For a particular year the simple interest at 10% is ₹ 800. The compound interest for the next year at the same rate is :

1. ₹ 880

2. ₹ 800

3. ₹ 720

4. ₹ 968

Given,

Interest = ₹ 800

T = 1 year

R = 10%

Let money on which interest is ₹ 800 be P.

By formula,

Interest = $\dfrac{P \times R \times T}{100}$

Substituting values we get :

$\Rightarrow 800 = \dfrac{P \times 10 \times 1}{100} \\[1em] \Rightarrow P = \dfrac{800 \times 100}{10} = 8000.$

We know that,

S.I. and C.I. for first year are equal.

C.I. for first year = ₹ 800

For second year :

P = ₹ 8000 + ₹ 800 = ₹ 8800

R = 10%

T = 1 year

I = $\dfrac{P \times R \times T}{100} = \dfrac{8800 \times 10 \times 1}{100}$ = ₹ 880.

Hence, Option 1 is the correct option.