Question

PLEASE SOLVE THIS ASAP!!!​

Answer 1

FORMULA TO USE :-

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

Question 1 :-

To find :-

Compound interest

Given :-

Principal = ₹1000

Time = 2 years

Rate = 10%

Substituting the values,

$a = 1000(1 + \dfrac{10}{100}) ^{2}$

$a = 1000( \dfrac{100 + 10}{100})^{2}$

$a = 1000( \dfrac{110}{100})^{2}$

$a = 1000( \dfrac{11\not0}{10\not0})^{2}$

$a = 1000( \dfrac{11}{10})^{2}$

$a = 1000 \times \dfrac{11}{10} \times \dfrac{11}{10}$

$a = 10\not0\not0 \times \dfrac{11}{1 \not0} \times \dfrac{11}{1 \not0}$

$a = 10 \times 11 \times 11 = 1210$

Compound interest = Amount - Principal

CI = ₹1210 - ₹1000

CI = ₹210

FORMULA TO USE :-

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

Question 2 :-

To find :-

Simple interest

Given :-

Principal = ₹2000

Rate = 5%

Time = 2 years

substituting the values,

$si = \dfrac{2000 \times 2 \times 5}{100}$

$si = \dfrac{2000 \times 10}{100}$

$si = \dfrac{20\not0\not0 \times 10}{1 \not0 \not0}$

$si = 20 \times 10 = 200$

Question 3 :-

Given :-

Principal = ₹10,000

Amount = ₹13924

Time = 2 years

To find :-

Compound interest

Compound interest = Amount - Principal

Substituting the values,

CI = ₹13924 - ₹10,000

CI = ₹3924

Some more formulas :-

When interest is compounded half-yearly,

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

When interest is compounded quarterly,

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