Question

9.If x:y=3:4 the value of x^(2)y+xy^(2):x^(3)+y^(3) is (a) 13:12 (b) 12:13 (c) 21:31 (d) none of these​

Answer 1

ANSWER:

• The value of x²y + xy² : x³ + y³ = 12 : 13

GIVEN:

• x : y = 3 : 4.

TO FIND:

• The value of x²y + xy² : x³ + y³.

EXPLANATION

$\implies\sf \dfrac{x}{y} = \dfrac{3}{4}\\ \\ \\ \implies\sf x = \dfrac{3y}{4}\\ \\ \\ \implies\sf \dfrac{x^2y + xy^2}{x^3 + y^3}=\dfrac{ \left(\dfrac{3y}{4}\right)^2 \times y +\left(\dfrac{3y}{4}\right)y^2}{ \left(\dfrac{3y}{4}\right)^3+ y^3}\\ \\ \\ \implies \sf \dfrac{x^2y + xy^2}{x^3 + y^3} = \dfrac{ \left(\dfrac{9y^2}{16}\right) \times y +\left(\dfrac{3y^3}{4}\right)}{ \left(\dfrac{27y^3}{64}\right)+ y^3}\\ \\ \\ \sf \implies\dfrac{x^2y + xy^2}{x^3 + y^3} = \dfrac{ \left(\dfrac{9y^3}{16}\right) +\left(\dfrac{3y^3}{4}\right)}{ \left(\dfrac{27y^3}{64}\right)+ y^3}$

$\implies\sf \dfrac{x^2y + xy^2}{x^3 + y^3} = \dfrac{ \left(\dfrac{9y^3}{16}\right) +\left(\dfrac{12y^3}{16}\right)}{ \left(\dfrac{27y^3}{64}\right)+ \dfrac{64 {y}^{3} }{64}}\\ \\ \\ \implies\sf \dfrac{x^2y + xy^2}{x^3 + y^3} = \dfrac{ \left(\dfrac{21y^3}{16}\right)}{ \left(\dfrac{91y^3}{64}\right)}\\ \\ \\ \implies \sf \dfrac{x^2y + xy^2}{x^3 + y^3} = \dfrac{ \left(\dfrac{3}{1}\right)}{ \left(\dfrac{13}{4}\right)}\\ \\ \\ \implies\sf \dfrac{x^2y + xy^2}{x^3 + y^3} = \left(\dfrac{3}{1}\right) \times \left(\dfrac{4}{13}\right)\\ \\ \\ \implies \sf \dfrac{x^2y + xy^2}{x^3 + y^3} = \dfrac{12}{13}$

x²y + xy² : x³ + y³ = 12 : 13.

Answer 2

Answer:

Given :-

• It x : y = 3 :4.

To Find :-

• What is the value of x²y + xy² : x³ + y³.

Solution :-

Given :

$\implies \sf x : y =\: 3 : 4$

$\implies \sf \dfrac{x}{y} =\: \dfrac{3}{4}$

$\implies \sf \bold{\purple{x =\: \dfrac{3y}{4}}}$

Now,

$\mapsto \sf\bold{x^2y + xy^2 : x^3 + y^3}$

$\longrightarrow \sf \dfrac{x^2y + xy^2}{x^3 + y^3}$

Now, we have :

$\leadsto \sf x =\: \dfrac{3y}{4}$

So, by putting the value of x we get,

$\longrightarrow \sf \dfrac{\bigg\lgroup \dfrac{3y}{4}\bigg\rgroup^2 y + \bigg\lgroup \dfrac{3y}{4}\bigg\rgroup\: y^2}{\bigg\lgroup \dfrac{3y}{4}\bigg\rgroup^3 + \bigg\lgroup y^3\bigg\rgroup}$

$\longrightarrow \sf \dfrac{\bigg\lgroup \dfrac{9y^3}{16}\bigg\rgroup + \bigg\lgroup \dfrac{3y^3}{4}\bigg\rgroup}{\bigg\lgroup \dfrac{27y^3}{64}\bigg\rgroup + \bigg\lgroup y^3\bigg\rgroup}$

$\longrightarrow \sf \dfrac{\bigg\lgroup \dfrac{9y^3}{16}\bigg\rgroup + \bigg\lgroup \dfrac{12y^3}{16}\bigg\rgroup}{\bigg\lgroup \dfrac{27y^3}{64}\bigg\rgroup + \bigg\lgroup \dfrac{64y^3}{64}\bigg\rgroup}$

$\longrightarrow \sf \dfrac{\bigg\lgroup \dfrac{9y^3 + 12y^3}{16}\bigg\rgroup}{\bigg\lgroup \dfrac{27y^3 + 64y^3}{64}\bigg\rgroup}$

$\longrightarrow \sf \dfrac{\bigg\lgroup \dfrac{21y^3}{16}\bigg\rgroup}{\bigg\lgroup \dfrac{91y^3}{64}\bigg\rgroup}$

$\longrightarrow \sf \bigg\lgroup \dfrac{21y^3}{16}\bigg\rgroup \times \bigg\lgroup \dfrac{64}{91y^3}\bigg\rgroup$

$\longrightarrow \sf \bigg\lgroup \dfrac{1344\cancel{y^3}}{1456\cancel{y^3}}\bigg\rgroup$

$\longrightarrow \sf \bigg\lgroup \dfrac{\cancel{1344}}{\cancel{1456}}\bigg\rgroup$

$\longrightarrow \sf \bigg\lgroup \dfrac{\cancel{672}}{\cancel{728}}\bigg\rgroup$

$\longrightarrow \bigg\lgroup \dfrac{12}{13}\bigg\rgroup$

$\longrightarrow \sf\bold{\red{12 : 13}}$

${\small{\bold{\underline{\therefore\: The\: value\: of\: x^2y + xy^2 : x^3 : y^3\: is\: 12 : 13\: .}}}}$

Hence, the correct options is option no (b) 12 : 13.