Question

The digit at the tens place of two digit number is 4 times that is the unit place . If the digits are reversed the new number will be 54 less than original number . Find the number .

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Answer 1

Answer: 82

Step-by-step explanation:

Let the digit at the ten's place as x and digit at one's place as y.

x=4y (given)

Original number= 10x+y

                          = 40y+y (x=4y)

                           =41y

Reversed number = 10y +x

                              = 14y

41y - 14y = 54  (reversed number = original number + 54 ----given)

27y = 54

y = 2

x=8 (x=4y)

Thus the number is 82

[Reverse number is 28. We can verify the solution by subtracting the two:

82-28 = 54]

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

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