Simple Interest for 4 yr = Rs( 1360 - 1040 ) = Rs 320
So, Simple Interest for 3 yr = Rs [(320/4) x 3]= Rs 240

Correct Option: B

Simple Interest for 4 yr = Rs(1360 - 1040 ) = Rs 320
so, Simple Interest for 3 yr= Rs [(320/4) x 3]= Rs 240

Sum = Rs (1040-240)= Rs 800

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

Correct Option: A

Let principal amount = P
As amount =3P, T=25 yr
∴ SI=3P-P=2P
∴ Rate R = (100 x SI)/(P x T)
where, SI= Simple Interest T = Time
= (100 x 2P)/(P x 25)=8%

Let the sum be 'y'
then, [(y x13 x 1)/100]-[(y x 12 x 1)/100]=110

Correct Option: C

Let the sum be 'y'
then, [(y x13 x 1)/100 ]- [( y x 12 x 1) / 100]=110
since ( y / 100 ) = Rs 110
∴ Y = Rs 11000

Simple in 3 year = Rs( 1350 - 1260) = Rs 90
Simple in 2 year = (2/3) &time; 90= Rs 60
Rate, R= (100 x SI)/(P x T)
where , SI= Simple Interest
P= Principal
T= Time

Correct Option: C

Simple in 3 year = Rs( 1350 - 1260) = Rs 90
Simple in 2 year = (2/3) &time; 90= Rs 60
Principal = Rs (1260 -60)= Rs 1200
Rate, R= (100 x SI)/(P x T)= (100 x 60)/(1200 x 2)=60/24= 25 %
where , SI= Simple Interest
P= Principal
T= Time

Let the sum be Rs 'y' , so amount = 2y
Simple interest =Rs y
Let R be the rate of interest,
R= (100 x SI)/(P x T)
where SI= Simple Interest
P = Principal
T = Time

Correct Option: A

Let the sum be Rs 'y' , so amount = 2y
Simple interest =Rs y
Let R be the rate of interest,
R= (100 x SI)/(P x T) = (100 x y) / (y x 8) = 12.5 %
where SI= Simple Interest
P = Principal
T = Time

now, the needed amount = Rs 4y
since SI = Rs (4x-x)= Rs 3y
since T= (100 x SI)/(P x R)
= (100 x 3y )/(y x 125)= 24 yr