Let sum = p
Then, after 15 yr
Sum = 8p

∴ SI = 8p - P = 7P
Now, 7P = (P x R x 15)/100

Correct Option: B

Let sum = p
Then, after 15 yr
Sum = 8p

∴ SI = 8p - P = 7P
Now, 7P = (P x R x 15)/100
⇒ 7 = 15R/100 = 3R/20
∴ R = (20 x 7)/3 = 140/3
= 46²/₃%

Let sum be P.
∴ 50% of P = P/2 = SI
Now, P/2 = (P x 5 x T) / 100 [ as time = 10 yr]

Correct Option: A

Let sum be P.
∴ 50% of P = P/2 = SI
Now, P/2 = (P x 5 x T) / 100 [ as time = 10 yr]
⇒ P/2 = 5PT / 100
⇒ 1/2 = T/20
∴ = T = 10 yr

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

Correct Option: D

Total interest needed in a year = Rs 400 × 12
= Rs 4800
Principal = (100 × SI)/R × T
where, R = Rate
T = Time
SI= Simple Interest

According to the formula. SI = (P x R x T) /100

Correct Option: D

The required difference in amount and principal is SI = A - P
Here, P = ₹ 4000, R = 5% T = 4 yr
According to the formula.
SI = (P x R x T) /100
= (4000 x 5 x 4)/100
= ₹ 800

Here, P = ₹ 2668, T = 5 yr, A = ₹ 4669
We know that,
Amount (A) = Principal (P) + (Simple Interest) SI
4669 = 2668 + SI
⇒ SI = 4669 - 2668 = ₹ 2001
Again, SI = (P x R x T)/100

Correct Option: D

Here, P = ₹ 2668, T = 5 yr, A = ₹ 4669
We know that,
Amount (A) = Principal (P) + (Simple Interest) SI
4669 = 2668 + SI
⇒ SI = 4669 - 2668 = ₹ 2001
Again, SI = (P x R x T)/100
∴ 2001 = (2668 x R x 5)/100
⇒ R= (2001 x 100) / (2668 x 5)
= (2001 x 5)/667
= 15%