Let P = Rs. 100. Then, S.I. Rs. 60 and T = 6 years.
R = (100 x 60)/(100x6) = 10% p.a.
Now, P = Rs. 12000
T = 3 years and R = 10% p.a.
So, C.I. = Rs.[12000 * [(1 + 10/100)^3- 1]]
= Rs. 12000 * 331/1000 = Rs. 3972
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A certain sum of money amounts to Rs. 1008 in 2 years and to Rs.1164 in 3 1/2 years. Find the sum and rate of interests.
S.I. for 1 and 1/2 years = Rs.(1164-1008) = Rs.156.
SI for 2 years = Rs.(156*(2/3)*2)=Rs.208.
Principal = Rs. (1008 - 208) = Rs. 800.
Now, P = 800, T = 2 and S.l. = 208.
Rate =(100 * 208)/(800 * 2)% = 13%
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What sum of money will accumulate to Rs.5300 at 8% rate of interest in 9 months?
Given 1st year depreciation amount = $10000
next 7 year's depreciation amount = 7 * 8% original cost price
= 7*8/100 * 52500 = $29400
Total depreciation amount = ($10000 + $29400) = $39400
Value after 8 years = $(52500-39400) = $13100
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simple interest on an amount at 4% per annum for 13 months is more than the simple interest on the same sum for 8 months at 6% per annum by Rs. 40. what is the principle amount
Given that the interest is only earned on principal, if an investment of Rs.1000.00 amount to Rs.1440.00 in two years, then what is the rate of interest earned?
As per scheme a car loan of Rs. 4 lakh at 12.5 p.c.p. rate of simple interest can be borrowed on a repayment term of lump sum amount at the end of 3 years.
As per scheme (b), the amount can be repaid at the end of 2 years, but compound interest (compounded annually) would be charged at the same rate. What would be the difference in amount of interest between the two schemes?
Given Intrest rate r = 8%
Deposited for year n = 2 years
Formula: SI = p*n*r/100
=> p*2*8/100 = 4p/25
CI = Amount - p
Amount = p(1 + r/100)^n
=> p(1 + 8/100)^2
CI = p(27/25)^2 - p
p(27/25)^2- p - 4p/25 = 8
p[27*27 - 625 - 100]/625 = 8
4p = 5000 => p = Rs. 1250
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