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 14 of 18.
Nipan said:
8 years ago
Thanks for explaining.
Sase m said:
8 years ago
Why should we apply like this;
s.i for 1 year = Rs 854-815=39
1yr = 39 then 4y = 156
then 854-156=698.
s.i for 1 year = Rs 854-815=39
1yr = 39 then 4y = 156
then 854-156=698.
Sase M said:
8 years ago
@Divya.
Its a initial amount.
Its a initial amount.
Madhu said:
8 years ago
Basic formula of to find r% and p%.
When give two amounts and two years to find p formula,
(A1*T2-A2*T1)/T2-T1.
R% FORMULA.
(A2-A1)/(A1*T2-A2*T1).
When give two amounts and two years to find p formula,
(A1*T2-A2*T1)/T2-T1.
R% FORMULA.
(A2-A1)/(A1*T2-A2*T1).
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.
Pramod nepali said:
8 years ago
Let p=x nd after 3year A1=815 nd4year A2=854... A( is total amount)=p+i we no P=(A*100)/(100+TR) .......p nd R is equal for both A1 nd A2 .
so P1 =p2-------(eq1) we have formula P1=(A1*100)/(100+T1R)--------(2)
And P2=(A2*100)/(100+T2R)---------(3)
put eq2 and eq3 on eq 1 and putting all value we go Rate (R)=5.58 value of R input on eq 2 or eq 3 we got the value of p.
so P1 =p2-------(eq1) we have formula P1=(A1*100)/(100+T1R)--------(2)
And P2=(A2*100)/(100+T2R)---------(3)
put eq2 and eq3 on eq 1 and putting all value we go Rate (R)=5.58 value of R input on eq 2 or eq 3 we got the value of p.
Pacy said:
8 years ago
You can use the formula, SI= PRT.
Here, R= (854-815)/815= 0.0475, T= 3 years, & P= 815.
Therefore, SI= 815*0.0475*3= 117.
Final answer= 815-117= 698.
You can replace T= 4 years & P= 854, the answer is same.
Here, R= (854-815)/815= 0.0475, T= 3 years, & P= 815.
Therefore, SI= 815*0.0475*3= 117.
Final answer= 815-117= 698.
You can replace T= 4 years & P= 854, the answer is same.
Nidhi Chouhan said:
8 years ago
Since principal with the interest is given for two consecutive years therefore its easy to get the S.I. Without any formula. You can also subtract from 854 given that you multiply 39 with 4 (years).
Hari said:
8 years ago
What is the percentage of the interest?
Madhav zayn said:
8 years ago
I am not getting it, Please explain me, anyone.
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