Question

$\huge \sf \color{blue}{\underline{\underline{Question}}}$
Sum of two numbers is 15. if the sum of their reciprocal is 3/10, find the two numbers​

Answer 1

$\huge \sf \color{purple}{\underline{\underline{Answer}}} :$

Two numbers are 5 and 10

$\huge \sf \color{purple}{\underline{\underline{Solution}}}:$

✯ Let the number be x .

✯ Then the other number is 15 - x.

$\sf \color{brown}{Using \: the \: given \: information \: , we\: get }$

$\sf \frac{1}{x}+\frac{1}{15-x}=\frac{3}{10} \\ \\ \sf \frac{15-x+x}{x(15-x)}=\frac{3}{10} \\ \\ \sf \frac{15}{(15x-x²)}=\frac{3}{10} \\ \\ \sf 150=45x-3x² \\ \\ \sf 3(x²-15+50)=0 \\\\ \sf x²-15+50=0 \\ \\ \sf x²-5x-10x+50 =0 \\ \\ \sf (x-10)(x-5)=0 \\ \\ \sf x-10=0 \: \: \ \ \ OR \ \ \ \ \sf x-5 \\ \\ \sf \boxed{ \colorbox{lightgreen}{x=10}} \ \ \ \ or \ \ \ \ \boxed{\colorbox{lightgreen}{ x = 5 }}$$\color{red}━━━━━━━━━━━━━━━━━━━━━━━━━$

When x = 10 , then the other number is 15-5 = 10

When x = 5, then the other number is 15-5 = 10

$\color{red}━━━━━━━━━━━━━━━━━━━━━━━━━$

$\\ \\ \sf \color{blue}{Hence, \ the \ two \ numbers \ are \ 5 \ and \ 10 }$

$\color{red}━━━━━━━━━━━━━━━━━━━━━━━━━$$\color{red}━━━━━━━━━━━━━━━━━━━━━━━━━$

Answer 2

AnswEr-:

• $\underline {\red{ \star {\mathrm { \:The\:two\:the\:numbers\:are\:10\:and\:5\:or\:5\:and\:10\:.}}}}$

Explanation-:

$\mathrm {\bf{ Given-:}}$

• The Sum of two numbers is 15.

• If the sum of their reciprocal is 3/10,

$\mathrm { \bf{To\:Find\;-:}}$

• The two numbers .

$\sf{\bf{Solution \:of\:Question \:-:}}$

$\mathrm {\bf{ Let's \:Assume \;-:}}$

• The number be x .

Then Given That ,

• Sum of two numbers is 15.

Then ,

• The second number is 15 -x .

Therefore,

• $\mathrm{\bf{\purple{The\:Number \:\: -:}}} \begin{cases} \sf{\blue{The\:First \:number \:\:is\:= \frak{x}}} & \\\\ \sf{\red{The\:Second \:number \:\:is \:=\:\frak{15-x}}}\end{cases}$

• Also , Given that ,

• If the sum of their reciprocal is 3/10,

• $\mathrm{\bf{\purple{Their\:Reciprocals\:\: -:}}} \begin{cases} \sf{\blue{The\:First \:number \:\:is\:= \frak{\dfrac{1}{x}}}} & \\\\ \sf{\red{The\:Second \:number \:\:is \:=\:\frak{\dfrac{1}{15-x}}}}\end{cases}$

Then ,

• There sum is 3/10

$\underline{\bf{\sf{\star{According \:to\:Question \:-:}}}}$

• $\longmapsto { \mathrm { \dfrac{1}{x} + \dfrac{1}{15-x}= \dfrac{3}{10} }}$

• $\longmapsto { \mathrm { \dfrac{15-x + x }{x(15 -x)} = \dfrac{3}{10} }}$

• $\longmapsto { \mathrm { \dfrac{15\cancel {-x }\cancel{+ x} }{x(15 -x)} = \dfrac{3}{10} }}$

• $\longmapsto { \mathrm { \dfrac{15 }{x(15 -x)} = \dfrac{3}{10} }}$

• $\mathrm {\bf{ By\:Cross \:Multiplication -:}}$

• $\longmapsto { \mathrm { 15 \times 10 = 3 x(15 -x) }}$

• $\longmapsto { \mathrm { 150 =45x - 3x^{2} }}$

• $\longmapsto { \mathrm { 3x^{2} -45x + 150 =0 }}$

$\mathrm {Taking\:3\:as\:common\:in\:each\:Term-:}$

• $\longmapsto { \mathrm { 3x^{2} -45x + 150 =0 }}$

• $\longmapsto { \mathrm { x^{2} -15x + 50 =0 }}$

• $\longmapsto { \mathrm { x^{2} -10x - 5x + 50 =0 }}$

• $\longmapsto { \mathrm { x(x - 10) -5 ( x - 10) =0 }}$

• $\pink{\underline{\boxed { \mathrm { x \:= 10 \:, \:x = 5 \: }}}}$

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When x = 5 -:

• $\mathrm{\bf{\purple{Putting \:x=5\:\: -:}}} \begin{cases} \sf{\blue{The\:First \:number \:\:is\:= \frak{x= 5}}} & \\\\ \sf{\red{The\:Second \:number \:\:is \:=\:\frak{15-x= 15 - 5 = 10 }}}\end{cases}$

Therefore,

• $\underline {\pink{ \star {\mathrm { When \: x= 5\:then\:the\:two\:the\:numbers\:are\:5\:and\:10.}}}}$

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• When x = 10 -:

• $\mathrm{\bf{\purple{Putting \:x=10\:\: -:}}} \begin{cases} \sf{\blue{The\:First \:number \:\:is\:= \frak{x= 10}}} & \\\\ \sf{\red{The\:Second \:number \:\:is \:=\:\frak{15-x= 15 - 10 = 5 }}}\end{cases}$

Therefore,

• $\underline {\pink{ \star {\mathrm { When \: x= 10\:then\:the\:two\:the\:numbers\:are\:10\:and\:5.}}}}$

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• $\underline {\red{ \star {\mathrm { \:The\:two\:the\:numbers\:are\:10\:and\:5\:or\:5\:and\:10\:.}}}}$

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