A sum was put at simple interest at a certain rate for 2 years . Had it been put at 3% higher rate, it would have fetched Rs 300 more. The sum is

Let sum = P
Then, according to the question.
SI = P/2
? P/2 = (P x 8 x 6)/100
? It is clear that data is inadequate.

Let the sum be P.
? SI = P/2
? P/2 = (P x 9 x 5)/100
Clearly data is inadequate.

If the sum be ? P, then
(2240 - P) = (P x 4 x 3)/100
? 2240 = 12P/100 + P
? 2240 = 112P/100
? P = (2240 x 100)/112 = ? 2000

Now, required interest,
SI = PRT/100 = [{2000 x (7/2) x (1/2)} /100]
= ? 35

Since, the two simple interest are equal.
Then, (4000 x 3 x R)/100 = (5000 x 12 x 2)/100
? R = 10%

SI = 26350 - 21250 = ? 5100
? Rate = (SI x 100) / (Principal x Time)
= (5100 x 100) / (21250 x 6)
= 4%

Let the sum be Rs 'y'
Since Simple Interest = Rs ( y / 2)
and, T = 6 yr , R = 10% per annum
So Simple Interest, SI = (P x R x T)/100
where, R = Rate
T = Time
SI= Simple Interest
now, According to problem, ( y / 2 ) = ( y x 10 x 6)/100
? (1/2)=(6/10)
? which is not true, so it is not a possible case.

As Sum = [(100 x SI)/(Time x Rate)]
here, let R =x%, T=x yr, and, SI=Rs x
? Sum=[(100 × x)/(x × x)]
=(100/x)

Let the sum be Rs 'y' and , rate of interest = R%
Simple Interest for 2 yr =Rs( 625 - 575 ) = Rs 50
? Sum of money, y=Rs( 575 - 75)= Rs 500

? R= [(100 x SI)/(Sum x Time)]
=[(100x 75)/(500 x 3)]=5 %

Let principle = Rs. P
Then S.I = P/9
Let Rate = R% per annum and time = R years

Then, as we know SI = (P x R x T) / 100.
? P/9 = ( P x R x R) / 100
? R² = 100/9
? R = 10/3 = 3¹/₃ % per annum

Let capital = Rs . P
Then, SI₁ - SI₂ = 104.
? (P x 13 x 1)/100 - (P x 25/2 x 1) /100 = 104
? 13P/100 - P/8 = 104
? 26P -25P = (104 x 200)
? P = 20800
? Capital = Rs. 20800