(kx5x1)/100 + [(1500 - k)x6x1]/100 = 85
5k/100 + 90 ? 6k/100 = 85
k/100 = 5
=> k = 500
Let the principal be P and rate of interest be R%.
Required ratio =
Let the sum be Rs. 100. Then,
S.I. for first 6 months = Rs.[ (100 x 10 x 1)/(100 x 2) ]= Rs.5
S.I. for last 6 months =Rs.[(102 x 10 x 1)/(100 x 2) ] = Rs.5.25
So, amount at the end of 1 year = Rs. (100 + 5 + 5.25) = Rs. 110.25
Effective rate = (110.25 - 100) = 10.25%
Let man invested Rs. A And, after two years amount invested = (A +
As the interest rate increases by 2%
=> (7000x3x2)/100 = 420
9200
--------
9620
We know that I = PTR/100
=> P = 20250/4.5 = 4500
Now, new Interest at 5% = 4500x1x5/100 = 225
Now the additional amount = 225 - 202.5 = Rs. 22.5
Let sum = S. Then, amount = 7S/6
S.I. = 7S/6 - S = S/6; Time = 3 years.
Rate = (100 x S) / (S x 6 x 3) = 5 5/9 = 50/9 %.
We know that,
I = PTR/100
ATQ,
15800 = 14800 x T x 6/100
T = 15800/148x7
T = 15800/1036
T = 15.25 yrs.
From the information provided we know that,
Principal + 8% p.a. interest on principal for n years = 180 ??.. (1)
Principal + 4% p.a. interest on principal for n years = 120 ??? (2)
Subtracting equation (2) from equation (1), we get
4% p.a. interest on principal for n years = Rs.60.
Now, we can substitute this value in equation (2),
i.e Principal + 60 = 120
= Principal = Rs.60.
We know that SI = , where p is the principal, n the number of years and r the rate percent of interest.
In equation (2), p = Rs.60, r = 4% p.a. and the simple interest = Rs.60.
Therefore, 60 =(60*n*4)/100
=> n = 100/4 = 25 years.
From the given data,
3500x7xt/100 = 500
=> t = 100/49 years
Now, in the second case
The interest per year = 49/100 x 800 = 392
=> 4900 x 1 x r/100 = 392
=> r = 8%
Let the rate be R% p.a.
Then,
Rate = 10%
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