Question

By cross multiplication method find such a two digit number such that the digit at unit place is twice the digit at tens place and the the number is 36 more than the original number​

Answer 1

Step-by-step explanation:

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Answer 2

❏ SolutioN :

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$\blacksquare\:\:\footnotesize{\underline{\underline{Given}}}$

$\footnotesize{\text{digit at unit place is twice the digit at tens place}}$

$\footnotesize{\text{the reverse number is 36 more than the original number}}$

$\blacksquare\:\:\footnotesize{\underline{\underline{To\: Find}}}$

$\footnotesize{\text{find the two digits and the number }}$

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$\footnotesize{\text{Let , the 1st place digit is = X}}$

$\footnotesize{\text{And , the 2nd place digit is = Y}}$

$\therefore\:\:\footnotesize{\text{The Original Number is = 10Y+X}}$

And

$\therefore\:\:\footnotesize{\text{The Reverse Number is = 10X+Y}}$

$\blacksquare\:\:\footnotesize{\underline{\underline{\red{Condition:1}}}}$

$\footnotesize{\text{digit at unit place is twice the digit at tens place}}$

$\therefore\:\:\footnotesize{X=2Y}$

$\implies\footnotesize{X-2Y+0=0}$ ...................(i)

$\blacksquare\:\:\footnotesize{\underline{\underline{\red{Condition:2}}}}$

$\footnotesize{\text{the reverse number is 36 more than the original number}}$

$\therefore\:\:\footnotesize{(10X+Y)=(10Y+X)+36}$

$\implies\footnotesize{10X+Y-10Y-X=36}$

$\implies\footnotesize{9X-9Y=36}$

$\implies\footnotesize{9(X-Y)=36}$

$\implies\footnotesize{X-Y=\dfrac{\cancel{36}}{\cancel9}}$

$\implies\footnotesize{X-Y=4}$

$\implies\footnotesize{X-Y-4=0}$ ......................(ii)

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Cross Multiplication process

$\footnotesize{\text{for the two linear equations in 2 variables , }}$

$\footnotesize{a_1x+b_1+c_1=0}$

$\footnotesize{a_2x+b_2+c_2=0}$

$\setlength{\unitlength}{1.6mm}\begin{picture}(30,20)\linethickness{0.08mm}\put(10,10){ c_1\:\:\:\:\:\:a_1\:\:\:\:\:\:b_1\:\:\:\:\:\:c_1 }\put(10,6){ c_2\:\:\:\:\:\:a_2\:\:\:\:\:\:b_2\:\:\:\:\:\:c_2 }\multiput(12,10)(5,0){3}{\vector(3,-4){2.6}}\multiput(12,7)(5,0){3}{\vector(3,4){2.6}}\put(0,10){.}\end{picture}$

$\therefore\:\:\footnotesize{\boxed{\dfrac{x}{b_1c_2-b_2c_1}=\dfrac{y}{c_1a_2-c_2a_1}=\dfrac{1}{a_1b_2-a_2b_1}}}$

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$\implies\footnotesize{X-2Y+0=0}$ ...................(i)

$\implies\footnotesize{X-Y-4=0}$ ..................(ii)

$\setlength{\unitlength}{1.6mm}\begin{picture}(30,20)\linethickness{0.08mm}\put(10,10){ 0\:\:\:\:\:\:\:1\:\:\:\:\:\:(-2)\:\:\:\:\:\:0 }\put(7,6){ (-4)\:\:\:\:\:\:1\:\:\:\:\:\:(-1)\:\:\:\:\:\:(-4) }\multiput(11.5,10)(4.5,0){2}{\vector(3,-4){2.6}}\multiput(11.5,7)(4.5,0){2}{\vector(3,4){2.6}}\put(23,10){\vector(3,-4){2.6}}\put(23,7){\vector(3,4){2.6}}\put(0,10){.}\end{picture}$

$\therefore\:\:\footnotesize{\dfrac{x}{(-2)(-4)-(-1)(0)}=\dfrac{y}{(0)(1)-(-4)(1)}=\dfrac{1}{(1)(-1)-(1)(-2)}}$

$\implies\footnotesize{\dfrac{x}{8-0}=\dfrac{y}{0+4}=\dfrac{1}{-1+2}}$

$\implies\footnotesize{\dfrac{x}{8}+\dfrac{y}{4}=\dfrac{1}{1}}$

$\implies\footnotesize{\dfrac{x}{8}=\dfrac{y}{4}=1}$

$\blacksquare\:\:\footnotesize{\dfrac{x}{8}=1}$

$\implies\:\:\footnotesize{\boxed{x=8}}$

$\blacksquare\:\:\footnotesize{\dfrac{y}{4}=1}$

$\implies\:\:\footnotesize{\boxed{y=4}}$

$\therefore\:\:\footnotesize{\text{The digits are 8 and 4 }}$

And

$\therefore\:\:\footnotesize{\text{The Original Number is = 10Y+X}}$

$\implies\:\:\footnotesize{\text{The Original Number is = 10(4)+8}}$

$\implies\:\:\footnotesize{\text{The Original Number is = 40+8}}$

$\implies\:\:\footnotesize{\boxed{\text{The Original Number is = 48}}}$

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