Aptitude - Simple Interest - Discussion
Discussion Forum : Simple Interest - General Questions (Q.No. 9)
9.
A sum of Rs. 725 is lent in the beginning of a year at a certain rate of interest. After 8 months, a sum of Rs. 362.50 more is lent but at the rate twice the former. At the end of the year, Rs. 33.50 is earned as interest from both the loans. What was the original rate of interest?
Answer: Option
Explanation:
Let the original rate be R%. Then, new rate = (2R)%.
Note:
Here, original rate is for 1 year(s); the new rate is for only 4 months i.e. 
year(s).
 |
 |
725 x R x 1 |  |
+ | |
362.50 x 2R x 1 | |
= 33.50 |
| 100 | 100 x 3 |
(2175 + 725) R = 33.50 x 100 x 3
(2175 + 725) R = 10050
(2900)R = 10050
| R = |
10050 | = 3.46 |
| 2900 |
Original rate = 3.46%
Discussion:
97 comments Page 3 of 10.
ANU said:
1 decade ago
This is how I did it.
[725*1*i/100]+[362.50*4/12*i/100] = 33.50.
7.25i+2.416i = 33.50.
9.66i = 33.50.
i = 33.50/9.66.
i = 3.46%.
Here I have taken i as rate of interest(R).
[725*1*i/100]+[362.50*4/12*i/100] = 33.50.
7.25i+2.416i = 33.50.
9.66i = 33.50.
i = 33.50/9.66.
i = 3.46%.
Here I have taken i as rate of interest(R).
Mahalaksmi said:
1 decade ago
I am not understanding this problem please explain me clearly,
725*r*1/100+31.50*2r*1/100*3.
What does this 100*3 mean?
Please help me guys.
725*r*1/100+31.50*2r*1/100*3.
What does this 100*3 mean?
Please help me guys.
Kasinath said:
1 decade ago
Look @Maha :).
Let P=principal, I=Interest, R=Rate of Interest, T=time.
Let us consider two loans as Loan1 and Loan2.
In Loan 1: P1=725, R1=x, T1=1year.
In loan 2: P2=362.50 R2=2x, T2=After 8months it means(12-8)4MONTHS i.e., 4/12=1/3YEARS.
Given Interest I=33.50Rs of both Loan1 and Loan2.
Since I=PTR/100.
=>(P1*T1*R1/100)+(P2*T2*R2/100) = I ----EQUATION 1.
Now Insert the values a/c to equation 1 then you will get result.
Let P=principal, I=Interest, R=Rate of Interest, T=time.
Let us consider two loans as Loan1 and Loan2.
In Loan 1: P1=725, R1=x, T1=1year.
In loan 2: P2=362.50 R2=2x, T2=After 8months it means(12-8)4MONTHS i.e., 4/12=1/3YEARS.
Given Interest I=33.50Rs of both Loan1 and Loan2.
Since I=PTR/100.
=>(P1*T1*R1/100)+(P2*T2*R2/100) = I ----EQUATION 1.
Now Insert the values a/c to equation 1 then you will get result.
Dolagobinda behera said:
1 decade ago
I agreed to @Tarun as the way he approached the question. I think the answer might be wrong as explained in the answer.
Amith sourya said:
1 decade ago
The solution he given is wrong. Because for first case he took 8 months and for second month he took the remaining months i.e. 12-8 = 4. But he took in solution as 1 year.
Abhi said:
1 decade ago
Ya you should calculate for 8 months and 4 months.
Asif said:
1 decade ago
It should be:
725*r*8/(12*100) + 362.5*2r*4/(12*100) = 33.50.
R will be = 4.62%
725*r*8/(12*100) + 362.5*2r*4/(12*100) = 33.50.
R will be = 4.62%
Sonu said:
1 decade ago
Hello sir,
Please let me know the right answer please.
Please let me know the right answer please.
Agnes said:
1 decade ago
Please why did you multiply 33.50*100*3?
Venu said:
1 decade ago
Hey guys its correct.
Here 725 is given to the person for 1 year. Now you calculate the SI for that, later after 8 months, he again burrows again but the catch is rate is doubled.
Now you know :).
Here 725 is given to the person for 1 year. Now you calculate the SI for that, later after 8 months, he again burrows again but the catch is rate is doubled.
Now you know :).
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers