Question

2.
A number is divided into two parts such that one part is 10 more than the other
parts are in the ratio 5:3, find the number and the two parts​

Answer 1

Answer:

Given :-

• Two number are in ratio 5:3
• One of them is 10 more than other

To Find :-

Numbers

SoluTion :-

Let the number be 5y and 3y respectively

$\large \sf \: 5y - 3y = 10$

$\large \sf \: 2y = 10$

$\large \sf \: y \: = \cancel \dfrac{10}{2}$

${ \boxed{ \pink{ \frak{y = 5}}}}$

Numbers are :-

$\large \bf \: 5y = 5(5) = 25$

$\large \bf \: 3y = 3(5 ) = 15$

Answer 2

Question :-

• A number is divided into two parts such that one part is 10 more than the other parts are in the ratio 5:3, find the number.

Given :-

• The Ratio of Numbers = 5:3
• One Number is 10 more then the other number.

To Find :-

• What are the Numbers ?

Solution :-

Let First Number be 5x

Let Second Number be 3x + 10

According To Question :-

$\longmapsto \sf{\bf{ \:5x \: = \: 3x \: + \: 10 }}$

$\longmapsto \sf{\sf{ \: 5x \: - \: 3x \: = \: 10 }}$

$\longmapsto \sf{\sf{ \: 2x \: = \: 10 }}$

$\longmapsto \sf{\sf{ \: x \: = \: \cancel\dfrac{10}{2} }}$

$\longmapsto \sf\boxed{\bf{ \: x \: = \: 5 }}$

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First Number = 5x = 5 × 5 = 15

Second Number = 3x + 10 = 3 × 5 + 10 = 25

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$\begin{gathered}\begin{gathered}\qquad \therefore\: \sf{ First \: Number \: = \underline {\underline{ 15}}}\\\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\qquad \therefore\: \sf{ Second \: Number \: = \underline {\underline{ 25}}}\\\end{gathered}\end{gathered}$

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