Question

math please answer fast ​

Answer 1

Step-by-step explanation:

1) A= P(1+r/100)n

A= 45000* 116/100*116/100*116/100

A= 70240.35

CI = A-P

CI = 70240.32-45000

CI = 25000.32

Answer 2

FORMULA TO USE :-

$amount = principal(1 + \dfrac{rate}{100})^{time}$

Question 1 :-

Given :-

Principal = ₹45000

Time = 3 years

Rate = 16% compounded annually.

To find :-

Compound interest

By substituting the values,

$a = 45000(1 + \dfrac{16}{100})^{3}$

$a = 45000( \dfrac{116}{100})^{3}$

By reducing to the lowest terms,

$a = 45000( \dfrac{29}{25})^{3}$

$a = 45000 \times \dfrac{29}{25} \times \dfrac{29}{25} \times \dfrac{29}{25}$

$a = \dfrac{ 45000 \times 24,389}{15625}$

$a = 70,240.32$

Compound interest :- Amount - Principal

CI = ₹70240.32 - ₹45000

CI = ₹25,240.32

Question 2 :-

Given :-

Compound interest = ₹287

Time = 2 years

Rate = 5% per annum

To find :-

Amount.

Let the principal be ₹x.

As we know that Compound interest = Amount - Principal,

Amount = Principal + Compound interest

substituting the values,

$x + 287 = x(1 + \dfrac{5}{100})^{2}$

$x + 287 = x( \dfrac{105}{100})^{2}$

By reducing to the lowest terms,

$x + 287 = x( \dfrac{21}{20})^{2}$

By transposing x to the LHS (Left Hand Side),

$\dfrac{x + 287}{x} = \dfrac{21 \times 21}{20 \times 20}$

$\cfrac{x + 287}{x} = \dfrac{441}{400}$

By cross multiplication,

$400(x + 287) = 441(x)$

$400x + 126,567 = 441x$

By transposing 400x to the RHS (Right Hand Side),

$126,567 = 441x - 400x$

$126,567 = 41x$

Transposing 41 to the LHS (Left Hand Side),

$\dfrac{126567}{41} = x$

$3,087 = x$

Therefore,

Amount = ₹3087 + ₹287

Amount = ₹ 3,374

Some more forumulas :-

$simple \: interest = \dfrac{principal \times time \times rate}{100}$

When interest is compounded half-yearly :-

$amount = principal(1 + \dfrac{rate}{200})^{2 \times time}$

When interest is compounded quarterly :-

$amount = principal(1 + \dfrac{rate}{400})^{4 \times time}$