rate=r%
1200 (1+r/100)^2=1348.32
r=6%
Let sum = P and original rate = R. Then
[(P * (R+2) * 3)/100] - [ (P * R * 3)/100] = 360
3P*(R+2) - 3PR = 36000
3PR + 6P - 3PR = 36000
6P = 36000
P = 6000
Let Rs.x be the amount that the elder daughter got at the time of the will. Therefore, the younger daughter got (3,500,000 - x).
The elder daughter?s money earns interest for (21 - 16) = 5 years @ 10% p.a simple interest.
The younger daughter?s money earns interest for (21 - 8.5) = 12.5 years @ 10% p.a simple interest.
As the sum of money that each of the daughters get when they are 21 is the same,
=>
(1500 x R1 x 3)/100
=> 4500 (R1-R2) = 1350
=> (R1-R2)= 1350/4500 = 0.3 %
We need to know the S.I, principal and time to find the rate. Since the principal is not given, so data is inadequate.
Let the original rate be R%. Then, new rate = (2R)%.
Note: Here, original rate is for 1 year(s); the new rate is for only 4 months i.e.1/3 year(s).
=> (2175 + 725) R = 33.50 x 100 x 3
=> (2175 + 725) R = 10050
=> (2900)R = 10050
=>
Original rate = 3.46%
At 5% more rate, the increase in S.I for 10 years = Rs.600 (given)
So, at 5% more rate, the increase in SI for 1 year = 600/10 = Rs.60/-
i.e. Rs.60 is 5% of the invested sum
So, 1% of the invested sum = 60/5
Therefore, the invested sum = 60 × 100/5 = Rs.1200
Let the sum be Rs. P. Then,
= 510
P[ - 1] = 510.
Sum = Rs. 1920
So, S.I. = (1920 x 25 x 2) / (2 x 100) = Rs. 480
Amount to be paid = = Rs. 115.
Simple interest is given by the formula SI = (pnr/100), where p is the principal, n is the numberof years for which it is invested, r is the rate of interest per annum
In this case, Rs. 1250 has become Rs.10,000.
Therefore, the interest earned = 10,000 ? 1250 = 8750.
8750 = [(1250 x n x 12.5)/100]
=> n = 700 / 12.5 = 56 years.
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