Question

The interest on a certain principal for 1 year
12.500. The compound interest on the same 2 year s
same rate of interest is 26.250. What is the rate s
1) 8p.c.p.a.
2)10 p.
cs
3) 12 pcpa 14) The informations suffin Show step by step​

Answer 1

Answer:

The rate of interest applied is 10%  

Step-by-step explanation:

Given as :

The principal amount = Rs P

The compound interest for 1 year = $CI_1$ = Rs 12,500

Let The Amount = Rs $A_1$

The time period = 1 year

Let The rate of interest = r%

From Compound Interest

Amount = principal × $(1+\frac{rate}{100})^{time}$

Or, $A_1$ = p × $(1+\frac{r}{100})^{1}$

∵ Interest = Amount - principal

So, Rs 12500 = $A_1$ - p

Or, 12500 = p $(1+\frac{r}{100})^{1}$ - p

Again

The principal amount = Rs P

The compound interest for 2 years = $CI_2$ = Rs 26,250

The time period = 2 years

Let The rate of interest = r%

Let The Amount = Rs $A_2$

From Compound Interest

Amount = principal × $(1+\frac{rate}{100})^{time}$

Or, $A_2$ = p × $(1+\frac{r}{100})^{2}$

∵ Interest = Amount - principal

So, Rs 26250 = $A_2$ - p

Or, 26250 = p $(1+\frac{r}{100})^{2}$ - p

let $(1 + \dfrac{r}{100)}$ = a

So, 12500 = p a - p                    ........A

And , 26250 = p a² - p             .........B

Solving eq A and eq B

$\dfrac{26250}{12500}$ = $\dfrac{p a^{2} - p}{pa - p}$

Or, $\dfrac{21}{10}$ = $\dfrac{a^{2} - 1 }{a - 1}$

Or, 2.1 = $\dfrac{(a + 1) ( a - 1)}{(a - 1)}$

Or, a + 1 = 2.1

Or, a = 2.1 - 1

∴  a = 1.1

Now, put the value of a

So, $(1 + \dfrac{r}{100)}$ = 1.1

Or, $\dfrac{r}{100}$ = 1.1 - 1

Or, r % = 0.1

i.e r = 10%

So, The rate of interest = r = 10%

Hence, The rate of interest applied is 10%  Answer

Answer 2

Answer:

THE INTEREST APPLIED IS 10 P.C.P.A

Step-by-step explanation:

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