Question

Find the amount of R 36,000 after 2 years, compounded annually, the rate of interest being 10% for the first year and 12% for the second year.​

Answer 1

Given

• Principal(P)= Rs. 36,000
• Time(T) = 2 Year's
• Rate(R) = 10% for 1st and 12% For 2nd Year's

To Find

• Amount(A)

Formula Used

$\boxed{ \bf{Amount = P\left( 1+\dfrac{R}{100}\right) ^{n} }}$

Solution

___________________

$\sf{Amount= 36000\left( 1+\dfrac{10}{100}\right)\left( 1+\dfrac{12}{100}\right) }$

$\sf{Amount= 36000\left( \dfrac{100 + 10}{100}\right)\left( \dfrac{100 + 12}{100}\right) }$

$\sf{Amount= 36000 \times \dfrac{1 10}{100} \times \dfrac{112}{100} }$

$\sf{Amount= 36 \cancel0 \cancel{00 }\times \dfrac{1 1 \cancel0}{ \cancel{100}} \times \dfrac{112}{ 1 \cancel0 \cancel0} }$

$\sf{Amount= 36 \times 11 \times 112 }$

$\sf{Amount= 44352 }$

Hence the Amount is Rs.44352.

Answer 2

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GIVEN::-

• Principal(P)= Rs. 36,000

• Time(T) = 2 Year's

• Rate(R) = 10% for 1st and 12% For 2nd Year's

To find::-

• Amount(A)

Formula Used::-

$\huge\boxed{ \bf{Amount = P\left( 1+\dfrac{R}{100}\right) ^{n} }}$

Solution::-

$\tiny \boxed{Amount= 36000\left( 1+\dfrac{10}{100}\right)\left( 1+\dfrac{12}{100}\right) }$

$\tiny\boxed{Amount= 36000\left( \dfrac{100 + 10}{100}\right)\left( \dfrac{100 + 12}{100}\right) }$

$\tiny\boxed{Amount= 36000 \times \dfrac{1 10}{100} \times \dfrac{112}{100} }$

$\boxed{Amount= 36 \cancel0 \cancel{00 }\times \dfrac{1 1 \cancel0}{ \cancel{100}} \times \dfrac{112}{ 1 \cancel0 \cancel0} }$

$\bold{Amount= 36 \times 11 \times 112 }$

$\bold{Amount= 44352 }$

Hence the Amount is Rs.44352.