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 3 of 18.
Nishan said:
1 decade ago
In my view the interest between the 3rd & 4th year of Rs. 39/- cannot determine the same interest component for the first three years as the interest will be paid on top of the capital?
As such the interest component should reduce proportionately over the previous years?
As such the interest component should reduce proportionately over the previous years?
Abdulwaheed said:
1 year ago
Simple interest for 1 year = 854 - 815 = 39.
Simple interest for 3 year = 39 * 3 = 117.
Simple interest for 4 year = 39 * 4 = 156.
Now,
Subtract the given amount of each simple interest amount
A = 815-117 = 698
B = 854-156 = 698
Both are same, so;
So, the answer is 698.
(15)
Simple interest for 3 year = 39 * 3 = 117.
Simple interest for 4 year = 39 * 4 = 156.
Now,
Subtract the given amount of each simple interest amount
A = 815-117 = 698
B = 854-156 = 698
Both are same, so;
So, the answer is 698.
(15)
(47)
K mahendra naidu said:
8 years ago
Hi, guys this problem can be solved in 2 ways i.e,
Sum of 3 years ------------ 815
Sum of 4 years------------- 854
So, the difference is 1 year.
Amount 39.
So,
solution1:
39*3=117(difference*years).
815-117=698.
Sol2:
39 * 4 = 156(difference*years),
854 - 156 = 698.
Sum of 3 years ------------ 815
Sum of 4 years------------- 854
So, the difference is 1 year.
Amount 39.
So,
solution1:
39*3=117(difference*years).
815-117=698.
Sol2:
39 * 4 = 156(difference*years),
854 - 156 = 698.
MICHAEL THIMMARAYAN said:
2 years ago
Simple interest for 1 year = 854 - 815 = 39.
Simple interest for 3 year = 39 * 3 = 117.
Simple interest for 4 year = 39 * 4 = 156.
Now,
Subtract the given amount of each simple interest amount
A = 815-117 = 698
B = 854-156 = 698
Both are same, so;
So, the answer is 698.
Simple interest for 3 year = 39 * 3 = 117.
Simple interest for 4 year = 39 * 4 = 156.
Now,
Subtract the given amount of each simple interest amount
A = 815-117 = 698
B = 854-156 = 698
Both are same, so;
So, the answer is 698.
(25)
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.
Ranjit Das said:
5 years ago
@Kannan.
As we say here that we have found out the interest for 3 years i.e, 117. That's why we subtracted 117 from 815.
If, we calculate the interest for 4 years..the amount will be, (39*4)= 156, then we should subtract 156 from 854, the will be same as above.
As we say here that we have found out the interest for 3 years i.e, 117. That's why we subtracted 117 from 815.
If, we calculate the interest for 4 years..the amount will be, (39*4)= 156, then we should subtract 156 from 854, the will be same as above.
(2)
Farru_faiz said:
6 years ago
P = sum.
SI = simple interest.
P+SI=815 ; P + SI=854
SI=815-P ; SI=854-P
SI=P*T*R/100
(815-P) =P*3*R/100 ------>1
(854-P) =P*4*R/100 ------>2
---- >>(1/2)
(815-p)/3 = (854-P)/4
(815-P)*4=3*(854-p)
3260-4P = 2562-3P
3260-2562 = 4 P-3P
698=P.
SI = simple interest.
P+SI=815 ; P + SI=854
SI=815-P ; SI=854-P
SI=P*T*R/100
(815-P) =P*3*R/100 ------>1
(854-P) =P*4*R/100 ------>2
---- >>(1/2)
(815-p)/3 = (854-P)/4
(815-P)*4=3*(854-p)
3260-4P = 2562-3P
3260-2562 = 4 P-3P
698=P.
(1)
Pavithrasai said:
8 years ago
Here, si=p*r*t/100.
So for 3 yrs, 815=p*r*3/100 => r=81500/3p and for 4yrs, 854=p*r*4/100 => r=85400/4p. Verify options by placing each option as p values and option which gives same values of r in both the equations is the answer. Hence, option c.
So for 3 yrs, 815=p*r*3/100 => r=81500/3p and for 4yrs, 854=p*r*4/100 => r=85400/4p. Verify options by placing each option as p values and option which gives same values of r in both the equations is the answer. Hence, option c.
Aquarius said:
8 years ago
Can't this question be solved using formula
p=100*si/r*t.
Total amt = p+si.
815 = p+si.
si = 815-p.
854 = p+si.
si = 854-p.
Substituting in formula,
p = 100*815-si/r*3.
Can anyone please help how to solve using this method of substitution.
p=100*si/r*t.
Total amt = p+si.
815 = p+si.
si = 815-p.
854 = p+si.
si = 854-p.
Substituting in formula,
p = 100*815-si/r*3.
Can anyone please help how to solve using this method of substitution.
Vaishali said:
1 year ago
S.I for 1 year = 854 - 815 = 39.
S.I for 3 year = 39 * 3 = 117.
S.I for 4 year = 39 * 4 = 156.
Now,
Subtract the given amount of each simple interest amount
A = 815 - 117 = 698.
B = 854 - 156 = 698.
Both are same, so;
So, the answer is 698.
S.I for 3 year = 39 * 3 = 117.
S.I for 4 year = 39 * 4 = 156.
Now,
Subtract the given amount of each simple interest amount
A = 815 - 117 = 698.
B = 854 - 156 = 698.
Both are same, so;
So, the answer is 698.
(105)
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