming? Please explain me.
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.
Shub said:
7 years ago
Amount after 1st year =1200.
After 2nd year=1200 + 120.
After 3rd year= 1200 + 120+120+12.
Total = 3972.
After 2nd year=1200 + 120.
After 3rd year= 1200 + 120+120+12.
Total = 3972.
Bhushan said:
7 years ago
SI for 6 yrs increase by 60%.
Means;
For 1st yr 10%.
1st yr , 10% of 12000 = 1200 - - - >(1)
Add 1200 ... ie. = 12000+1200= 13200.
For 2nd yr, 10% of 13200 = 1320 - - - > (2)
So, 13200+1320= 14520.
For 3rd yr, 10% of 14520 = 1452. - - - > (3)
So add the interest amount of all 3 equations ie....
1200 + 1320 + 1452 = 3972 is the right answer.
Means;
For 1st yr 10%.
1st yr , 10% of 12000 = 1200 - - - >(1)
Add 1200 ... ie. = 12000+1200= 13200.
For 2nd yr, 10% of 13200 = 1320 - - - > (2)
So, 13200+1320= 14520.
For 3rd yr, 10% of 14520 = 1452. - - - > (3)
So add the interest amount of all 3 equations ie....
1200 + 1320 + 1452 = 3972 is the right answer.
(2)
Ganesh said:
7 years ago
@All.
1st find rate R of 6 yr Assume 100rs, for 6 yr.
R=(100 * s.i/p * n) = 10%.
After that find C.I using ratio proposition of 3 yr is 3:3:1 is fixed and 2yr is (2:1).
Then 1st yr c.i. is 3* 10% of 12000,
2nd yr C.I is 3* 10% of 1200,
3rd yr C.I is 1* 10% of 120 add all the yrs.
3600 + 360 + 12 = 3972 is the answer.
1st find rate R of 6 yr Assume 100rs, for 6 yr.
R=(100 * s.i/p * n) = 10%.
After that find C.I using ratio proposition of 3 yr is 3:3:1 is fixed and 2yr is (2:1).
Then 1st yr c.i. is 3* 10% of 12000,
2nd yr C.I is 3* 10% of 1200,
3rd yr C.I is 1* 10% of 120 add all the yrs.
3600 + 360 + 12 = 3972 is the answer.
(1)
Sangavi said:
7 years ago
When will we use -P in the compound interest formula?
Rahul krishna said:
6 years ago
Let p=100 and si=60(as amount increase to 60%).
Si=(p*r*t)/100.
Find r=?.
Then, with the help of r find ci.
Si=(p*r*t)/100.
Find r=?.
Then, with the help of r find ci.
Rahul krishna said:
6 years ago
@Sangavi we use - p because.
CI = Amount - principal.
So, CI=p(1+r/100)^n-p.
CI = Amount - principal.
So, CI=p(1+r/100)^n-p.
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