Let the money interest at 8% interest be ? P .
Then, the money interest at 10% interest = ?(4000 - P)
According to the question,
(P x 8 x 1)/100 + [(4000 - P) x 10 x 1]/100 = 352
? 8P + 40000 - 10P = 35200
? 40000 - 35200 = 2P
? P = 4800/2 = ? 2400
Let the amount be P, then amount after 12 yr = 2 P
SI = 2P - P = P
SI = (P x R x T)/ 100
P = (P x R x 12)/ 100
? R = 100/12 = 81/3%
Let the required period of time be T yr.
Then, (1680 x 4 x 15) /100 = (1200 x T x 7)/100
T = (1680 x 2 x 12)/(1200 x 7)
= 6 yr
Hence, the required time period is 6 yr.
A = ? 3400 T = 3 yr , R = 1% per month = 12 % per annum Let the principle be ? P.
SI = (PTR/100) = (P x 3 x 12)/100 = ? 36P/100
A = (P + SI) = P + 36P/100 = ? 136P/100
? 136P/100 = 3400
? P = (3400 x 100)/136 = 2500
? Sum = ? 2500
If interest is 40% of the principal then time = 5 years.
So, when interest would be equal to 100% of the principal time would be
= (100/40) x 5 years = 12.5 years
= 12 yr 6 months
? [ (2000 x 8 x 1) /100] + [ (4000 x 15/2 x 1)/100] + [(1400 x 17/2 x 1) / 100] + [(2600 x R x 1)/100] = (10000 x 8.13 x 1) / 100
? 160 + 300 + 119 + 26R = 813
? 26R = 234
? R = 9%
Let the extra money invested = ? P
According to the question .
(3000 x 5 x 1)/100 + (P x 8 x 1)/100 = [(3000 + P) x 6 x 1)]/100
? 15000 + 8P = 18000 + 6P
? 2P = 18000 - 15000
? P = 3000/2 = ? 1500
Initially ,
let P = ? A, R = 13/2% per annum and T = 1yr
SI = PRT/100 = (A x 1 x 13/2)/100 = ? 13A/200
Now, new deposit = ? (A - 600), R = 11/2 % per annum and T = 1yr
SI = PTR/100 = [(A - 600) x 1 x 11/2)/100] = ? 11(A - 600)/200
By the given condition ,
13A/200 - 11(A - 600)/200 = 73
? 13A - 11(A - 600 ) = 200 x 73
? 2A = 200 x 73 - 600 x 11
? 2A = 14600 - 6600
? 2A = 8000
? A = 4000
Hence, the initial investment is ? 4000.
Amount after 5 yr, 4200 + (4200 x 10 x 5)/100
= (4200 + 2100) = Rs 6300
? Cost of the cow = (6300 - 5000) = ? 1300
Let the principle and rate of interest be P and R%
Then, (P x R x 10)/100 = 600
? PR = 6000
Total interest at the end of year = 300 + (3P x R x 5)/100 = 300 + (15 x 6000)/100 = ? 1200
Let each sum be ? P then,
(P x 10 x 5)/100 - (P x 10 x 4)/100 = 220
? P/10 = 220
? P = 2200
Hence, each sum is ? 2200.
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