On a sum of money, the simple interest for 2 years is Rs.660, while compound interest is Rs.696.30, the rate of interest being the same in both the cases. The rate of interest is?
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Heya friend,
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Let ₹P be the principal and R be rate of interest. Then,
S.I. = P×R×T/100
=> 660 = P×R×2/100
=> 660 = P×R/50
=> 660×50 = P×R
=> 33,000 = P×R
=> P = 33,000
C.I. = P{(1+R/100)^n - 1}
=> 696.30 = 33,000/R{(100+R/100)^2 - 1}
=> 69,630R/100×33,000 = {(100)^2 + (R)^2 + 2×100×R/(100)^2 - 1}
=> 211R/10,000 = {10,000 + (R)^2 + 200R - 10,000/10,000}
=> 211R×10,000/10,0000 = (R)^2 + 200R
=> 211R = R(R+200)
=> 211R/R = R+200
=> 211 = R+200
=> 211-200 = R
=> R = 11 % per annum
Hence, the rate of interest is 11 % per annum.
Thanks
With regards@
Tanisha
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Let ₹P be the principal and R be rate of interest. Then,
S.I. = P×R×T/100
=> 660 = P×R×2/100
=> 660 = P×R/50
=> 660×50 = P×R
=> 33,000 = P×R
=> P = 33,000
C.I. = P{(1+R/100)^n - 1}
=> 696.30 = 33,000/R{(100+R/100)^2 - 1}
=> 69,630R/100×33,000 = {(100)^2 + (R)^2 + 2×100×R/(100)^2 - 1}
=> 211R/10,000 = {10,000 + (R)^2 + 200R - 10,000/10,000}
=> 211R×10,000/10,0000 = (R)^2 + 200R
=> 211R = R(R+200)
=> 211R/R = R+200
=> 211 = R+200
=> 211-200 = R
=> R = 11 % per annum
Hence, the rate of interest is 11 % per annum.
Thanks
With regards@
Tanisha
indupriya123:
The answer is correct but I didn’t understand how did u get p=33,000 at the starting
Answered by
0
Given:
SI of two years
Formula for SI is
where P is principle, T is time and R is rate of interest
So,
Now, Compound Interest for two years
Solving the above equation, we get
So the rate of interest will be
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