2/3 part of my sum is lent out at 3%, 1/6 part is Lent out at 6% and remaining part is lent out 12% All the three parts are lent out at simple interest. If the annual income is ? 25, what is the sum?

Let rate is R%
According to the question.
[(R x 400 x 2)/100] + [(R x 100 x 4)/100] = 60
? 12R = 60
? R = 60/12 = 5%

Let the two rates be R₁ and R₂

According to the question,
[(1000 x 2 x R₁)/100] - [(1000 x 2 x R₂)/100] = 20
? [2000 x (R₁ - R₂)] / 100 = 20
? R₁ - R₂ = 20/20 = 1%

Here, n = 2, m =10, T₁ = 6 yr, T₂ = ?
? T₂ = [(m - 1) / (n - 1)] x T₁
= [(10 - 1)/(2 - 1)] x 6
= 54 yr

SI at 5% = 6P - P = 5P
? 5P = (P x 5 x T) /100
? T = 100 yr
Now, for new rate (R),
11P = (P x R x 100)/100
? R =11%

According to the question
[P x (R + 2) x 4]/100 - [P x R x 4]/100 = 56
? (4PR + 8P - 4PR)/100 = 56
? 8P/100 = 56
? P = (56 x 100)/8 = ? 700

Given, R₁ = 6% R₂ = 10%
According to the question,
800 + [(800 x 6 x T)/100] = 600 + [(600 x 10 x T)/100]
800 + 48T = 600 + 60T
? 12T = 200
? 3T = 50
? T = 50/3 = 16²/₃ yr

Given, T₁ = 5 yr, R₁ = 10% and
T₂ = 10 yr, R₂ = 8%
Let the first part = P
Then, second part = (1521 - P)

Now, according to the question.
[(P x 5 x 10)/100] = [{(1521 - P) x 10 x 8}/100]
? 5P = 12168 - 8P
? 13P = 12168
? P = ? 936

So second part = 1521 - 936 = ? 585

Given. T₁ = 2 yr and T₂ = 4 yr,
P₁ = 600, P₂ = 150.

According to the question,
SI₁ + SI₂ = 80
[(600 x R x 2)/100] + [(150 x R x 4)/100] = 80
? 120R + 60R = 800
? 180R = 800
? R = 800/180 = 80/18 = 40/9
= 4⁴/₉%

Let the principal be P.
Then, according to the question,
(P x T x T) / 100 = P/144 [? time and rate are equal]
? T² = 100/144
? T = 10/12 = 5/6 %

Let principal = P, time = T and rate = T
According to the question,
(P x T x T) / 100 = P/16 [? time and rate are equal]
? T² = 100/16
or T = 10/4 = 5/2 = 2¹/₂ %