Question

A two digit number is such that the product of the digits is 20. If 9 is subtracted from the number, the digits interchange their places. Find the number. ​

Answer 1

$\huge{\blue{\bold{\underline{\underline{Answer :}}}}}$

$\:\:$

$\large{\green{\underline \bold{\tt{Given :-}}}}$

$\:\:$

• Product of digits of a two digit number is 20

$\:\:$

• If 9 is subtracted from the number, the digits interchange their places

$\:\:$

$\large{\red{\underline \bold{\tt{To \: Find :-}}}}$

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• The original number

$\:\:$

$\large{\orange{\underline{\tt{Solution :-}}}}$

$\:\:$

• Let the ten's digit be "x"

• Let the one's digit be "y"

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$\purple{\underline \bold{According \: to \: the \ question :}}$

$\:\:$

〚 Product of digits is 20 〛

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➠ x × y = 20

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➠ $\sf x = \dfrac { 20 } { y }$ ------- (1)

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$\underline{\bold{\texttt{Original number :}}}$

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➠ 10x + y -------- (a)

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$\underline{\bold{\texttt{Reversed number :}}}$

$\:\:$

➠ 10y + x

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〚 When 9 is subtracted from the number, the digits interchange their places 〛

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So,

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➜ 10x + y - 9 = 10y + x

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➜ 10x - x + y - 10y = 9

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➜ 9x - 9y = 9

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〚 Dividing the above equation by "9"〛

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➜ x - y = 1 ------ (2)

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$\underline{\bold{\texttt{Putting the value of "x" from (1) to (2)}}}$

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➜ x - y = 1

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➜ $\sf \dfrac { 20 } { y } - y = 1$

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➜ $\sf \dfrac { 20 - y^2} { y } = 1$

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➜ y = 20 - y²

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➜ y² + y - 20 = 0

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➜ y² + 5y - 4y - 20 = 0

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➜ y(y + 5) -4(y + 5) = 0

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➜ (y + 5)(y - 4) = 0

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Hence,

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• y = -5

• y = 4

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Case I [ y = -5 ]

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$\underline{\bold{\texttt{Putting y = -5 in (1) }}}$

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➜ $\sf x = \dfrac { 20 } { y }$

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➜ $\sf x = \dfrac { 20 } { -5}$

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➨ x = -4

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Case II [ y = 4 ]

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$\underline{\bold{\texttt{Putting y = 4 in (2) }}}$

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➜ $\sf x = \dfrac { 20 } { y }$

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➜ $\sf x = \dfrac { 20 } { 4}$

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➨ x = 5

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Thus we got 2 pairs

• x = -4 , y = -5

• x = 5 , y = 4

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$\underline{\bold{\texttt{Putting these values in (a) }}}$

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Original number

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➠ 10x + y

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➜ 10(-4) + (-5)

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➜ -40 -5

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➨ -45

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Original number

$\:\:$

➠ 10x + y

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➜ 10(5) + 4

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➜ 50 + 4

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➨ 54

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• Hence the original number is 54 or -45

Answer 2

$\huge{\blue{\bold{\underline{\underline{Answer :}}}}}$

$\:\:$

$\large{\green{\underline \bold{\tt{Given :-}}}}$

$\:\:$

• Product of digits of a two digit number is 20

$\:\:$

• If 9 is subtracted from the number, the digits interchange their places

$\:\:$

$\large{\red{\underline \bold{\tt{To \: Find :-}}}}$

$\:\:$

• The original number

$\:\:$

$\large{\orange{\underline{\tt{Solution :-}}}}$

$\:\:$

Let the ten's digit be "x"

Let the one's digit be "y"

$\:\:$

$\purple{\underline \bold{According \: to \: the \ question :}}$

$\:\:$

〚 Product of digits is 20 〛

$\:\:$

➠ x × y = 20

$\:\:$

➠ $\sf x = \dfrac { 20 } { y }$ ------- (1)

$\:\:$

$\underline{\bold{\texttt{Original number :}}}$

$\:\:$

➠ 10x + y -------- (a)

$\:\:$

$\underline{\bold{\texttt{Reversed number :}}}$

$\:\:$

➠ 10y + x

$\:\:$

〚 When 9 is subtracted from the number, the digits interchange their places 〛

$\:\:$

So,

$\:\:$

➜ 10x + y - 9 = 10y + x

$\:\:$

➜ 10x - x + y - 10y = 9

$\:\:$

➜ 9x - 9y = 9

$\:\:$

〚 Dividing the above equation by "9"〛

$\:\:$

➜ x - y = 1 ------ (2)

$\:\:$

$\underline{\bold{\texttt{Putting the value of "x" from (1) to (2)}}}$

$\:\:$

➜ x - y = 1

$\:\:$

➜ $\sf \dfrac { 20 } { y } - y = 1$

$\:\:$

➜ $\sf \dfrac { 20 - y^2} { y } = 1$

$\:\:$

➜ y = 20 - y²

$\:\:$

➜ y² + y - 20 = 0

$\:\:$

➜ y² + 5y - 4y - 20 = 0

$\:\:$

➜ y(y + 5) -4(y + 5) = 0

$\:\:$

➜ (y + 5)(y - 4) = 0

$\:\:$

Hence,

$\:\:$

y = -5

y = 4

$\:\:$

Case I [ y = -5 ]

$\:\:$

$\underline{\bold{\texttt{Putting y = -5 in (1) }}}$

$\:\:$

➜ $\sf x = \dfrac { 20 } { y }$

$\:\:$

➜ $\sf x = \dfrac { 20 } { -5}$

$\:\:$

➨ x = -4

$\:\:$

Case II [ y = 4 ]

$\:\:$

$\underline{\bold{\texttt{Putting y = 4 in (2) }}}$

$\:\:$

➜ $\sf x = \dfrac { 20 } { y }$

$\:\:$

➜ $\sf x = \dfrac { 20 } { 4}$

$\:\:$

➨ x = 5

$\:\:$

Thus we got 2 pairs

x = -4 , y = -5

x = 5 , y = 4

$\:\:$

$\underline{\bold{\texttt{Putting these values in (a) }}}$

$\:\:$

Original number

$\:\:$

➠ 10x + y

$\:\:$

➜ 10(-4) + (-5)

$\:\:$

➜ -40 -5

$\:\:$

➨ -45

$\:\:$

Original number

$\:\:$

➠ 10x + y

$\:\:$

➜ 10(5) + 4

$\:\:$

➜ 50 + 4

$\:\:$

➨ 54

$\:\:$

• Hence the original number is 54 or -45