Question

A man invested 36,000, a part of it at 12% and the rest at 15% per annum simple
interest. If he recieved a total annual interest of 4,890, how much did he invest at
each rate?​

Answer 1

Answer:

Step-by-step explanation:

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Answer 2

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

Let us assume that

• The amount invested at the rate of 12 % per annum = Rs x

and

• The amount invested at the rate of 12 % per annum = Rs y

So,

According to statement,

$\red{\bf :\longmapsto\:x + y = 36000 - - - - (1)}$

We know that,

☆ Simple Interest on a certain sum of money Rs p invested at the rate of r % per annum for t years is

$\bf :\longmapsto\:SI = \dfrac{p \times r \times t}{100}$

Case 1 :-

• Amount invested, p = Rs x

• Rate of interest, r = 12 % per annum

• Time = 1 year

Thus,

☆ Simple Interest in this case is

$\rm :\longmapsto\:SI_1 = \dfrac{x \times 12 \times 1}{100}$

$\rm :\longmapsto\:SI_1 = \dfrac{12x}{100} - - - (2)$

Case 2 :-

• Amount invested, p = Rs y

• Rate of interest, r = 15 % per annum

• Time = 1 year

Thus,

☆ Simple Interest in this case is

$\rm :\longmapsto\:SI_2 = \dfrac{y \times 15 \times 1}{100}$

$\rm :\longmapsto\:SI_2 = \dfrac{15y}{100} - - - (3)$

According to statement,

☆ The annual interest received = Rs 4890

$\rm :\longmapsto\:SI_1 + SI_2 = 4890$

$\rm :\longmapsto\:\dfrac{12x}{100} + \dfrac{15y}{100} = 4890$

$\rm :\longmapsto\:\dfrac{4x}{100} + \dfrac{5y}{100} = 1630$

$\rm :\longmapsto\:4x + 5y = 163000$

$\rm :\longmapsto\:4(36000 - y) + 5y = 163000$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \red{\bigg \{ \because \:using \: (1) \bigg \}}$

$\rm :\longmapsto\:144000 - 4y + 5y = 163000$

$\rm :\longmapsto\:y = 163000 - 144000$

$\bf\implies \:y = 19000$

☆ On substituting the value of y = 19000 in equation (1),

$\rm :\longmapsto\:x + 19000 = 36000$

$\bf\implies \:x = 17000$

Hence,

• Amount invested in first case = Rs 17000

And

• Amount invested in second case = Rs 19000