Let the sum of money be Rs y
So Amount = y +[( y x 5 x 4 )/100]

Correct Option: A

Let the sum of money be Rs y
So Amount = y +[( y x 5 x 4 )/100]
But Amount = Rs 900
⇒ 900 = y +(20y)/100
⇒ 900 = 6y / 5
⇒ y = ( 900 x 5) /6
= Rs 750

Simple Interest in 1 year= Rs (1729 - 1586)
= Rs 143

And R= (100 × SI)/(P × T)
where, R = rate
SI= Simple Interest
P = Principal
T= Time

Correct Option: D

Simple Interest in 1 year= Rs (1729 - 1586)
= Rs 143
now, SI in 2 year = Rs 286

Principal P= Rs(1586 - 286)
= 1300
And R= (100 × SI)/(P × T)
= (100 × 143)/(1300 × 1)
= 11 %
where, R = rate
SI= Simple Interest
P = Principal
T= Time

Let the amount instalment be Rs ' x '
Then According to question,
(Amount of 'x' for 3 yrs) + (Amount of 'x' for 2 yrs) + (Amount of 'x' for 1 yrs) + x =3220

Correct Option: C

Let the amount instalment be Rs ' x '
Then According to question,
(Amount of 'x' for 3 yrs) + (Amount of 'x' for 2 yrs) + (Amount of 'x' for 1 yrs) + x =3220
or, [x+(x × 10 × 3)/100] + [x+(x × 10 × 2)/100] + [x+(x × 10 × 1)/100] + x=3220
⇒ 4x+ (30x/100)+(20x/100)+(10x/100)=3220
⇒ 460x=322000
⇒ x=Rs 700

∴ Each Instalment= Rs 700

Amount A = 2 × P
where , P = Principal

Rate R = (100 × SI)/(P × T)
where , SI= Simple Interest
T= Time

Correct Option: B

Let Sum = P, Then SI=P
As Amount A = 2 × P
where , P = Principal

Rate R = (100 × SI)/(P × T)
= (100 × P)/(P × 8) %
= 12.5 %
where , SI= Simple Interest
T= Time

Simple Interest = (Principal × Rate × Time)/100

Correct Option: C

Here , P= Rs 1000
T= 4 yrs
R= 4 %
where, P= Principal
T= Time
R= Rate
Since , Simple Interest on Rs 1000=(1000 × 4 × 4)/100
= Rs 160

now, simple interest=Rs 160
P = Rs 400
R = 10 %
then, T=(100 × SI)/P × R
= (100 × 160)/(400 × 10)
= 4 yr