Let the sum be Rs. `x` , Then ,

C.I.  = `x(1 + 10/100)^2 - x =  (21x)/(100),`

S.I. = `((x xx 10 xx 2)/(100)) =  x/5`.

`:.`      (C.I.) -  (S.I.) =  `((21x)/(100) - x/5) = x/100`

`:.`     ` x/100 =   631`

`hArr      x`= 63100

Hence , the sum is Rs. 63,100.

Let the sum be Rs.  `x`. Then

C.I.  = `[x xx (1 + (50)/(3 xx 100)^3 - x] = ((343x)/(216) - x) =  (127x)/(216)`

`:.`      ` (127x)/(216) = 1270`

or    `x = (1270 xx 216)/(127)`

or  `x = 2160`

Thus , he sum is Rs. 2160.

`:.`       S.I. = Rs.`(2160 xx 50/3 xx 3 xx 1/100)` = Rs. 1080

Principal = Rs. 500      Amount =  Rs. 583.20     , Time = 2 years

Let the rate  be  R%  per annum , Them

`[ 500  (1 + R/100)^2] = 583.20`

or `(1 + R/100)^2 = 583.2/500`

or   `(1 + R/100)^2 = 5832/5000`

or      `(1 + R/100)^2 = 11664/10000`

or      `(1 + R/100)^2 =  (108 /100)^2`

or      ` 1 + R/100 =  108/100`

or      `(100 + R)/(100) = 108/100`

or    R = 108 - 100 = 8

or   R = 8

So , rate  =  8% p.a.

Principal  =  Rs. 1000      amount  = Rs. 1331       Rate =   10%  p.a.

Let the time be n years , Then ,

=    `[1000 (1 + 10/1000)^n]` = 1331

or    ` (11/10)^n =  (1331/1000)`

or     `(11/10)^n = (11/10)^3`

or   n = 3

`:.`      n = 3 years.

Clearly rate =  5% p.a.     Time =  3 years,         S.I.  = Rs. 1200

So principal  = Rs. `((100 xx 1200)/(3 xx 5))` =  Rs. 8000

Amount = Rs`[8000 xx (1 + 5/100)^3]`

= Rs.` (8000 xx 21/20 xx 21/20 xx 21/20)`  

= Rs. 9261

`:.`      C.I.  = Rs. 9261 - 8000) =  Rs. 1261.

Principal  = Rs. 16000      ,   Time  = 9 months = 3 quarters.

Rate = 20% per annum = 5% per annum

`:.`     Amount = Rs.` [16000 xx (1 + 5/100)^3]`

= Rs. `(16000 xx 21/20 xx 21/20 xx 21/20)`

=Rs. 18522

`:.`     C.I. = Rs.  (18522 - 16000)

=Rs. 2522

principal = Rs. 10000,    Rate = 2%  per half  year ,     Time =  2  years =  4 half years

`:.`    Amount = Rs.` [10000 xx (1 + 2/100)^4]`

= Rs.  ` (10000 xx 51/50 xx 51/50 xx 51/50 xx 51/50)`

= Rs. 10824.32.

`:.`        C.I.    = Rs.  (10824.32 -  10000) =   Rs. 824.32

Time  = 2 years 4 months =  `2 4/12` years =  `2 1/3` years.

Amount  = Rs` [8000 xx (1 + 15/100)^2 xx (1 + (1/3 xx 15)/(100))]`

=Rs.` (8000 xx 23/20 xx 23/20 xx 21/20)`

= Rs. 11109

`:.`       C.I. = Rs. (11109 - 8000) = Rs. 3109.

Amount = Rs.` [7500 xx(1 + 4/100)^2]`

= Rs. `(7500 xx 26/25 xx 26/25)` =  Rs. 8112.

`:.`       C.I.   = Rs. (8112 - 7500) = Rs. 612.

Answer : Option A

Explanation :

Let the sum be P
$MF#%\text{Time, T} = 1\dfrac{1}{2}\text{ year} = \dfrac{3}{2}\text{ year}\\\\ \text{Amount after }1\dfrac{1}{2}\text{ years} = \text{P}\left(1 + \dfrac{\text{(R/2)}}{100}\right)^\text{2T} = \text{P}\left(1 + \dfrac{(4/2)}{100}\right)^{2 \times \frac{3}{2}} = \text{P}\left(1 + \dfrac{2}{100}\right)^3\\\\= \text{P}\left(\dfrac{102}{100}\right)^3=\text{P}\left(\dfrac{51}{50}\right)^3$MF#%
Given that amount after 1¹â„₂ years = 13265.10
$MF#%=> \text{P}\left(\dfrac{51}{50}\right)^3 = 13265.10\\\\ => \text{P} = 13265.10\left(\dfrac{50}{51}\right)^3 = \dfrac{13265.10 \times 50 \times 50 \times 50}{51 \times 51 \times 51} = \dfrac{260.1 \times 50 \times 50 \times 50}{51 \times 51}= \dfrac{5.1 \times 50 \times 50 \times 50}{51}\\\\ = 0.1 \times 50 \times 50 \times 50 = \text{Rs. 12500}$MF#%