Question

) A sum of money triples itself at compound interest in 3 years. In 9 years it will be

Answer 1

Answer:

27 times the principle

Step-by-step explanation:

we know that A=p(1+r/100)^n

substitute A =3A and n=3 as per question

now cubing on both sides, we get

27A=(1+r/100)^9

hence proved!!

Answer 2

The sum of money will be 27 times in 9 years.

Given:

• A sum of money triples itself at compound interest in 3 years.

To find:

• In 9 years it will be ?

Solution:

Formula to be used:

Let A be sum of money, Pbe principal amount, r be rate and n be years.

Compound interest

$\bf A=P \left( {1 + \frac{r}{100} } \right)^{n}$

Step 1:

Write the equation according to the statement.

ATQ,

Sum of money triples in 3 years.

A=3P

So,

$3P=P \left( {1 + \frac{r}{100} } \right)^{3}$

$3= \left( {1 + \frac{r}{100} } \right)^{3}$

taking cube root both sides.

$\bf \sqrt[3]{3} = 1 + \frac{r}{100}...eq1$

Step 2:

Find the sum of money in 9 years.

$A=P \left( {1 + \frac{r}{100} } \right)^{9}...eq2$

put the value of eq1 in eq2.

$A=P( \sqrt[3]{3} )^{9}$

$A=P ( {3} )^{ \frac{9}{3} }$

$A=P ( {3} )^{ 3 }$

$\bf A=27P$

Thus,

The sum of money will be 27 times in 9 years.

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