on a certain sum of money,invested at the rate of 10 percent per annum compounded annually,the interest for the first year plus the interest for the third year is 2652.find the sum.
Answers
Answered by
24
Let the sum be P
As , A = P(1 +R/100)^t
⇒A = P(1+10/100)
Interest = P/10
now, for 3rd year interest, the amount at the end of 2nd year will be sum.
i.e.
A = P(1 +10/100)^2
=121P/100
now amount for 3rd year will be
A = (121P/100)(1 +10/100)
=1331P/1000
interest = 331P/1000
A/q
P/10 + 331P/1000 = 2652
⇒431P/1000 = 2652
⇒P ≈ Rs 6153
As , A = P(1 +R/100)^t
⇒A = P(1+10/100)
Interest = P/10
now, for 3rd year interest, the amount at the end of 2nd year will be sum.
i.e.
A = P(1 +10/100)^2
=121P/100
now amount for 3rd year will be
A = (121P/100)(1 +10/100)
=1331P/1000
interest = 331P/1000
A/q
P/10 + 331P/1000 = 2652
⇒431P/1000 = 2652
⇒P ≈ Rs 6153
Answered by
20
Let the principal be x
R =10%
T= 3rd year
For 1st year
I = x*10*1÷100. = x÷10
A =x+x÷10 = 11x/10
For 2nd year
p=11x/10
r=10%
t=1year
I=p*r*t/100
=11x*10*1/10*100
Amount=11x/10+11x/100
=121x/100
For 3rd year
p=121x/100
r=10%
t=1
I=121x*10*1/100*100
=121x/1000
Amount=121x/100+121x/1000
=1331x1000
x/10+121x/1000=2652
=100x+121x/1000=2652
221x/1000=2652
221x =2652*1000
x=2652*1000/221
x=12,000
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