Question

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Answer 1

Answer:

Given:-

Govind Has 2 sons

Ageing 10 and 8 each

Invests 1.22 lakh in names of both sons at compound interest of 20% p.a

At age of 18,they would receive equal amounts

To Find:- amount he planned to invest in his older son,that is of 10 years now

Solution:-

We know

$A=P(1 + R \div 100 )^{n}$

A= amount

P=Principal

R= rate

n= Time

Let the sum be x (in lakhs)

and 1.22-x

For second kid (8 years old)

A=x(1+20/100)⁸ ( eq. 1 )

For first kid (10 years old)

A= (1.22-x) (1+20/100)¹⁰ ( eq. 2)

from both (eq.1) & (eq.2)

x(1+20/100)⁸=(1.22-x)(1+20/100)¹⁰

as Govind wanted to give equal amounts

$\frac{x}{1.22 - x} = \frac{( \frac{12}{10}) ^{10}}{ (\frac{12}{10}) ^{8} } \\ \frac{x}{1.22 - x} = \frac{1.2 ^{10} }{1.2 ^{8} } = {1.2}^{2} \\ \frac{x}{1.22 - x} = 1.44 \\ 1.44(1.22 - x) = x \\ 1.756 - 1.44x = x \\ x + 1.44x = 1.756 \\ 2.44x = 1.756 \\ x = \frac{1.756}{2.44} \\ x = 0.72(rounded \: off)$

therefore the son will get Rs 0.72 lakhs

that is Rs 72,000

Answer 2

Answer:

Given :-

• Ages of Gobind are given as 10 years and 8 years.

• Gobind plans to invest a total of 1.22 lakh in the names of the two sons at 20% per annum compound interest

• When 2 sons attained the age of 18 years, would receive equal amounts.

To find :-

How much does he plan to invest in the name of his older son

Let's begin

So let the total amount be x lacks

If Gobind gives one son x lacks so 2nd son will get (1.22-x)

Amount for first son

As we know

$A = P {(1 + \frac{R}{100} )}^{n}$

we're

A is Amount

P is principal which is x

n is time period which is 8 years

R is rate which is 20%

$A = x {(1 + \frac{20}{100} )}^{8} - - - -\sf \: equation \: \: 1$

Amount for second son

Using same formula

$A = P {(1 + \frac{R}{100} )}^{n}$

we're

A is Amount

P is principal which is (1.22-x)

n is time period which is 10 years

R is rate which is 20%

$A = (1.22 - x ){(1 + \frac{20}{100} )}^{10} - - - - - -\sf \: equation \:2$

When we compare both the equation we get

$x {(1 + \frac{20}{100} )}^{8} = (1.22 - x ){(1 + \frac{20}{100} )}^{10}$

$x = { ( \frac{100 + 20 }{100} )}^{8} = (1.22 - x)({ \frac{100 + 20}{100} })^{10} \: \: \sf Done \: By \: LCM$

$x = { ( \frac{120}{100} )}^{8} = (1.22 - x)({ \frac{120}{100} })^{10} \:$

$\frac{x}{1.22 - x} = \frac{ \frac{ {120}^{10} }{10} }{ { \frac{120}{100} }^{8} } \: \sf Done \: By \: Transposing$

$\frac{x}{1.22 - x} = \frac{ {1.2}^{10} }{{1.2}^{8} } \: \sf Done \: By \: cancelling$

Using exponential law which is

when bases are same powers are subtracted

$\frac{x}{1.22 - x} = {1.2}^{10 - 8}$

$\frac{x}{1.22 - x} = {1.2}^{2}$

$\frac{x}{1.22 - x} = 1.44$

By Transposing

$x = 1.44(1.22 - x)$

$x = 1.756 - 1.44x$

$2.44x = 1.756$

$x = \frac{1.756}{0.72}$

$x = 0.72 \: \: approx$

So Govind invests 0.72 on his older son.