Question

Difference between two positive integer is 30 ratio of these integer is 2 is to 5 find the integer

Answer 1

Suppose the integers are 2x and 5x
then... from the given information...
5x-2x=30
3x=30
x=10

then the integers are 20 and 50 respectively
2*10=20
5*10=50 which has the ratio of 2:5



Hope it helps you...

Answer 2

$\underline{\underline{\Large{\mathfrak{Solution : }}}} </p><p>.$

$\begin{gathered}\textsf{Let the integers are x and y . } \\ \\ \underline{\textsf { According to the question , }} \\ \\ \end{gathered}$

Let the integers are x and y .

According to the question ,

$\begin{gathered} < br / > \mathsf{\implies x \: - \: y \: = \: 30 } \\ \\ \mathsf{ \implies x \: - \: y \: - \: 30 \: = \: 0 \qquad...(1)}\\ \\ \underline{\textsf{Now,}} \\ \\ \mathsf{\implies x \: : \: y \: = \: 2 \: : \: 5 } \\ \\ \mathsf{ \implies \: \dfrac{x}{y} \: = \: \dfrac{2}{5} } \\ \\ \mathsf{ \implies 5x \: = \: 2y} \\ \\ \mathsf{ \: \: \therefore \: 5x \: - \: 2y \: = \: 0 \qquad...(2)}\end{gathered} </p><p>$

$\begin{gathered}\underline{\textsf{Using Cross Multiplication Method : }} \\ \\ \textsf{Coe. of x \quad Coe. of y \quad Constant term } \\ \\ \textsf{ \: \: \: 1 \quad \: \: \: \: \: \: - 1 \quad \: \: \: \: \: \: - 30} \\ \\ \textsf{ \: \: \: 5 \quad \: \: \: \: \: \: - 2 \quad \: \: \: \: \: \: \: 0} \\ \\ \mathsf{ \implies \dfrac{x}{( - 1 \: \times \: 0) \: - \: ( - 30 \: \times \: - 2)} \: = \: \dfrac{y}{( - 30 \: \times \: 5) \: - \: ( 1 \: \times \: 0)} \: =} \\ \mathsf{ \: \: \: \: \: \: \: \: \: \: \: \: \dfrac{1}{(1 \: \times \: - 2) \: - \: ( - 1 \: \times \: 5)} }\end{gathered}$

$\begin{gathered} \mathsf{ \implies \dfrac{x}{ \: - \: 60} \: = \: \dfrac{ \: \: \: \: y \: }{ - 150} \: = \: \dfrac{1}{ - 2 \: + \: 5} } \\ \\ \\ \mathsf{ \implies \dfrac{x}{ - 60} \: = \: \dfrac{y}{ - 150} \: = \: \dfrac{1}{3} }\end{gathered}$

$\begin{gathered} < br / > \textsf{Now,} \\ \\ \mathsf{\implies \dfrac{x}{-60} \: = \: \dfrac{ 1}{3}} \\ \\ \mathsf{\implies 3x \: = \: -60 } \\ \\ \mathsf{\implies x \: = \: \dfrac{-60}{3}} \\ \\ \mathsf{\therefore \: \: x \: = \: -20} < br / > \end{gathered} </p><p> Now,$

$\begin{gathered} < br / > \textsf{And,} \\ \\ \mathsf{\implies \dfrac{y}{-150} \: = \: \dfrac{1}{3}} \\ \\ \mathsf{\implies 3y \: = \: -150} \\ \\ \mathsf{\implies y \: = \: \dfrac{-150}{3}} \\ \\ \mathsf{\therefore \: y \: = \: -50} \end{gathered} </p><p> And,$

$\Large{\boxed{\mathsf{Hence, integers \: are \: -20 \: and \: -50.}}}$