RishebhKumar (Studying) (128 Points)
16 May 2012Simple Interest
Q1. The sum of required to earn a monthly interest of Rs.1200 at 18% per annum SI is……….?
Q2. A sum of money amount to Rs.6200 in 2 years and Rs.7400 in 3 years.The principal and rate of interest are………..?
Q3. A sum of money double itself in 10 years. The number of years it would triple itself is………..?
Explain with solution plz....
intermediate(ipc)course
(no)
(1460 Points)
Replied 16 May 2012
2. 6200 = p*2/12*i----------------------------1
7400 = p*3/12*i
therefore,
6200+7400 = p*2/12*i+p*3/12*i
13600 = 1/6 pi +1/4 pi
=10/24 pi
pi = 13600*24/10
=32640(principal 32000 + interest 640)
intermediate(ipc)course
(no)
(1460 Points)
Replied 16 May 2012
3. 2*p*10/12*i = 3*p*n*i
or,5/3 *p*i = 3*p*i*n
or,9*n = 5
or,n=5/9=0.56=0.6.
Tejaswi Kasturi
(student-cpt)
(427 Points)
Replied 16 May 2012
1. The interest rate is 18% per annum as such per month it is 18/12% i.e. 1.5%. as the interest per month is 1.5% i = p*r*n/100;
1200 = p*1.5/100 implies p = 1200*100/1.5 =80,000
2.the question doesn't specify whether the interest is compound or simple
if simple interest
A = P+I = P + P*r*n/100
for 2 years
6200 = P + P*r*2/100
for 3 years
7400 = P + P*r*3/100
subtracting first from second we get P*r/100 = 1200
substituting it in the first equation we get 6200-2*(P*r/100)=P
i.e. P = 6200 - 2*(1200) = 6200 - 2400 = 3800
r = 1200*100/3800 = 31.5% (approx)
if compound interest
P(1+r)^2 = 6200; P(1+r)^3 = 7400 dividing second by first we get (1+r) = 7400/6200
implies r = 1200/6200 = 19.35%
P = 6200/(1.1935)^2 = 4352.22
3. Again the question doesn't specify whether the Interest is simple or compound.
Simple Interest:
P+P*r*10/100 = 2P implies r= 10% (cancel P on both sides)
to triple the amount P+ P*10*n/100 = 3P implies n = 20 i.e. 20 years
Compound Interest
P(1+r)^10 = 2P implies r = 7.17% (take logarithms on both sides)
(1.0717)^n = 3 n = 15.85 years
Usman
(Accounting)
(24 Points)
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DKG Dexterous Zone
(Institute)
(32 Points)
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