Question

a sum of money amounts to 2100 in a year and 2205 in two years . find the rate of interest and sum.

Answer 1

Year 1 $=$ 2100

Year 2 = 2205

r $=$ ?

p $=$ ?

Difference $=$ 2205 $-$ 2100

$=$ 105

I $=$ p $\times$ R $\times$ T/100

105$=$ $\frac {2100\times R\times1}{100}$

R$=$ 5%

Part b

Let the principal for first year be x

So we'll use this formula;

I $=$ P $\times$ R$\times$ T/100

Whereby

$\frac{5x}{100}$ $=$ $\frac{x}{20}$

Amount for first year $=$ x$+\frac{x}{20}$ $=\frac{21x}{20}$

For second year p $=$ $\frac{21x}{20}$

R= 5%p.a T = 1 year

So we solve for this;

I $=$ $[\frac{21x}{20}\times5\times1 ]/$ 100

$=$ $\frac{21x}{4}/$ 100

From the equation above we'll get

$\frac{21x}{4}$ /100

=2100

400$\times$ $\frac{21x}{400}=$ 2100$\times$ 400

$\frac{21x}{21}$ $=$ $\frac{840000}{21}$

X$=$ 40000

So the original sum of money is

40,000