Question

divide rs 3600 into two parts such that if one part be lent at 9% per annum and the other at 10% per annum ,the total annual income is rs 333.

Answer 1

Question :-

Divide Rs 3600 into two parts such that if one part be lent at 9% per annum and the other at 10% per annum ,the total annual income is Rs 333.

$\large\underline{\sf{Solution-}}$

Given that Rs 3600 into two parts such that if one part be lent at 9% per annum and the other at 10% per annum, the total annual income is Rs 333.

Let assume that

First part be Rs x

Second part be Rs 3600 - x

Now, According to statement, Rs x is invested at rate of 9 % per annum.

So, annual income received,

$\rm \: A_1 = x \times \dfrac{9}{100} = \dfrac{9x}{100} - - - (1)$

According to statement again, Rs 3600 - x is invested at the rate of 10 % per annum.

So, annual income received,

$\rm \: A_2 = (3600 - x) \times \dfrac{10}{100} = \dfrac{36000 - 10x}{100} - - - (2)$

According to statement again,

$\rm \: A_1 \: + \: A_2 = 333$

$\rm \: \dfrac{9x}{100} + \dfrac{36000 - 10x}{100} = 333$

$\rm \: \dfrac{9x + 36000 - 10x}{100} = 333$

$\rm \: \dfrac{36000 - x}{100} = 333$

$\rm \: 36000 - x = 33300$

$\rm\implies \:x = 2700$

Hence,

First part invested at the rate of 9 % = Rs 2700

Second part invested at the rate of 10 % = Rs 900

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Additional Information

1. Amount received on a certain sum of money of Rs P invested at the rate of r % per annum compounded annually for n years is given by

$\boxed{\sf{ \:Amount = P {\bigg[1 + \dfrac{r}{100} \bigg]}^{n} \: }}$

2. Amount received on a certain sum of money of Rs P invested at the rate of r % per annum compounded semi - annually for n years is given by

$\boxed{\sf{ \:Amount = P {\bigg[1 + \dfrac{r}{200} \bigg]}^{2n} \: }}$

3. Amount received on a certain sum of money of Rs P invested at the rate of r % per annum compounded quarterly for n years is given by

$\boxed{\sf{ \:Amount = P {\bigg[1 + \dfrac{r}{400} \bigg]}^{4n} \: }}$

4. Amount received on a certain sum of money of Rs P invested at the rate of r % per annum compounded monthly for n years is given by

$\boxed{\sf{ \:Amount = P {\bigg[1 + \dfrac{r}{1200} \bigg]}^{12n} \: }}$

5. Simple Interest received on a certain sum of money of Rs P invested at the rate of r % per annum for n years is given by

$\boxed{\sf{ \: \:SI = \frac{P \times r \times n}{100} \: \: }}$

Answer 2

Answer:

Given :-

• One part be lent at 9% per annum and the other at 10% per annum, the total income is Rs 333.

To Find :-

• Divide Rs 3600 into two parts.

Formula Used :-

$\clubsuit$ Simple Interest or S.I Formula :

$\dashrightarrow \sf\boxed{\bold{\pink{S.I =\: \dfrac{P \times r \times t}{100}}}}$

where,

• S.I = Simple Interest
• P = Principal
• r = Rate of Interest
• t = Time Period

Solution :-

Let,

$\mapsto \bf First\: Part =\: Rs\: x$

$\mapsto \bf Other\: Part =\: Rs\: (3600 - x)$

$\sf\bold{\green{\underline{\bigstar\: In\: the\: first\: case\: :-}}}$

$\leadsto$ One part of lent at 9% per annum.

Given :

• Principal = Rs x
• Rate of Interest = 9% per annum
• Time Period = 1 year

According to the question by using the formula we get,

$\implies \sf S.I =\: \dfrac{x \times 9 \times 1}{100}$

$\implies \sf S.I =\: \dfrac{x \times 9}{100}$

$\implies \sf\bold{\purple{S.I =\: \dfrac{9x}{100}\: ------\: (Equation\: No\: 1)}}$

$\sf\bold{\green{\underline{\bigstar\: In\: the\: second\: case\: :-}}}$

$\leadsto$ The other part at 10% per annum.

Given :

• Principal = Rs (3600 - x)
• Rate of Interest = 10% per annum
• Time Period = 1 year

According to the question by using the formula we get,

$\implies \sf S.I =\: \dfrac{(3600 - x) \times 10 \times 1}{100}$

$\implies \sf S.I =\: \dfrac{(3600 - x) \times 1\cancel{0}}{10\cancel{0}}$

$\implies\sf\bold{\purple{S.I =\: \dfrac{3600 - x}{10}\: ------\: (Equation\: No\: 2)}}$

Now, according to the question,

$\leadsto$ The total income is Rs 333.

$\implies \sf \dfrac{9x}{100} + \dfrac{3600 - x}{10} =\: 333$

$\implies \sf \dfrac{9x + 36000 - 10x}{100} =\: 333$

By doing cross multiplication we get,

$\implies \sf 9x + 36000 - 10x =\: 333(100)$

$\implies \sf 9x - 10x + 36000 =\: 33300$

$\implies \sf - x + 36000 =\: 33300$

$\implies \sf - x =\: 33300 - 36000$

$\implies \sf {\cancel{-}} x =\: {\cancel{-}} 2700$

$\implies \sf\bold{\red{x =\: 2700}}$

Hence, the required parts are :

✯ First Part :

➳ First Part = Rs x

➳ First Part = Rs 2700

✯ Other Part :

➳ Other Part = Rs (3600 - x)

➳ Other Part = Rs (3600 - 2700)

➳ Other Part = Rs 900

$\therefore$ The two parts of Rs 3600 are 2700 and 900 .