A sum of money amounts to ? 2240 at 4% per annum simple interest in 3 yr. The interest on the same sum for 6 months at 3.5% per annum is

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 sum = P, then SI = P/5 time = 10 yr.
? Rate = (100 x P)/(P x 5 x 10) = 2%

Given, SI₂ = 60, SI₁ = 30, T₁ = 4 yr, T₂ = 8 yr
According to the question,
[(1500 x R x 8)/100] - [(1500 x R x 4)/100] = 60 - 30
? (6000 x R)/100 = 30
? R = 30/60 = 1/2 = 0.5%

Here, n = 3, m = 2, T₁ = 20 yr
? T₂ = [(m -1) / (n - 1)] x T₁
= (2 - 1) / (3 - 1) x 20 = 10 yr

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

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.
And the original rate be y% per annum.
Then new rate=(y+3)% per annum
According to question, [(P × (y+3) × 2)/100]=[(P × y × 2)/100]=300
? [(Py + 3P)/100]=[Py/100] = 150
? Py+ 3P - Py=15000
? 3P=15000
? P= 5000
Thus, the sum is Rs 5000

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)