Aptitude - Simple Interest - Discussion
Discussion Forum : Simple Interest - General Questions (Q.No. 1)
1.
A sum of money at simple interest amounts to Rs. 815 in 3 years and to Rs. 854 in 4 years. The sum is:
Answer: Option
Explanation:
S.I. for 1 year = Rs. (854 - 815) = Rs. 39.
S.I. for 3 years = Rs.(39 x 3) = Rs. 117.
Principal = Rs. (815 - 117) = Rs. 698.
Discussion:
171 comments Page 6 of 18.
Imran syed said:
9 years ago
For simple interest, this formula is very helpful.
That is I = Prt.
That is I = Prt.
Salomi said:
9 years ago
@Gayathri.
Thank you for explaining the solution clearly.
Thank you for explaining the solution clearly.
Kabali said:
9 years ago
@Gayathri nice explanation.
Mick sudama said:
9 years ago
Here you can do it by another S. I 1 yr = 39, in 4 years =156, & principle=854 - 156 = 698.
Aparna said:
9 years ago
Crystal clear explanation, Thanks @Gayathri.
Kitty said:
9 years ago
@Appu.
Your explanation is good, Thank you.
Your explanation is good, Thank you.
Sachin said:
9 years ago
Thanks @Vikram and @Gayatri.
Sahid Hussain said:
9 years ago
Let the sum is Rs. X and rate of interest is y%.
We know A = P + SI where SI = (P * R * T) /100 Therefore, A = P + (P * R * T) /100.
Case 1 : A = 815, P = x, T = 3 years, R = y%.
A = x + (x * y * 3) /100->1
Case 2 : A = 854, P = x, T = 4 years, R = y%.
A = x + (x * y * 4) /100 ->2
Subtracting equation (1) from equation (2), we get x * y = 3900.
Now by putting this value of x * y = 3900 in any one of above two equation, we'll get sum = x = 698 Answer.
We know A = P + SI where SI = (P * R * T) /100 Therefore, A = P + (P * R * T) /100.
Case 1 : A = 815, P = x, T = 3 years, R = y%.
A = x + (x * y * 3) /100->1
Case 2 : A = 854, P = x, T = 4 years, R = y%.
A = x + (x * y * 4) /100 ->2
Subtracting equation (1) from equation (2), we get x * y = 3900.
Now by putting this value of x * y = 3900 in any one of above two equation, we'll get sum = x = 698 Answer.
Rakesh.h said:
9 years ago
I will explain in a simple way.
We have A = P + I.
815 = P + I -----> Eq 1.
854 = P + I -----> Eq 2.
Subtract Eq 1 and 2.
Simple interest I = 815 - P - 854 + P
= 39 for 1year.
Then take for 3 years or 4 years.
You will get the same answer.
For 3 years:
39 * 3 = 117.
Or For 4 years:
39 * 4 = 156.
Then use formula A = P + I.
For 3 years:
815 = P + 117,
P = 815 - 117,
= 698.
For 4 years:
854 = P + 156,
P = 854 - 155,
= 698.
Take for 3 or 4 years answer will be same. I think the explanation is very useful.
We have A = P + I.
815 = P + I -----> Eq 1.
854 = P + I -----> Eq 2.
Subtract Eq 1 and 2.
Simple interest I = 815 - P - 854 + P
= 39 for 1year.
Then take for 3 years or 4 years.
You will get the same answer.
For 3 years:
39 * 3 = 117.
Or For 4 years:
39 * 4 = 156.
Then use formula A = P + I.
For 3 years:
815 = P + 117,
P = 815 - 117,
= 698.
For 4 years:
854 = P + 156,
P = 854 - 155,
= 698.
Take for 3 or 4 years answer will be same. I think the explanation is very useful.
(3)
Aryan said:
9 years ago
Let principal = x,
Principal + simple interest = total sum (after any number of years),
For 3 years,
X + x * r * 3 = 815 ----> Eqution 1,
For 4 years,
X + x * r * 4 = 854 ----> Equation 2,
Now there are 2 unknown and 2 equations solve for x and we get.
X = 698.
Principal + simple interest = total sum (after any number of years),
For 3 years,
X + x * r * 3 = 815 ----> Eqution 1,
For 4 years,
X + x * r * 4 = 854 ----> Equation 2,
Now there are 2 unknown and 2 equations solve for x and we get.
X = 698.
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