Find 2 consecutive integers such that two-fifth of the smaller exceeds two- ninth of the grater by 4 About the author Cora
Step-by-step explanation: \large{ \bold{ \color{blue}{ \underline{Question : – }}}} Question:− Find two consecutive odd integers such that two – fifth of the smaller exceeds two – ninth of the greater by 4. \large{ \bold{ \color{blue}{ \underline{ \color{blue}{Solution : – }}}}} Solution:− Let the smaller odd integer be x and greater odd integer be ( X + 2 ) \begin{gathered} \frac{2}{5} \\ \end{gathered} 5 2 of smaller exceedo \begin{gathered} \frac{2}{5th} \\ \end{gathered} 5th 2 of greater odd by 4. \begin{gathered} = > \frac{2x}{5} = \frac{2}{5}(x + 2) + 4 \\ \end{gathered} => 5 2x = 5 2 (x+2)+4 \begin{gathered} = > \frac{2x}{5} = \frac{2x + 4 + 36}{9} \\ \end{gathered} => 5 2x = 9 2x+4+36 = > 18x \: = 5(2x + 40)=>18x=5(2x+40) = > 18x \: = 10x \: + 200=>18x=10x+200 = > 8x \: = 200=>8x=200 = > X \: = 25=>X=25 \begin{gathered}= > X + 2 \: = 25 \: + 2 \\ \: \: \: \: \: \: \: \: \:\: \: \: = 27\end{gathered} =>X+2=25+2 =27 So, two consecutive odd integers are 25 and 27 . ________________________________ Reply
[tex] \large{ \bold{ \color{blue}{ \underline{Question : – }}}}[/tex] Find two consecutive odd integers such that two – fifth of the smaller exceeds two – ninth of the greater by 4. [tex] \large{ \bold{ \color{blue}{ \underline{ \color{blue}{Solution : – }}}}}[/tex] Let the smaller odd integer be x and greater odd integer be ( X + 2 ) [tex] \frac{2}{5} \\ [/tex] of smaller exceedo [tex] \frac{2}{5th} \\ [/tex] of greater odd by 4. [tex] => \frac{2x}{5} = \frac{2}{5}(x + 2) + 4 \\ [/tex] [tex] => \frac{2x}{5} = \frac{2x + 4 + 36}{9} \\ [/tex] [tex] = > 18x \: = 5(2x + 40)[/tex] [tex] = > 18x \: = 10x \: + 200[/tex] [tex] => 8x \: = 200[/tex] [tex] => X \: = 25 [/tex] [tex]=> X + 2 \: = 25 \: + 2 \\ \: \: \: \: \: \: \: \: \:\: \: \: = 27[/tex] So, two consecutive odd integers are 25 and 27 . ___________________________________ Hope it’s helps uhh…. Be Brainly♡~ Reply
Step-by-step explanation:
\large{ \bold{ \color{blue}{ \underline{Question : – }}}}
Question:−
Find two consecutive odd integers such that two – fifth of the smaller exceeds two – ninth of the greater by 4.
\large{ \bold{ \color{blue}{ \underline{ \color{blue}{Solution : – }}}}}
Solution:−
Let the smaller odd integer be x and greater odd integer be ( X + 2 )
\begin{gathered} \frac{2}{5} \\ \end{gathered}
5
2
of smaller exceedo \begin{gathered} \frac{2}{5th} \\ \end{gathered}
5th
2
of greater odd by 4.
\begin{gathered} = > \frac{2x}{5} = \frac{2}{5}(x + 2) + 4 \\ \end{gathered}
=>
5
2x
=
5
2
(x+2)+4
\begin{gathered} = > \frac{2x}{5} = \frac{2x + 4 + 36}{9} \\ \end{gathered}
=>
5
2x
=
9
2x+4+36
= > 18x \: = 5(2x + 40)=>18x=5(2x+40)
= > 18x \: = 10x \: + 200=>18x=10x+200
= > 8x \: = 200=>8x=200
= > X \: = 25=>X=25
\begin{gathered}= > X + 2 \: = 25 \: + 2 \\ \: \: \: \: \: \: \: \: \:\: \: \: = 27\end{gathered}
=>X+2=25+2
=27
So, two consecutive odd integers are 25 and 27 .
________________________________
[tex] \large{ \bold{ \color{blue}{ \underline{Question : – }}}}[/tex]
Find two consecutive odd integers such that two – fifth of the smaller exceeds two – ninth of the greater by 4.
[tex] \large{ \bold{ \color{blue}{ \underline{ \color{blue}{Solution : – }}}}}[/tex]
Let the smaller odd integer be x and greater odd integer be ( X + 2 )
[tex] \frac{2}{5} \\ [/tex] of smaller exceedo [tex] \frac{2}{5th} \\ [/tex] of greater odd by 4.
[tex] => \frac{2x}{5} = \frac{2}{5}(x + 2) + 4 \\ [/tex]
[tex] => \frac{2x}{5} = \frac{2x + 4 + 36}{9} \\ [/tex]
[tex] = > 18x \: = 5(2x + 40)[/tex]
[tex] = > 18x \: = 10x \: + 200[/tex]
[tex] => 8x \: = 200[/tex]
[tex] => X \: = 25 [/tex]
[tex]=> X + 2 \: = 25 \: + 2 \\ \: \: \: \: \: \: \: \: \:\: \: \: = 27[/tex]
So, two consecutive odd integers are 25 and 27 .
___________________________________
Hope it’s helps uhh….
Be Brainly♡~