8.
fü) If(10x+3y):(5x+2y)=9:5, let us show that, (2x+y): (x+2y)=11:13
Let us calculate what term should be added to both

2 thoughts on “8.<br />fü) If(10x+3y):(5x+2y)=9:5, let us show that, (2x+y): (x+2y)=11:13<br />Let us calculate what term should be added to both”

1. Answer:

   Question No 1 :-

   [tex]\longrightarrow[/tex] If (10x + 3y) : (5x + 2y) = 9 : 5, let us show that (2x + y) : (x + 2y) = 11 : 13.

   Given :–

   [tex]\mapsto[/tex] [tex]\sf (10x + 3y) : (5x + 2y) = 9 : 5.[/tex]

   Show That :

   [tex]\mapsto[/tex] [tex]\sf (2x + y) : (x + 2y) = 11 : 13.[/tex]

   Solution :-

   [tex]\implies[/tex] [tex]\sf 10x + 3y : 5x + 2y = 9 : 5[/tex]

   Then, we can write as :

   [tex]\implies[/tex] [tex]\sf \dfrac{10x + 3y}{5x + 2y} =\: \dfrac{9}{5}[/tex]

   By doing cross multiplication we get,

   [tex]\implies[/tex] [tex]\sf 5(10x + 3y) =\: 9(5x + 2y)[/tex]

   [tex]\implies[/tex] [tex]\sf 50x + 15y =\: 45x + 18y[/tex]

   [tex]\implies[/tex] [tex]\sf 50x – 45x =\: 18y – 15y[/tex]

   [tex]\implies[/tex] [tex]\sf 5x =\: 3y[/tex]

   [tex]\implies[/tex] [tex]\sf \dfrac{x}{y} =\: \dfrac{3}{5}[/tex]

   Let,

   [tex]\longmapsto[/tex] [tex]\sf \dfrac{x}{y} =\: \dfrac{3}{5} =\: k\: [where\: k \neq 0][/tex]

   Then,

   ➲ [tex]\sf x =\: 3k[/tex]

   ➲ [tex]\sf y =\: 5k[/tex]

   Now by taking L.H.S = (2x + y) : (x + 2y)

   ↦ [tex]\sf (2 \times 3k + 5k) : (3k + 2 \times 5k)[/tex]

   ↦ [tex]\sf (6k + 5k) : (3k + 10k)[/tex]

   ↦ [tex]\sf 11{\cancel{k}} : 13{\cancel{k}}[/tex]

   ➠ [tex]\bold{\red{11 : 13}}[/tex] = R.H.S

   [tex]\leadsto[/tex] [tex]\sf\boxed{\bold{\pink{(PROVED)}}}[/tex]

   _______________________________________

   Question No 2 :-

   [tex]\longrightarrow[/tex] Let us calculate what term should be added to both terms of the ratio 2 : 5 to make the ratio 6 : 11.

   Solution :-

   Let, the number should be added be x

   Now, according to the question,

   ⇒ [tex]\sf \dfrac{2 + x}{5 + x} =\: \dfrac{6}{11}[/tex]

   By doing cross multiplication we get,

   ⇒ [tex]\sf 11(2 + x) =\: 6(5 + x)[/tex]

   ⇒ [tex]\sf 22 + 11x =\: 30 + 6x[/tex]

   ⇒ [tex]\sf 11x – 6x =\: 30 – 22[/tex]

   ⇒ [tex]\sf 5x =\: 8[/tex]

   ➠ [tex]\sf\bold{\red{x =\: \dfrac{8}{5}}}[/tex]

   [tex]\therefore[/tex] [tex]\sf\bold{\dfrac{8}{5}}[/tex] is should be added to the number.

   Reply

2. Answer:

   11/13 and 8/5

   Step-by-step explanation:

   [tex]\implies \sf{\dfrac{10x+3y}{5x+2y} = \dfrac{9}{5} }\\\\\\\implies \sf{\dfrac{10x\frac{y}{y} +3y}{5x\frac{y}{y}+2y}=\dfrac{9}{5} }\\\\\\\implies \sf{\dfrac{y\big(10\frac{x}{y}+3\big)}{y\big(5\frac{x}{y}+2\big)}=\dfrac{9}{5} }[/tex]

   Let [tex]\dfrac{x}{y}=k[/tex],

   [tex]\implies\sf{\dfrac{10k+3}{5k+2}=\dfrac{9}{5}} \\\\\implies\sf{5(10k+3)=9(5k+2) }\\\\\implies\sf{k=\dfrac{3}{5}}[/tex]

   Hence,

   [tex]\implies \sf{\dfrac{2x+y}{x+2y} }\\\\\\\implies \sf{\dfrac{2x\frac{y}{y} +y}{x\frac{y}{y}+2y} }\\\\\\\implies \sf{\dfrac{y\big(2\frac{x}{y}+1\big)}{y\big(\frac{x}{y}+2\big)} }[/tex]

   [tex]\implies\sf{\dfrac{2k+1}{k+2} =\dfrac{2\big(\frac{3}{5}\big)+1}{\frac{3}{5}+2} }\\\\\implies\sf{ \dfrac{11}{13} }[/tex]

   Question 2:

   Let the x should be added,

   ⇒ (2 + x)/(5 + x) = 6/11

   ⇒ 11(2 + x) = 6(5 + x)

   ⇒ 22 + 11x = 30 + 6x

   ⇒ 11x – 6x = 30 – 22

   ⇒ 5x = 8

   ⇒ x = 8/5

   8/5 should be subtracted

   Reply