Question

The difference of two natural number is 3 and the difference of their reciprocals is 3\28, find the numbers.

Answer 1

$\bf{\underline{\underline{\rm{\red{Given:}}}}}$

• The difference of two natural numbers is 3
• the differences of their reciprocals is $\large\frac{3}{28}$

$\bf{\underline{\underline{\rm{\pink{To\:find:}}}}}$

• The numbers

$\bf{\underline{\underline{\rm{\orange{Solution:}}}}}$

• Let the greater number be x.

• Let the smaller number be y.

$\bf{\underline{\underline{\rm{\red{As\:per\:first\:condition:}}}}}$

The difference of two natural numbers is 3

Representing the condition mathematically.

=> x - y = 3

=> x = y + 3 -----> 1

$\bf{\underline{\underline{\rm{\red{As\:per\:second\:condition:}}}}}$

the differences of their reciprocals is $\large\frac{3}{28}$

Reciprocal of greater number = $\bf\large\frac{1}{x}$

Reciprocal of smaller number = $\bf\large\frac{1}{y}$

Representing second condition mathematically.

=> $\large\sf\frac{1}{x}$

=> $\large\sf\frac{1}{y}$

=> $\large\sf\frac{3}{28}$

Cross multiplying and multiplying on LHS,

=> $\large\sf\frac{y-x}{xy}$

= $\large\sf\frac{3}{28}$

Cross multiplying,

=> 28 ( y - x) = 3 ( xy )

Substitute the value of x from equation 1,

=> 28 ( y - y + 3) = 3 ( y + 3 ) ( y )

=> 28y - 28y + 84 = 3 ( y² + 3y )

=> 84 = 3y² + 9y

=> 3y² + 9y = 84

Converting the equation to a quadratic equation,

=> 3y² + 9y - 84 = 0 ---->2

Solving the equation further using factorization method,

Divide equation 2 by 3,

=> y² + 3y - 28 = 0

=> y² + 7y - 4y - 28 = 0

=> y ( y + 7 ) - 4 ( y + 7) = 0

=> (y + 7) ( y - 4) = 0

=> y + 7 = 0 OR y - 4 = 0

=> y - 7 OR y = 4

y = - 7 is not acceptable since - 7 is not a natural number.

•°• y = 4

Substitute y = 4 in equation 1,

=> x = y + 3

=> x = 4 + 3

=> x = 7

$\bf{\large{\underline{\boxed{\rm{\red{Greater\:number\:x\:=\:7}}}}}}$

$\bf{\large{\underline{\boxed{\rm{\pink{Smaller\:number\:y\:=\:4}}}}}}$

Answer 2

Given:

The difference of two natural numbers is 3

the differences of their reciprocals is \large\frac{3}{28}

28

3

\bf{\underline{\underline{\rm{\pink{To\:find:}}}}}

Tofind:

The numbers

\bf{\underline{\underline{\rm{\orange{Solution:}}}}}

Solution:

Let the greater number be x.

Let the smaller number be y.

\bf{\underline{\underline{\rm{\red{As\:per\:first\:condition:}}}}}

Asperfirstcondition:

The difference of two natural numbers is 3

Representing the condition mathematically.

=> x - y = 3

=> x = y + 3 -----> 1

\bf{\underline{\underline{\rm{\red{As\:per\:second\:condition:}}}}}

Aspersecondcondition:

the differences of their reciprocals is \large\frac{3}{28}

28

3

Reciprocal of greater number = \bf\large\frac{1}{x}

x

1

Reciprocal of smaller number = \bf\large\frac{1}{y}

y

1

Representing second condition mathematically.

=> \large\sf\frac{1}{x}

x

1

=> \large\sf\frac{1}{y}

y

1

=> \large\sf\frac{3}{28}

28

3

Cross multiplying and multiplying on LHS,

=> \large\sf\frac{y-x}{xy}

xy

y−x

= \large\sf\frac{3}{28}

28

3

Cross multiplying,

=> 28 ( y - x) = 3 ( xy )

Substitute the value of x from equation 1,

=> 28 ( y - y + 3) = 3 ( y + 3 ) ( y )

=> 28y - 28y + 84 = 3 ( y² + 3y )

=> 84 = 3y² + 9y

=> 3y² + 9y = 84

Converting the equation to a quadratic equation,

=> 3y² + 9y - 84 = 0 ---->2

Solving the equation further using factorization method,

Divide equation 2 by 3,

=> y² + 3y - 28 = 0

=> y² + 7y - 4y - 28 = 0

=> y ( y + 7 ) - 4 ( y + 7) = 0

=> (y + 7) ( y - 4) = 0

=> y + 7 = 0 OR y - 4 = 0

=> y - 7 OR y = 4

y = - 7 is not acceptable since - 7 is not a natural number.

•°• y = 4

Substitute y = 4 in equation 1,

=> x = y + 3

=> x = 4 + 3

=> x = 7

\bf{\large{\underline{\boxed{\rm{\red{Greater\:number\:x\:=\:7}}}}}}

Greaternumberx=7

\bf{\large{\underline{\boxed{\rm{\pink{Smaller\:number\:y\:=\:4}}}}}}

Smallernumbery=4