Question

The digit at the tens place of a two digit number is 4 times that of the unit place if the digits are reversed the new number 54 less than the original number find the number

Answer 1

Answer:

Let us take the digit at the ten's place as X and digit at one's place as Y.

given: x = 4y

original number= 10x+y or 41y (x=4y)

reversed number = (10y +x)  or 14 y

as 10y +x is 54 less than 10x+y, we can say

41y - 14y = 54

27y = 54

therefore, y = 54/27

                    = 2

therfore, y = 2

since x = 4y, x=8

therefore the original number is 82 and the reverse is 28.

we can check the solution by subtracting the two:

82-28 = 54

as the condition holds, answer is verif

Answer 2

☯ Let's consider the ten's digit and unit digit of original number be x and y respectively.

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Therefore,

The two digit number is = 10x + y

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$\qquad\quad\boxed{\bf{\mid{\overline{\underline{\bigstar\: According\: to \: the \: Question :}}}}\mid}$

The digit at tens place of two digit number is 4 times that in the unit place.

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$:\implies\sf Ten's\:digit = 4 \times unit\:digit\\ \\ \\:\implies\sf x = 4 \times y$

$\qquad\qquad:\implies\sf x = 4y\qquad\qquad\bigg\lgroup\bf eq\:(1)\bigg\rgroup$

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Also,

If the digits are reversed the new number will be 54 less than original number.

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Number after reversing digit = 10y + x

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Therefore,

$:\implies\sf 10y + x = (10x + y) - 54\\ \\ \\ :\implies\sf 10y - y + x - 10x = - 54\\ \\ \\ :\implies\sf 9y - 9x = - 54\\ \\ \\ :\implies\sf 9(y - x) = - 54\\ \\ \\:\implies\sf y - x = \cancel{\dfrac{-54}{9}}\\ \\ \\ :\implies\sf y - x = - 6\\ \\ \\ \dag\;{\underline{\frak{Substituting\:value\:of\:'x'\:from\:eq\:(1),}}}\\ \\ \\ :\implies\sf y - 4y = -6\\ \\ \\ :\implies\sf -3y = -6\\ \\ \\ :\implies\sf y = \cancel{\dfrac{-6}{-3}}\\ \\ \\:\implies{\underline{\boxed{\frak{\purple{y = 2}}}}}\;\bigstar$

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$\dag\;{\underline{\frak{Now\:putting\;value\:of\:y\;in\;eq\:(1),}}}\\ \\ \\ :\implies\sf x = 4 \times 2\\ \\ \\:\implies{\underline{\boxed{\frak{\pink{x = 8}}}}}\;\bigstar$

$\therefore\:{\underline{\sf{Hence,\:the\:original\:number\:is\;{\textsf{\textbf{82}}}.}}}$