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:
174 comments Page 14 of 18.
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.
Monisha Lodhi said:
8 years ago
Why we find out only for three years. Is principle is found on only for the first year money?
Divya said:
8 years ago
Is 698 is principal amount?
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.
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